Srinivas

Regular practice with AP Inter 2nd Year Chemistry Study Material Chapter 1 Solutions Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Chemistry 1st Lesson Solutions Questions and Answers

I. Multiple Choice Questions

Question 1.
Which of the following concentration term is independent of temperature? [ ]
1. Molarity
2. Molality
3. Mass-Volume percentage
4. Volume -Volume percentage
Answer:
2. Molality
Molality depends on mass (temperature independent)

Question 2.
The boiling point of an azeotropic mixture of water and ethanol is less than that of water. This mixture shows [ ]
1. negative deviation from Raoult’s law
2. positive deviation from Raoult’s law
3. no deviation from Raoult’s law
4. deviation which cannot be predicted
Answer:
2. positive deviation from Raoult’s law
Lower boiling point → mixture vaporizes more easily. This means intermolecular forces are weaker than ideal. So vapour pressure is higher than expected.

Question 3.
The van’t Hoff factor for 0.1 m Ba(NO₃)₂ solution is 2.74. Its degree of dissociation (α) is [ ]
1. 0.74
2. 0.92
3.0.87
4. 0.78
Answer:
3.0.87
α = \(\frac{i-1}{n-1}\) = \(\frac{2.74-1}{3-1}\) = \(\frac{1.74}{2}\) = 0.87
[Ba(NO₃)₂→Ba⁺²+2NO₃ n=3]

Question 4.
The solubility of gases in liquids increases with [ ]
1. increase in pressure and increase in temperature
2. decrease in pressure and decrease in temperature
3. decrease in pressure and increase in temperature
4. increase in pressure and decrease in temperature
Answer:
4. increase in pressure and decrease in temperature
Gas solubility follows: Increase in Pressure → more gas dissolves (Henry’s law)
Increase in Temperature → gas escapes→ solubility decreases

Question 5.
Which of the following is an example of an ideal solution?
1. CS₂ + CH₃COCH₃
2. C₆H₅OH + C₆H₅NH₂
3. C₆H₆ + C₆H₅CH₃
4. C₂H₅OH + CH₃COCH₃
Answer:
3. C₆H₆ + C₆H₅CH₃
C₆H₆ + C₆H₅CH₃ is an ideal solution.

Question 6.
The mole fraction of benzene in solution containing 30% by mass in CCl₄ is [ ]
1. 0.5611
2. 0.9102
3. 0.4586
4. 0.0214
Answer:
3. 0.4586
30% by mass of benzene means 30 g of benzene is present in 100g of solution.
Mass of benzene = 30 g; Mass of CCl₄= 100-30-70g;
Molar mass of benzene = 78, Molar mass of CCl₄ = 154
No. of moles of benzene (n₁) = \(\frac{\text { weight }}{\text { GMM }}\) = \(\frac{30}{78}\) = 0.385
No. of moles of CCl₄ (n₂) = \(\frac{70}{154}\) = 0.455
Mole fraction of Benzene = \(\frac{\mathrm{n}_1}{\mathrm{n}_1+\mathrm{n}_2}\) = \(\frac{0.385}{0.385+0.455}\) = \(\frac{0.385}{0.84}\) = 0.458

GMM of CCl4
(1 × 12) + (4 × 35.5) = 154
GMM of C₆H₆
(6 × 12) + (6 × 1) = 78

Question 7.
Which of the following solutions has the highest boiling point?
1.0.1 M KNO₃
2. 0.1 M Na₃PO₄
3. 0.1 M BaCl₂
4. 0.1 M K₂SO₄
Answer:
2. 0.1 M Na₃PO₄
Formula: Elevation of boiling point ∆T_b = ik_b × m;
∆T_b is directly proportional to Van’t Hoff factor (i). If i is more ∆T_b is more.
From the given options we have

Question 8.
For a dilute solution containing non-volatile solute Raoult’s law states that [ ]
1. the relative lowering of vapour pressure is equal to the mole fraction of the solute
2. the relative lowering of vapour pressure is equal to the mole fraction of the solvent
3. the lowering of vapour pressure is equal to the mole fraction of the solute
4. the lowering of vapour pressure is equal to the mole fraction of the solvent
Answer:
1. the relative lowering of vapour pressure is equal to the mole fraction of the solute
For dilute solutions with non-volatile solute: \(\frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}\) = χ_solute.
RLVP = mole fraction of solute

Question 9.
45 g of ethylene glycol molar mass (62 g mol^-1) is mixed with 600 g of water. The freezing point depression is (K_f = 1.86 K kg mol^-1)
1. 5.66 K
2. 8.33 K
3. 2.25 K
4. 4.81 K
Answer:
3. 2.25 K

Given mass of ethylene glycol = 45 g.
Given molar mass of ethylene glycol = 62 g
Mass of water = 600g, Molar mass of water = 18
Freezing point constant K = 1.86 K kgmol^-1
We know that ∆T_ƒ = K_ƒ × m, m = molality
∆T_ƒ = 1.8 × \(\frac{\text { Mass of solute }}{\text { GMM of solute }}\) × \(\frac{1000}{\text { Mass of solvent (in g) }}\) = 1.8 × \(\frac{45}{62}\) × \(\frac{100}{600}\) = 2.32 K
The freezing point depression ∆T_ƒ = ?

Question 10.
Which law explains the solubility of gases in liquids ?
1. Boyle’s law
2. Charles’ law
3. Henry’s law
4. Raoult’s law
Answer:
3. Henry’s law
Henry’s law explains the solubility of gases in liquids.

Question 11.
At 283 K, the osmotic pressure of 2% (w/v) solution of a substance X is 7.87 × 104 N.m^-2. The molar mass of X (in g mol^-1) is
1. 280
2. 660
3. 598
4. 300
Answer:
3. 598
Given T = 283K, Osmotic pressure П = 7.87 × 10⁴ Nm^-2
2% (w/v) means 2 g of solute present in 100 mL of solution,
w_s = 2g, V = 100 mL = 10^-4m³, Molar mass of solute (M_s)=?
Formula: Osmotic pressure П = CRT
П = \(\frac{\mathrm{w}_{\mathrm{s}} \mathrm{RT}}{\mathrm{M}_{\mathrm{s}} \mathrm{~V}}\)
⇒ M_s = \(\frac{\mathrm{w}_{\mathrm{s}} \mathrm{RT}}{\Pi \mathrm{~V}}\) = \(\frac{2 \times 8.314 \times 283}{7.87 \times 10^4 \times 10^{-4}}\) = 597.9 = 598 g

Question 12.
The vapour pressure of an aqueous solution obtained by adding 18 g of glucose to 180 g of water at 373 K is (Vapour pressure of pure water at 373 K is 760 mm Hg) [ ]
1. 752.4 mm Hg
2. 741.4 mm Hg
3. 759 mm Hg
4. 745.3 mm Hg
Answer:
1. 752.4 mm Hg
Given P^o = 760 mm, we have to find P^s
Mass of glucose (w_s) = 18 g, Molar mass of glucose (M_s) = 180 g.mol^-1
Mass of water (w_o) = 180 g, Molar mass of water (M) = 18 g.mol^-1
According to Raoult’s law \(\frac{\mathrm{P}^0-\mathrm{P}^{\mathrm{s}}}{\mathrm{P}^0}\) = \(\frac{\mathrm{w}_{\mathrm{s}}}{\mathrm{M}_{\mathrm{s}}}\) × \(\frac{\mathrm{M}_0}{\mathrm{w}_0}\)
⇒ \(\frac{760-\mathrm{P}^{\mathrm{S}}}{760}\) = \(\frac{18}{180}\) × \(\frac{18}{180}\) ⇒ 1 – \(\frac{\mathrm{P}^{\mathrm{s}}}{760}\) = \(\frac{1}{100}\) ⇒ 1 – \(\frac{1}{100}\) = \(\frac{\mathrm{P}^{\mathrm{s}}}{760}\)
⇒ 1 – 0.01 = \(\frac{\mathrm{P}^{\mathrm{s}}}{760}\) ⇒ P^s = 0.99 × 760 = 752.4

Question 13.
The value of Henry’s law constant, K_H is [ ]
1. greater for gases with higher solubility
2. constant for all gases
3. not related to the solubility of gases
4. greater for gases with lower solubility
Answer:
4. greater for gases with lower solubility
Henry’s law: Higher K_H → gas is less soluble (needs more pressure).
So solubility ∝ < 1/K_H. Greater for gases with lower solubility

Question 14.
The mass of solute (in g) required to be added to 100 g of water so as to observe an elevation of boiling point of 0.052 °C is
(Given: K_b (H₂O) = 0.52 K kg mol^-1 molar mass of solute = 100 g mol^-1)
1.0.30
2. 0.05
3. 2.00
4. 1.00
Answer:
4. 1.00
Given Molar mass of solute M_s = 100,
Mass of solvent H₂O = 100
ΔT_b = 0.052 °C, K_b = 0.52 K kg mol^-1, Mass of solute w_s=?
Elevation in boiling point ΔT_b = k_b × m, molarity
⇒ 0.052 = 0.52 × \(\frac{\mathrm{w}_{\mathrm{s}}}{\mathrm{M}_{\mathrm{s}}}\) × \(\frac{1000}{\text { Mass of solvent }(\mathrm{g})}\) ⇒ 0.052 = 0.52 × \(\frac{w_s}{100}\) × \(\frac{1000}{100}\)
0.052 × 10 = 0.52 × w_s ⇒ w_s = 1

Question 15.
P°_A and P°_B are the vapour pressures of two pure liquids A and B respectively of an ideal solution. If X_A is the mole fraction of liquid A, the total pressure of the solution will be
1. P°_B + X_A (P°_A – P°_B)
2. P°_A + X_A (P°_B – P°_A)
3. P°_A + X_A (P°_A – P°_B)
4. P°_B + X_A (P°_B – P°_B)
Answer:
1. P°_B + X_A (P°_A – P°_B)

II. Fill in the Blanks

Question 1.
The units of boiling point elevation constant K_b are ____
Answer:
K kg mol^-1

Question 2.
Solubility of a gas in liquid ____ with increase in pressure.
Answer:
increases

Question 3.
For an ideal solution, Δ_mixH is equal to ____.
Answer:
Zero

Question 4.
Desalination plants work on the basis of the phenomenon called ____.
Answer:
reverse osmosis

Question 5.
Human blood is isotonic with _____. solution (or) Saline.
Answer:
0.9% (w/v) NaCl

III. One Word Answer Questions

Question 1.
What is the term used for a solution that does not follow Raoult’s law?
Answer:
Non-ideal solution does not follow the Raoult’s law.

Question 2.
In which type of a solution, dynamic equilibrium exists between the dissolved solute and undissolved solute?
Answer:
In saturated solution dynamic equilibrium exists.

Question 3.
What is the name given to binary liquid mixtures which have same composition in liquid and vapour phase and boil at a constant temperature?
Answer:
Azeotropic mixtures have same composition in liquid and vapour phase.

Question 4.
In the expression ΔT_f = K_fm, what is the name given to the constant K_f? Answer:
K_f is called Cryoscopic constant (or) Molal depression constant.

Question 5.
What is the name given to the solutions, when they have same osmotic pressure at a given temperature?
Answer:
Isotonic solutions have same osmotic pressure at a given temperature.

IV. Very Short Answer Questions

Question 1.
Define molarity.
Answer:
Molarity (M): It is the number of moles of the solute dissolved per litre of the solution.
Formula: Molarity(M) \(=\frac{\text { No.of moles of solute }}{\text { Volume of solution in Litre }}\)

Question 2.
Define molality.
Answer:
Molality(m): It is the number of moles of the solute dissolved in one kilogram (Kg) of the solvent
Formula: Molality(m) \(=\frac{\text { No.of moles of solute }}{\text { Mass of solvent in Kg }}\)
No. of moles of solute Mass of solvent in Kg

Question 3.
Define mole fraction.
Answer:
Mole Fraction(χ): It is the ratio of the number of moles of one component to the total number of moles of all the components present in the solution.
Formula: Mole fraction χ \(=\frac{\text { No. of moles of one component }}{\text { Total no. of moles of all the components }}\)
No. of moles of one component Total no. of moles of all the components

Question 4.
State Henry’s law.
Answer:
Henry’s law: The partial pressure (p) of a gas in vapour phase is directly proportional to the mole fraction (χ) of the gas in the solution.
Formula: Partial pressure of a gas P_gas = K_Hχ_gas.

Question 5.
What is Ebullioscopic constant?
Answer:
Ebullioscopic constant(K_b): It is the elevation in boiling point of a solution when one mole of non-volatile solute is added to 1kg of solvent. Formula: ΔT_b = K_bm

Question 6.
What is Cryoscopic constant?
Answer:
Cryoscopic constant (K_f): It is the depression in the freezing point of a solution when one mole of non-volatile solute is dissolved in 1kg of solvent. Formula: ΔT_f = K_f×m

Question 7.
Define osmotic pressure.
Answer:
Osmotic pressure(\(\Pi\)): It is the minimum pressure required to stop the flow of solvent particles through a semipermeable membrane from a pure solvent into a solute. (or)
The pressure required just to stop the Osmosis is called Osmotic pressure.
Formula: \(\Pi\) = CRT

Question 8.
What are isotonic solutions?
Answer:
Isotonic solutions: Solutions which have the same osmotic pressure at a given temperature are called isotonic solutions.
Ex: Blood is isotonic with 0.9% (w/v) NaCl.

Question 9.
Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution of benzoic acid in methanol.
Answer:
Molarity(M)=0.15M, Volume (V)=250mL,
GMM (C₆H₅COOH) = 122, Weight of C₆H₅COOH (w) =?

Question 10.
What is van’t Hoff’s factor?
How is it related to ‘a’ in the case of a binary electrolyte (1:1)?
Answer:
Van’t Hoff’s factor(i): It is the ratio of the experimental value of the colligative property to the calculated value of the colligative property.
Incase of a binary electrolyte (1:1), the relation between a and i is as follows:
i = \(\frac{1+(n-1) \alpha}{1}\), Degree of dissociation, α = \(\frac{i-1}{n-1}\)
The Van’t Hoff factor i accounts for dissociation or association of solute.

V. Short Answer Questions

Question 1.
How many types of solutions are formed? Give an example for each type of solution.
Answer:

Question 2.
Define mass percentage, volume percentage and mass to volume percentage solutions.
Answer:

1. Mass percentage (w/w): Mass of the component (in g) present in 100g of solution. Mass % of a component \(=\frac{\text { Mass of the component in the solution }}{\text { Total mass of the solution }}\) × 100
2. Volume percentage (V/V): Volume of the component (in mL) present in 100 mL of solution.
   Volume % (V/V) of a component \(=\frac{\text { Volume of the component }}{\text { Total volume of the solution }}\) × 100
3. Mass by volume % (w/v): The mass of solute in (g) present in 100 mL of solution.
   Mass by volume % (w/v) of solute \(=\frac{\text { Mass of the solute in the solution }}{\text { Total volume of the solution }}\) × 100

Question 3.
A solution of glucose in water is labeled as 10% w/w. What would be the molality of the solution ?
Answer:
10% w/w glucose solution means 10g of glucose is dissolved in 100g of solution.
Weight of Glucose solute (w_s) = 10 g, GMM of glucose = 180,
Moles of glucose n \(=\frac{\text { weight }}{\text { GMM }}\) = \(\frac{10}{180}\), Weight of solvent (W) = 100 – 10 = 90g
Molality of the solution m = n × \(\frac{1000}{W(g)}\) = \(\frac{10}{180} \times \frac{1000}{90}\) = 0.617 m
∴ Molality of the given solution = 0.617m
GMM of C₆H₁₂O₆
(6 × 12) + (12 × 1) + (6 × 16) + (16 × 2) + (1 × 1) = 122

Question 4.
A solution of sucrose in water is labeled as 20% w/w. What would be the mole fraction of each component in the solution?
Answer:
20% w/w sucrose solution means 20g of sucrose is present in 100g of solution.
Weight of sucrose solute = 20g, GMM of sucrose (C₁₂H₂₂O₁₁)= 342 g mol^-1.
Moles of sucrose, n_A = \(\frac{\text { weight }}{\text { GMM }}\) = \(\frac{20}{342}\) = 0.0585
Weight of solution = 100g, Mass of solvent = 100 – 20 = 80g, GMM of water (H₂O) = 18 g mol^-1.

GMM of C₁₂H₂₂O₁₁
(12×12)+(22×1)+(11×16)
= 342

Moles of water, n_B = \(=\frac{\text { weight }}{\mathrm{GMM}}\) = \(\frac{80}{18}\) = 4.4445
Mole fraction of sucrose (χ_A) \(=\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{A}}+\mathrm{n}_{\mathrm{B}}}\) = \(\frac{0.0585}{0.0585+4.4445}\) = \(\frac{0.0585}{4.503}\) = 0.013
Mole fraction of sucrose (χ_B) = 1 – χ_A = 1 – 0.013 = 0.987 [∵ χ_A + χ_B = 1]
Mole fraction of water (χ_B) = 1 − χ_A = 1 – 0.013 = 0.987

Question 5.
What is meant by positive deviations from Raoult’s law and how is the sign of Amix H related to positive deviation from Raoult’s law ?
Answer:

1. When the vapour pressure of a solution is higher than the predicted value by Raoult’s law, it is called positive deviation.
2. In such cases, intermolecular interactions between solute and solvent particles (A and B) are weaker than those between solute-solute (A-A) and solvent-solvent (B-B).
3. Hence, the molecules of (A or B) will escape more easily from the surface of solution than in their pure state. Therefore, the vapour pressure of the solution will be higher.
4. Characteristics of a solution showing positive deviation
   1. P_total > P_A + P_B
   2. ΔH_mix > 0 (+Ve)
   3. ΔV_mix > 0 (+Ve)
5. Ex:
   1. Ethyl alcohol and water
   2. Acetone and benzene.

Question 6.
What is meant by negative deviation from Raoult’s law and how is the sign of Δ_mixH related to negative deviation from Raoult’s law?
Answer:

1. When the vapour pressure of a solution is lower than the predicted value by Raoult’s law, it is called negative deviation.
2. In such cases, intermolecular interactions between solute and solvent particles (A and B) are stronger than those between solute-solute (A-A) and solvent-solvent (B-B).
3. It leads to decrease in vapour pressure resulting in negative deviation.
4. Characteristics of a solutions showing negative deviation
   1. P_total < P_A + P_B;
   2. ΔН_mix < 0, (-Ve)
   3. ΔV_mix < 0, (-Ve)
5. Ex:
   1. Nitric acid and water
   2. Hydrochloric acid and water.

Question 7.
Calculate the mass of a non-volatile solute (molar mass 40g mol^-1) which should be dissolved in 114g Octane to reduce its vapour pressure to 80%.
Answer:
Vapour pressure is reduced 80% means vapour pressure of the solution reduces to 80.
Vapour pressure of solvent P^o=100, Vapour pressure of solution P^s=80
Mass of solvent (octane) (w_o)=114g, Molar mass of Octane (C₈H₁₈) is M_o=114 g/mol
Molar mass of solute is M_s=40 g/mol, Mass of solute is w_s =?

GMM of C₈H₁₈
=96+18=122

From Raoult’s law \(\frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}^{\mathrm{S}}}{\mathrm{P}^{\mathrm{o}}}\) = \(\frac{w_s}{M_s} \times \frac{M_o}{w_o}\)
⇒ \(\frac{100-80}{100}\) = \(\frac{w_s}{40} \times \frac{114}{114}\) ⇒ \(\frac{20}{100}\) ⇒ \(\frac{w_s}{40}\) ⇒ w_s = \(\frac{20 \times 40}{100}\) = 8 g

Question 8.
If the osmotic pressure of glucose solution is 1.52 bar at 300K. What would be its concentration if R=0.083 L bar mol^-1 K^-1 ?
Answer:
Given osmotic pressure Π = 1.52, R = 0.083 L bar mol^-1 K^-1, T= 300K, Concentration C = ?
Formula: П=CRT ⇒ C = \(\frac{\Pi}{\mathrm{RT}}\) = \(\frac{1.52}{0.083 \times 300}\)
= 0.061 M

Question 9.
What is relative lowering of vapour pressure? How is it useful to determine the molar mass of a solute?
Answer:
When a non-volatile solute is dissolved in a volatile solvent, the vapour pressure decreases.
LVP (Δp):This difference between the vapour pressure of pure solvent (P^o) and the vapour pressure of the solution (PS). Thus, Δp = p^o – p^s
RLVP: It is the ratio of lowering of vapour pressure (ΔP) to the vapour pressure of pure solvent (P^o) Also, RVP is equal to the molefraction of the solute (χ₂)

Question 10.
How is molar mass related to the depression in freezing point of a solution?
Answer:
Depression in Freezing point ΔTƒ=Kƒ×m, where m= molality of the solution
⇒ ΔT_f = K_f×\(\frac{\mathrm{w}_{\mathrm{s}}}{\mathrm{M}}\)×\(\frac{1000}{w_0(g)}\) ⇒ M_s = \(\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{w}_{\mathrm{s}} \times 1000}{\Delta \mathrm{~T}_{\mathrm{f}} \times \mathrm{w}_{\mathrm{o}}}\)
Here, M_s = molar mass of solute, K_f = molal depression constant
W_s = Mass of solute in grams, ΔT_f = depression in freezing point, w_o = Mass of solvent in grams

VI. Long Answer Questions

Question 1.
Calculate the depression in the freezing point of water when 10g of CH₃CH₂ CHClCOOH is added to 250g of water. K_a = 1.4 × 10^-3, K_f = 1.86 K kgmol^-1.
Answer:
We know depression in the freezing point ΔT_f = iK_fm
So we have to find
(i) molality (m)
(ii) Degree of dissociation (α)
(iii) Van’t Hoff factor (i)

(i) To find molality(m):

(ii) To find Degree of dissociation (α) :

(iii) To find Van’t Hoff factor (i)
Calculation of Van’t Hoff factor: ”

Question 2.
19.5g of CH₂FCOOH is dissolved in 500g of water. The depression in freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
Answer:
(i) Calculation of Van’t Hoff factor(i):

(ii) Calculation of dissociation constant, (K1):

Question 3.
Determine the osmotic pressure of a solution prepared by dissolving 25mg of K₂SO₄ in two litre of water at 25°C assuming that it is completely disassociated.
Answer:
Weight of K₂SO₄ dissolved (W_s) = 25 mg = 0.025 g, Volume of solution = 2L, T=25+273=298K
GMM of K₂SO₄ (M_s)= 174, Gas constant R= 0.083
K₂SO₄ dissociates completely: K₂SO₄ → 2K⁺ + SO₂

Question 4.
Benzene and Toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300K are 50.71mm of Hg and 32.06mm of Hg respectively. Calculate the mole fraction of benzene in vapour phase of 80g of benzene is mixed with 100g of toluene.
Answer:

Textual Solved Questions

Question 1.
Calculate the mole fraction of ethylene glycol (C₂H₆O₂) in a solution containing 20% of C₂H₆O₂ by mass.
Answer:
Assume that we have 100 g of solution.
20% of C₂H₆O₂ means 20 g of ethylene glycol and 80 g of water.
Weight of ethylene glycol = 20g, GMM of C₂H₆O₂ = 12 × 2 + 1 × 6 + 16 ×
Moles of C₂H₆O₂, n₁ \(=\frac{\text { weight }}{\text { GMM }}\) = \(\frac{20}{62}\) = 0.322
Weight of water H₂O = 80g, GMM of H₂O = 18
Moles of water H₂O, n₂ \(=\frac{\text { weight }}{\mathrm{GMM}}\) = \(\frac{80}{18}\) = 4.444
Mole fraction of ethylene glycol (χ₁) = \(\frac{\mathrm{n}_1}{\mathrm{n}_1+\mathrm{n}_2}\) = \(\frac{0.322}{0.322+4.444}\) = \(\frac{0.322}{4.766}\) = 0.068
Mole fraction of water (χ₂) = 1 – χ₁ = 1 – 0.068 = 0.932

Question 2.
Calculate the molarity of a solution containing 5 g of NaOH in 450 ml solution.
Answer:
Given weight of solute (NaOH) is w = 5 g; GMM of NaOH = 40
Moles of NaOH, n = \(\frac{\text { weight }}{\text { GMM }}\) = \(\frac{5}{40}\)
Volume of the solution in mL = 450 mL, Molarity=?
Molarity M = n × \(\frac{1000}{V(\mathrm{~mL})}\) = \(\frac{5}{40} \times \frac{1000}{450}\) = 0.278 mol dm ^-3

Question 3.
Calculate molality of 2.5 g of ethanoic acid (CH₃COOH) in 75g of benzene.
Solution:
Given weight of ethanoic acid = 2.5g
GMM of ethanoic acid C₂H₄O₂ = (12 × 2) + (1 × 4) + (16 × 2)= 60
Moles of C₂H₄O₂, n = \(\frac{\text { weight }}{\text { GMM }}\) = \(\frac{2.5}{60}\)
weight of solvent (benzene)= 75 g Molality (m)=?
Molality(m) = n × \(\frac{1000}{\text { Mass of solvent (in g) }}\) = \(\frac{2.5}{60} \times \frac{1000}{75}\) = 0.556 mol.kg^-1

Question 4.
If N₂ gas is bubbled through water at 293 K, how many millimoles of N₂ gas would dissolve in 1 litre of water? Assume that N₂ exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N₂ at 293 K is 76.48 kbar.
Answer:
Weight of water 1 liter = 1000 mL, GMM of H₂O =
No.of moles of water (n₂) = \(\frac{\text { weight }}{\text { GMM }}\) = \(\frac{1000}{18}\) = 55.5
Given that p = 0.987 bar. K_н = 76.48 k bar = 76,480 bar
From Henry’s law p = K_Hχ₁ ⇒ χ₁ = \(\frac{\mathrm{p} \text { (nitrogen) }}{\mathrm{K}_{\mathrm{H}}}\) = \(\frac{0.987}{76480}\) = 1.29 × 10^-5
We have n₁ <<<<55.5 ⇒ n₁ + n₂ = 55.5
Mole fraction of nitrogen, χ₁ = \(\frac{n_1}{n_1+n_2}\) ⇒ 1.29 × 10^-5 = \(\frac{\mathrm{n}_1}{55.5}\)
⇒ n₁ = 1.29 × 10^-5 × 55.5 = 7.16 × 10^-4 = 0.716 moles

Question 5.
Vapour pressure of chloroform (CHCl₃) and dichloromethane (CH₂Cl₂) at 298 K are 200mm Hg and 415 mm Hg respectively.
(i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at 298 K and,
(ii) mole fractions of each component in vapour phase.
Solution:
(i) Weight of dichloromethane = 40 g,
GMM of CH₂Cl₂ = 12 × 1 + 1 × 2 + 35.5 × 2 = 85
Moles of CH₂Cl₂, n₁ = \(\frac{\text { weight }}{\text { GMM }}\) = \(\frac{40}{85}\) = 0.47 mol
Weight of chloroform = 25.5 g
GMM of CHCl₃ = 12 × 1 + 1 × 1 + 35.5 × 3 = 119.5
Moles of CHCl₃, n₂ = \(\frac{25.5}{119.5}\) = 0.213 mol
Total number of moles is n₁ + n₂ = 0.47 + 0.213 = 0.683 mol
Mole fraction of CH₂Cl₂(χ₁) = \(\frac{n_1}{n_1+n_2}\) = \(\frac{0.47}{0.683}\) = 0.688
Mole fraction of CHCl₃ (χ₂) = 1 – χ₁ = 1.00 0.688 = 0.312
Given \(\mathrm{p}_1{ }^0\) = 200mm Hg \(\mathrm{p}_2{ }^0\) = 415 mm Hg
According to Raoult’s law, P_total = \(\mathrm{p}_1{ }^0\) + (\(\mathrm{p}_2{ }^0\) – \(\mathrm{p}_1{ }^0\))χ₁
= 200 + (415 – 200) × 0.688 = 200 + 147.9 = 347.9 mm Hg
In vapour phase p₁ = χ₁ × Ptotal ⇒ χ₁ = P₁/P_total
∴ p₁(CH₂Cl₂) = 0.688 × 415 = 285.5 mm Hg; p₂(CHCl₂) = 0.312 × 200 = 62.4 mm Hg

(ii) Mole fraction of CH₂Cl₂(χ₁) = 285.5 /347.9 = 0.82 [∵ χ₁ = p₁/_total]
Mole fraction of CHCl₃ (χ₂) = 62.4/347.9= 0.18 [∵ χ₂ = p₂/_total]

Question 6.
The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non- volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g mol^-1). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?
Answer:
Given that p_o = 0.850bar, M₀ = 78 g mol^-1, w₀ = 39g
P_s = 0.845 bar, w_s = 0.5g, M_s = ?
From Raoult’s law \(\frac{\mathrm{P}^{\mathrm{O}}-\mathrm{P}^{\mathrm{S}}}{\mathrm{P}^{\mathrm{O}}}\) = \(\frac{\mathrm{w}_{\mathrm{s}}}{\mathrm{M}_{\mathrm{s}}}\) × \(\frac{\mathrm{M}_0}{\mathrm{w}_0}\) ⇒ M_s = \(\frac{\mathrm{P}^{\mathrm{O}}}{\mathrm{P}^{\mathrm{O}}-\mathrm{P}^{\mathrm{S}}}\) × \(\frac{w_s \times \mathbf{M}_o}{\mathbf{w}_o}\)
= \(\frac{0.850}{0.850-0.845}\) × \(\frac{0.5 \times 78}{39}\) = 170 g mol^-1
∴ M_s = 170 g mol^-1

Question 7.
18 g of glucose, C₆H₁₂O₆, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? K_b for water is 0.52 K kg mol^-1.
Answer:
Given K_b = 0.52 K kg mol^-1.
Weight of glucose w = 18 g, GMM of glucose M_o = 180, Mass of solvent W₀ = 1000g,
Moles of glucose n \(=\frac{\text { weight }}{\mathrm{GMM}}\) = \(\frac{18}{180}\)
ΔT_b = K_b × m = K_b × n × \(\frac{1000}{\mathrm{~W}_0(\mathrm{~g})}\) = 0.52 × \(\frac{18}{180}\) × \(\frac{1000}{1000}\) = 0.052
Water boils at 373.15 K at 1.013 bar pressure,
∴ the boiling point of solution is 373.15 + 0.052 = 373.202 K.

Question 8.
The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_b for benzene is 2.53 K kg mol^-1.
Answer:
Given boiling point of the solution (T_b) = 354.11k, boiling point of pure solvent (T_o) = 353.23K
Weight of solute (w) = 1.80 g, Weight of solvent (w)=90 g, GMM of solute M_s=?
The rise in the boiling point ΔT_b = T_b – T₀ = 354.11 K – 353.23 K = 0.88 K
Elevation in boiling point ΔT_b = K_bm, m = molality
ΔT_b = K_b × \(\frac{\mathrm{W}_{\mathrm{s}}}{\mathrm{M}_{\mathrm{s}}}\) × \(\frac{1000}{w_{\mathrm{o}}(\mathrm{~g})}\) ⇒ 0.88 = 2.53 × \(\frac{1.8}{\mathrm{M}_{\mathrm{s}}}\) × \(\frac{1000}{90}\)
⇒ M_s = \(\frac{2.53 \times 1.8 \times 1000}{0.88 \times 90}\) = 58g mol^-1

Question 9.
45g of ethylene glycol (C₂H₆O₂) is mixed with 600g of water. Calculate (a) the freezing point depression and
(b) the freezing point of the solution.
Answer:
Given weight of ethylene glycol = 45g, weight of solvent (H₂O) = 600g
GMM of ethylene glycol (C₂H₆O₂) = 62, K_f = 1.86 K kgmol^-1 (for H₂O)

(a) Depression in Freezing point ΔT_f = K_fm
ΔT_f = K_f\(\left(\frac{\mathrm{w}_{\mathrm{s}}}{\mathrm{M}_{\mathrm{s}}} \times \frac{1000}{\mathrm{~W}_0(\mathrm{~g})}\right)\) = 1.86 × \(\frac{45}{62}\) × \(\frac{1000}{600}\) = 2.2K

(b) We know ΔT_f = T₀ – T_f·
∴ Freezing point of aqueous solution T_f – T₀ – ΔT_f = 273.15 K -2.2 K= 270.95 K

Question 10.
1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol^-1. Find the molar mass of the solute.
Answer:
Given K_f = 5.12 K kg mol^-1, ΔT_f = 0.40 K, Weight of solute (w_s) = 1.00 g
Weight of solvent (w) = 50 g, GMM of solute (M_s)=?
Depression in Freezing point ΔT_f = K_fm, m=molality
ΔT_f = K_f\(\left(\frac{\mathrm{w}_{\mathrm{s}}}{\mathrm{M}_{\mathrm{s}}} \times \frac{1000}{\mathrm{~W}_0(\mathrm{~g})}\right)\) ⇒ M_s = \(\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{w}_{\mathrm{s}} \times 1000}{\Delta \mathrm{~T}_{\mathrm{f}} \times \mathrm{w}_0}\) = \(\frac{5.12 \times 1.00 \times 1000}{0.40 \times 50}\) = 256 g mol^-1
Thus, GMM of the solute = 256 g mol^-1

Question 11.
200 cm³ of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10^-3 bar. Calculate the molar mass of the protein.
Answer:
Given data: П = 2.57 × 10^-3 bar, Weight of protein W= 1.26g.
V = 200 cm³ = 0.200 litre, T = 300 K, R= 0.083 L bar mol^-1 K^-1,
C = molality, Molar mass of protein M=?
П = CRT = \(\left(\frac{\text { mass }}{\text { molar mass }} \times \frac{1000}{V(\mathrm{~mL})}\right)\)RT ⇒ 2.57 × 10^-3 = \(\frac{1.26}{\text { molar mass }}\) × \(\frac{1000}{200}\) × 0.083 × 300
Molar mass = \(\frac{1.26 \times 5 \times 0.083 \times 30}{2.57 \times 10^{-3}}\) = 61.022 g.mol^-1

Question 12.
2 g of benzoic acid (C₆H₅COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9K kg mol. What is the percentage association of acid if it forms dimer in solution?
Answer:
Given data: Weight of solute w_s = 2 g; GMM of solute(C₆H₅COOH) = 122,
moles of solute n \(=\frac{\text { weight }}{\mathrm{GMM}}\) = \(\frac{2}{122}\) = 0.01639 mol
Weight of solvent W = 25 g = 0.025kg
molality, m = \(\frac{\mathrm{n}}{\mathrm{~W}(\mathrm{Kg})}\) = \(\frac{0.01639}{0.025}\) = 0.6556 mol kg^-1
Given K_f = 4.9 K kg mol^-1; ΔT_f = 1.62 K
Depression in freezing point ΔT_f = iK_fm ⇒ i = \(\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{~K}_{\mathrm{f}} \cdot \mathrm{~m}}\) = \(\frac{1.62}{4.9 \times 0.6556}\) = \(\frac{1.62}{3.212}\) = 0.504
Now consider the following equilibrium for the benzoic acid:

If x is a degree of association, 1-α moles of benzoic acid are left undissociated and the corresponding α/2 are associated moles of benzoic acid at equilibrium.
∴ 1 – \(\frac{\alpha}{2}\) = i ⇒ 1 – \(\frac{\alpha}{2}\) = 0.504 ⇒ \(\frac{\alpha}{2}\) = 0.4958 ⇒ α = 0.9916
Percentage of association of acid if it forms dimer = α × 100 = 0.9916 × 100 = 99.16%
∴ degree of association of benzoic acid in benzene is 99.2%.

Question 13.
0.6 mL of acetic acid (CH₃COOH), having density 1.06 g mL^-1, is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205°C. Calculate the van’t Hoff factor and the dissociation constant of acid.
Answer:
Given data: density d = 1.06 g mL^-1, V = 0.6 mL
Mass = V × d = 0.6 × 1.06 = 0.6408, Kf = 1.86 K kg mol^-1
GMM of CH₃COOH
(2 × 12) + (4 × 1) + (2 × 16)
= 24 + 4 + 32 = 60
No.of moles, n \(=\frac{\text { mass }}{\text { molar mass }}\) = \(\frac{0.6408}{60}\)
Molality(m) = n × \(\frac{1000}{\mathrm{~W}(\mathrm{~g})}\) = \(\frac{0.6408}{60}\) × \(\frac{1000}{1000}\) = 0.0106 mol kg^-1
Depression in freezing point ΔT_f = K_f × m = 1.86 K × 0.0106 = 0.0197 K
Van’t Hoff Factor, i \(=\frac{\text { Observed freezing point }}{\text { Calculated freezing point }}\) = \(\frac{0.0205 \mathrm{~K}}{0.0197 \mathrm{~K}}\) = 1.041

Thus, total moles of particles at equilibrium: n (1 – α + α + α) = n(1 + α)
∴ i = \(\frac{\mathrm{n}(1+\alpha)}{\mathrm{n}}\) = 1 + α
⇒ i = 1 + α ⇒ α = i – 1 = 1.041 – 1 = 0.041(Degree of dissociation)
For [CH₃COOH]= n(1 – α) = 0.0106(1 – 0.041)
For [CH₃COO ̄]̄]= nα = 0.0106 × 0.041,[H⁺] = nα = 0.0106 × 0.041
K_a = \(\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}\) = \(\frac{0.0106 \times 0.041 \times 0.0106 \times 0.041}{0.0106(1-0.041)}\)
= 1.86 × 10^-5

Objective Questions

Question 1.
On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following cases dissolution of sugar will be most rapid?
1) Sugar crystals in cold water.
2) Sugar crystals in hot water.
3) Powdered sugar in cold water.
4) Powdered sugar in hot water.
Answer:
4) Powdered sugar in hot water.

Question 2.
A beaker contains a solution of substance ‘A’. Precipitation of substance ‘A’ takes place when small amount of ‘A’ is added to the solution. The solution is _____
1) saturated
2) supersaturated
3) unsaturated
4) concentrated
Answer:
2) supersaturated

Question 3.
Maximum amount of,a solid solute that can be dissolved ¡n a specified amount of a given liquid solvent does not depend upon ___
1) Temperature
2) Nature of solute
3) Pressure
4) Nature of solvent
Answer:
3) Pressure

Question 4.
Concentrated aqueous sulphuric acid is 98% H₂SO₄ by mass and has a density of 1.80 g mL^-1. Volume of add acid required to make one litre of MH₂SO₄ solution is
1) 11.10 mL
2) 16.65 mL
3) 22.20 mL
4)5.55 mL
Answer:
4)5.55 mL

Question 5.
At equilibrium, the rate of dissolution of a solid solute in a volatile liquid solvent ____
Is _______
1) less than the rate of crystallisation
2) greater than the rate of crystallisation
3) equal to therate of crystallisation
4) zero
Answer:
3) equal to therate of crystallisation

Question 6.
4L of 0.02 M aqueous solution of NaCH was diluted by adding one litre of water. The molality of the resultant solution is
1) 0.004
2) 0.008
3) 0.012
4) 0.016
Answer:
4) 0.016

Question 7.
Which of the following is dependent on temperature?
1) Molarity
2) Mole fraction
3) Weight percentage
4) Molality
Answer:
1) Molarity

Question 8.
Which of the following units is useful in relating concentration of solution with its vapour pressure?
1) molefraction
2) partspermillion
3) mass percentage
4) molality
Answer:
1) molefraction

Question 9.
Colligative properties depend on ____
1) the nature of the solute particles dissolved in solution.
2) the number of solute particles in solution.
3) the physical properties of the solute particles dissolved in solution.
4) the nature of solvent particles.
Answer:
2) the number of solute particles in solution.

Question 10.
At a given temperature, osmotic pressure of a concentrated solution of a substance
1) is higher than that at a dilute solution.
2) is lower than that of a dilute solution.
3) is same as that of a dilute solution.
4) cannot be compared with osmotic pressure of dilute solution.
Answer:
1) is higher than that at a dilute solution.

Question 11.
Value of Henry’s constant K ____
1) increases with increase in temperature.
2) decreases wìth increase in temperature.
3) remains constant.
4) first increases then decreases
Answer:
1) increases with increase in temperature.

Question 12.
The value of Henry’s constant K_H is
1) greater for gases with higher solubility.
2) greater for gases with lower solubility.
3) constant for all gases.
4) not related to the solubility of gases.
Answer:
2) greater for gases with lower solubility.

Question 13.
A solution containing components A and B follows Raoult’s law, when
1) A-B attraction force is greater than A-A and B-B
2) A-B attraction force is less than A-A and B-B
3) A-B attraction force remains same as A-A and B-B
4) Volume of solution is different from sum of volumes of solute and solvent.
Answer:
3) A-B attraction force remains same as A-A and B-B

Question 14.
For an ideal solution, the correct option is
1) Δ_mix G=0 at constant T and P
2) Δ_mixS=0 at constant T and P
3) Δ_mixV#0 at constant T and P
4) Δ_mixH=0 at constant T and P
Answer:
4) Δ_mixH=0 at constant T and P

Question 15.
An ideal solution is formed when its components
1) have no volume change on mixing
2) have no enthalpy change on mixing
3) have both the above characteristics
4) have high solubility
Answer:
3) have both the above characteristics

Question 16.
A solution of acetone in ethanol
1) shows a negative deviation from Raoult’s law
2) shows a positive deviation from Raoult’s law
3) behaves like a near ideal solution
4) obeys Raoult’s law
Answer:
2) shows a positive deviation from Raoult’s law

Question 17.
Which of the following aqueous solutions should have the highest boiling point?
1) 1.0 M NaOH
2) 1.0 M Na₂SO₄
3) 1.0 M NH₄NO₃
4) 1.0 M KNO₃
Answer:
2) 1.0 M Na₂SO₄

Question 18.
Pure water can he obtained from sea water by
1) centrifugation
2) plasmolysis
3) reverse osmosis
4) sedimentation
Answer:
3) reverse osmosis

Question 19.
The unit of ebulioscopic constant is
1) K kg mol^-1 or K (molality)^-1
2) mol kg K^-1 or K^-1(molality)
3) kg mol^-1 K^-1 or K^-1(molality)^-1
4) K mol kg^-1 or K (molality)
Answer:
1) K kg mol^-1 or K (molality)^-1

Question 20.
In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl₂ solution is
1) the same
2) about twice
3) about three times
4) about six times
Answer:
3) about three times

Question 21.
Blood cells retain their normal shape in solutions which are
1) hypotonic to blood
2) isotonic to blood
3) hypertonic to blood
4) equinormal to blood
Answer:
2) isotonic to blood

Question 22.
An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because ____
1) it gains water due to osmosis.
2) it loses water due to reverse osmosis.
3) it gains water due to reverse osmosis.
4) it loses water due to osmosis.
Answer:
4) it loses water due to osmosis.

Question 23.
Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?
1) KCl
2) C₆H₁₂O₆
3) Al₂(SO₄)₃
4) K₂SO₄
Answer:
3) Al₂(SO₄)₃

Question 24.
In water saturated air, the mole fraction of water vapour is 0.02. If the total pressure of the saturated air is 1.2 atm, the partial pressure of dry air is
1) 1.18 atm
2) 1.76 atm
3) 1.176 atm
4) 0.98 atm
Answer:
3) 1.176 atm

Question 25.
The values of Van’t Hoff factors for KCl, NaCl and K₂SO₄, respectively, are ____ water by
1) 2, 2 and 2
2) 2, 2 and 3
3) 1, 1 and 2
4) 1, 1 and 1
Answer:
2) 2, 2 and 3

Regular practice with AP Inter 2nd Year Chemistry Study Material Chapter 7 Alcohols, Phenols and Ethers Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Chemistry 7th Lesson Alcohols, Phenols and Ethers Questions and Answers

I. Multiple Choice Questions

Question 1.
Glycerol is a:
1) dihydric alcohol
2) monohydric alcohol
3) hexahydric alcohol
4) trihydric alcohol
Answer:
4) trihydric alcohol
Count number of-OH groups in glycerol. It has three —OH groups attached to different carbons.Alcohol classification depends on number of —OH groups. Trihydric alcohol

Question 2.
Primary and secondary alcohols on action of red-hot copper give
1) aldehydes and ketones respectively
2) ketones and aldehydes respectively
3) only aldehydes
4) only ketones
Answer:
1) aldehydes and ketones respectively
Red-hot Cu causes dehydrogenation (removal of H₂).
Primary alcohol → aldehyde. Secondary alcohol → ketone. Because of different carbon environments. Aldehydes and ketones respectively

Question 3.
Which of the following can work as a dehydrating agent for alcohols?
1) Conc. H₂SO₄
2) Anhydrous Al₂O₃
3) H₃PO₄
4) All of these
Answer:
4) All of these
Dehydrating agents remove water. Cone. H₂SO₄, Al₂O₃, and H₃PO₄ all promote elimination of water. So all act as dehydrating agents.

Question 4.
Which of the following shows strong hydrogen bonding?
1) Ethyl amine
2) Ammonia
3) Ethyl alcohol
4) Diethyl ether
Answer:
3) Ethyl alcohol
Strong hydrogen bonding requires -OH group (more polar than -NH).
Alcohols form stronger H-bonds than amines/ethers.

Question 5.
What is the IUPAC name of the compound given below?

1) 5, 5-dimethyl-hexan-2-ol
2) 5,5,5-trimethyl-pentan-2-ol
3) 2,2-dimethyl-hexan-5-ol
4) 2,2-dimethyl-pentan-5-ol
Answer:
1) 5, 5-dimethyl-hexan-2-ol
Choose longest chain containing -OH. Number from end nearest to -OH.
Identify substituents correctly. Applying rules → hexane backbone, -OH at C-2, two methyl at C-5. 5,5-dimethyl-hexan-2-ol

Question 6.
Sodium phenoxide on reaction with carbon dioxide followed by acidification gives
1) benzoic acid
2) salicylic acid
3) phenylacetic acid
4) phthalicacid
Answer:
2) salicylic acid
Sodium phenoxide + CO₂ → Kolbe’s reaction.
CO₂ attaches at ortho position → after acidification → salicylic acid

Question 7.
What is the major product formed when ethanol is dehydrated with concentrated H₂SO₄ at 413K? .
1) Ethene
2) Methoxymethane
3) Methoxyethane
4) Ethoxyethane
Answer:
4) Ethoxyethane
Dehydration of alcohol depends on temperature: 413 K → ether formation (intermolecular) 443 K → alkene. So ethanol gives Ethoxyethane at 413 K .

Question 8.
The reaction between tert-Butyl chloride and sodium ethoxide gives
1) tert-butyl ethyl ether
2) tert-butyl methyl ether
3) 2-methylprop-1-ene
4) butene
Answer:
3) 2-methylprop-1-ene
tert-butyl chloride forms stable carbocation → elimination favored.
Strong base (ethoxide) → E₂ elimination → alkene formation Gives most substituted alkene. 2-methylprop-1 -ene

Question 9.
One molecule of dialkyl ether produces how many molecules of alkyl halides with excess of halogen acid?
1) 1
2) 2
3) 3
4) 4
Answer:
2) 2
Ether + excess HX → cleavage of both C-O bonds (2 molecules)

Question 10.
On heating aqueous solution of benzene dia/onium chloride, which of the following is formed?
1) benzene
2) chlorobenzene
3) phenol
4) aniline
Answer:
3) phenol
Benzene diazonium salt + water (heat) → hydrolysis.
\(\mathrm{N}_2^{+}\) replaced by -OH group. Forms phenol

Question 11.
When phenol is treated with excess bromine water it gives
1) m-bromophenol
2) o- and p-bromophenol
3) 2,4-dibromophenol
4) 2,4,6-tribromophenol
Answer:
4) 2,4,6-tribromophenol
Phenol is strongly activating (-OH group). Directs substitution to ortho & para positions.
Excess Br₂ → substitution at 2,4,6 positions. 2,4,6-tribromophenol

Question 12.
The compound obtained by the reaction of ethene with diborane followed by hydrolysis with alkaline H₂O₂ is
1) ethanol
2) propanol
3) ethanal
4) triethyl bromide
Answer:
1) ethanol
Reaction is hydroboration-oxidation: Alkene +. diborane → alcohol (anti-Markovnikov addition)
Ethene gives ethanol (-OH adds to less substituted carbon).

Question 13.
Which of the following is formed when phenol is exposed to air?
1) Benzoquinone
2) Benzyl quinone
3) Acetophenone
4) Benzene
Answer:
1) Benzoquinone
Phenol on exposure to air gets oxidized slowly. Forms quinone (benzoquinone).

Question 14.
The alcohol which does not react with Lucas reagent at room temperature is
1) isobutyl alcohol
2) n-butanol
3) tert-butyl alcohol
4) sec-butyl alcohol
Answer:
2) n-butanol
Lucas test: 3° alcohol → immediate reaction. 2° alcohol → slow. 1 ° alcohol → no reaction at room temp, n-butanol is primary alcohol.

Question 15.
Which statement from the following is true?
1) During dehydration reaction H₂O molecule is added to reactant.
2) Lucas test is used for detection of ethers.
3) During esterification OH^– from acid molecule and H⁺ from alcohol are removed.
4) During esterification H⁺ from acid molecule and OH^– from alcohol are removed.
Answer:
3) During esterification OH^– from acid molecule and H⁺ from alcohol are removed.
Esterification: Acid + Alcohol → Ester + Water. Water is formed by removal of: OH from acid.
H from alcohol. Check options carefully OH^– from acid and H⁺ from alcohol are removed

II. Fill in the Blanks

Question 1.
Isopropyl benzene is commonly called ________.
Answer:
cumene.

Question 2.
Boiling points of alcohols and phenols are higher in comparison to other classes of compounds due to __________.
Answer:
intermolecular hydrogen bonding

Question 3.
Alcohols and phenols react with carboxylic acids to form __________.
Answer:
esters

Question 4.
Lucas reagent is _________
Answer:
Cone. HCl/ZnCl₂ (anhydrous)

Question 5.
On treating phenol with chloroform in the presence of sodium hydroxide, salicylaldehyde is formed. This reaction is known as ___________.
Answer:
Reimer-Tiemann reaction

III. One Word Answer Questions

Question 1.
Phenol on heating with zinc dust gives a compound X. What is X?
Answer:
Phenol on heating with zinc dust gives Benzene (C₆H₆)

Question 2.
When an alkyl halide is allowed to react with sodium alkoxide, ether is formed. What is the name given to this reaction?
Answer:
Williamson synthesis: Alkyl halide + sodium alkoxide → Ether (R’-X + R-O-Na → R-O-R’ + NaX)

Question 3.
In the preparation of phenol from cumene, what is the byproduct obtained?
Answer:
The byproduct obtained in the preparation of phenol from cumene is Acetone (CH₃COCH₃)

Question 4.
What is the HJPAC name of glycerol?
Answer:
IUPAC name of glycerol is Propane-1,2,3-triol.

Question 5.
Which class of compounds are obtained when alkenes react with water in the presence of acid as catalyst?
Answer:
Alkenes when react with water in the presence of acid as catalyst gives Alcohols.

IV. Very Short Answer Questions

Question 1.
Explain why propanol has higher boiling point than that of the hydrocarbon butane.
Answer:
The boiling point of propanol is higher because of the presence of intermolecular hydrogen bonding in the molecules. But, it is not present in butane due to the absence of polar -OH group.
The boiling point (391K) of propanol is more than that of butane (309K).

Question 2.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable j molecular masses. Explain this fact.
Answer:
Alcohols can form H-bonds with water and break the H-bonds already existing between water molecules. Hence, they are soluble in water.
But hydrocarbons cannot form H-bonds with water and hence they are insoluble in water.

Question 3.
Give the reagents used for the preparation of phenol from chlorobenzene.
Answer:

Question 4.
Give the reagents used for the preparation of phenol from chlorobenzene.
Answer:
Phenol can be obtained by heating chlorobenzene in the presence of a catalyst, with 10% NaOH solution and HC1 at 350°C under 200-300 atm pressure.

Question 5.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:
Ethers cannot be prepared by the dehydration of secondary (or) tertiary alcohols, because alkenes are formed easily in these reactions. This happens due to steric hinderance.

Question 6.
Write the mechanism of the reactions of HI with methoxymethane.
Answer:
The ether molecule is initially protonated by the halogen acid (HI) to form protonated ether.

The protonated ether is then attacked by the halide ion (I^–) which acts as nucleophile as follows:

Question 7.
Name the reagents used in the following- reactions:
i) Oxidation of primary alcohol to carboxylic acid
ii) Oxidation of primary alcohol to aldehyde
Answer:
i) Acidified KMnO₄(or)acidified K₂Cr₂O₇ is used in the oxidation of primary alcohol to carboxylic acid
ii) CrO₃ (or) pyridine chlorochromate (pcc) is used as a reagent in the oxidation of primary alcohol to aldehyde.

Question 8.
Write the equations for the following reactions
i) Broniination of phenol to 2.4,6-trihromophcnil
ii) Benzl alcohol to benzoic acid
Answer:
i) Bromination of phenol to 246-rihron1ophenol:
Phenol when treated with excess of aqueous bromine solution forms 2,4,6-tribromophenol.

ii) Benzl alcohol to benzoic acid:
Benzyl alcohol oxidises into Benzaldehyde and then to Benzoic acid in the presence of alkaline KMnO₄.

Question 9.
Identify the reactant needed to form t-butyi alcohol from acetone.
Answer:
CH₃MgBr (Methyl magnesium bromide) is needed to form t-butyl alcohol from acetone.

Question 10.
Write the struclures for the foIIoshig compounds
1) Ethoxy Ethane
2) Ethoxybutane
3) Phenoy ethane
Answer:
1) Elbow Elhane : C₂H₅ — O — C₂H₅
2) Ethow butane : C₂H₅ — O — CH₂ — CH₂ — CH₂ — CH₃
3) Pheuoy etliane or Elhoy henzene or Ethyl phcnyl ether : C₆H₅ — O — C₂H₅

V. Short Answer Questions

Question 1.
Give the equations for the preparation of phenol from Cumene.
Answer:
Cumene is the common name for Isopropylbenzene .
Preparation: When Cumene is oxidised in the presence of air, it is converted into cumene hydroperoxide. When treated with dilute acid, it is then converted into phenol and acetone.

Question 2.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
Ethene is converted into alcohol by direct addition of water in the presence of phosphoric Acid (H₃PO₄).

Mechanism(S_N¹): The mechanism of the reaction involves the following three steps:
Step 1: Protonation of alkene to form carbocation by electrophilic attack of H₃O⁺

Step 2: Nucleophilic attack of water on carbocation.

Step 3: Deprotonation to form an alcohol

Question 3.
Explain the acidic nature of phenols and compare with that of alcohols.
Answer:
I) Acidic nature of phenols: Phenols are weakly acidic
a) Phenols turn blue litmus to red
b) Phenols react with sodium metal and NaOH to form H₂ and H₂O.

II) Acidic nature of phenols Vs alcohols:
Phenols are relatively more acidic than alcohols and also water.
a) The hydroxyl group in phenol is directly attached to SP² carbon on benzene ring. The carbon acts as an electron withdrawing group and Phenol exhibits resonance.
In these structures oxygen carries positive charge. This oxygen attracts electron pair of O-H bond strongly towards itself and facilitates the release of H⁺ ion and C₆H₅O^–.
The formed phenoxide ion is more stable than phenol.

Question 4.
Write the products formed by the reduction and oxidation of phenol.
Answer:
a) Reduction: Phenols on distillation with zinc dust undergo reduction to form benzene

b) Oxidation: Phenol undergo oxidation with chromic acid to produce conjugated diketone which is known as p-benzoquinone.

Question 5.
Write the products of the following reactions.

Answer:

Question 6.
Draw the structures of all isomeric alcohols of molecular formula C₅H₁₂O and give their IUPAC names and classify them as primary, secondary, and tertiary alcohols.
Answer:
The structures, IUPAC names and the classification of all isomeric alcohols of molecular formula C₅H₁₂O are eight and follows as

Question 7.
Ethanol reacts with H₂SO₄ at 443K forms ethene while at 413 K it forms Ethoxy Ethane. Explain the mechanism.
Answer:
I) Formation of Ethene at 443K:
Ethanol undergoes dehydration when it is heated it with concentrated H₂SO₄ at 443 K..

S<sub>N¹ Mechanism: Dehydration of ethanol involves the following steps.
Step 1: Formation of protonated alcohol (oxonium ion)

Step 2: Formation of carbocation: It is the slowest step and hence it has become the rate determining step of the reaction.

Step 3: Formation of ethene by elimination of proton</sub>

II) Formation of ethoxy ethane at 413 K:
Ethanol is dehydrated to ethoxyethane in the presence of sulphuric acid at 413 K.

S_{N² Mechanism: The formation of ether is a nucleophilic bimolecular reaction and it involves the attack of alcohol molecule on a protonated alcohol, as indicated below}

Question 8.
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Cke reason.
Answer:
The ortho and para isomers can be separated by steam distillation. The ortho nitro phenol is steam volatile due to intra molecular hydrogen bonding. The para-nìtrophenol is less volatile due to intermolecular hydrogen bonding which results in the association of molecules.
o- nitrophenol shows intra molecular hydrogen bonding. So, it is relatively more volatile than p-isomer.

p-nitrophenol shows inter molecular hydrogen bonding.

So, this is less volatile than 0-isomer.

Question 9.
Account for the statement: Alcohols boil at highest temperature than hydrocarbons and ethers of comparable molecular masses.
Answer:
Alcohols have intermolecular attractions due to intermolecular hydrogen bonds. This type of intermolecular hydrogen bonding do not exist in hydrocarbons and ethers. Hence alcohols boil at higher temperature than hydrocarbons and ethers of comparable molecular masses.

Question 10.
Explain why in anisole electrophilic substitution takes place at ortho and para positions and not at meta position.
Answer:
Anisole is the common name for Methoxy benzene
Structure of Anisole:

It is the example of alkyl-aryl ethers.
In anisole, Methoxy group (+M group) is electron releasing group and activates the benzene ring at ortho and para positions. Thus the aromatic ethers undergo electrophilic substitution in the benzene ring at the ortho and para positions.
Due to resonance electron density is more at ortho and para positions.
Hence Anisole exhibits electrophilic substitution at ortho and para positions.
Resonance structures of Anisole:

VII. Long Answer Questions

Question 1.
Illustrate hydroboration-oxidation reaction with a suitable example.
Answer:
Hydroboration-oxidation reaction: It is the addition of diborane to an alkene followed by
reaction with hydrogen peroxide (H₂O₂) in the presence of OH^– giving an alcohol.
Ex: Alkene reacts with Diborane [B₂H₆ = (BH₃)₂] to give trialkyl borane as addition product. This is oxidised to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.

Question 2.
Explain why-
i) ortho-nitrophenol is more acidic than ortho-methoxyphenol
ii) -OH group attached to benzene ring activates it towards electrophilic substitution.
Answer:
i) ortho-nitrophenol is more acidic than ortho- methoxyphenol
The structures of the two compounds are here

Nitro group(-NO₂)is an electron withdrawing group. So,nitrophenol is more aeidic than phenol. Methoxy group (-OCH₃) is an electron releasing group. So, methoxyphenol is less acidic than phenol. Thus, ortho nitrophenol is more acidic than ortho-methoxy phenol.

ii) The -OH group is an electron releasing group. It increases the electron density at ortho and para positions on the benzene ring. Hence it activates benzene ring towards electrophilic substitution reactions.

Question 3.
With a suitable example write equations for the following:
i) Kolbe’s reaction
ii) Reimer-Tiemann reaction
iii) Williamsons ether synthesis
iv) Oxidation of alcohol with PCC
Answer:
i) Kolbe’s reaction: When sodium salt of phenol is heated with carbon dioxide at 135°C and under 4-7 atm pressure, a carboxyl group, mainly in the ortho position is introduced.

ii) Riemer-Tiemann reaction: On treating phenol with chloroform in the presence of sodium hydroxide, a -CHO group is introduced at ortho position of benzene ring.

iii) Williamson’s synthesis:
Reaction of Alkyl halides with sodium alkoxide forms ethers.

Ex: When methyl iodide reacts with sodium Ethoxide it forms methoxyethane.
CH₃I + C₂H₅ONa ⟶ CH₃O – C₂H₅ + NaBr

iv) Oxidation of alcohol with PCC:
A better reagent for oxidation of primary alcohols to aldehydes in good yield is pyridinium chlorochromate (PCC), a complex of chromium trioxide with pyridine and HCl.

Question 4.
Explain why phenol with bromine water forms 2,4,6-tribromophenol, while on reaction with bromine in CS₂ at low temperatures forms para-bromophenol as the major product.
Answer:
1) Formation of 2,4,6-tribromophenol: Phenol when treated with excess of aqueous bromine solution yields 2,4,6 tribromopehnol. Because in water, phenol ionises to give phenoxide ion which is strongly activating and gives trisubstituted product.

2) Formation of para-bromophenol: However, ifbromination is carried out in a solvent of low dielectric constant, such as chloroform, carbontetrachloride (or) carbon disulphide, ortho and para bromophenols are formed.This is because in these solvents ionisation is less and ring is slightly activated. Hence monosubstituted product is formed.

Bromination of phenol takes place even in the absence of Lewisacid, such as FeBr₃.
It is due to the highly activating effect of the -OH group on the benzene ring.

Objective Questions

Question 1.
The functional group present in alcohols is:
1. -CHO
2. -OH
3. -COOH
4. -NH₂
Answer:
2. -OH

Question 2.
The general formula of phenol is:
1. ArOH
2. RCHO
3. RCOOH
4. RNH₂
Answer:
1. ArOH

Question 3.
Which of the following is a primary alcohol?
1. Propan-2-ol
2. Butan-2-ol
3. Ethanol
4. tert-Butanol
Answer:
3. Ethanol

Question 4.
Phenol reacts with NaOH because phenol is:
1. Neutral
2. Basic
3. Acidic
4. Amphoteric
Answer:
3. Acidic

Question 5.
Which reagent converts alcohol into alkyl chloride?
1. NaOH
2. SOCl₂
3. H₂SO₄
4. KMnO₄
Answer:
4. KMnO₄

Question 6.
Ethanol on oxidation gives:
1. Methanoic acid
2. Ethanoic acid
3. Propanoic acid
4. Acetone
Answer:
2. Ethanoic acid

Question 7.
Phenol reacts with Bromine in CS₂ at lower temperature to give
1. p-bromophenol(80%)
2. o-Bromophenol (80%)
3. Tribromophenol
4. di bromophenol
Answer:
1. p-bromophenol(80%)

Question 8.
Which alcohol gives turbidity immediately with Lucas reagent?
1. Primary alcohol
2. Secondary alcohol
3. Tertiary alcohol
4. Phenol
Answer:
3. Tertiary alcohol

Question 9.
Phenol gives violet colour with:
1. NaOH
2. FeCl₃
3. HCl
4. NaCl
Answer:
2. FeCl₃

Question 10.
Ether contains the functional group:
1. -OH
2. -CO-
3. -O-
4. -CHO
Answer:
3. -O-

Question 11.
The IUPAC name of CH₃-O-CH₃ is:
1. Methoxy methane
2. Ethoxy methane
3. Dimethyl ether
4. Methanol
Answer:
1. Methoxy methane

Question 12.
Phenol is more acidic than alcohol because:
1. Phenoxide ion is resonance stabilized
2. Alcohol is basic
3. Phenol is neutral
4. Alcohol forms hydrogen bonds
Answer:
1. Phenoxide ion is resonance stabilized

Question 13.
Which compound undergoes Kolbe reaction?
1. Ethanol
2. Phenol
3. Ether
4. Methanol
Answer:
2. Phenol

Question 14.
Which compound undergoes Reimer- Tiemann reaction?
1. Ether
2. Alcohol
3. Phenol
4. Alkene
Answer:
2. Alcohol

Question 15.
Diethyl ether on heating with excess HI gives:
1. Ethene
2. Ethanol
3. Ethyl iodide
4. Acetaldehyde
Answer:
3. Ethyl iodide

Question 16.
Alcohols show hydrogen bonding due to:
1. Presence of carbon
2. Presence of oxygen
3. Presence of hydrogen attached to oxygen
4. Double bond
Answer:
3. Presence of hydrogen attached to oxygen

Question 17.
Which has the highest boiling point?
1. Ether
2. Alkane
3. Alcohol
4. Alkene
Answer:
3. Alcohol

Question 18.
Which compound is commonly called carbolic acid?
1. Methanol
2. Ethanol
3. Phenol
4. Ether
Answer:
3. Phenol

Question 19.
Williamson synthesis is used for preparation of:
1. Alcohols
2. Phenols
3. Ethers
4. Aldehydes
Answer:
3. Ethers

Question 20.
Methanol is also called:
1. Wood spirit
2. Grain alcohol
3. Vinegar
4. Carbolic acid
Answer:
1. Wood spirit

Question 21.
Which compound is most acidic?
1. Ethanol
2. Water
3. Phenol
4. Methanol
Answer:
3. Phenol

Question 22.
Lucas reagent contains:
1. anhydrous ZnCl₂ +Conc.HCl
2. NaOH + HCl
3. FeCl₃ + HCl
4. H₂SO₄ + Zn
Answer:
1. anhydrous ZnCl₂ +Conc.HCl

Question 23.
Phenol reacts with bromine water to give:
1. Bromobenzene
2. Tribromophenol
3. Dibromophenol
4. Bromoethane
Answer:
2. Tribromophenol

Question 24.
The major product formed when ethanol is heated with cone. H₂SO₄ at 443 K is:
1. Ethane
2. Ethene
3. Ethanal
4. Ether
Answer:
2. Ethene

Question 25.
Which alcohol gives ketone on oxidation?
1. Primary alcohol
2. Secondary alcohol
3. Tertiary alcohol
4. Methanol
Answer:
2. Secondary alcohol

Question 26.
Which reagent is used to distinguish phenol from ethanol?
1. NaOH
2. FeCl₃
3. H₂SO₄
4. NaCl
Answer:
2. FeCl₃

Question 27.
Tertiary alcohols resist oxidation because:
1. They are acidic
2. They lack±-hydrogen
3. They are unstable
4. They are volatile
Answer:
2. They lack±-hydrogen

Question 28.
Anisole is:
1. Alcohol
2. Ether
3. Phenol
4. Ester
Answer:
2. Ether

Question 29.
The product formed when phenol reacts with dilute nitric acid is:
1. Picric acid
2. Nitrobenzene
3. o- and p-nitrophenol
4. Benzoic acid
Answer:
3. o- and p-nitrophenol

Question 30.
Which compound forms explosive peroxides on standing in air?
1. Alcohols
2. Phenols
3. Ethers
4. Aldehydes
Answer:
3. Ethers

Question 31.
Which alcohol is least soluble in water?
1. Methanol
2. Ethanol
3. Propanol
4. Butanol
Answer:
4. Butanol

Question 32.
Hvdroboration-oxidation of alkenes gives:
1. Alcohols
2. Ketones
3. Aldehydes
4. Phenols
Answer:
1. Alcohols

Question 33.
Sodium metal reacts with alcohol to liberate:
1. Oxygen
2. Nitrogen
3. Hydrogen
4. Chlorine
Answer:
3. Hydrogen

Question 34.
Phenol on distillation with zinc dust gives:
1. Cyclohexane
2. Benzene
3. Toluene
4. Benzoic acid
Answer:
2. Benzene

Question 35.
Which compound gives iodoform test?
1. Methanol
2. Ethanol
3. Phenol
4. Ether
Answer:
2. Ethanol

Question 36.
The ether formed by dehydration of ethanol at 413 K is:
1. Methoxy methane
2. Diethyl ether
3. Phenol
4. Ethanal
Answer:
2. Diethyl ether

Question 37.
Which compound does NOT react with NaOH?
1. Phenol
2. Ethanoic acid
3. Ethanol
4. Picric acid
Answer:
3. Ethanol

Question 38.
The hybridisation of oxygen atom in alcohol is:
1. sp
2. sp²
3. sp³
4. dsp²
Answer:
3. sp³

Question 39.
Which is the strongest acid?
1. Phenol
2. p-Nitrophenol
3. Ethanol
4. Methanol
Answer:
2. p-Nitrophenol

Question 40.
Cleavage of ethers by HI proceeds through:
1. Free radical substitution
2. Electrophilic substitution
3. Nucleophilic substitution
4. Addition reaction
Answer:
3. Nucleophilic substitution

Regular practice with AP Inter 2nd Year Chemistry Study Material Chapter 8 Aldehydes, Ketones and Carboxylic Acids Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Chemistry 8th Lesson Aldehydes, Ketones and Carboxylic Acids Questions and Answers

I. Mutliple Choice Questions

Question 1.
With which of the following reagents both ethanal and propanone react?
1) Tottens’
2) Setoff’s
3) Fehling
4) Grignard
Answer:
4) Grignard
Tollens ’, Fehling, Schiff ’ —»react mainly with aldehydes
Grignard reacts with both aldehydes & ketones (adds to C=0) Grignard reagent

Question 2.
Cannizzaro reaction is not given by
1) CH₃CHO
2) PhCHO
3) HCHO
4) (CH₃)₃C-CHO
Answer:
1) CH₃CHO
annizzaro reaction occurs only with aldehydes without a-hydrogen.
CH₃CHO has α-H does not give Cannizzaro

Question 3.
A new C—C bond is formed ¡n
1) Cannizzaro reaction
2) Rosenmund reduction
3) Clemmensen reduction
4) AIdol condensation
Answer:
4) AIdol condensation
New C-C bond formation happens when molecules combine.
Aldol condensation forms C-C bond between two aldehydes/ketones.

Question 4.
Fehling reagent oxidised an aliphatic aldehyde to carboxylic acid. The compound responsible for reddish brown precipitate is
1) CuO
2) Cu₂O
3) CU(OH)₂
4) (R-COO)₂Cu
Answer:
2) Cu₂O
Fehling’s solution contains Cu²⁺ ions. On reduction ⟶ CU₂O (reddish-brown ppt).

Question 5.
Clemmensen and Wolff-Kishner reductions are used to convert
1) R-Cl → R-H
2) R-CHO → R-CH₂OH
3) >C=O → >CH₂
4) R-COOH → R-CH₂OH
Answer:
3) >C=O → >CH₂
Clemmensen & Wolff-Kishner reduce carbonyl group
>C=O → – CH₂

Question 6.
Which of the following is a dicarboxvlic acid?
1) oxalic acid
2) benzoic acid
3) salicylic acid
4) picric acid
Answer:
1) oxalic acid
Dicarboxylic acid = 2 COOH groups.Oxalic acid = HOOC-COOH

Question 7.
CH₃CHO and C₆H₅CH₂-CHO can be distinguished chemically by
1) Benedict test
2) Iodoform test
3) Tollens’reagent
4) Fehling’s reagent
Answer:
2) Iodoform test
Iodoform test detects -COCH₃ group. CH₃CHO gives iodoform, C₅H₅CH₂CHO does not.

Question 8.
Conversion of CH₃COOH to CH₃COCI cannot be achieved by
1) SOCl₂
2) PCl₅
3) PCl₃
4) Cl₂/ red P
Answer:
4) Cl₂/ red P
Carboxylic acid → acid chloride requires chlorinating agent.
Cl₂ /red P forms PCl₃ in situ → converts COOH → COCl

Question 9.
Common reagent used to detect aldehydes and ketones is
1) Lucas reagent
2) 2, 4-DNP
3) Baeyer’s reagent
4) Tollens’ reagent
Answer:
2) 2, 4-DNP
2,4-DNP reacts with carbonyl group (C=O).Gives yellow/orange ppt → common test.

Question 10.
Benzaldehyde can be prepared from toluene by reacting with
1) CrO₂Cl₂ / CS₂
2) KMnO₄ / KOH
3) Pd / BaSO₄
4) CO+HCl/AlCl₃
Answer:
1) CrO₂Cl₂ / CS₂
Controlled oxidation of toluene gives benzaldehyde.
Chromyl chloride (CrO₂Cl₂) → Etard reaction.

Question 11.
Choose the weakest acid among the following
1) FCH₂COOH
2) Cl₂CCOOH
3) CH₂COOH
4) CH₃CH₂COOH
Answer:
4) CH₃CH₂COOH
Electron-donating groups decrease acidity. Alkyl group (+I effect) → weakest acid.
Propionic acid weakest here. CH₃CH₂COOH

Question 12.
Which one of the following is not correctly matched?
1) Formic acid – methanoic acid
2) acetic acid – ethanoic acid
3) malonic acid – propanedioic acid
4) adipic acid – ethanedioic acid
Answer:
4) adipic acid – ethanedioic acid
Adipic acid formula = hexanedioic acid, not ethanedioic. So mismatch.
adipic acid – ethanedioic acid

Question 13.
NaHSO₃ forms an adduct with all compounds except
1) CH₃CHO
2) H₃CCOCH₃
3) PhCHO
4) glucose
Answer:
4) glucose
NaHSO₃ forms addition compounds with aldehydes/ketones.
Glucose is not typical carbonyl (exists mainly cyclic).

Question 14.
Which one of the following is more reactive towards nucleophilic addition?
1) CH₃CHO
2) CH₃CH₂CHO
3) CH₃COCH₃
4) H₃CCH₂COCH₃
Answer:
1) CH₃CHO
Reactivity in nucleophilic addition depends on: +I effect (alkyl groups) →decreases reactivity
Steric hindrance → decreases reactivity. Order: Aldehyde > Ketone, and smaller
alkyl groups → more reactive. CH₃CHO has: only one alkyl group, less steric hindrance

Question 15.
The correct trend of boiling point is
1) CH₃COOH > CH₃COCH₃ > CH₃CH₂OH > CH₃CH₂Cl
2) CH₃CH₂OH > CH₃COOH > CH₃COCH₃ > CH₃CH₂Cl
3) CH₃CH₂OH > CH₃COOH > H₃CCH₂Cl > CH₃COCH₃
4) CH₃COOH > CH₃CH₂OH > H₃CCOCH₃ > CH₃CH₂Cl
Answer:
4) CH₃COOH > CH₃CH₂OH > H₃CCOCH₃ > CH₃CH₂Cl
Boiling point depends on intermolecular forces:Carboxylic acid → strongest (H-bonded dimers)
Alcohol → H-bonding. Ketone → dipole-dipole.
Alkyl halide → weak van der Waals
So order: Carboxylic acid > Alcohol > Ketone > Alkyl halide
Acid > Alcohol > Ketone > Alkyl chloride

II. Fill in the Blanks

Question 1.
IUPAC name of formaldehyde is _________
Answer:
methanal (HCHO)

Question 2.
Chromium trioxide (CrO₃) in acidic (H₂SO₄) media is called _______ reagent
Answer:
Jones

Question 3.
Tollens’ reagent is ________
Answer:
freshly prepared ammonical silver nitrate solution

Question 4.
Carboxylic acids having an α-hydrogen are halogenated at the α-position on treatment with chlorine or bromine in presence of small amount of phosphorus to give α- halocarboxylic acids. This reaction is known as _________
Answer:
Hell-Volhard-Zellnskv reaction (HVZ reaction)

Question 5.
IUPAC name of Mesityl oxide is ___________
Answer:
4-methvlnent-3-en-2-one

III. One Word Answer Questions

Question 1.
What is name of the product formed when benzene is reacted with CO and HCl in the presence of anhydrous AlCl₃? .
Answer:
Benzaldehyde is obtained when benzene is reacted with CO and HCl.

Question 2.
What is Rochelle salt?
Answer:
Rochelle salt: Sodium potassium tartarate (KNaC₄H₄O₆.4H₂O) .
( A double salt of Tartaric acid)

Question 3.
Which acid is used as solvent in food industry as vinegar?
Answer:
Vinegar: Ethanoic acid (CH₃COOH or acetic acid>

Question 4.
Aldehydes and ketones undergo what type of addition reactions?
Answer:
Aldehydes and ketones undergo Nucleophilic addition reactions.

Question 5.
Phthalic acid on reaction with ammonia followed bv strong heating gives a compound ‘X’ along with elimination of NH₃. What is the name of that compound ‘X’?
Answer:
phthalimide

IV. Very Short Answer Questions

Question 1.
Arrange the following compounds in increasing order of their property indicated.
i) acetaldehyde, acetone and t-butylmethyl ketone reactivity towards HCN
ii) fluoroacetic acid, monochloroacetic acid, acetic acid and dichloroacetic acid (acid strength)
Answer:
i) Reactivity towards HCN: t-butylmethyl ketone < acetone < acetaldehyde.
ii) Acid strength: dichloroacetic acid > fluoroaceticacid > monochloroacetic acid > acetic acid.

Question 2.
Write the reaction showing a-halogenation of carboxylic acid and give its name.
Answer:
HVZ Reaction: Carboxylic acids having an a-hydrogen are halogenated at the a-position on treatment with chlorine (or) bromine in the presence of small amount of red phosphorus to give a-halo carboxylic acids. The reaction is known as Hell-Volhard-Zelinsky reaction.
Reaction of a-halogenation of carboxylic acid (HVZ reaction):

Question 3.
Although the phenoxide ion has a greater number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Answer:
A carboxylic acid is a stronger acid than phenol due to the following reasons:
a) The conjugate base of carboxylic acid is carboxylate ion (RCOO^–), which is stabilised by two equivalent resonance structures in which the negative charge is at the more electronegative oxygen atom. But, the conjugate base of phenol, is phenoxide ion, has nonequivalent resonance structures in which the negative charge is at the less electro negative carbon atom. Hence, resonance in phenoxide ion is not as important as it is in carboxylate ion.

b) In a carboxylate ion, the negative charge is delocalised over two electro negative oxygen atoms, whereas it is less effectively delocalised on less electronegative carbon atoms in the phenoxide ion. So, a carboxylic acid is a stronger acid than a phenol.

Question 4.
How do you distinguish acetophenone and benzophenone?
Answer:
Acetophenone undergoes iodoform reaction due to the presence of acetyl group CH₃-CO .
But benzophenone does not undergo iodoform reaction due to the absence of CH₃-CO group

Question 5.
Explain the position of electrophilic substitution in benzoic acid.
Answer:
COOH group on benzene, deactivates the ring at ortho and para positions.
Hence substitution takes place at meta position.

Question 6.
Write equations showing the conversion of
i) acetic acid to acetyl chloride
ii) benzoic acid to benzamide
Answer:
i) acetic acid to acetyl chloride:
3CH₃COOH + PCl₃ ⟶ 3CH₃C0Cl + H₃PO₃
CH₃COOH + PCl₅ ⟶ CH₃COCl + POCl₃ + HCl
CH₃COOH + SOCl₂ ⟶ CH₃COCl + SO₂ + HCl

ii) benzoic acid to benzamide

Question 7.
An organic acid with molecular formula C₈H₈O₂ on decarhoylation forms Toluene. Identify the organic acid.
Answer:
The organic acid formed is phenyl acetic acid (2-Phenyl ethanoic acid)

Question 8.
List the reagents suitable to reduce carhoxy lic acid to alcohol.
Answer:
LiAlH₄ (or) copper chromate are the needed reducing reagents.

Question 9.
Write the mechanism of esterification.
Answer:

Question 10.
Compare the acidic strength of acetic acid, Chloroacetic acid, benzoic acid and Phenol.
Answer:
The relative values of k_a and pk_a indicate the acidic strength:
Higher k_a value and lower pk_a value indicates that the acid is strong.
Order of acidic strength: Chloroacetic acid > benzoic acid > acetic acid > phenol
The acidic strength of acetic acid, chloroactic acid, benzoic acid and phenol:

V. Short Answer Questions

Question 1.
Write the equations of any aldehyde with Fehling’s reagent.
Answer:
Fehling’s solution is an alkaline solution of copper sulphate containing Rochelle salt(sodium potassium tartrate) .
Fehling’s test: When Fehling’s solution is heated with an aldehyde a reddish brown precipitate of (Cu₂O) is formed.
RCHO + 2Cu⁺² + 5OH^– ⟶ RCOO^– + Cu₂O + 3H₂O

Question 2.
What is Tollen’s reagent? Explain its reaction with Aldehydes.
Answer:
1) Tollen’s reagent: Ammonical silver nitrate solution is called Tollen’s reagent.
Its formula is [Ag(NH₃)₂]⁺OH^–

2) Reaction of toulene with aldehydes ( Silver mirror test):
When Tollen’s reagent is warmed with an aldehyde, it gets reduced to metallic silver.
This metallic silver gets deposited on the inner wall of the test tube to form a silver mirror.
RCHO + 2 [Ag(NH₃)₂]⁺ + 3OH^– ⟶ RCOO^–Silver mirror + 2Ag ↓ + 4NH₃ + 2H₂O

Question 3.
Explain why Aldehydes and ketones undergo nucleophilic addition while alkenes undergoes electrophilic addition though both are unsaturated compounds.
Answer:
1) Nucleophilic addition: Aldehydes and ketones undergo nucleophilic addition because the intermediate anion formed by the attack of nucleophile on carbonyl group is more stable than the cation formed by attack of electrophile on carbonyl group.

2) Electrophilic addition: Alkenes undergoes electrophilic addition beacuse in alkenes there is double bond between two carbon atoms and is electron rich. This double bond is highly reactive. So it is attacked by electrophilic first. So alkenes undergo electrophilic addition reactions.

Question 4.
Arrange the following in the increasing order of their acidic strength:
benzoic acid, 4-methoxybenzoic acid, 4-nitrobenzoic acid and 4-nethylbenzoic acid.
Answer:
Acidic strength of electron with drawing -NO₂ group increases
Acidic strength of electron with releasing -OCH₃, -CH₃ groups decreases
But releasing power of -OCH₃ is more than that of -CH₃.
Lower Pka values indicate higher acidity.
Hence the increasing order (low to high) acidic strengths are given below:
4-methoxy benzoic acid < 4-methyl benzoic acid < benzoic acid < 4-nitrobenzoic acid.
P^ka values: (4.46) < (4.36) < (4.19) < (3.41)

Question 5.
Describe the following:
i) Crossed aldol condensation
ii) Decarboxylation
Answer:
i) Crossed aldol condensation: It is the condensation of two different carbonyl compounds (one of which must have one a-hydrogen) in the presence of alkali. It is the condensation of an aldehyde with a ketone in the presence of an alkali

ii) Decarboxylation: It removes a carboxyl group and releases CO₂.
Heating of anhydrous potassium salt of carboxylic acid with sodalime (CaO + NaOH) forms Alkane with the liberation of CO₂. The formed alkane contains one carbon less than that of parent acid.

Question 6.
Write the oxidation products of acetaldehyde, acetone and acetophenone.
Answer:

Question 7.
Write the IUPAC names of the following
(i) CH₃CH₂CH(Br)CH₂COOH
(ii) PhCH₂COCH₂COOH
(iii) CH₃CH(CH₃)CH₂COOC₂H₅
Answer:
i) CH₃CH₂CH(Br)CH₂COOH: 3-Bromol-pentanoicacid
ii) PhCH₂COCH₂COOH: 4-Phenyl-3-oxo butanoic acid
iii) CH₃CH(CH₃)CH₂COOC₂H₅: Ethyl-3-Methyl butanote

Question 8.
Explain the role of electron withdrawing and electron releasing groups on the acidity of carboxylic acids.
Answer:
1) The role of electron withdrawing group on the acidity of carboxylic acids:
The electron withdrawing substituents tend to withdraw electrons away from the carboxyl carbon. This favours declocalisation of the negative charge. Delocalisation of negative charge stabilizes the carboxylate anions and makes the release of H atom as H⁺ easier. Hence, the presence of an electron withdrawing substituent increases the acid strength of the acid.

2) The role of electron releasing groups on the acidity of carboxylic acids:
Alkyl groups, such as -CH₃, -C₂H₅ etc., are electron releasing groups. The presence of an electron releasing group tends to increase the negative charge on the oxygen atom of the anion. This localisation of negative charge on the oxygen of the carboxylic group destabilizes the anion. Due to the increased electrostatic effects the release of H atom of the —COOH group as a proton (H⁺) becomes more difficult. As a result, the strength of the acid decreases.

Question 9.
Draw the Structures of the following derivatives:
i) Acetaldehydedimethylacetal
ii) The ethylene ketal of hexan-3-one
iii) The methyl hemiacetal of formaldehyde
Answer:

Question 10.
An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It doesn’t reduce Tollens’ reagent but forms sodium hydrogen sulphite adduct and gives positive iodoform test. On vigorous oxidation forms ethanoic and propanoic acids. Write the possible structure of the compound.
Answer:
Determination of molecule formulae:

Empirical formula = C₅H₁₀O
Empirical formula mass = (5 × 12) + (10 × 1) + (1 × 16) = 86 Given molecular mass) = 86
n = \(\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{86}{86}\) = 1
∴ Molecular formula of the compound = C₅H₁₀O
The compound does not reduce Fehling’s solution, but forms a bisulphite addition compound.
So, the compound is a ketone. This ketone compound gives idoform test.
So, the ketone is a methyl ketone. Then the possible compound is CH₃-CO-C₃H₇.
This compound on oxidation gives mixture of ethanoic acid and propanoic acid.
Therefore the possible structure is

VI. Long Answer Questions

Question 1.
Explain the following terms. Give an example of the reaction in each case.
i) Cyanohydrin
ii) Acetal
iii) Semicarbazone
iv) Aldol
v) Hemiacetal
vi) Oxime
Answer:
i) Cyanohydrin: Cyanohydrins are very useful compounds for organic synthesis. Hydrogen cyanide (HCN) adds to aldehydes and ketones to form cyanohydrin

ii) Acetal: Aldehydes react with alcohols in the presence of dry hydrogen chloride to form gem- dialkoxy compounds which are known as acetals.

iii) Semicarbozone: Aldehydes and ketones react with semicarbazide (NH₂NHCONN₂) to form semicarbazone.

iv) Aldol: The β-hydroxy aldehydes (or) β-hydroxy ketones are called aldol.
Aldol reaction: Aldehydes and ketones having at least one α-hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form β-hydroxy aldehyde (aldol) or β-hydroxy ketones (ketol) respectively.

v) Hemi Acetal: When one molecule of alcohol is added to one molecule of aldehyde, it forms a hemiacetal. It is an unstable compound. It contains functional groups of both alcohol and ether.

Hemiacetal reacts with one more alcohol molecule to form an acetal.

vi) Oxime: Aldehydes and ketones reacts with hydroxylamine to form oximes.

Question 2.
how do you distinguish the following pairs of compounds?
i) Propanal and propanone
ii) Acetophvnone and benzophennne
iii) Phenol and benìok acid
iv) Pentan-2-one and Pentan-3-one
Answer:
i) Propanal Vs propanone: Propanal(CH₃CH₂CH₂OH) does not undergo idoform reaction, where as propanone (CH₃—CO—CH₃) gives idoform when warmed with iodine in the presence of alkali.

ii) Acetophenone Vs benzopheiionc: Acetophenone is a methyl ketone (C₆H₅COCH₃) while benzophenone (C₆H₅COC₃₆H₅) is a diphenyl ketone. So acetophenone undergoes Idoform reaction. But Bezophenone does not undergo iodoform reaction.

iii) Phenol Vs benzoic acid:
a) Phenol gives coupling reactions with diazonium compounds, while benzoic acid does not. Phenol gives azodyes when reacted with benzenediazonium chloride (C₅H₅N₂Cl) Benzoic acid does not give any reaction with diazonium salts.

b) With neutral FeCl₃ phenols give reddish colour,, while benzoic acid gives buff-coloured precipitate.

iv) Pentan-2-one and Pentan-3-one: Pentan-2-one is a methyl ketone. So, it undergoes idoform test. So, when heated with iodine and alkali pentan-2-one, it gives idoform.
But pentan-3-one does not give this test.

Question 3.
Describe the following:
i) Acetylation
ii) Cannizzaro reaction
iii) Cross aldol condensation
iv) Decarboxylation
Answer:
i) Acety lation: Alcohols when treated with an acyl chloride (or) acid anhydride in the presence of pyridine give esters. In this reaction, H of the -OH group of alcohol is replaced by acyl (CH₃-CO-) group. So, this reaction is called acylation.

ii) Cannizzaro reaction:Aldehydes having no ‘a’ hydrogen undergo Cannizzaro’s reaction. Such aldehydes in the presence of concentrated alkaline solution undergo self oxydation-reduction to give a mixture of alcohol and a salt of carboxylic acid.
Ex: 1) 2 molecules of formaldehyde undergo self oxydation in presence of conc.NaOH to form methanol and Sodium formate.

2) Benzaldehyde gives benzyl alcohol and sodium benzoate.

iii) Cross aldol condensation: The condensation of two different carbonyl compounds (one of which must have one a-hydrogen) in the presence of alkali is called cross aldol condensation.

iv) Decarboxylation: It removes a carboxyl group and releases CO₂. Heating of anhydrous potassium salt of carboxylic acid with sodalime (CaO+NaOH) forms Alkane with the liberation of C02. The formed alkane contains one carbon less than that of parent acid.

Question 4.
Complete each synthesis by giving the missing starting material, reagent or product.

Answer:

Question 5.
Explain the following reactions
a) Aldol reaction
b) Esterification
c) H.V.Z reaction
d) Broniinaition of Benzoic acid
Answer:
a) Aldol reaction : Aldehydes and ketones having at least one α-hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form β-hydroxy aldehyde.
This is also called Aldol condensation.

b) Esterification: Carboxylic acid reacts with alcohol in the presence of acidic medium (HCl, H₂SO₄) to give Ester.

c) H.V.Z Reaction: Carboxylic acids having an a-hydrogen are halogenated at the a-position on treatment with chlorine (or) bromine in the presence of small amount of red phosphorus to give a-halo carboxylic acids.

d) Bromination of Benzoic acid:
Bromination of Benzoic acid is an electrophilic aromatic substitution (EAS).
Bromine reacts with FeBr₃ to form a strong electrophile.The aromatic ring of benzoic acid attacks Br⁺ at the meta position, forming a sigma complex (arenium ion).
Loss of H⁺ restores aromaticity and gives the final product meta-bromobenzoic acid.

Objective Questions

Question 1.
Give IUPAC name of the compound given below.

1) 2-Chloro-5-hydroxyhexane
2) 2-Hydroxy-5-chlorohexane
3) 5-ChIorohexan-2-ol
4) 2-Chlorohexan-5-ol
Answer:
3) 5-ChIorohexan-2-ol

Question 2.
IUPAC name of
is
1) 1 -methoxy-1 -methylethane
2) 2-methoxy-2-methylethane
3) 2-methoxypropane
4) isopropylmethyl ether
Answer:
3) 2-methoxypropane

Question 3.
IUPAC name of m-cresol is
1) 3-methylphenol
2) 3-chlorophenol
3) 3-methoxyphenol
4) benzene-1,3-diol
Answer:
1) 3-methylphenol

Question 4.
How many alcohols with molecular formula C₄H₁₀O are chiral in nature?
1) 1
2) 2
3) 3
4)4
Answer:
1) 1

Question 5.
What is the correct order of reactivity of alcohols in the following reaction?

1) 1° > 2° > 3°
2) 1° < 2° > 3°
3) 3° > 2° > 1°
4) 3° > 1° > 2°
Answer:
3) 3° > 2° > 1°

Question 6.
The process of converting alkyl halides into alcohols involves ________
1) addition reaction
2) substitution reaction
3) dehydrohalogenation reaction
4) rearrangement reaction
Answer:
2) substitution reaction

Question 7.
Which of the following compounds will react with sodium hydroxide solution in water?
1) C₆H₅OH
2) C₆H₅CH₂OH
3) (CH₃)₃COH
4) C₂H₅OH
Answer:
1) C₆H₅OH

Question 8.
In the following sequence of reactions.

the end product is
1) acetone
2) methane
3) acetaldehyde
4) ethylalcohol
Answer:
4) ethylalcohol

Question 9.
Consider the following reaction,

The product Z is
1) toluene
2) benzaldehyde
3) benzoic acid
4) benzene
Answer:
3) benzoic acid

Question 10.
Monochlorination of toluene in sunlight followed by hydrolysis with aq. NaOH yields.
1) o-Cresol
2) m-Cresol
3) 2, 4-Dihydroxytoluene
4) Benzyl alcohol
Answer:
4) Benzyl alcohol

Question 11.
CH₃CH₂OH can be converted into CH₃CHO by _________
1) catalytic hydrogenation
2) treatment with LiAlH₄
3) treatment with pyridinium chlorochromate
4) treatment with KMnO₄
Answer:
3) treatment with pyridinium chlorochromate

Question 12.
The correct order of increasing acidic strength is ________
1) Phenol < Ethanol < Chloroacetic acid < Acetic acid
2) Ethanol < Phenol < Chloroacetic acid < Acetic acid
3) Ethanol < Phenol < Acetic acid < Chloroacetic acid
4) Chloroacetic acid < Acetic acid < Phenol < Ethanol
Answer:
3) Ethanol < Phenol < Acetic acid < Chloroacetic acid

Question 13.
Which of the following is most acidic?
1) Benzyl alcohol
2) Cyclohexanol
3) Phenol
4) m-Chlorophenol
Answer:
4) m-Chlorophenol

Question 14.
Which of the following compounds is aromatic alcohol?

1) A, B, C, D
2) A, D
3) B, C
4) A
Answer:
3) B, C

Question 15.
Compound can be prepared by the reaction of _________
1) Phenol and benzoic acid in the presence of NaOH
2) Phenol and benzoyl chloride in the presence of pyridine
3) Phenol and benzoyl chloride in the presence of ZnC₂
4) Phenol and benzaldehyde in the presence of palladium.
Answer:
2) Phenol and benzoyl chloride in the presence of pyridine

Question 16.
Among the following sets of reactants which one produces anisole?
1) CH₃CHO, RMgX
2) C₆H₅OH, NaOH, CH₃I
3) C₆H₅OH, neutral FeCl₂
4) C₆H₅-CH₃, CH₃COCl, AlCl₃
Answer:
2) C₆H₅OH, NaOH, CH₃I

Question 17.
The reagent which does not react with both, acetone and benzaldehyde.
1) Sodium hydrogensulphite
2) Phenyl hydrazine
3) Fehling’s solution
4) Grignard reagent
Answer:
3) Fehling’s solution

Question 18.
Which of the following compounds will give butanone on oxidation with alkaline KMnO₄ solution?
1) Butan-1 -ol
2) Butan-2-ol
3) Both of these
4) None of these
Answer:
2) Butan-2-ol

Question 19.
Phenol is less acidic than _______
1) ethanol
2) o-nitrophenol
3) o-methylphenol
4) o-methoxyphenol
Answer:
2) o-nitrophenol

Question 20.
Cannizaro’s reaction is not given by _______

3) H CHO
4) CH₃CHO
Answer:
4) CH₃CHO

Question 21.
In Ciemmensen Reduction carbonyl compound is treated with _________
1) Zinc amalgam + HCl
2) Sodium amalgam + HCl
3) Zinc amalgam + nitric acid
4) Sodium amalgam + HNO₃
Answer:
1) Zinc amalgam + HCl

Question 22.
Ciemmensen reduction of a ketone is carried out in the presence of which of the following?
1) Zn-Hg with HCl
2) LiAlH₄
3) H and Pt as catalyst
4) Glycol with KOH
Answer:
1) Zn-Hg with HCl

Question 23.
Reduction of aldehydes and ketones into hydrocarbons using zinc amalgam and conc.HCl is called
1) Ciemmensen reduction
2) Cope reduction
3) Dow reduction
4) Wolff-Kishner reduction
Answer:
1) Ciemmensen reduction

Question 24.
Which one of the following on treatment with 50% aqueous solution hydroxide yields the corresponding alcohol and acid?
1) C₆H₅CH₂CHO
2) C₆H₅CHO
3) CH₃CH₂CH₂CHO

Answer:
2) C₆H₅CHO

Question 25.
Iodoform test is not given by
1) 2-pentanone
2) ethanol
3) ethanal
4) 3-pentanone
Answer:
4) 3-pentanone

Reviewing AP Inter 2nd Year Economics Study Material Chapter 10 Economy of Andhra Pradesh Questions and Answers can help students prepare confidently for exams.

AP Inter 2nd Year Economics 10th Lesson Economy of Andhra Pradesh Questions and Answers

Very Short Answer Questions

Question 1.
Explain P4 approach initiated by Andhra Pradesh.
Answer:
1) The P4 approach (Public-Private-People Partnership) is a flagship initiative of Andhra Pradesh aimed at achieving “Zero Poverty by 2029”.

2) It involves a mentorship model where each poor family (Bangaru Kutumbam) is adopted by a mentor (Margadarsi) who provides guidance and support. The government acts as a facilitator, ensures access to services and schemes, and uses real-time monitoring, while encouraging community participation to create jobs and reduce poverty.

Question 2.
Briefly mention the export potential of Andhra Pradesh.
Answer:

1. Andhra Pradesh has strong export potential across agriculture, marine, industrial, and services sectors. It is the sixth largest exporting state, contributing about 4.5% of India’s exports.
2. The state exports rice (25% of India’s exports), marine products, chillies, tobacco, pharmaceuticals and software services, with a focus on improving quality and becoming a global logistics hub under Swarna Andhra Vision 2047.

Question 3.
Which are the backward districts in AP eligible for Central assistance as per the Reorganisation Act?
Answer:
As per the Andhra Pradesh Reorganisation Act, 2014, seven backward districts are identified for Central assistance:

1. North Coastal region: Srikakulam, Vizianagaram, Visakhapatnam
2. Rayalaseema region: Kurnool, Anantapur, YSR Kadapa, Chittoor
   These districts receive Central assistance under a 90:10 (Centre: State) cost-sharing pattern.

Question 4.
Mention any two schemes of school education in A.P.
Answer:
Any two school education schemes in Andhra Pradesh are:

1. Talliki Vandanam – Provides financial assistance of ₹ 15,000 per year to support students’ education and development.
2. Mana Badi – Mana Bhavishyathu – Focuses on improving school infrastructure and providing better learning facilities.

Question 5.
Comment on the Global Capability Centres (GCC) in Andhra Pradesh.
Answer:

1. Global Capability Centres (GCCs) are offshore units of multinational companies providing IT, business, analytics and R&D services. Andhra Pradesh promotes GCCs through its IT & GCC Policy (2024-29) to make the state a major hub.
2. Cities like Visakhapatnam and Vijayawada are emerging GCC hubs, which help in job creation, skill development and attracting investment, strengthening the state’s knowledge economy.

Question 6.
Mention any two central institutions established in AP after state reorganisation, with their locations.
Answer:
Two central institutions established after the Andhra Pradesh Reorganisation Act, 2014 are:

1. Indian Institute of Technology (IIT), Tirupati
2. Indian Institute of Management (IIM), Visakhapatnam

Short Answer Questions

Question 1.
Write about social welfare schemes of A.P.
Answer:
The Government of Andhra Pradesh is implementing several welfare schemes under the “Super Six”, along with other initiatives aimed at poverty alleviation and social support.

1) “Super Six” Schemes: These schemes provide direct financial and material assistance:

• Talliki Vandanam – ₹ 15,000 per year to support children’s education (Classes I – XII).
• Monthly Financial Aid for Women -₹ 1,500 per month to eligible women (19-59 years).
• Annadata Sukhibhava – ₹ 20,000 per year to farmers for agricultural expenses.
• Unemployment Allowance – Short Answer Questions3,000 per month for educated youth with job creation goal.
• Deepam 2.0 – Three free LPG gas cylinders annually.
• Free Bus Travel for Women – Free travel in APSRTC buses.

2) Other Key Welfare Initiatives

• NTR Bharosa Pension – ₹ 4,000 monthly support to vulnerable groups.
• Anna Canteens – Meals at ? 5 for the poor.
• Dr. NTR Vaidya Seva – Cashless healthcare for about 1.43 crore families.
• Dokka Seethamma Mid-Day Meal & School Kits – Improve nutrition and education.
• PMAY, Jal Jeevan Mission, Saubhagya- Housing, drinking water, and electrification.
• P4 Approach – Targets “Zero Poverty by 2029” through mentorship support.

Question 2.
Explain the impact of AP reorganisation on the resource endowment in Andhra Pradesh.
Answer:
The reorganisation of Andhra Pradesh in 2014 significantly altered the state’s economic landscape and resource endowment.

1. Overall Resource Share: After bifurcation, Andhra Pradesh inherited only 46% of the total resources of the combined state.
2. Forestry and Biodiversity: Forests became fragmented, as Telangana retained richer forest cover, reducing forest revenue, jobs, and eco-tourism potential.
3. Energy and Minerals: Loss of Singareni coalfields affected energy security. Though some minerals like barytes and uranium remained, key resources like limestone, granite, and iron ore were lost, reducing revenue and employment.
4. Water Resources: Godavari and Krishna rivers became interstate rivers, causing water disputes and delays in irrigation projects.
5. Economic Infrastructure: Loss of Hyderabad created a revenue gap and fiscal strain, forcing the state to rebuild institutions and adopt new development strategies.

Question 3.
Examine the Govt, initiatives for agricultural sector development in Andhra Pradesh.
Answer:
The Government of Andhra Pradesh has adopted a multi-faceted approach to develop agriculture, focusing on sustainability, financial security, and infrastructure modernization.

1) Sustainable and Natural Farming: Through APCNF & ZBNF, the state promotes natural farming to reduce input costs, improve climate resilience, and reduce dependence on chemicals. Support includes interest-free loans, subsidized seeds, and tools.

2) Financial and Legal Support: Schemes like Annadata Sukhibhava (₹ 20,000) and PM-Kisan (₹ 6,000) provide financial aid. Crop Cultivator Rights Cards (CCRCs) help tenant farmers access loans.

3) Institutional and Infrastructure Support: Rythu Seva Kendralu (RSKs) provide inputs and guidance, while farm mechanization, cold storage, and irrigation projects (Polavaram) improve productivity and reduce losses.

4) Market and Future Strategy: Market Intervention Schemes (MIS) ensure fair prices, and focus on seed quality and Agri-Tech (Al, loT, blockchain) under Swarna Andhra Vision 2047 supports long-term agricultural growth.

Question 4.
List out any four approaches and initiatives of the AP Industrial Policy.
Answer:
The Andhra Pradesh Industrial Development Policy (4.0) 2024-29 outlines key approaches to make the state a globally competitive industrial hub.

1. Prioritizing High-Growth Industries: Focus on sectors such as electronics, pharmaceuticals, food processing, green energy, semiconductors, and aerospace to drive industrial growth.
2. Infrastructure and Logistics: Provision of land consolidation, affordable infrastructure, and efficient logistics to support industries.
3. Ease of Doing Business: Implementation of investor-friendly policies including a Single Window Mechanism for quick approvals.
4. Support for MSMEs: Establishment of 175 MSME parks with financial assistance, capital subsidies, tax rebates, and employment-linked incentives.

Question 5.
Write a brief note on role of the service sector in Andhra Pradesh.
Answer:
The service sector is a crucial pillar of the Andhra Pradesh economy, acting as a major driver of growth and modernization.

1. Dominant Economic Contribution: The service sector is the largest contributor, accounting for 54.2% of GSDP (2024-25), with rapid growth in trade, hotels, and banking.
2. Significant Employer: About 31.6% of the workforce is employed in services such as trade, transport, education, health, and government jobs, with focus on IT and logistics.
3. Surge in Exports: Service exports, especially computer software, increased sharply from $3.9 billion (2023) to $26 billion (2024), strengthening the state’s global presence.
4. Emerging IT and Tourism Hub: Cities like Visakhapatnam and Vijayawada are growing as IT and GCC hubs, while tourism (over 254 million visits) boosts the hospitality sector.

Question 6.
Explain the employment generation strategy of Andhra Pradesh.
Answer:
The employment generation strategy of Andhra Pradesh is a multi-pronged approach focusing on skill development, industrial growth, and support for emerging sectors.

1) Strategic Vision and Skill Development: Under Swarna Andhra Vision 2047, the state promotes employment through education reforms and skill training. APSDC and skill hubs enhance technical, digital, and entrepreneurial skills.

2) Industrial and Infrastructure Expansion: Development of SEZs, industrial parks, and logistics hubs creates jobs in sectors like textiles, electronics, and pharmaceuticals. MSMEs and IT & GCC policy aim to generate large-scale employment.

3) Public Employment and Livelihood Support: Schemes like MGNREGA provide rural employment, while support to fisheries, handloom, and handicrafts ensures inclusive growth. Unemployment allowance (₹ 3,000) supports youth.

4) Future-Oriented Jobs: Focus on green energy, startups, and innovation hubs promotes jobs in renewable energy and digital sectors.

Long Answer Questions

Question 1.
Discuss the characteristics of A.P economy.
Answer:
Characteristics of A.P economy:
1. Strong Macroeconomic Performance

• The Gross State Domestic Product (GSDP) is estimated at ₹16.06 lakh crore (2024-25), with a growth rate of 12.94%, indicating a rapidly expanding economy.
• The Per Capita Income of ₹ 2,68,653 is higher than the national average, reflecting improved living standards. This demonstrates economic resilience and steady development.

2. Structural Shift in Sectoral Composition

• The economy has undergone significant structural transformation:
  • Services – 54.2 %
  • Industry – 27.6 %
  • Agriculture – 18.2 %
• This shows a transition from an agriculture-based economy to a service-oriented economy.

3. Employment Pattern and Challenges

• Agriculture contributes only 18.2% to GSDP but employs 45.6% of the workforce, indicating over-dependence on agriculture.
• Services employ 31.6% and industry 22.8%, reflecting gradual diversification.
• The unemployment rate is 4.1%, while graduate unemployment is around 24%, highlighting skill mismatch. Government established 192 Skill Hubs and aims to create 20 lakh jobs.

4. Rich Natural Resources and Export Strength

• Andhra Pradesh ranks 5th in mineral wealth, with resources such as barytes, limestone, quartz, and uranium.
• It has the 3rd longest coastline (1,053 km), contributing 40% of marine exports and 30% of fish production.
• Known as the “Rice Bowl of India,” it contributes significantly to rice exports along with chillies and tobacco.
• Service exports are also growing strongly, with software exports reaching $26 billion in 2024.

5. Demographic and Human Development Features

• The population is about 53.59 million, with a low growth rate of 0.35%, indicating demographic stability.
• Poverty has declined to 9.2%, reflecting effective welfare measures.
• Better health indicators and institutions such as NT Tirupati and IIM Visakhapatnam contribute to human capital development.

6. Impact of Reorganisation (2014)

• Andhra Pradesh retained only 46% of total resources, creating a weaker economic base.
• The loss of Hyderabad resulted in revenue and fiscal challenges.

7. Infrastructure and Development Focus

• The state is investing in industrial corridors, ports, logistics networks, and irrigation projects.
• Focus on MSMEs, industrial parks, and IT hubs aims to promote employment and industrialization.

8. Future Vision – Swarna Andhra Vision 2047

• Andhra Pradesh aims to become a $2.4 trillion economy by 2047.
• Key focus areas include: Zero Poverty (P4 approach),Farmer and Agri-Tech advancement (ZBNF) Deep-Tech integration (Al, Blockchain, IoT), Global logistics and infrastructure development
• The vision seeks to ensure sustainable, inclusive, and long-term economic growth.

Question 2.
Examine the employment generation strategy of A.P.
Answer:
Employment Generation Strategy of Andhra Pradesh: The employment generation strategy of Andhra Pradesh is a multi-dimensionai approach that combines long-term planning, skill development, industrial growth, and livelihood support to create sustainable employment opportunities.

I) Strategic Vision and Skill Development:

1. Swarna Andhra Vision 2047 places employment generation at the core of development, targeting a $2.4 trillion GSDP by 2047.
2. It focuses on preparing youth for a technology-driven economy through education reforms and skill-based training.
3. AP State Skill Development Corporation (APSDC) and 192 Skill Hubs provide technical, digital, and vocational training.
4. These centres also promote entrepreneurship, encouraging self-employment along with wage employment.
5. This helps create a skilled workforce required for industrial and service sector growth.

II) Public Employment Schemes and Industrial Expansion:

1. MGNREGA and other livelihood programmes provide large-scale wage employment, particularly in rural areas, reducing seasonal unemployment.
2. Urban livelihood programmes create employment opportunities in towns and cities.
3. The government promotes industrial growth through SEZs, industrial parks, and logistics hubs, attracting investment and generating jobs.
4. Sectors such as textiles, food processing, electronics, and pharmaceuticals are prioritized because of their high employment potential.
5. MSMEs are supported through subsidies, incentives, and easier credit access, creating local employment opportunities.

III) Startups, Green Jobs, and Inclusive Livelihoods:

1. Innovation hubs and incubation centres encourage youth participation in IT, digital services, and startups.
2. This promotes modern employment opportunities aligned with global trends.
3. Renewable energy projects, organic farming, and sustainable practices generate green jobs.
4. Traditional sectors such as fisheries, handloom, handicrafts, and agriculture-related activities continue to support rural and coastal livelihoods.
5. These measures promote inclusive growth and reduce regional disparities.

IV) Outcomes and Continuing Challenges:

1. Government initiatives helped reduce the unemployment rate from 6.8% (2023) to 5.9% (2024).
2. However, challenges remain:
   1. Regional imbalances in employment opportunities
   2. Underemployment in agriculture
   3. Need for continuous skill upgradation to meet industry requirements.

Question 3.
Explain the main features of Swarna Andhra Vision 2047.
Answer:
Swarna Andhra Vision 2047 is a comprehensive long-term development roadmap launched on December 13,2024, aiming to transform Andhra Pradesh into a “Wealthy, Healthy and Happy State” by 2047. It aligns with the national goal of Viksit Bharat 2047 and targets a Gross State Domestic Product (GSDP) of $2.4 trillion, reflecting an ambitious plan for rapid and inclusive growth.

1) Ten Sutras – Core Strategic Framework:
i) Zero Poverty: Focuses on complete eradication of poverty through inclusive growth strategies like welfare schemes and the P4 approach, ensuring that all sections benefit from development.

ii) Population Management and Human Resource Development: Aims at balanced population growth while strengthening education, healthcare, and skill levels, thereby improving the quality of human capital.

iii) Employment and Skilling: Targets creation of 20 lakh jobs and emphasizes skill development programs, ensuring youth are equipped for modern, technology-driven employment.

iv) Water Security: Ensures efficient irrigation, water conservation, and Al-based water management systems, which are essential for agriculture and sustainable development.

v) Farmer & Agri-Tech Advancement: Focuses on modernizing agriculture through smart farming techniques, digital support systems, and sustainable practices, improving productivity and farmer incomes.

vi) Global-Best Logistics: Aims to develop world-class infrastructure, ports, and logistics networks, leveraging the state’s coastline to become a global trade and logistics hub.

vii) Cost Optimization in Energy & Fuel: Promotes renewable energy sources and efficient power generation, reducing costs and ensuring energy security.

viii) Product Perfection: Focuses on improving quality standards of goods and exports, making Andhra Pradesh competitive in global markets.

ix) Swachh Andhra: Emphasizes sanitation, clean environment, waste management, and urban infrastructure, improving living conditions and public health.

x) Deep-Tech in All Sectors: Encourages integration of advanced technologies such as Al, Blockchain, and loT across agriculture, industry, and services to drive innovation and efficiency.

2) Focus on inclusive and Balanced Growth

• The vision ensures inclusive development, addressing regional disparities and improving opportunities for all sections of society.
• It promotes citizen participation, ensuring that development is people-centered.

3) Emphasis on Sustainability and Future Readiness

• Strong focus on green energy, sustainable agriculture, and efficient resource utilization.
• Encourages technology-driven growth, preparing the state for future economic challenges.

4) Improvement in Quality of Life

• The vision prioritizes better healthcare, education, housing, sanitation, and infrastructure, ensuring overall well-being of citizens.
• It aims at creating not just economic growth, but also a healthy and happy society.

Multiple Choice Questions

Question 1.
The forest cover in Andhra Pradesh approximately accounts for:
1) 19 %
2) 23 %
3) 25 %
4) 27 %
Answer:
2) 23 %

Question 2.
Under AP Reorganized Act, 2014, All MS is established in ____________
1) Visakhapatnam
2) Tirupathi
3) Mangalagiri
4) Tadepalligudem
Answer:
3) Mangalagiri

Question 3.
The share of industry and manufacturing sector in Andhra Pradesh GSDP: (As per AP Socio-economic survey 2024-25)
1) 26 %
2) 21 %
3) 16.8%
4) 22.8%
Answer:
4) 22.8%

Question 4.
What does the Society of Elimination of Rural Poverty (SERP) promote through SHGs?
1) Self-employment
2) Handicrafts production
3) Rural Housing
4) Microfinancing
Answer:
1) Self-employment

Question 5.
The GSDP target of Swara Andhra Vision 2047:
1) $ 2.0 trillion
2) $ 2.2 trillion
3) $ 2.4 trillion
4) $ 2.6 % trillion
Answer:
3) $ 2.4 trillion

Question 6.
The following is true regarding the impact of rorganisation of Andhra Pradesh in 2014
1) AP inherited 46 % of resources
2) AP’s forest-based revenue increased
3) The income from mines were not affected
4) Lost deposits of Barytes and Uranium
Answer:
1) AP inherited 46 % of resources

Question 7.
Which of the following statements is not correct?
1) AP IT policy aims to create 10 lakh jobs
2) AP aims to double the share in software exports by 2047
3) AP contributes about 45% of India’s rice exports
4) AP’s graduate unemployment rate is less than the national average.
Answer:
1) AP IT policy aims to create 10 lakh jobs

Question 8.
MEPMA is nodal agency that aims to:
1) Improve municipal governance
2) Eliminate rural poverty
3) Eliminate poverty in urban areas
4) Improve sanitation in urban slums
Answer:
1) Improve municipal governance

Question 9.
The rank of the population growth rate of Andhra Pradesh among the states in 2025 is:
1) 2nd lowest
2) 3rd lowest
3) 4th lowest
4) 6th lowest
Answer:
1) 2nd lowest

Question 10.
Andhra Pradesh is among India’s top exporters of:
1) Cotton, tobacco
2) Chillis, oil seeds
3) Fish and mangoes
4) Chillis and tobacco
Answer:
4) Chillis and tobacco

Fill in the Blanks

Question 1.
Andhra Pradesh is the ____________ largest producer of raw silk in India.
Answer:
second

Question 2.
The amount given annually under Talliki Vandanam scheme is Rs. ____________.
Answer:
15,000

Question 3.
The growth rate of Agriculture sector is ____________ than other two sectors in AP in 2024-25.
Answer:
higher

Question 4.
Under Deepam-2 scheme eligible families get ____________ LPG gas cylinders per year.
Answer:
3

Question 5.
Under Annadata Sukhibhava Scheme, the Annual financial assistance of ____________ will be given to eligible farmers.
Answer:
Rs. Rs. 20,000

One Word Answers

Question 1.
What is the scheme under which farmers can have subsidized drones?
Answer:
RKVY (Rashtriya Krishi Vikas Yojana)- The Kisan Drone Yojana 2025

Question 2.
What does Sankalpam scheme focus on?
Answer:
Life skills (training in core life skills to school children of 6-8 classes)

Question 3.
When was Dokka Seethamma Madhyahna Badi Bhojanam extended to Junior Colleges?
Answer:
January 2025.

Question 4.
Where was the Indian Institute of Science Education and Reserch (USER) was located in Andhra Pradesh?
Answer:
Tirupati

Question 5.
Expand ASHA.
Answer:
Accredited Social Health Activist.

Regular practice with AP Inter 2nd Year Chemistry Study Material Chapter 9 Amines Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Chemistry 9th Lesson Amines Questions and Answers

I. Multiple Choice Questions

Question 1.
Which of the following amines cannot be prepared by Gabriel phthalimide reaction?
1) Benzylamine
2) Aniline
3) Ethylamine
4) Methylamine
Answer:
2) Aniline
Gabriel synthesis gives only aliphatic primary amines.
Aromatic amines (like aniline) cannot be prepared.

Question 2.
Which of the following reaction does not produce amine?
1) Gabriel phthalimide synthesis
2) Hoffmann bromamide degradation reaction
3) Carbylamine reaction
4) Reduction of nitriles
Answer:
3) Carbylamine reaction
Gabriel, Hofmann, Reduction → all give amines.
Carbylamine reaction is only a test (does not prepare amine)

Question 3.
Classification of amines into 1°, 2″, 3″ is based on the
1) number of amino groups
2) nature of carbon atom
3) degree of substitution at nitrogen
4) degree of unsaturation
Answer:
3) degree of substitution at nitrogen
Classification (1°, 2°, 3°) depends on how many groups are attached to nitrogen.
Not number of -NH₂ groups. Degree of substitution at nitrogen

Question 4.
Which of the following amines does not react with Hinsberg reagent?
1) CH₃CH₂NH₂
2) (CH₃CH₂)₂NH
3) (CH₃CH₂)₃N
4) all will react
Answer:
3) (CH₃CH₂)₃N
Hinsberg reagent reacts with: 1 ° amine → soluble product
2° amine → insoluble product; 3° amine → no reaction (CH₃CH₂)₃N
Tertiary amines do not react.

Question 5.
Identify the product Y in the given reaction sequence.

Answer:

Sequence: Nitrobenzene → (Sn/HCl) → aniline (X)
Aniline → diazonium salt → coupling with phenol (in basic medium)
Gives azo dye (-N=N-) with -OH group on ring.
Azo compound with -OH substituted benzene ring

Question 6.
Propanoic acid on reaction with NH₃ followed by heating gives a compound X, which on reaction with Br₂ and aqueous KOH gives.
1) Propylamine
2) Ethanamine
3) Ethanenitrile
4) Propanenitrile
Answer:
2) Ethanamine
Propanoic acid + NH₃ → amide
Amide + Br₂/KOH → Hofmann bromamide reaction
Gives amine with one less carbon → ethylamme

Question 7.
Benzene diazonium chloride salt solution is heated at about 283 K the salt gets hydrolysed to compound “A”: A can be identified with
1) Lucas reagent
2) Neutral ferric chloride
3) Tollens’reagent
4) Hinsberg reagent
Answer:
2) Neutral ferric chloride
Diazonium salt + water (heat) → phenol
Phenol gives violet color with FeCl₃

Question 8.
The number of primary amines possible for a compound with molecular formula C₄H₁₁N are
1) 1
2) 2
3) 3
4) 4
Answer:
4) 4
Count possible structures of primary amines (—NH₂) for C₄H₁₁N
Different carbon skeletons give multiple isomers. Total = 4

Question 9.
Which of the following is most volatile?
1) CH₃CH₂CH₂NH₂
2) (CH₃)₃N

4) CH₃OH
Answer:
2) (CH₃)₃N
Volatility ∝ lower intermolecular forces. H-bonding decreases volatility
Tertiary amine has no N—H → least H-bonding → most volatile (CH₃)₃N

Question 10.
Aliphatic primary amine on heating with chloroform and ethanolic KOH gives
1) an alcohol
2) an alkanediol
3) an alkyl isocyanide
4) an alkyl cyanide
Answer:
3) an alkyl isocyanide
Reaction with CHCl₃ + alcoholic KOH → CarIylamine reaction.
Only primary amines give isocyanides (foul smell).

Question 11.
Which of the following does not reduce C₃H₅NO₂ to C₆H₅NH₂?
1) Fe/ HCl
2) Zn/ HCl
3) LiAlH₄
4) SnCl₂ in HCl
Answer:
3) LiAlH₄
Reduction of nitrobenzene to aniline is done by metal + acid (Fe/HCl, Zn/HCl, SnCl₂/HCl).
LiAlH₄ generally reduces different functional groups, not used here.

Question 12.
The end product in the given reaction sequence is

1) CH₃-CH₂-NH₂
2) CH₃CONH₂
3) CH₃-NH-CH₃
4) CH₃CH₂OH
Answer:
4) CH₃CH₂OH

Question 13.
Which among the following has the lowest pKb value in aqueous medium?
1) NH₃
2) CH₃NH₂
3) (CH₃)₂NH
4) (CH₃)₃N
Answer:
3) (CH₃)₂NH
Basicity in aqueous medium depends on +I effect + solvation.
Secondary amines are most basic due to balance of both.
Order: 2° > 1° > 3° > NH₃ (CH₃)₂NH

Question 14.
Formation of benzene from benzene diazonium chloride can be carried out with
1) H₃PO₃
2) H₃PO₄
3) H₃PO₂
4) HPO₃
Answer:
3) H₃PO₂
Benzene diazonium salt → benzene requires reduction.
H₃PO₂ acts as reducing agent (replaces N₂⁺ with H)

Question 15.
Aniline reacts with the following to give benzanilide
1) Benzyl chloride
2) Benzal chloride
3) Benzoyl chloride
4) Benzo tri chloride
Answer:
3) Benzoyl chloride
Aniline + acyl chloride → benzoylation (Schotten—Baumann reaction).
Gives benzanilide. Beuzoyl chloride

II. Fill in the Blanks

Question 1.
The shape of trimethyl amine is _________
Answer:
pyramidal

Question 2.
Benzenesulphonyl chloride (C₆H₅SO₂Cl), is also known as _________
Answer:
Hinsbere’s reagent

Question 3.
Carbylamine reaction or isocyanide test is used for the identification of ___________
Answer:
primary amines or 1° amines

Question 4.
The conversion of primary aromatic amines into diazonium salts is known as ___________
Answer:
diazotisation

Question 5.
Coupling reaction of aryl diazonium salts with phenols or aryl amines gives rise to the formation of __________
Answer:
azodyes

III. One Word Answer Questions

Question 1.
What is the IUPAC name of ethyl methyl amine?
Answer:
IUPAC name of ethyl methyl amine is N-methyl ethanamine.

Question 2.
The process of cleavage of the C-X bond of alkyl halides by ammonia molecule is known as
Answer:
Ammonolysis is the cleavage of the C-X bond of alkyl halides by ammonia molecule.

Question 3.
Aniline and other arylamines are usually colourless but get coloured on storage why?
Answer:
Aniline and other arylamines get coloured on storage due to Atmospheric oxidation.

Question 4.
Tertiary amines like trimethylamine are used as
Answer:
Tertiary amines are used to attract flies or insects.

IV. Short Answer Questions

Question 1.
Write the IUPAC names of the following compounds and classify into primary, secondary and tertiary amines.
(i) (CH₃)₂CHNH₂
(ii) CH₃(CH₂)₂NH₂
(iii) (CH₃CH₂)₂NCH₃.
Answer:

Question 2.
Explain why ethylamine is more soluble in water where as aniline is sparingly soluble.
Answer:
Ethylamine forms hydrogen bonds with water molecules and hence dissolves in water. Aniline, due to its large hydrocarbon part, develops hydrophobic character. As a result it does not dissolve in water.

Question 3.
Why aniline does not undergo Friedel-Craft reactions?
Answer:
Aniline doesnot undergo Friedel-Crafts reaction. Actually, aniline being a Lewis base forms a complex with AlCl₃ which is a Lewis acid. The amino group is not in a position to activate the benzene ring towards electrophilic substitution. Hence, alkylation (or) acylation which lead to Fridel Crafts reaction is not possible.

Question 4.
Gabriel phthalimide synthesis exclusively forms primary amines . Explain.
Answer:
Gabriel phthalimide synthesis is preferred for synthesizing aliphatic primary amines because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

Question 5.
Arrange the following bases in decreasing order of pK_b values C₂H₅NH₂, C₆H₅NHCH₃, (C₆H₅)₂NH and C₆H₅NH₂.
Answer:
Decreasing order of pK_b : C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₆H₅)₂NH
pK_b values: 9.38 > 9.30 > 3.29 > 3.00
[Decreasing order of Basicity: (C₆H₅)₂NH > C₂H₅NH₂ > C₆H₅NHCH₃ > C₆H₅NH₂

Question 6.
Arrange the following bases in increasing order of their basic strength. Aniline, p-nitroaniline and p-toluidine.
Answer:
Increasing order (low’ to high) of basic strength:
p-nitroaniline < Aniline < p-toluidine.

Question 7.
Write equations for carbylamine reaction of an aliphatic amine.
Answer:
When a primary amine is heated with alcoholic caustic potash and chloroform, an offensive
smelling compound called carbylamine (Ethylisocyanide) is formed. The reaction is used as a test for chloroform (or) for a primary amine.

Question 8.
Give structures of A, B and C in the following reaction.

Answer:

Question 9.
Accomplish the following conversions:
(i) Benzoic acid to benzamide
(ii) Aniline to p-bromoaniline
Answer:

Question 10.
Why aromatic primary amines cannot be prepared by Gabriel phthalimide?
Answer:
Aryl halides does not react with potassium phthalimide. Because C-X bond in haloarene is difficult to be cleaved due to partial double bond character. They do not undergo nucleophilic substitution with the anion formed by phthalimide. So, aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis.

V. Short Answer Questions

Question 1.
Give one chemical test to distinguish between the following pairs of compounds:
i) Methylamine and dimethvlamine
ii) Aniline and N-methylaniline
iii) Ethylamine and aniline
Answer:
i) Methylamine is a primary amine, while dimethyl amine is a secondary amine.
So, methyl amine undergoes carbylamine test, while dimethylamine does not.
When methylamine is heated with chloroform and alcohalic KOH, an offensive smelling compound methyl carbylamine is formed.

ii) Aniline is a primary amine, while N-methylaniline is a secondary amine.
So aniline undergoes carbylamine test, while N-methylaniline doesnot.

iii) Ethylamine and aniline are primary aromatic amines.
Aniline undergoes azodye test, while ethylamine doesnot.

Question 2.
How do you prepare the following?
i) N, N-Dimethyl propanamine from ammonia
ii) Propanamine from chloroethane
Answer:

Question 3.
Compare the basicity of the following in gaseous state and in aqueous state and arrange them in increasing order of basicity.
CH₃NH₂, (CH₃)₂NH, (CH₃)₃N and NH₃
Answer:
P^kb values of NH₃, CH₃NH₂, (CH₃)₂NH, (CH₃)₃N are 4.74, 3.35, 3.27, 4.22 respectively.
“Smaller is the value of Pkb, greater is the basic nature of amine.
Hence, increasing order of basicity: NH₃ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH
In gaseous phase, basic character of amines increases with the increase in number of electron releasing groups(+I groups). Due to +1 effect, the order of basicity of amines is as follows
[(CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃]
In aqueous state, basicity depends on stability of protonated amines.
Greater the strength of H-bonding greater will be stability and more is the basic strength.
Hence order of basicity : NH₃ > CH₃NH₂ >(CH₃)₂NH > (CH₃)₃N
But overall basic strength of Amines depends on combined effect of steric, solvation and Inductive effect. Hence overall order is (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃

Question 4.
How do you carry out the following conversions?
i) N-ethylamine to N,N-diethvl propanamine
ii) Aniline to Benzene sulphonamide
Answer:
i) N-ethylamine to N,N-diethyl propanamine:

ii) Aniline to Benzene sulphonamide

Question 5.
Explain with a suitable example, how benzene sulphonylchloride can distinguish primary, secondary and tertiary amines.
Answer:
Benzene Sulphonyl chloride (C₆H₅SO₂Cl) is used to distinguish 1 °, 2°, 3° Amines.
This is also known as Hinsberg’s reagent.
i) 1 ° Amines (Ethylamine) react with Benzene sulphonyl chloride to form N-Ethyl benzene sulphonamide.

The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence it is soluble in alkali.

ii) 2° Amines (Diethylamine) react with Benzene sulphonyl chloride to form N,N-Diethyl- benzene sulphonamide.

Since N,N-Diethylbenzene sulphonamide doesnot contain any hydrogen atom attached to nitrogen atom, it is not acidic and hence insoluble in alkali.

iii) Tertiary amines do not react with benzene sulphonyl chloride. Hence Hinsberg reagent is useful to distinguish primary, secondary and tertiary amines.

Question 6.
Write the reactions of
(i) aromatic and (ii) aliphatic primary amines with nitrous acid, highlighting the differences.
Answer:
i) Aromatic amines react with nitrous acid at low temperatures (273-278K) to form diazonium salts.

ii) Aliphatic primary amines react with nitrous acid to form aliphatic diazonium salts which being unstable, liberate nitrogen gas and alcohols. Quantitative evolution ofnitrogenis used for the estimation of amino acids and proteins.

Question 7.
Explain why amines are less acidic than alcohols of comparable molecular masses.
Answer:
Electronegativity of oxygen is more than that of nitrogen. Thus the oxygen in alcohols tends to pull electrons twoards itself and releases H+ more easily. On the other hand, the presence of alkyl group on the nitrogen atom in amines increases electron density to make it more basic than alcohol. Therefore, amines are less acidic than alcohols.

Question 8.
Write the equations involved in the reaction of Nitrous acid with ethylamine and aniline.
Answer:
1) Primary aliphatic amines like C₂H₅NH₂ react with nitrous acid and form aliphatic diazonium salts. But it being unstable, dissociates to give ethyl alcohol and N₂ gas.

2) Aromatic amines like C₆H₅NH₂ react with nitrous acid at low temperatures (273-278K) and form diazonium salts.

Question 9.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’, which on heating with Br2 and KOH forms compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.
Answer:
From the given data, compound B upon heating with Br2 and KOH forms a compound C .
This indicates that the compound B is an acid amide and C is amine.
Since B has been formed upon heating compound A which is an aromatic acid and it is benzoic acid.
The reactions involved are given as follows:

Question 10.
Account for the following:
pK_b of aniline is more than thit of methyla mine
Answer:
In aniline due to resonance, the lone pair of electrons on the N atom are delocalized over the benzerie ring. Due to this, electron density on the nitrogen decreases.
But in CH₃NH₂, +1-effect of CH₃ increases the electron density on the N-atom.
Thus, aniline is a weaker hase than methylamine and hence its pK_b value is more than that of methylamine.

VI. Long Answer Questions

Question 1.
Explain why the order of basicity for methylamine, N,N-dimethyl amine and N,N,N-trimethylamine changes in gaseous and aqueous medium.
Answer:
In gaseous phase, due to +1 effect, basic character of amines increases with increase in number of electron releasing groups.
So the order of basicity: [(CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃]
In aqueous state, basicity depends on stability of protonated amines. Greater the strength of El- bonding greater will be stability and more is the basic strength.
So order of basicity :NH₃ > CH₃NH₂ >(CH₃)₂NH > (CH₃)₃N
But overall basic strength of Amines depends on combined effect of steric, solvation and Inductive effect.
Hence overall order : (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃

Question 2.
Complete the following conversions.
Aniline to (i) Fluorobenzene (ii) Cyanobenzene (iii) Benzene and (iv) Phenol.
Answer:
i) Aniline to Fluorobenzene: This conversion involves two stages:
a) Preparation of Benzenediazonium chloride: Benzene diazonium chloride is prepared by the reaction of aniline with nitrous acid at 273-278K.

b) Preparation of Fluorobenzene: Benzene diazonium chloride on heating with fluoroboric acid gives fluorobenzene.

The reaction is called Balz-Schiemann reaction.

Question 3.
Explain the following name reactions:
i) Sandmeyer reaction
ii) Gattermann reaction
Answer:
i) Sandmeyer reaction: This reaction is used to convert aniline into an alkyl halide (Cl^–, Br^–) or Cyanide (CN^–). It is used in the synthesis of chlorobenzene, bromobenzene and benzionitris.
This process involves forming a dizaonium salt at 0-5°C, which undergoes copper catalysed radical substitution releasing nitrogen gas.
The Cl^–, Br^– and CN^– nucleophiles can be introduced in the benzene ring in the presence of Cu₂Cl₂ / HCl (or) Cu₂Br₂ / HBr (or) CuCN/KCN.

ii) Gattermann reaction: This reaction is used for obtaining chiorobenzene (or) bromobenzene from benzene diazonium chloride by treating it with Cu/HCl (or) Cu/HBr respectively.

Question 4.
Write the steps involved in the coupling of benzene diazonium chloride with aniline and phenol.
Answer:
Coupling reactions: Benzene diazonium salt reacts with certain aromatic compounds having an electron rich group (-OH, -NH₂ etc.,) to form azo compounds. Azo compounds are highly coloured and are used as dyes. This reaction is known as coupling reaction and is carried by controlling the pH of the medium.
Coupling reactions is an electrophilic substitution reaction.
Ex 1: Benzene diazonium chloride reacts with phenol. In this reaction diazonium group is coupled with phenol at para position by electrophilic substitution reaction to form p-hydroxy azobenzene.

Ex 2: Reaction of diazonium salt with aniline yields p-amino azobenzene.

Objective Questions

Question 1.
Which of the following is a 3° amine?
1) 1 -methylcyclohexylamine
2) Triethylamine
3) tert-butylamine
4) N-methylaniline
Answer:
2) Triethylamine

Question 2.
The correct 1UPAC name for CH₂==CHCH₂ NHCH₃ is
1) Allylmethylamine
2) 2-amino-4-pentene
3) 4-aminopent-1-ene
4) N-methylprop-2-en-1-amine
Answer:
4) N-methylprop-2-en-1-amine

Question 3.
Acid anhydrides on reaction with primary amines give _________
1) amide
2) imide
3) secondary amine
4) imine
Answer:
1) amide

Question 4.
Amongst the following, the strongest base in aqueous medium is _______
1) CH₃NH₂
2) NCCH₂NH₂
3) (CH₃)₂ NH
4) C₆H₅NHCH₃
Answer:
3) (CH₃)₂ NH

Question 5.
Benzylamine may be alkylated as shown in the following equation :
C₆H₅CH₂NH₂ + R—X → C₆H₅CH₂NHR Which of the following alkylhalides is best suited for this reaction through S_N1 mechanism?
1) CH₃Br
2) C₆H₅Br
3) C₆H₅CH₂Br
4) C₂H₅Br
Answer:
3) C₆H₅CH₂Br

Question 6.
Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine?
1) H_2(excess) /Pt
2) LiAlH₄ in ether
3) Fe and HCl
4) Sn and HCl
Answer:
2) LiAlH₄ in ether

Question 7.
Best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain is
1) Hoffmann Bromamide reaction
2) Gabriel.phthalimide synthesis
3) Sandmeyer reaction
4) Reaction with NH₃
Answer:
2) Gabriel.phthalimide synthesis

Question 8.
The gas evolved when methylamine reacts with nitrous acid is ______
1) NH₃
2) N₂
3) H₂
4) C₂H₆
Answer:
2) N₂

Question 9.
Electrolytic reduction of nitrobenzene in weakly acidic medium gives
1) aniline
2) nitrosobenzene
3) N-phenyl hydroxylamine
4) p-hydroxyaniline
Answer:
1) aniline

Question 10.
Amongst the given set of reactants, the most appropriate for preparing 2°amine is ______
1) 2° R—Br + NH₃
2) 2° R—Br + NaCN followed by H₂/Pt
3) 1° R—NH₂ + RCHO followed by H₂/Pt
4) 1° R—Br (2 mol) + potassium phthalimide followed by H₃O⁺/heat
Answer:
3) 1° R—NH₂ + RCHO followed by H₂/Pt

Question 11.
The source of nitrogen in Gabriel synthesis of amines is ________
1) Sodium azide, NaN₃
2) Sodium nitrite, NaNO₂
3) Potassium cyanide, KCN
4) Potassium phthalimide, C₆H₄(CO)₂N^–K⁺
Answer:
4) Potassium phthalimide, C₆H₄(CO)₂N^–K⁺

Question 12.
The best reagent for converting, 2-phenylpropanamide into 1- phenylethanamine is __________
1) excess H₂/Pt
2) NaOH/Br₂
3) NaBH₄/methanol
4) LiAlH₄/ether
Answer:
2) NaOH/Br₂

Question 13.
The best reagent for converting 2-phenylpropanamide into 2-phenylpropanamine is _______
1) excess H₂
2) Br₂ in aqueous NaOH
3) iodine in the presence of red phosphorus
4) LiAlH₄ in ether
Answer:
4) LiAlH₄ in ether

Question 14.
Hoffmann Bromamide Degradation reaction is shown by _______
1) ArNH₂
2) ArCONH₂
3) ArNO₂
4) ArCH₂NH₂
Answer:
2) ArCONH₂

Question 15.
Acetamide is treated with the following reagents separately. Which one of these would yield niethylamine?
1) NaOH+Br₂
2)Sodalime
3) Hot Cone. H₂SO₄
4) PCl₅
Answer:
1) NaOH+Br₂

Question 16.
Methylamine reacts with HNO₂ to form ________
1) CH₃—O—N=0
2) CH₃—O—CH₃
3) CH₃OH
4) CH₃CHO
Answer:
3) CH₃OH

Question 17.
Reduction of aromatic nitro compounds using Fe and HCl gives _________
1) aromatic oxime
2) aromatic hydrocarbon
3) aromatic primary amine
4) aromatic amide
Answer:
3) aromatic primary amine

Question 18.
The reaction

is named as ________
1) Sandmeyer reaction
2) Gatterman reaction
3) Claisen reaction
4) Carbylamine reaction
Answer:
2) Gatterman reaction

Question 19.
Which of the following will be most stable diazonium salt \(\mathrm{RN}_2^{+} \mathrm{X}^{-}\)?
1) \(\mathrm{CH}_3 \mathrm{~N}_2^{+} \mathrm{X}^{-}\)
2) \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2^{+} \mathrm{X}^{-}\)
3) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{~N}_2^{+} \mathrm{X}^{-}\)
4) \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{~N}_2^{+} \mathrm{X}^{-}\)
Answer:
2) \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2^{+} \mathrm{X}^{-}\)

Question 20.
Which of the following statements about primary amines is false?
1) Alkyl amines are stronger bases than aryl amines
2) Alkyl amines react with nitrous acid to produce alcohols.
3) Aryl amines react with nitrous acid to produce phenols
4) Alkyl amines are stronger bases than ammonia
Answer:
3) Aryl amines react with nitrous acid to produce phenols

Question 21.
Which of the following compound will not undergo azo coupling reaction with benzene diazonium chloride.
1) Aniline
2) Phenol
3) Anisole
4) Nitrobenzene
Answer:
4) Nitrobenzene

Question 22.
The correct decreasing order of basic strength of H₂O, NH₃,OH , \(\mathrm{NH}_2^{-}\) species is ________
1) \(\mathrm{NH}_2^{-}\) >OH^–>NH₃ >H₂O
2) OH^– > \(\mathrm{NH}_2^{-}\) > H₂O > NH₃
3) NH₃ > H₂O > \(\mathrm{NH}_2^{-}\) > OH^–
4) H₂O > NH₃ > OH^– >\(\mathrm{NH}_2^{-}\)
Answer:
1) \(\mathrm{NH}_2^{-}\) >OH^–>NH₃ >H₂O

Reviewing AP Inter 2nd Year Economics Study Material Chapter 9 Industry and Service Sectors Questions and Answers can help students prepare confidently for exams.

AP Inter 2nd Year Economics 9th Lesson Industry and Service Sectors Questions and Answers

Very Short Answer Questions

Question 1.
Describe Ratna categories of PSUs
Answer:
The Ratna categories are a classification system for Central Public Sector Enterprises (CPSEs) in India, designed to grant well-performing companies greater financial and operational autonomy based on specific criteria.
The three main categories are:

• Maharatnas (introduced in 2010) -14 companies
• Navaratnas (introduced in 1997) – 25 companies
• Miniratnas (introduced in 1997) – 73 companies, further divided into Miniratna I and II.

Question 2.
State the objectives of Atma Nirbhar Bharat Abhiyan
Answer:
Objectives of Atma Nirbhar Bharat Abhiyan:
The Atma Nirbhar Bharat Abhiyan (2020) aims to support MSMEs during the pandemic through collateral-free loans, equity infusion and support to stressed units. It also focuses on modernising MSME classification and boosting domestic production through schemes like PLI, with the broader goal of achieving self-reliance and transforming India into a global manufacturing hub.

Question 3.
Explain the forms of FDI.
Answer:
Foreign Direct Investment (FDI) in India can take several forms:

1. Buying Shares – Purchasing shares in an Indian company to gain ownership or control.
2. Acquisition-Taking over an existing Indian company.
3. Partnership – Entering into a partnership with an Indian firm.
4. lnvestment Instruments – Investing through debentures, bonds, or venture capital funds.

Question 4.
Explain Maritime Waterways with examples
Answer:
Maritime waterways refer to sea or ocean routes used for domestic and international shipping, carrying about 95% of India’s trade volume and 70% of its value. They are managed by the Ministry of Ports, Shipping and Waterways, with major and minor ports handling trade.
Ex: Transportation of coal, iron ore, crude oil, LPG, machinery and food grains, along with activities like cruise shipping.

Question 5.
How does UDAN scheme achieve its objective?
Answer:
The Ude Desh ka Aam Nagrik (UDAN) scheme 2017, achieves its objective of affordable regional air connectivity by capping airfares at ₹ 2,500 for one-hour flights and providing viability gap funding to airlines. It also focuses on developing unserved and underserved airports and expanding routes, thereby connecting remote areas to the national network.

Question 6.
Differentiate between General Insurance and Life Insurance.
Answer:

Short Answer Questions

Question 1.
Mention any four aspects of Indian industry.
Answer:
Aspects of Indian Industry:
1) Broad Scope of Activities: Industry involves the processing of raw materials and the production of goods, encompassing sectors such as manufacturing, construction, and the supply of electricity, gas, and water. This shows that industry is not limited to factories alone but includes a wide range of productive activities.

2) Economic Support and Growth: It plays a vital role in economic development by supporting agriculture, boosting trade, and creating significant employment opportunities. Thus, industry acts as a key driver of overall economic progress.

3) Foundational Requirements: The success and growth of the industrial sector are heavily dependent on technology, innovations, and competitiveness, as well as access to resources and a workforce of skilled labor and efficient technicians. These factors ensure higher productivity and efficiency.

4) Infrastructure-Based Core industries: A critical segment of the industrial landscape consists of the eight core industries—coal, crude oil, natural gas, refinery products, fertilizers, steel, cement, and electricity, which are considered the backbone of the country’s infrastructure.

Question 2.
Write any four measures taken to promote MSME in India.
Answer:
Measures to Promote MSMEs in India:

1) Credit and Financial Support: The government established specialized institutions and schemes to improve access to finance, including the Small Industries Development Bank of India (SIDBI) (1990) and the Credit Guarantee Scheme (CGTMSE) (2000), which provides collateral-free bank loans up to ₹2 crores. This ensures MSMEs get adequate financial assistance.

2) Digital Registration (Udyam Registration): In 2020, the registration process was simplified and digitalized through the Udyam Registration System, which includes a self-declaration facility, making it easier for MSMEs to register and access government incentive schemes.

3) Public Procurement and E-Market Reforms: The Public Procurement Policy (2012) mandates that 25% of government procurement must be sourced from MSMEs. The Government e-Marketplace (GeM) enables MSMEs to sell products directly to government departments, improving market access.

4) Atmanirbhar Bharat Package (2020): This package supports MSMEs through ₹3 lakh crore collateral-free automatic loans, equity infusion via a Fund of Funds, and support for stressed MSME units, helping them sustain and grow.

Question 3.
Explain the significance of Communications sector in Indian economy.
Answer:
Significance of Communication Sector in Indian Economy:
1) Foundation for Development: The communications sector is essential for areas like education, governance, trade, and commerce, as it facilitates the transmission of information through enhanced connectivity. Information and Communications Technology (ICT) has made communication faster and more accessible, transforming how individuals and organization’s function.

2) Support for Other Sectors: It acts as a driving force for agriculture, manufacturing, and services by providing necessary infrastructure. It enables activities such as marketing and efficient logistical coordination, thereby improving overall productivity.

3) Modernization and Technology Adoption: India’s telecom sector has shifted from analog to digital transmission, leading to increased tele-density and widespread use of mobile phones and broadband. It is now adopting advanced technologies like 5G, cloud computing, and virtualization.

4) Global Leadership and Social Development: India’s IT services sector is a major global player, contributing significantly to exports. Additionally, communication technologies improve governance (transparency and efficiency) and support social services like telemedicine, e- health records, and digital education, along with growth in media and broadcasting.

Question 4.
Comment on Make-in India programme.
Answer:
1) The Make in India programme, launched in 2014, is a major initiative aimed at transforming India into a global manufacturing and innovation hub. It encourages both domestic and multinational companies to expand manufacturing in India, covering 25 focus sectors (17 manufacturing and 8 services).

2) The primary objectives are to boost manufacturing output and generate employment opportunities, with targets such as increasing manufacturing’s share in GDP to 25% and creating 100 million jobs.

3) The programme also aims to attract FDl, develop infrastructure, promote innovation and technology, improve ease of doing business, and achieve self-reliance and sustainable development.

4) It is supported by initiatives like Skill India, Startup India, Digital India, PLI Scheme, and infrastructure proiects like NIP and PM Gati Shakti, which together strengthen industrial growth.

Question 5.
Explain the importance of service sector in India.
Answer:
Importance of Service Sector in India:

1. Significant Contribution to GDP: The service sector, also known as the tertiary sector, is a dominant contributor, accounting for more than 55% of India’s GVA. Its growth rate often exceeds the overall economic growth, making it a key driver of the economy.
2. Employment Generation: Although its share in employment is less than 30%, it provides a substantial number of jobs, with strong potential for further employment in areas like IT, tourism, and healthcare.
3. Supporting Other Sectors: The service sector supports agriculture and manufacturing by providing essential services such as logistics, banking, communication, and credit, thereby improving efficiency and productivity.
4. Boosting Exports and Development: It contributes significantly to export earnings, attracts FDI inflows, promotes technological advancement, and improves quality of life through services like education, healthcare, and transport, while also integrating India into the global economy.

Question 6.
Explain India’s Tourism at a glance.
Answer:
India’s Tourism-At a Glance:
1) Growth and Economic Importance: India’s tourism sector is a fast-growing part of the economy that generates direct and indirect employment, foreign exchange, and promotes trade in logistical services. Its growth depends on the availability of high-quality infrastructure such as transport, hotels, and hospitality services.

2) Performance and Key Data: India improved its global ranking from 54th (2021) to 39th (2024) in the Travel and Tourism Development Index. In 2022, foreign tourist arrivals were 6.19 million, domestic visits reached 1,731 million, and foreign exchange earnings were US $16.928 billion, placing India 14th in world tourism receipts.

3) Major Destinations and Contributors: Top source countries include the USA, Bangladesh, and the UK, while Indians travel mainly to the UAE, Saudi Arabia, and USA. Domestically, Uttar Pradesh, Tamil Nadu, and Andhra Pradesh receive the highest visits, and popular monuments include the Taj Mahal, Agra Fort, and Fatehpur Sikri.

4) Government Initiatives: The sector is promoted through measures like ITDC (1966), Tourism Policy (2002), Incredible India campaign, Swadesh Darshan Yojana (2015), E-Visa facilities, railway tourism packages, and tourism awards, which enhance infrastructure and global promotion.

Long Answer Questions

Question 1.
Explain the Industrial Policy, 1991.
Answer:
Industrial Policy, 1991: The New Industrial Policy, announced on July 24,1991, was a landmark reform that shifted India’s economy towards Liberalization, Privatization, and Globalization (LPG).

It aimed to reduce government control, increase efficiency, and integrate India with the global economy.

I) Objectives of the Policy: The policy was designed to achieve the following goals:

1. Building on past gains already made within the industrial sector.
2. Correcting distortions or weaknesses in the pattern of industrial growth.
3. Maintaining sustained growth in productivity and gainful employment.
4. Attaining technological dynamism and international competitiveness.

II) Key Features and Decisions

1. Delicensing: Industrial licensing was abolished for almost all industries, except a few such as alcohol, explosives, hazardous chemicals, and defence-related equipment, reducing government control.
2. De-reservation for Public Sector: Industries reserved for the public sector were drastically reduced, with only Atomic Energy and Railways remaining, allowing greater private sector participation.
3. Removal of MRTP Restrictions: Pre-approval under the MRTP Act was removed, and it was later replaced by the Competition Act, with the Competition Commission of India (CCI) ensuring fair competition.
4. Foreign Investment & Technology: FDI limits were increased (from 40% to 51 % and later up to 100% in some sectors), along with automatic approval for foreign technology agreements, encouraging global integration.
5. Public Sector Reforms: Efficient PSUs were given greater autonomy (leading to Ratna categories), while sick units were referred to BIFR for revival or closure.
6. Financial & Location Policy: Industrial location restrictions were relaxed, and the mandatory convertibility clause of loans into equity was abolished, improving financial flexibility.

III. Impact of the Policy (1991-2000):

1. Accelerated Growth: Industries expanded rapidly, especially in sectors like automobiles, pharmaceuticals, IT, and consumer goods, leading to higher production and modernization.
2. Increased Investment: There was a sharp rise in FDI and FII inflows, bringing capital, technology, and global business practices into India.
3. Diversification: New industries such as software and IT services emerged, and private participation increased in sectors like telecom, banking, and insurance, transforming the industrial landscape.
4. Challenges: Despite its success, the policy led to some issues such as jobless growth (growth without sufficient employment), regional imbalances, and unequal distribution of income and wealth.

Question 2.
Explain problems of MSMEs in India.
Answer:
Challanges Faced by MSMEs

1. Financial Constraints

• Limited Access to Credit: MSMEs often struggle to obtain loans from formal institutions due to lack of collateral, strict lending norms, and complex procedures. Many depend on costly informal sources, limiting growth and modernization.
• Delayed Payments: Delays in payments from large companies and government agencies disrupt cash flow and create working capital shortages.
• High Cost of Credit: Even when loans are available, higher interest rates increase financial burden and reduce profitability.

2. Infrastructure Deficiencies

• Poor Infrastructure: Inadequate roads, irregular power supply, water shortages, and weak internet connectivity increase production costs and delay deliveries.
• Limited Access to Resources: MSMEs often face shortages of quality raw materials, skilled labour, and modern technology, restricting expansion.

3. Technological Backwardness

• Outdated Technologies: Use of old machinery and traditional methods results in low efficiency, poor quality, and higher wastage.
• Limited Adoption of New Technologies: High costs and lack of technical knowledge hinder modernization and competitiveness.

4. Skills Gap

• Lack of Skilled Labour: Shortages of workers trained in advanced machinery, digital tools, and modern techniques affect productivity.
• Training and Development Issues: Limited access to skill-development programmes restricts workforce upgradation.

5. Regulatory and Compliance Issues

• Complex Regulations: Compliance with multiple laws is often costly and time-consuming for small enterprises.
• Bureaucracy: Excessive paperwork and delays in approvals hinder business operations and growth.

6. Marketing and Sales Challenges

• Limited Marketing Capabilities: Lack of funds and expertise in branding, advertising, and market research limits market reach.
• Intense Competition: MSMEs face strong competition from large industries, multinational companies, and online retailers.

7. Other Challenges

• Low Productivity: Small-scale production often leads to higher costs and lower efficiency.
• Lack of Standardization: Inconsistent quality control affects product reliability and operational efficiency.
• Limited Information Access: Lack of awareness about market trends, technologies, and government schemes restricts growth.
• External Shocks: Economic crises, pandemics, and natural disasters can severely disrupt MSME operations.

Question 3.
Evaluate the post 2000 scenerio of manufacturing sector.
Answer:
Manufacturing Sector Reforms in India Since 2000: The post-2000 period witnessed a major shift in India’s manufacturing strategy from restrictive policies to infrastructure development, policy reforms, and incentive-based growth.

1. Shift from EPZs to Special Economic Zones (SEZs)

• Export Processing Zones (EPZs) achieved limited success due to rigid regulations. To address this, the government introduced the SEZ Policy (2000) and SEZ Act (2005).
• SEZs offered better infrastructure, tax incentives, and simplified procedures, attracting domestic and foreign investment.
• As a result, 276 operational SEZs generated exports worth 4.47 lakh crore (2024-25) and provided employment to about 31 lakh people.

2. Large-Scale Infrastructure:

• Industrial corridors integrate transport, power, and logistics networks to improve efficiency and reduce costs.
• The National Industrial Corridors Development Corporation (NIDDC) was established in 2016. By 2023, 11 industrial corridors had been announced, including Delhi-Mumbai (DMIC) and Vizag-Chennai (VCIC).
• The PM MITRA Scheme (2021) promotes world-class textile parks with modern infrastructure.

3. National Manufacturing Policy (NMP), 2011

• The NMP aimed to raise manufacturing growth to 12-14%, increase its GDP share to 25%, and create 100 million jobs.
• It introduced National Investment and Manufacturing Zones (NIMZs) with advanced infrastructure.
• These zones encourage industrial investment and employment generation.

4. Make in India and NIIF

• Make in India (2014) seeks to transform India into a global manufacturing hub by promoting investment, innovation, and skill development across 25 sectors.
• National Investment and Infrastructure Fund (NIIF) (2015) was established to finance long-term infrastructure projects.
• Together, they strengthen industrial capacity and attract global investors.

5. PLI Scheme and Infrastructure Support

• Production Linked Incentive (PLI) Scheme (2020-21) provides financial incentives for increased production in 14 key sectors, including electronics and pharmaceuticals.
• Initiatives such as the National Infrastructure Pipeline (NIP) and PM Gati Shakti improve infrastructure and logistics.
• These measures reduce costs and enhance global competitiveness.

6. Current Standing and Challenges

• Despite reforms, India’s share in global manufacturing is about 2.8%, much lower than China’s.
• Infrastructure gaps, skill shortages, and employment challenges continue to limit the sector’s full potential.

Multiple Choice Questions

Question 1.
Industry Sector includes;
1) Manufacturing
2) Construction
3) Electricity, Gas and Water Supply
4) All the above
Answer:
4) All the above

Question 2.
Which of the following industry does not require to get License:
1) Ammunition
2) Hazardous Chemicals
3) Small Scale Industries
4) Industrial Explosives
Answer:
3) Small Scale Industries

Question 3.
Which of the following Authority regulates Insurance in Indias:
1) RBI
2) IRDA
3) NABARD
4) SIDBI
Answer:
2) IRDA

Question 4.
Make in India Scheme was launched in the year:
1) 2014
2) 2018
3) 2021
4) 2024
Answer:
1) 2014

Question 5.
Limit of MUDRA loan amount to SISHU category:
1) Rs.0.5 lakhs
2) Rs.2.0 lakhs
3) Rs.5.0 lakhs
4) Rs. 10.00 lakhs
Answer:
1) Rs.0.5 lakhs

Question 6.
In which year Golden Quadrangle Programme was lauched?
1) 1999
2) 2001
3) 2014
4) 2024
Answer:
2) 2001

Question 7.
Which is not a core industry?
1) Refinery
2) Cement
3) Fertilisers
4) Textiles
Answer:
4) Textiles

Question 8.
Which of the following statements is correct?
1) MRTP Act was repealed by FEMA
2) The functions of MRTP Act were taken over by CCI
3) Prior to MRTPA restricted Public Sector expansion
4) MRTP companies limits were enhanced in 2002
Answer:
2) The functions of MRTP Act were taken over by CCI

Question 9.
Who are the beneficiaries of Stand-up Scheme?
1) Women entrepreneurs
2) SC and ST entrepreneurs
3) Both (1) and (2)
4) Backward entrepreneurs
Answer:
3) Both (1) and (2)

Question 10.
Number of sectors covered by Production Linked Incentive Scheme covers
1) 8
2) 10
3) 12
4) 14
Answer:
4) 14

Fill in the Blanks

Question 1.
PM MITRA is related to ___________ parks
Answer:
Industrial textile

Question 2.
The investment limit for Micro enterprises under 2025 definition is Rs. ?___________ crores.
Answer:
2.5

Question 3.
The newest Railway zone, South Coast Railways, established with as ___________ headquarters.
Answer:
Visakhapatnam

Question 4.
According to Travel and Tourism Index (TTDI) 2024 Report, published by World Economic Forum, India has ranked ___________ among 119 countries
Answer:
39th

Question 5.
Expansion of MUDRA is ___________ Development and Refinance Agency.
Answer:
Micro Unit Development and Refinance Agency

One Word Answers

Question 1.
What is the purpose of Special Economic Zones?
Answer:
Export promotion

Question 2.
In which year P.M Jandhan Yojana introduced?
Answer:
2014

Question 3.
In which fund are the disinvestment proceeds deposited?
Answer:
National Investment Fund

Question 4.
What does Start-up scheme aims to,promote?
Answer:
Promotion of innovations

Question 5.
What is the function of DIPAM?
Answer:
Management of disinvestment