Book Back Questions

• Ln11 : Q19 , 20
• Ln12 : Q2 5 Mark, Q5 3 Mark
• Ln13 : Q5(All) || 6th subdivision very important (Ph Based Question)
• Ln13 : Electrophilic substitution reaction(Imp) see other properties in box

Q1.A compound ‘A’ of molecular formula C₂H₃N on reduction with Na(Hg)/C₂H₅OH gives ‘B’ of molecular formula C₂H₇N which undergoes carbylamine test. Compound ‘B’ on reaction with nitrous acid gives compound ‘C’ of molecular formula C₂H₆O by liberating nitrogen. Identify A, B and C and write the reactions involved. (Repeating)

Answer:

• A = Acetonitrile (CH₃CN)
• B = Ethylamine (C₂H₅NH₂)
• C = Ethanol (C₂H₅OH)

Reason:

1. Compound B gives carbylamine test, so it must be a primary amine. Hence B = Ethylamine.
2. Primary amine reacts with nitrous acid to give alcohol with liberation of N₂ gas, so C = Ethanol.
3. The compound A (C₂H₃N) which on reduction gives ethylamine is acetonitrile.

Reactions involved:

1. Reduction:

CH₃CN + 4[H] → CH₃CH₂NH₂

(Acetonitrile) (Ethylamine)

Reagent: Na(Hg) / C₂H₅OH

1. Reaction with nitrous acid:

CH₃CH₂NH₂ + HNO₂ → CH₃CH₂OH + N₂ ↑ + H₂O

(Ethylamine) (Ethanol)

Thus, A = CH₃CN (Acetonitrile), B = C₂H₅NH₂ (Ethylamine), C = C₂H₅OH (Ethanol).

Q2.An organic compound A with molecular formula C₇H₆O reduces Tollens’ reagent but does not reduce Fehling’s solution. A reacts with 50% NaOH to give B and C. C on treatment with soda lime gives D. Identify A, B, C and D. Write the equations.

Answer:

• A = Benzaldehyde (C₆H₅CHO)
• B = Benzyl alcohol (C₆H₅CH₂OH)
• C = Sodium benzoate (C₆H₅COONa)
• D = Benzene (C₆H₆)

Reason:

1. Compound A reduces Tollens’ reagent but not Fehling’s solution, which indicates it is an aromatic aldehyde. Hence A = Benzaldehyde.
2. Benzaldehyde with 50% NaOH undergoes Cannizzaro reaction producing benzyl alcohol and sodium benzoate.
3. Sodium benzoate on heating with soda lime undergoes decarboxylation to form benzene.

Reactions involved:

1. Cannizzaro reaction:

2C₆H₅CHO + NaOH → C₆H₅CH₂OH + C₆H₅COONa

(Benzaldehyde)  (Benzyl alcohol) (Sodium benzoate)

1. Decarboxylation with soda lime:

C₆H₅COONa + NaOH → C₆H₆ + Na₂CO₃

(Sodium benzoate) (Benzene)

Thus, A = Benzaldehyde, B = Benzyl alcohol, C = Sodium benzoate and D = Benzene.

3. Aldol reaction:
CH₃COCH₃ + HCHO
dil. NaOH
→ CH₃COCH₂CH₂OH → C₄H₈O₂ (β-hydroxy carbonyl compound)

Answer :

• A = Ethanal (CH₃CHO)
• B = 3-Hydroxybutanal
• C = Crotonaldehyde (CH₃CH=CHCHO).

Reason:

This reaction involves Rosenmund reduction followed by reactions with NaOH.

1. First step (Rosenmund reduction):
   CH₃COCl + H₂ —Pd/BaSO₄→ CH₃CHO

So,

A = Ethanal (CH₃CHO)

1. Second step (reaction with NaOH – Aldol reaction):
   2 CH₃CHO —NaOH→ CH₃CH(OH)CH₂CHO

So,

B = 3-Hydroxybutanal (Aldol)

1. Third step (heating / dehydration):
   CH₃CH(OH)CH₂CHO —Δ→ CH₃CH=CHCHO

So,

C = Crotonaldehyde (But-2-enal)

Q2.An organic compound (A) – C₃H₈O₃ used as a sweetening agent, on oxidation with Fenton’s reagent (H₂O₂ / Fe²⁺) gives a mixture of compounds B and C. Identify A, B and C. Write possible reactions.

Answer

• A = Glycerol (Propane-1,2,3-triol)
• B = Glyceraldehyde
• C = Dihydroxyacetone

Reason

The molecular formula C₃H₈O₃ and use as a sweetening agent indicate glycerol.
On oxidation with Fenton’s reagent, glycerol undergoes oxidation of alcohol groups forming glyceraldehyde and dihydroxyacetone.

Reactions

1. Oxidation to glyceraldehyde

HO–CH₂–CHOH–CH₂OH
(H₂O₂ / Fe²⁺)
→ HO–CH₂–CHOH–CHO
(Glyceraldehyde)

2. Oxidation to dihydroxyacetone

HO–CH₂–CHOH–CH₂OH
(H₂O₂ / Fe²⁺)
→ HO–CH₂–CO–CH₂OH
(Dihydroxyacetone)

Q3.In the following Question A : is Methyl Amine & B N- Methyl hydroxylamine (It wil be blank in the questions)

Q4.A (C₂H₅NO) —(Br₂ / KOH)→ CH₅N (B). Identify A and B and write the reaction.

Answer

• A = Acetamide (CH₃CONH₂)
• B = Methylamine (CH₃NH₂)

Reason

The reagent Br₂ / KOH indicates the Hofmann bromamide reaction, where an amide loses one carbon atom and forms a primary amine.
Thus acetamide (C₂H₅NO) gives methylamine (CH₃NH₂).

Reaction

CH₃CONH₂ + Br₂ + 4KOH
→ CH₃NH₂ + K₂CO₃ + 2KBr + 2H₂O

Q4.Identify A and B

A → (Na(Hg) / C₂H₅OH) → CH₃–CH₂–NH₂

B → (Na(Hg) / C₂H₅OH) → CH₃–NH–CH₃

Answer:

• A = CH₃–C≡N (Acetonitrile / Methyl cyanide)
• B = CH₃–N≡C (Methyl isocyanide)

Q5.Compound (A) → C₃H₈O₃ oxidised with sodium hypobromite to give a mixture of compounds (B) and (C). Identify A, B and C. Write the possible reactions. Name the mixture of (B) and (C).

Q6.Phenol is distilled with Zn dust gives compound A. A reacts with propene gives compound B, which on air oxidation gives compound C. Identify A, B and C.

Answer:

• A = Benzene (C₆H₆)
• B = Cumene (Isopropyl benzene)
• C = Phenol

Reactions:

1. Phenol + Zn → Benzene + ZnO
2. Benzene + Propene → Cumene
   (Friedel–Crafts alkylation)
3. Cumene + O₂ → Cumene hydroperoxide → Phenol + Acetone

Q7.Phenol is distilled with Zn dust gives compound A. A reacts with propene gives compound B, which on air oxidation gives compound C. Identify A, B and C.

Answer:

• A = Benzene (C₆H₆)
• B = Cumene (Isopropyl benzene)
• C = Phenol (C₆H₅OH)

Reason:

• Phenol on distillation with Zn dust forms benzene.
• Benzene + propene gives cumene (isopropyl benzene).
• Cumene on air oxidation forms phenol (cumene process).

Reactions:

1. C₆H₅OH + Zn → C₆H₆ + ZnO
2. C₆H₆ + CH₃–CH=CH₂ → C₆H₅–CH(CH₃)₂ (Cumene)
3. Cumene + O₂ → Cumene hydroperoxide → C₆H₅OH (Phenol) + Acetone.

Q8.Ethanoic acid —(SOCl₂)→ A —(Pd/BaSO₄)→ B. Identify A and B.

Answer:

• A = Acetyl chloride (CH₃COCl)
• B = Ethanal (CH₃CHO)

Reactions:

1. CH₃COOH + SOCl₂ → CH₃COCl + SO₂ + HCl
2. CH₃COCl + H₂ —Pd/BaSO₄→ CH₃CHO (Ethanal)

Q.9

Answer:

• A = Aniline (C₆H₅NH₂)
• B = Benzene diazonium chloride (C₆H₅N₂⁺Cl⁻)
• C = Phenol (C₆H₅OH)

Reactions:

1. Reduction of nitrobenzene:
   C₆H₅NO₂ + 6[H] —Fe/HCl→ C₆H₅NH₂ + 2H₂O
2. Diazotization:
   C₆H₅NH₂ + NaNO₂ + HCl (273 K) → C₆H₅N₂⁺Cl⁻ + 2H₂O
3. Hydrolysis:
   C₆H₅N₂⁺Cl⁻ + H₂O (283 K) → C₆H₅OH + N₂ + HCl