Q.Give the structures of A, B and C in the following reactions:

Here are the structures for each sequence, with the key reaction type noted.

(i) Concept: Nucleophilic substitution (SN2) followed by Hofmann rearrangement.

• $CH_3CH_2I \xrightarrow{NaCN} CH_3CH_2CN$ (A, propanenitrile)
• Partial hydrolysis gives $CH_3CH_2CONH_2$ (B, propanamide) …

(i) $CH_3CH_2I \xrightarrow{NaCN} A \xrightarrow{\text{partial hydrolysis}} B \xrightarrow{NaOH,\ Br_2} C$

Concept: This is a classic chain-extension via cyanide, followed by controlled hydrolysis to an amide, then the Hofmann rearrangement to an amine.

1. Step 1: $S_N2$ substitution. Iodide is a good leaving group. Cyanide ion ($^-CN$) is a strong nucleophile and attacks the primary carbon.

   $CH_3CH_2I + NaCN \rightarrow CH_3CH_2CN + NaI$

   A = ethyl cyanide (propanenitrile).

2. Step 2: Partial hydrolysis. A nitrile can be fully hydrolysed to a carboxylic acid, but partial hydrolysis (using controlled conditions, e.g., dilute acid or base at moderate temperature) stops at the amide.

   $CH_3CH_2CN + H_2O \xrightarrow{\text{partial}} CH_3CH_2CONH_2$

   B = propanamide.

3. Step 3: Hofmann rearrangement. Treating an amide with bromine in aqueous NaOH converts it to a primary amine with one fewer carbon. The mechanism: bromination of the amide nitrogen, then rearrangement to an isocyanate, which hydrolyses to the amine.

   $CH_3CH_2CONH_2 + Br_2 + 4NaOH \rightarrow CH_3CH_2NH_2 + 2NaBr + Na_2CO_3 + 2H_2O$

   C = ethylamine.

(ii) $C_6H_5N_2^+Cl^- \xrightarrow{CuCN} A \xrightarrow{H_2O/H^+} B \xrightarrow{NH_3,\ \Delta} C$

Concept: The diazonium group is replaced by cyanide (Sandmeyer reaction), then the nitrile is hydrolysed to a carboxylic acid, which is then converted to an amide.

1. Step 1: Sandmeyer reaction. Diazonium salts undergo substitution with cuprous cyanide to give aryl cyanides.

   $C_6H_5N_2^+Cl^- + CuCN \rightarrow C_6H_5CN + N_2 + CuCl$

   A = benzonitrile.

2. Step 2: Acid hydrolysis. The nitrile is fully hydrolysed under acidic conditions to benzoic acid.

   $C_6H_5CN + 2H_2O \xrightarrow{H^+} C_6H_5COOH + NH_4^+$

   B = benzoic acid.

3. Step 3: Amide formation. Heating a carboxylic acid with ammonia gives the ammonium salt, which on further heating dehydrates to the amide.

   $C_6H_5COOH + NH_3 \xrightarrow{\Delta} C_6H_5COONH_4 \xrightarrow{-\ H_2O} C_6H_5CONH_2$

   C = benzamide.

(iii) $CH_3CH_2Br \xrightarrow{KCN} A \xrightarrow{LiAlH_4} B \xrightarrow{HNO_2,\ 0^\circ C} C$

Concept: Cyanide substitution, then reduction to a primary amine, then diazotisation and replacement by hydroxyl.

1. Step 1: $S_N2$ substitution. Ethyl bromide reacts with KCN to give propanenitrile.

   $CH_3CH_2Br + KCN \rightarrow CH_3CH_2CN + KBr$

   A = propanenitrile.

2. Step 2: Reduction with $LiAlH_4$. Lithium aluminium hydride reduces nitriles to primary amines.

   $CH_3CH_2CN \xrightarrow{LiAlH_4} CH_3CH_2CH_2NH_2$

   B = propylamine (1-aminopropane).

3. Step 3: Diazotisation and replacement. At $0^\circ C$, nitrous acid ($HNO_2$) converts a primary aliphatic amine to a diazonium salt, which is unstable and immediately decomposes to a carbocation, then to an alcohol (via $S_N1$ with water).

   $CH_3CH_2CH_2NH_2 + HNO_2 \rightarrow CH_3CH_2CH_2N_2^+ \rightarrow CH_3CH_2CH_2OH + N_2$

   C = propan-1-ol.

(iv) $C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow{NaNO_2 + HCl,\ 273\,K} B \xrightarrow{H_2O/H^+,\ \Delta} C$

Concept: Reduction of nitrobenzene to aniline, diazotisation, then hydrolysis of the diazonium salt to phenol.

1. Step 1: Reduction. Nitrobenzene is reduced to aniline using Fe/HCl (or Sn/HCl).

   $C_6H_5NO_2 + 6[H] \xrightarrow{Fe/HCl} C_6H_5NH_2 + 2H_2O$

   A = aniline.

2. Step 2: Diazotisation. At $273\,K$, aniline reacts with $NaNO_2/HCl$ to form the diazonium salt.

   $C_6H_5NH_2 + NaNO_2 + 2HCl \rightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O$

   B = benzenediazonium chloride.

3. Step 3: Hydrolysis. Heating the diazonium salt with water replaces the $-N_2^+$ group with $-OH$, giving phenol.

   $C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2 + HCl$

   C = phenol.

(v) $CH_3COOH \xrightarrow{NH_3,\ \Delta} A \xrightarrow{NaOBr} B \xrightarrow{NaNO_2/HCl} C$ …

Method: Retrosynthetic Analysis with Reaction Mapping

This method works backwards from known reagents and conditions to identify each intermediate and final product. For each sequence, we identify the type of reaction at each step, then deduce the structure.

(i) $CH_3CH_2I \xrightarrow{NaCN} A \xrightarrow{\text{partial hydrolysis}} B \xrightarrow{NaOH,\ Br_2} C$

Step 1: $CH_3CH_2I + NaCN \rightarrow$ Nucleophilic substitution (SN2)

• CN⁻ replaces I⁻
• A = $CH_3CH_2CN$ (ethyl cyanide / propanenitrile)

Step 2: Partial hydrolysis of nitrile

• $RCN \xrightarrow{H_2O/OH^-} RCONH_2$ (amide)
• B = $CH_3CH_2CONH_2$ (propanamide)

Step 3: $RCONH_2 \xrightarrow{NaOH,\ Br_2}$ Hoffmann bromamide degradation

• Amide loses CO₂, gives primary amine with one less carbon
• C = $CH_3CH_2NH_2$ (ethylamine)

Final: A = $CH_3CH_2CN$, B = $CH_3CH_2CONH_2$, C = $CH_3CH_2NH_2$

(ii) $C_6H_5N_2^+Cl^- \xrightarrow{CuCN} A \xrightarrow{H_2O/H^+} B \xrightarrow{NH_3,\ \Delta} C$

Step 1: Diazonium salt + CuCN → Sandmeyer reaction

• CN⁻ replaces N₂⁺
• A = $C_6H_5CN$ (benzonitrile)

Step 2: Acid hydrolysis of nitrile

• $C_6H_5CN \xrightarrow{H_2O/H^+} C_6H_5COOH$
• B = $C_6H_5COOH$ (benzoic acid)

Step 3: Benzoic acid + $NH_3$ (heat) → Ammonium salt → amide

• $C_6H_5COOH \xrightarrow{NH_3,\ \Delta} C_6H_5CONH_2$
• C = $C_6H_5CONH_2$ (benzamide)

Final: A = $C_6H_5CN$, B = $C_6H_5COOH$, C = $C_6H_5CONH_2$

(iii) $CH_3CH_2Br \xrightarrow{KCN} A \xrightarrow{LiAlH_4} B \xrightarrow{HNO_2,\ 0^\circ C} C$

Step 1: $CH_3CH_2Br + KCN \rightarrow$ SN2

• A = $CH_3CH_2CN$ (propanenitrile)

Step 2: $LiAlH_4$ reduces nitrile to primary amine

• $RCN \xrightarrow{LiAlH_4} RCH_2NH_2$
• B = $CH_3CH_2CH_2NH_2$ (propan-1-amine)

Step 3: Primary aliphatic amine + $HNO_2$ at $0^\circ C$ → diazotisation, but the aliphatic diazonium ion is unstable — it decomposes at once, losing $N_2$, and water traps the resulting carbocation to give the alcohol

• $CH_3CH_2CH_2NH_2 \xrightarrow{HNO_2} [CH_3CH_2CH_2N_2^+] \xrightarrow{H_2O,\ -N_2} CH_3CH_2CH_2OH$
• C = $CH_3CH_2CH_2OH$ (propan-1-ol)

Final: A = $CH_3CH_2CN$, B = $CH_3CH_2CH_2NH_2$, C = $CH_3CH_2CH_2OH$

(iv) $C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow{NaNO_2 + HCl,\ 273\,K} B \xrightarrow{H_2O/H^+,\ \Delta} C$

Step 1: Nitrobenzene reduction → aniline

• $C_6H_5NO_2 \xrightarrow{Fe/HCl} C_6H_5NH_2$
• A = $C_6H_5NH_2$ (aniline)

Step 2: Aniline + $NaNO_2/HCl$ at $273\,K$ → diazotisation

• B = $C_6H_5N_2^+Cl^-$ (benzenediazonium chloride — aromatic, so stable at this temperature)

Step 3: Diazonium salt + $H_2O/H^+$ (heat) → hydrolysis

• $C_6H_5N_2^+ \xrightarrow{H_2O} C_6H_5OH + N_2$
• C = $C_6H_5OH$ (phenol)

Final: A = $C_6H_5NH_2$, B = $C_6H_5N_2^+Cl^-$, C = $C_6H_5OH$

(v) $CH_3COOH \xrightarrow{NH_3,\ \Delta} A \xrightarrow{NaOBr} B \xrightarrow{NaNO_2/HCl} C$

Step 1: Acetic acid + $NH_3$ (heat) → ammonium acetate → acetamide

• $CH_3COOH \xrightarrow{NH_3} CH_3COONH_4 \xrightarrow{\Delta} CH_3CONH_2$
• A = $CH_3CONH_2$ (acetamide) …

Here are the common mistakes students make in Nucleophilic Substitution reaction sequences, organized by the specific reaction step, along with how to avoid each.

1. Confusing Cyanide as a Nucleophile vs. a Reducing Agent

Mistake: Treating NaCN or KCN as a reducing agent (like NaBH₄).

Reality: Cyanide ion (:CN⁻) is a strong nucleophile and a weak base. It attacks the electrophilic carbon (in SN² or SN¹ fashion) to form a nitrile (–C≡N).

How to avoid:

• Always check the substrate:
  • Alkyl halide + NaCN → alkyl cyanide (nitrile).
  • Aryl diazonium salt + CuCN → aryl cyanide (Sandmeyer reaction).
• Never write –CN as –NC (isocyanide) unless specified (e.g., AgCN gives isocyanide).

Example (i):

CH₃CH₂I + NaCN → CH₃CH₂CN (A = propanenitrile), not CH₃CH₂NC.

2. Forgetting Partial Hydrolysis of Nitrile Stops at Amide

Mistake: Writing complete hydrolysis to carboxylic acid when the question says partial hydrolysis.

Reality: Partial hydrolysis (controlled H₂O, mild conditions) stops at the amide (–CONH₂), not the acid.

How to avoid:

• Remember the sequence: R–CN → (partial) R–CONH₂ → (full) R–COOH.
• If the reagent is H₂O/H⁺ (without “partial”), assume full hydrolysis to acid.
• If it says “partial hydrolysis” or “H₂O (limited)”, write amide.

Example (i):

CH₃CH₂CN → partial hydrolysis → CH₃CH₂CONH₂ (B = propanamide).

3. Misapplying the Hoffmann Bromamide Degradation

Mistake: Thinking NaOH/Br₂ on an amide gives a bromo compound or an amine directly without rearrangement.

Reality: This is the Hoffmann bromamide reaction:

R–CONH₂ + Br₂ + NaOH → R–NH₂ (primary amine with one less carbon).

How to avoid:

• Count carbons: the product amine has n–1 carbons relative to the amide.
• The mechanism involves rearrangement (migration of R from C=O to N), so the alkyl group stays intact but the carbonyl carbon is lost as CO₂.

Example (i):

CH₃CH₂CONH₂ → CH₃CH₂NH₂ (C = ethylamine), not CH₃CH₂CONHBr.

4. Forgetting the Sandmeyer Reaction Uses CuCN, Not NaCN

Mistake: Using NaCN directly on a diazonium salt (which would give a different product or no reaction).

Reality: Aryl diazonium salts require CuCN (Sandmeyer) to replace N₂⁺ with –CN. NaCN alone gives a diazo coupling product or tar.

How to avoid:

• Memorize:
  • ArN₂⁺X⁻ + CuCN → Ar–CN (Sandmeyer).
  • ArN₂⁺X⁻ + H₂O → Ar–OH (phenol).
• The copper salt is essential for the radical/redox mechanism.

Example (ii):

C₆H₅N₂⁺Cl⁻ + CuCN → C₆H₅CN (A = benzonitrile).

5. Writing the Wrong Product for Nitrile Reduction

Mistake: Reducing a nitrile with LiAlH₄ and writing an aldehyde or a secondary amine.

Reality: LiAlH₄ reduces –C≡N to primary amine (–CH₂NH₂), not aldehyde (that’s DIBAL-H).

How to avoid:

• LiAlH₄ on nitrile → 1° amine (adds 2 H atoms to the triple bond).
• NaBH₄ does not reduce nitriles effectively.
• DIBAL-H at low temperature gives aldehyde.

Example (iii):

CH₃CH₂CN → LiAlH₄ → CH₃CH₂CH₂NH₂ (B = propylamine).

6. Forgetting That HNO₂ at 0°C Converts 1° Amine to Diazonium Salt

Mistake: Writing a nitroso compound or an alcohol directly from a primary amine with HNO₂.

Reality:

• 1° aliphatic amine + HNO₂ (0°C) → diazonium salt (unstable, decomposes to carbocation → mixture of alkene, alcohol, etc.).
• 1° aromatic amine + HNO₂ (0°C) → stable diazonium salt.

How to avoid:

• For aliphatic: the diazonium salt is transient; the final product is often an alcohol (from H₂O attack) or alkene (elimination).
• For aromatic: the diazonium salt is stable and can be used for coupling or substitution.

Example (iii):

CH₃CH₂CH₂NH₂ → HNO₂ → unstable diazonium → CH₃CH₂CH₂OH (C = propanol, major).

Example (vi):

C₆H₅NH₂ → HNO₂ → C₆H₅N₂⁺ (stable, B).

7. Mixing Up Diazonium Coupling vs. Substitution

Mistake: When a diazonium salt reacts with a phenol, writing a substitution product (like chlorobenzene) instead of an azo compound.

Reality: Diazonium salts couple with activated aromatic rings (phenols, anilines) at the para position (or ortho if para blocked) to form azo dyes (–N=N–).

How to avoid:

• Coupling requires an electron-rich aromatic (phenol, aniline) in basic/neutral medium.
• The product has a –N=N– bridge between two aromatic rings.
• Substitution (like replacing N₂⁺ with Cl, Br, CN) requires CuX (Sandmeyer) or HX (without Cu, gives Ar–X but with side products).

Example (vi):

C₆H₅N₂⁺ + C₆H₅OH → HO–C₆H₄–N=N–C₆H₅ (C = p-hydroxyazobenzene).

8. Forgetting That NaOBr (or Br₂/NaOH) on Amide Is Hoffmann, Not Oxidation

Mistake: Thinking NaOBr oxidizes the amide to a nitro compound or a carboxylic acid.

Reality: NaOBr (sodium hypobromite) is the reagent for Hoffmann rearrangement of amides to amines (same as Br₂/NaOH).

How to avoid: …