Q.How will you convert:

Key Idea: Chain-length change reactions of amines

Chain-descending conversions go through the amide, then Hofmann bromamide degradation ($\ce{Br2/NaOH}$), which removes one carbon. Chain-extending conversions go through the cyanide route ($\ce{R-X ->[KCN] R-CN ->[H3O+] R-COOH}$), which adds one carbon. Amines and alcohols interconvert via diazotisation ($\ce{HNO2}$) and oxidation/reduction.

1. Ethanoic acid -> methanamine - form the amide, then Hofmann-degrade it: $\ce{CH3COOH ->[NH3] CH3CONH2 ->[Br2/NaOH] CH3NH2}$.
2. Hexanenitrile -> 1-aminopentane - hexanenitrile ($\ce{CH3(CH2)4CN}$) has 6 carbons; direct reduction would give hexan-1-amine (6 C), not the 5-carbon target. So hydrolyse the nitrile to the 6-carbon acid first, then Hofmann-degrade that amide to lose one carbon: $\ce{CH3(CH2)4CN ->[H2O/H+] CH3(CH2)4COOH ->[NH3] CH3(CH2)4CONH2 ->[Br2/NaOH] CH3(CH2)3CH2NH2}$.
3. Methanol -> ethanoic acid - extend the chain via the cyanide route: $\ce{CH3OH ->[HI] CH3I ->[KCN] CH3CN ->[H3O+] CH3COOH}$.
4. Ethanamine -> methanamine - Hofmann degradation only works on an amide, so the amine must first be converted to the corresponding acid (via diazotisation to the alcohol, then oxidation): $\ce{CH3CH2NH2 ->[HNO2] CH3CH2OH ->[K2Cr2O7/H+] CH3COOH ->[NH3] CH3CONH2 ->[Br2/NaOH] CH3NH2}$.
5. Ethanoic acid -> propanoic acid - extend by one carbon: reduce to the alcohol, convert to the halide, then run the cyanide route: $\ce{CH3COOH ->[LiAlH4] CH3CH2OH ->[PBr3] CH3CH2Br ->[KCN] CH3CH2CN ->[H3O+] CH3CH2COOH}$.
6. Methanamine -> ethanamine - same idea as (v) applied to an amine: go via the alcohol and halide to the nitrile, then reduce the nitrile (now one carbon longer) back to an amine: $\ce{CH3NH2 ->[HNO2] CH3OH ->[HI] CH3I ->[KCN] CH3CN ->[H2/Ni] CH3CH2NH2}$.
7. Nitromethane -> dimethylamine - reduce the nitro group, then methylate under controlled (1 : 1) conditions so the reaction stops at the secondary amine: $\ce{CH3NO2 ->[H2/Ni] CH3NH2 ->[CH3I (1 equiv.)] (CH3)2NH}$. …

(i) Ethanoic acid to methanamine

Descend by one carbon: form the amide, then apply Hofmann degradation.

1. $\ce{CH3COOH ->[NH3, \Delta] CH3CONH2}$ (ethanamide)
2. $\ce{CH3CONH2 ->[Br2/NaOH] CH3NH2}$ (Hofmann bromamide degradation)

(ii) Hexanenitrile to 1-aminopentane

Hexanenitrile, $\ce{CH3(CH2)4CN}$, has 6 carbons; 1-aminopentane has only 5, so a carbon must be removed. Direct reduction of the nitrile would give hexan-1-amine (6 C) - not what is wanted - so the nitrile is hydrolysed to the acid first, and that amide is degraded by Hofmann's method.

1. $\ce{CH3(CH2)4CN + 2H2O ->[H+] CH3(CH2)4COOH + NH3}$ (hexanoic acid)
2. $\ce{CH3(CH2)4COOH ->[NH3] CH3(CH2)4CONH2}$ (hexanamide)
3. $\ce{CH3(CH2)4CONH2 ->[Br2/NaOH] CH3(CH2)3CH2NH2}$ (pentan-1-amine)

(iii) Methanol to ethanoic acid

Extend the chain by one carbon via the cyanide route.

1. $\ce{CH3OH ->[HI] CH3I}$
2. $\ce{CH3I ->[KCN] CH3CN}$ (acetonitrile)
3. $\ce{CH3CN + 2H2O ->[H+] CH3COOH + NH3}$

(iv) Ethanamine to methanamine

Hofmann degradation only removes a carbon from an amide, not an amine directly, so the amine is first converted to the corresponding acid (via diazotisation to the alcohol, then oxidation), then taken through the amide/Hofmann sequence.

1. $\ce{CH3CH2NH2 ->[HNO2] CH3CH2OH}$
2. $\ce{CH3CH2OH ->[K2Cr2O7/H+] CH3COOH}$
3. $\ce{CH3COOH ->[NH3] CH3CONH2}$
4. $\ce{CH3CONH2 ->[Br2/NaOH] CH3NH2}$

(v) Ethanoic acid to propanoic acid

Extend by one carbon: reduce to the alcohol, convert to the halide, then run the cyanide route.

1. $\ce{CH3COOH ->[LiAlH4] CH3CH2OH}$
2. $\ce{CH3CH2OH ->[PBr3] CH3CH2Br}$
3. $\ce{CH3CH2Br ->[KCN] CH3CH2CN}$
4. $\ce{CH3CH2CN + 2H2O ->[H+] CH3CH2COOH + NH3}$

(vi) Methanamine to ethanamine

Same idea as (v), applied to an amine: go via the alcohol and halide to the nitrile, then reduce the nitrile back to an amine (now one carbon longer).

1. $\ce{CH3NH2 ->[HNO2] CH3OH}$
2. $\ce{CH3OH ->[HI] CH3I}$
3. $\ce{CH3I ->[KCN] CH3CN}$
4. $\ce{CH3CN ->[H2/Ni] CH3CH2NH2}$

(vii) Nitromethane to dimethylamine

1. $\ce{CH3NO2 ->[H2/Ni] CH3NH2}$ (reduction of the nitro group) …

Here are the clear solution methods for each conversion, named and stepwise. Two workhorse tools do almost all of it: Hofmann bromamide degradation (removes one carbon: amide → amine) and the cyanide route (adds one carbon: halide → nitrile → acid or amine). Count carbons first, then pick the tool.

(i) Ethanoic acid into methanamine

Method: Amide formation, then Hofmann bromamide degradation (descend by one carbon: 2 C → 1 C)

Steps:

1. Amide formation: heat ethanoic acid with ammonia — the ammonium salt dehydrates to the amide. $\ce{CH3COOH ->[NH3, \Delta] CH3CONH2}$ (ethanamide)
2. Hofmann bromamide degradation: the amide loses its carbonyl carbon (as carbonate) and gives the amine with one carbon less. $\ce{CH3CONH2 + Br2 + 4NaOH -> CH3NH2 + Na2CO3 + 2NaBr + 2H2O}$ Result: Methanamine ($\ce{CH3NH2}$)

(ii) Hexanenitrile into 1-aminopentane

Method: Hydrolysis to the acid, amide formation, then Hofmann degradation (6 C → 5 C)

Hexanenitrile ($\ce{CH3(CH2)4CN}$) has 6 carbons; 1-aminopentane has 5. Direct reduction ($\ce{H2/Ni}$ or $\ce{LiAlH4}$) keeps all 6 carbons and gives hexan-1-amine — the wrong product. A carbon must be removed, and Hofmann degradation is the tool.

Steps:

1. Hydrolysis of the nitrile: $\ce{CH3(CH2)4CN + 2H2O ->[H+] CH3(CH2)4COOH + NH3}$ (hexanoic acid)
2. Amide formation: $\ce{CH3(CH2)4COOH ->[NH3, \Delta] CH3(CH2)4CONH2}$ (hexanamide)
3. Hofmann bromamide degradation: $\ce{CH3(CH2)4CONH2 ->[Br2/NaOH] CH3(CH2)3CH2NH2}$ Result: 1-aminopentane / pentan-1-amine ($\ce{CH3(CH2)3CH2NH2}$)

(iii) Methanol to ethanoic acid

Method: Cyanide chain extension (1 C → 2 C), then hydrolysis

Direct oxidation of methanol can only ever give the 1-carbon methanoic acid — the chain must first grow by one carbon.

Steps:

1. Halide formation: $\ce{CH3OH ->[HI] CH3I}$
2. Cyanide substitution ($S_N2$): $\ce{CH3I ->[KCN] CH3CN}$ (ethanenitrile/acetonitrile — now 2 carbons)
3. Hydrolysis: $\ce{CH3CN + 2H2O ->[H+] CH3COOH + NH3}$ Result: Ethanoic acid ($\ce{CH3COOH}$)

(iv) Ethanamine into methanamine

Method: Diazotisation → oxidation → amide → Hofmann degradation (2 C → 1 C)

Hofmann degradation removes a carbon from an amide, not from an amine directly — so the amine is first carried to the corresponding acid.

Steps:

1. Diazotisation to the alcohol: $\ce{CH3CH2NH2 ->[HNO2] CH3CH2OH}$ (the aliphatic diazonium ion is unstable and is displaced by water)
2. Oxidation: $\ce{CH3CH2OH ->[K2Cr2O7/H+] CH3COOH}$
3. Amide formation: $\ce{CH3COOH ->[NH3, \Delta] CH3CONH2}$
4. Hofmann degradation: $\ce{CH3CONH2 ->[Br2/NaOH] CH3NH2}$ Result: Methanamine ($\ce{CH3NH2}$)

(v) Ethanoic acid into propanoic acid

Method: Reduction → halide → cyanide chain extension → hydrolysis (2 C → 3 C)

Steps:

1. Reduction to the alcohol: $\ce{CH3COOH ->[LiAlH4] CH3CH2OH}$
2. Halide formation: $\ce{CH3CH2OH ->[PBr3] CH3CH2Br}$
3. Cyanide substitution: $\ce{CH3CH2Br ->[KCN] CH3CH2CN}$ (propanenitrile — now 3 carbons)
4. Hydrolysis: $\ce{CH3CH2CN + 2H2O ->[H+] CH3CH2COOH + NH3}$ Result: Propanoic acid ($\ce{CH3CH2COOH}$)

(vi) Methanamine into ethanamine

Method: Diazotisation → halide → cyanide chain extension → reduction (1 C → 2 C)

Steps:

1. Diazotisation to the alcohol: $\ce{CH3NH2 ->[HNO2] CH3OH}$
2. Halide formation: $\ce{CH3OH ->[HI] CH3I}$ …

Here are the common mistakes students make on these functional-group interconversions, part by part, and how to avoid each. The single biggest habit that prevents them all: count the carbons before choosing a route.

(i) Ethanoic acid → Methanamine

Common Mistake:

Making the amide and then reducing it with $\ce{LiAlH4}$: $\ce{CH3CONH2 ->[LiAlH4] CH3CH2NH2}$. That reduction keeps both carbons — it gives ethanamine, not methanamine.

How to Avoid:

The target has one carbon fewer than the acid, so the carbonyl carbon must be lost, not reduced. That is exactly what Hofmann bromamide degradation does:

$\ce{CH3COOH ->[NH3, \Delta] CH3CONH2 ->[Br2/NaOH] CH3NH2}$

Key Concept: $\ce{LiAlH4}$ on an amide preserves the carbon count; Hofmann degradation removes one carbon.

(ii) Hexanenitrile → 1-Aminopentane

Common Mistake:

Reducing the nitrile directly ($\ce{H2/Ni}$ or $\ce{LiAlH4}$). Hexanenitrile has 6 carbons, so reduction gives hexan-1-amine (6 C) — not the 5-carbon target.

How to Avoid:

Descend by one carbon via Hofmann degradation:

$\ce{CH3(CH2)4CN ->[H2O/H+] CH3(CH2)4COOH ->[NH3, \Delta] CH3(CH2)4CONH2 ->[Br2/NaOH] CH3(CH2)3CH2NH2}$

Key Concept: "Nitrile → amine" by reduction never changes the carbon count; when the target is shorter, go nitrile → acid → amide → Hofmann.

(iii) Methanol → Ethanoic acid

Common Mistake:

Simply oxidising methanol — that gives methanoic acid (1 C), not ethanoic acid (2 C).

How to Avoid:

Grow the chain first with cyanide, then hydrolyse:

$\ce{CH3OH ->[HI] CH3I ->[KCN] CH3CN ->[H3O+] CH3COOH}$

Key Concept: Use cyanide to add one carbon; oxidation alone never changes the chain length.

(iv) Ethanamine → Methanamine

Common Mistake:

Using the carbylamine reaction and then "hydrolysing the isocyanide to methanamine". Hydrolysis of ethyl isocyanide gives back ethanamine (plus methanoic acid) — the ethyl group stays on the nitrogen, so no carbon is removed from the amine.

How to Avoid:

Only an amide loses a carbon (Hofmann), so first carry the amine to the acid:

$\ce{CH3CH2NH2 ->[HNO2] CH3CH2OH ->[K2Cr2O7/H+] CH3COOH ->[NH3, \Delta] CH3CONH2 ->[Br2/NaOH] CH3NH2}$

Key Concept: In $\ce{R-NC}$ hydrolysis, the R–N bond survives — the products are R–NH₂ and methanoic acid. It is not a chain-shortening tool.

(v) Ethanoic acid → Propanoic acid

Common Mistake:

Trying to bolt a methyl group on directly (e.g., with $\ce{CH3MgBr}$) — a Grignard reagent is simply protonated by the acid's –COOH; no chain extension happens.

How to Avoid:

Go down to the alcohol, over to the halide, then extend with cyanide:

$\ce{CH3COOH ->[LiAlH4] CH3CH2OH ->[PBr3] CH3CH2Br ->[KCN] CH3CH2CN ->[H3O+] CH3CH2COOH}$

Key Concept: Chain-extending an acid = reduce → halogenate → $\ce{KCN}$ → hydrolyse.

(vi) Methanamine → Ethanamine

Common Mistake (two versions):

• Direct alkylation ($\ce{CH3NH2 + CH3I}$) — this gives dimethylamine (a $2^\circ$ amine, still 1 C per chain), not ethanamine.
• Gabriel synthesis on $\ce{CH3I}$ — this gives methanamine back (Gabriel adds no carbon).

How to Avoid:

The target is one carbon longer, so the extra carbon must come from cyanide:

$\ce{CH3NH2 ->[HNO2] CH3OH ->[HI] CH3I ->[KCN] CH3CN ->[H2/Ni] CH3CH2NH2}$

Key Concept: $\ce{CH3CH2NH2}$ has a 2-carbon chain — only the nitrile route builds it from a 1-carbon start.

(vii) Nitromethane → Dimethylamine

Common Mistake:

Thinking reduction of nitromethane gives dimethylamine directly — it gives methylamine ($1^\circ$, one N–C bond).

How to Avoid:

Reduce first, then add the second methyl under controlled (1 equivalent) conditions:

$\ce{CH3NO2 ->[H2/Ni] CH3NH2 ->[CH3I (1 equiv.)] (CH3)2NH}$

(An excess of $\ce{CH3I}$ would over-alkylate to trimethylamine/the quaternary salt.) …