Q.In order to prepare a 1° amine from an alkyl halide with simultaneous addition of one $CH_2$ group in the carbon chain, the reagent used as source of nitrogen is ____.

The key idea is the Gabriel phthalimide synthesis, which gives a primary amine without over-alkylation, and the use of KCN to add one carbon via nitrile formation followed by reduction.

Reasoning:

1. To add one $CH_2$ group, the alkyl halide must first undergo nucleophilic substitution with a cyanide ion ($CN^-$), giving a nitrile ($R-CN$) — this extends the chain by one carbon.
2. Reduction of the nitrile (e.g., with $H_2$/Ni or $LiAlH_4$) yields a primary amine ($R-CH_2NH_2$). …

The question asks for a reagent that does two things at once: convert an alkyl halide into a primary amine while lengthening the carbon chain by exactly one carbon atom. This is a classic "chain extension" problem in organic synthesis.

Let’s think about what each option does. The goal is to go from $R-X$ to $R-CH_2-NH_2$. That means we need to insert a $CH_2$ group between the original carbon and the nitrogen. A direct substitution with ammonia ($NH_3$) would give $R-NH_2$ — no extra carbon. So we need a reagent that first adds a carbon-containing group, which can later be transformed into an amine.

1. Option (A) Sodium amide, $NaNH_2$ — This is a strong base, often used for deprotonation or in elimination reactions. It can also act as a nucleophile, replacing the halide with $NH_2^-$ to give $R-NH_2$. But that gives a primary amine without adding a carbon. So this does not satisfy the "simultaneous addition of one $CH_2$ group" condition. It’s a dead end here.

2. Option (B) Sodium azide, $NaN_3$ — This is a good nucleophile. It replaces the halide with an azide group ($-N_3$), giving $R-N_3$. Reduction of the azide (e.g., with $LiAlH_4$ or catalytic hydrogenation) yields $R-NH_2$. Again, no extra carbon. So this also fails the chain-extension requirement.

3. Option (C) Potassium cyanide, KCN — This is the classic reagent for adding one carbon. The cyanide ion ($CN^-$) is a strong nucleophile and displaces the halide to form an alkyl cyanide (nitrile): $R-X \rightarrow R-CN$. The nitrile has one more carbon than the original alkyl halide. Then, reduction of the nitrile (e.g., with $H_2$/Ni or $LiAlH_4$) gives $R-CH_2-NH_2$ — a primary amine with exactly one extra $CH_2$ group. This is exactly what the question describes. …

Concept: Gabriel Phthalimide Synthesis & Nitrile Reduction

The question asks for a method to prepare a 1° amine from an alkyl halide while adding one carbon atom to the chain. This means the nitrogen source must also supply an extra carbon.

Method: Nitrile formation followed by reduction

Reasoning:

• Sodium amide ($NaNH_2$) gives direct substitution → no carbon added.
• Sodium azide ($NaN_3$) gives alkyl azide → reduction gives 1° amine but no carbon added.
• Potassium phthalimide gives Gabriel synthesis → 1° amine but no carbon added.
• Potassium cyanide (KCN) reacts with alkyl halide to form alkyl cyanide (nitrile), which on reduction gives a 1° amine with one extra carbon.

Steps:

1. Nucleophilic substitution $R-X + KCN \xrightarrow{\text{SN2}} R-CN + KX$ (Cyanide ion, $:C \equiv N^-$, acts as nucleophile; carbon attaches to alkyl group) …

Here are the common mistakes students make on this concept and how to avoid each.

The Core Concept

This question tests the Gabriel phthalimide synthesis and the Hofmann rearrangement (or the nitrile route). The key requirement is:

1. Prepare a 1° amine.
2. Add one $CH_2$ group to the carbon chain (increase chain length by one carbon).

The correct reagent is Potassium cyanide (KCN). The alkyl halide ($R-X$) reacts with KCN to give $R-CN$ (nitrile, +1C). Reduction of the nitrile ($R-CN \rightarrow R-CH_2NH_2$) yields the 1° amine with one extra carbon.

Mistake 1: Choosing Sodium Azide ($NaN_3$) or Sodium Amide ($NaNH_2$)

Why students pick it: They remember that $NaN_3$ or $NaNH_2$ can be used to make amines from alkyl halides.

Why it’s wrong:

• $NaNH_2$ is a strong base; it causes elimination (alkene formation) rather than substitution with most alkyl halides.
• $NaN_3$ gives an alkyl azide ($R-N_3$), which on reduction yields a 1° amine ($R-NH_2$). However, no carbon is added — the chain length remains the same.

How to avoid: Always check the carbon count. If the question says “simultaneous addition of one $CH_2$ group,” the nitrogen source must contain a carbon atom that gets incorporated. $NaN_3$ and $NaNH_2$ contain no carbon.

Mistake 2: Choosing Potassium Phthalimide (Gabriel’s reagent)

Why students pick it: The Gabriel synthesis is the classic method for making pure 1° amines from alkyl halides.

Why it’s wrong here:

• The Gabriel synthesis gives $R-NH_2$ (same number of carbons as the alkyl halide). No carbon is added to the chain.

How to avoid: Remember that the Gabriel reaction is a substitution of the halide by the phthalimide nitrogen. The alkyl group $R$ remains unchanged. For chain elongation, you need a reagent that introduces a new carbon atom (like cyanide).

Mistake 3: Forgetting the Reduction Step After Cyanide Addition

Why students pick it: They might think KCN directly gives an amine.

Why it’s wrong: KCN gives a nitrile ($R-CN$), not an amine. The nitrile must be reduced (e.g., with $LiAlH_4$ or $H_2$/catalyst) to get the 1° amine ($R-CH_2NH_2$). The question implies the overall transformation, but students often miss that the cyanide is the nitrogen source for the two-step process. …