Q.Write chemical equations for the following conversions:

Concept: Nucleophilic Substitution Reactions (SN2) — using cyanide ion as a nucleophile to extend the carbon chain by one carbon, followed by reduction of the nitrile to a primary amine.

Reasoning:

1. Both starting materials are primary alkyl halides. Treat each with alcoholic $KCN$ (or $NaCN$) to perform an SN2 attack, replacing the chlorine with a cyano group ($-CN$). This adds one carbon atom.
2. The resulting nitrile ($-CN$) is then reduced to a primary amine ($-CH_2NH_2$). A common reducing agent for this step is $LiAlH_4$ in dry ether, or catalytic hydrogenation ($H_2$/Ni or $H_2$/Pd).

Stepwise equations:

1. $CH_3-CH_2-Cl \xrightarrow{KCN(alc.)} CH_3-CH_2-CN \xrightarrow{LiAlH_4/H_2O} CH_3-CH_2-CH_2-NH_2$
2. $C_6H_5-CH_2-Cl \xrightarrow{KCN(alc.)} C_6H_5-CH_2-CN \xrightarrow{LiAlH_4/H_2O} C_6H_5-CH_2-CH_2-NH_2$
   ✓Final answer

   The conversions are achieved via SN2 with $KCN$ followed by $LiAlH_4$ reduction.

The Core Idea: Nucleophilic Substitution + Reduction

You have an alkyl halide (a good electrophile) and you want a product whose carbon chain is one carbon longer, ending in $\text{CH}_2\text{NH}_2$. The way to do that is to replace the halogen with a carbon nucleophile that carries the nitrogen, then reduce.

The cyanide ion ($\text{CN}^-$) is perfect: it's a strong nucleophile, attacks the carbon bearing the halogen in an $\text{S}_\text{N}2$ reaction, and the resulting nitrile ($\text{R–CN}$) can be reduced to $\text{R–CH}_2\text{NH}_2$ — exactly the product you need, with the nitrile carbon supplying the extra $\text{CH}_2$.

Step-by-Step Solution

1. First conversion: $\text{CH}_3\text{CH}_2\text{Cl} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$

Step 1a: Nucleophilic substitution with $\text{KCN}$ (or $\text{NaCN}$)

The chlorine atom is a good leaving group. In ethanol (the NCERT solution writes "ethanolic NaCN" — ethanol is a polar protic solvent, and the reaction works well in it), the cyanide ion attacks the electrophilic carbon.

This is an $\text{S}_\text{N}2$ reaction — the cyanide approaches from the back, inverting the configuration (though here the carbon is not chiral, so no stereochemical consequence). The product is propanenitrile (ethyl cyanide).

Step 1b: Reduction of the nitrile to a primary amine

The nitrile group ($-\text{CN}$) can be reduced to a primary amine ($-\text{CH}_2\text{NH}_2$) using a strong reducing agent. The classic choice is lithium aluminium hydride ($\text{LiAlH}_4$) in dry ether, followed by hydrolysis. Alternatively, catalytic hydrogenation ($\text{H}_2$/Ni) works equally well — that is the reagent NCERT itself uses in part (ii).

The reduction adds two hydrogen atoms to the carbon and one to the nitrogen, converting the triple bond into a single bond.

Overall equation for (i):

2. Second conversion: $\text{C}_6\text{H}_5\text{CH}_2\text{Cl} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{NH}_2$

Step 2a: Nucleophilic substitution with $\text{KCN}$

Benzyl chloride ($\text{C}_6\text{H}_5\text{CH}_2\text{Cl}$) is even more reactive toward $\text{S}_\text{N}2$ than a simple primary halide — the adjacent aromatic ring stabilises the transition state, so cyanide attack is fast.

The product is phenylacetonitrile (phenylethanenitrile / benzyl cyanide).

Step 2b: Reduction of the nitrile

Same reduction as before ($\text{H}_2$/Ni, as NCERT writes, or $\text{LiAlH}_4$):

The product is 2-phenylethan-1-amine (phenethylamine).

Overall equation for (ii):

Method: Nucleophilic Substitution via Alkyl Cyanide (Nitrile → Amine)

This is a two-step chain elongation method using cyanide ion ($\text{CN}^-$) as a nucleophile, followed by reduction.

General Principle

• Step 1: Alkyl halide undergoes $\text{S}_\text{N}2$ with $\text{KCN}$ (or $\text{NaCN}$, in ethanol — NCERT writes "ethanolic NaCN") to form an alkyl cyanide (nitrile).
• Step 2: The nitrile is reduced (e.g., with $\text{LiAlH}_4$ or catalytic hydrogenation, $\text{H}_2$/Ni) to a primary amine with one extra carbon.

(i) $CH_3-CH_2-Cl \rightarrow CH_3-CH_2-CH_2-NH_2$

Step 1 — Nucleophilic substitution:

Step 2 — Reduction of nitrile:

Key point: The cyanide carbon becomes the extra $\text{CH}_2$ group next to the amine.

(ii) $C_6H_5-CH_2-Cl \rightarrow C_6H_5-CH_2-CH_2-NH_2$

Step 1 — Nucleophilic substitution:

Step 2 — Reduction of nitrile:

Key point: Benzyl chloride ($C_6H_5CH_2Cl$) is especially reactive in $\text{S}_\text{N}2$ — the adjacent aromatic ring stabilises the transition state — so the substitution proceeds smoothly.

Summary of the Method

Final result: Both conversions increase the carbon chain by one and introduce a primary amine at the terminal position.

Here are the common mistakes students make when solving these nucleophilic substitution conversions, along with how to avoid each.

Mistake 1: Choosing a Route That Doesn't Change the Carbon Count

The Mistake:

Students reach for a standard amine preparation — direct ammonolysis or the Gabriel phthalimide synthesis — without counting carbons:

• $CH_3CH_2Cl + NH_3 \rightarrow CH_3CH_2NH_2$ ✗ (ethylamine — only 2 carbons)
• $CH_3CH_2Cl$ + potassium phthalimide → $N$-ethylphthalimide → hydrolysis → $CH_3CH_2NH_2$ ✗ (still ethylamine)

Why it's wrong:

Both routes attach nitrogen to the existing carbon skeleton. The target of (i) is $CH_3CH_2CH_2NH_2$ (propan-1-amine, 3 carbons) — one carbon longer than the starting halide — so any route that doesn't add a carbon cannot give it.

How to Avoid:

Count carbons first. A one-carbon extension means the cyanide route: the $CN^-$ nucleophile supplies the extra carbon, and reduction turns $-C \equiv N$ into $-CH_2NH_2$.

✓ Correct approach for (i):

1. $CH_3CH_2Cl + KCN \xrightarrow{\text{ethanol}} CH_3CH_2CN + KCl$
2. Reduction ($LiAlH_4$ or $H_2$/Ni) → $CH_3CH_2CH_2NH_2$

Mistake 2: Forgetting the Carbon Chain Length in (ii)

The Mistake:

Students write the product as $C_6H_5CH_2NH_2$ (benzylamine) instead of $C_6H_5CH_2CH_2NH_2$ (2-phenylethan-1-amine).

Why it's wrong:

The target has two carbons between the benzene ring and the amino group. The starting material has only one carbon. You must increase the chain length by one carbon.

How to Avoid:

Always count the carbon atoms in the product vs. starting material. If the product has one more carbon, you need a cyanide ion ($CN^-$) as the nucleophile first, then reduce.

✓ Correct approach for (ii):

1. $C_6H_5CH_2Cl$ + $KCN$ (ethanolic) → $C_6H_5CH_2CN$ (benzyl cyanide)
2. Reduction: $H_2$/Ni or LiAlH₄ → $C_6H_5CH_2CH_2NH_2$

Mistake 3: Inventing "Better" Solvent Conditions Than the Standard Ones

The Mistake:

Insisting the substitution must be run in an anhydrous polar aprotic solvent (acetone, DMF) and marking the ethanol route wrong.

Why it's wrong:

The standard (and NCERT's own printed) condition for this reaction is ethanolic NaCN/KCN — cyanide is a strong enough nucleophile that the $S_N2$ displacement works well in ethanol. Aprotic solvents can accelerate $S_N2$ reactions, but they are not required here, and "correcting" the printed conditions loses marks.

How to Avoid:

Write the reagent the syllabus uses: ethanolic $KCN$ (or $NaCN$), with heating. Mention $S_N2$ as the mechanism.

Mistake 4: Choosing a Reducing Agent That Doesn't Reduce Nitriles to Primary Amines

The Mistake:

Using $NaBH_4$ (which does not reduce nitriles), or DIBAL-H (which stops at the aldehyde stage), and expecting a primary amine.

How to Avoid:

For $R-CN \rightarrow R-CH_2NH_2$, use $LiAlH_4$ in dry ether (then water) or catalytic hydrogenation ($H_2$/Ni) — the reagent NCERT itself uses. ($H_2$/Raney Ni with ammonia suppresses secondary-amine coupling by-products, a useful refinement but not required.)

✓ Correct reduction:

$R-CN \xrightarrow{LiAlH_4/\text{ether, then } H_2O \text{ (or } H_2/Ni)} R-CH_2NH_2$

Mistake 5: Writing Incomplete or Unbalanced Equations

The Mistake:

Writing only the organic product and forgetting byproducts (like $KCl$) or not balancing atoms.

Example of wrong:

$CH_3CH_2Cl + KCN \rightarrow CH_3CH_2CN$ (missing $KCl$)

How to Avoid:

Always write complete, balanced equations with all products. Check that the number of atoms of each element is the same on both sides.

✓ Correct:

$CH_3CH_2Cl + KCN \xrightarrow{\text{ethanol}} CH_3CH_2CN + KCl$

Quick Summary Table

Final Tip: For primary amine preparation from alkyl halides, remember two standard routes:

• No chain extension → Gabriel phthalimide
• Chain extension by one carbon → $KCN$ followed by reduction

Both targets in this question are one carbon longer than their halides — so both must go through the cyanide route.