Q.Amongst the given set of reactants, the most appropriate for preparing 2° amine is ____.

The key idea is that direct alkylation of ammonia (option A) gives a mixture of primary, secondary, and tertiary amines, so it is not suitable for preparing a pure secondary amine. Option B yields a primary amine after reduction of the nitrile. Option D is the Gabriel phthalimide synthesis, which gives a pure primary amine, not secondary. …

To prepare a secondary amine cleanly, you need a method that avoids mixtures of primary, secondary, and tertiary products. Let's see why each option works or fails.

Nucleophilic Substitution Reactions — the core concept here — involve a nucleophile attacking an electrophilic carbon. When you use an alkyl halide with ammonia, the product (a primary amine) is itself a better nucleophile than ammonia. So it immediately attacks another alkyl halide molecule, giving a cascade of over-alkylation. That's the fundamental problem with direct alkylation.

Let's examine each option:

1. Option (A): 2° R–Br + NH₃

   Ammonia attacks the secondary alkyl bromide to give a primary amine salt. But the free primary amine formed is more nucleophilic than ammonia, so it reacts with another molecule of R–Br to give a secondary amine, then a tertiary amine, and finally a quaternary ammonium salt. You end up with a mixture — not a clean route to a secondary amine.

2. Option (B): 2° R–Br + NaCN followed by H₂/Pt

   This gives a nitrile (R–CN) which on reduction yields a primary amine (R–CH₂–NH₂). The carbon skeleton gains one extra carbon, and the product is a primary amine, not secondary. So this is outright wrong for preparing a secondary amine.

3. Option (C): 1° R–NH₂ + RCHO followed by H₂/Pt

   This is reductive amination. The aldehyde and primary amine first form an imine (Schiff base) by condensation. Catalytic hydrogenation then reduces the C=N bond to a C–N single bond, giving a secondary amine.

   The beauty: no free alkyl halide is present to cause over-alkylation. The imine intermediate is less nucleophilic than the starting amine, so further reaction is suppressed. This is the most controlled, high-yield method.

4. Option (D): 1° R–Br (2 mol) + potassium phthalimide followed by H₃O⁺/heat …

Concept: Preparation of Secondary Amines via Reductive Amination

The most appropriate method for preparing a secondary (2°) amine is option (C).

Method Name: Reductive Amination

Steps:

1. Condensation reaction A primary amine ($1° R{-}NH_2$) reacts with an aldehyde ($RCHO$) to form an imine (Schiff base):

2. Catalytic hydrogenation The imine is reduced using $H_2$ in the presence of a Pt catalyst:

Final product: A secondary amine ($R{-}NH{-}CH_2{-}R$).

Why other options fail:

• (A) $2° R{-}Br + NH_3$ → gives a mixture of primary, secondary, and tertiary amines (not selective). …

Here are the common mistakes students make on this question, along with how to avoid each.

Mistake 1: Confusing the product of the Hoffmann Ammonolysis (Option A)

• The Mistake: Students see $2^\circ R{-}Br + NH_3$ and think it directly gives a pure $2^\circ$ amine. They forget that $NH_3$ is a nucleophile and the product ($1^\circ$ amine) is also a nucleophile.
• Why it's wrong: The reaction doesn't stop. The $1^\circ$ amine product reacts with more alkyl halide to give a $2^\circ$ amine, then a $3^\circ$ amine, and finally a quaternary ammonium salt. You get a mixture.
• How to avoid: Remember the Hoffmann Ammonolysis rule: Alkylation of ammonia always gives a mixture of $1^\circ$, $2^\circ$, $3^\circ$ amines, and quaternary salt. It is never a clean method for preparing a specific $2^\circ$ amine.

Mistake 2: Misidentifying the product of the Nitrile Route (Option B)

• The Mistake: Students think $2^\circ R{-}Br + NaCN$ followed by reduction gives a $2^\circ$ amine.
• Why it's wrong: The $2^\circ$ alkyl halide undergoes $S_N2$ with $CN^-$ to give a nitrile ($R{-}CN$). Reduction of a nitrile ($H_2$/Pt) gives a $1^\circ$ amine ($R{-}CH_2{-}NH_2$). The carbon skeleton increases by one, but the amine is still primary.
• How to avoid: Trace the carbon chain. The $CN^-$ adds a carbon, but the $NH_2$ group ends up on that new carbon. The product is always $R{-}CH_2{-}NH_2$ — a $1^\circ$ amine, regardless of the starting halide's degree.

Mistake 3: Forgetting the Reductive Amination Mechanism (Option C)

• The Mistake: Students see $1^\circ R{-}NH_2 + RCHO$ and think it forms a $2^\circ$ amine directly, or they confuse it with simple imine formation.
• Why it's correct (and why students miss it): The $1^\circ$ amine reacts with the aldehyde to form an imine ($R{-}N{=}CH{-}R'$). The $H_2$/Pt then reduces the $C{=}N$ bond to a $C{-}N$ bond, giving a $2^\circ$ amine ($R{-}NH{-}CH_2{-}R'$). This is reductive amination — the most reliable method.
• How to avoid: Memorize the two-step logic:
  1. Condensation: $1^\circ$ amine + aldehyde/ketone → imine.
  2. Reduction: Imine + $H_2$/catalyst → $2^\circ$ amine. This avoids over-alkylation (unlike option A).

Mistake 4: Misapplying the Gabriel Phthalimide Synthesis (Option D)

• The Mistake: Students think using 2 moles of $1^\circ R{-}Br$ with potassium phthalimide gives a $2^\circ$ amine. …