Q.The best reagent for converting 2-phenylpropanamide into 2-phenylpropanamine is ____.

The key idea is Nucleophilic Substitution Reactions — specifically, the Hofmann rearrangement (for converting an amide to an amine with one fewer carbon) versus direct reduction (which preserves the carbon skeleton).

Reasoning:

1. 2-phenylpropanamide has the structure $\ce{C6H5-CH(CH3)-CONH2}$. To get 2-phenylpropanamine $\ce{C6H5-CH(CH3)-CH2NH2}$, the carbonyl carbon must be reduced to $\ce{CH2}$ without changing the carbon count.
2. Option (D) $\ce{LiAlH4}$ in ether is a strong reducing agent that directly reduces amides to amines, keeping the same number of carbons. …

This is a classic functional group transformation: turning an amide into an primary amine. The key is to recognise that the amide group ($-CONH_2$) must be reduced to a $-CH_2NH_2$ group, while leaving the benzene ring and the alkyl side chain untouched.

Let’s look at each option carefully.

1. Option (A): excess $H_2$

   Hydrogen gas alone is not a reducing agent for amides. Even with a metal catalyst (like Pd/C or Ni), $H_2$ typically reduces alkenes, alkynes, nitro groups, and nitriles — but not amides. The amide carbonyl is quite stable toward catalytic hydrogenation under normal conditions. So this won’t work.

2. Option (B): $Br_2$ in aqueous NaOH

   This is the Hoffmann bromamide degradation reaction. It converts an amide into a primary amine with one fewer carbon atom — because the carbonyl carbon is lost as $CO_2$.

   For 2-phenylpropanamide ($C_6H_5-CH(CH_3)-CONH_2$), this would give 1-phenylethanamine ($C_6H_5-CH(CH_3)-NH_2$), which is not 2-phenylpropanamine. The carbon skeleton changes. So this is wrong for the target product.

3. Option (C): iodine in the presence of red phosphorus

   This combination is used for the conversion of alcohols to alkyl iodides (the HI/P method). It has no application in amide reduction. Red phosphorus and iodine generate $HI$ in situ, which can reduce some functional groups, but amides are not affected in this way. This is a distractor.

4. Option (D): $LiAlH_4$ in ether …

Concept: Hofmann Rearrangement (Hofmann Bromamide Degradation)

This reaction converts a primary amide into a primary amine with one fewer carbon atom. The reagent used is bromine in aqueous NaOH (or KOH).

Method: Hofmann Rearrangement

Steps:

1. Identify the functional group

   The starting compound is 2-phenylpropanamide — a primary amide ($\ce{R-CONH2}$).

2. Recall the reagent for Hofmann rearrangement

   The classic reagent is $\ce{Br2}$ in aqueous $\ce{NaOH}$ (or $\ce{KOH}$). This converts $\ce{R-CONH2}$ to $\ce{R-NH2}$ with loss of one carbon (as $\ce{CO2}$).

3. Apply to the given molecule

   • 2-phenylpropanamide: $\ce{C6H5-CH(CH3)-CONH2}$
   • After Hofmann rearrangement: $\ce{C6H5-CH(CH3)-NH2}$ (2-phenylpropanamine)
   • The carbon chain length remains the same because the carbonyl carbon is lost, but the alkyl group attached to the carbonyl becomes attached to the nitrogen.

4. Check other options …

Let’s break this down step-by-step — first the concept, then the common mistakes, and finally how to avoid each.

🧪 The Core Concept

You are converting an amide into an amine.

• 2-phenylpropanamide is a primary amide ($\ce{R-CONH2}$).
• 2-phenylpropanamine is a primary amine ($\ce{R-CH2NH2}$).

The reaction needed is reduction of the amide to the amine.

✓ Correct Reagent

The best reagent here is:

(D) $\ce{LiAlH4}$ in ether

Why?

$\ce{LiAlH4}$ reduces amides cleanly to amines.

The reaction is:

No other option does this transformation correctly.

✗ Common Mistakes & How to Avoid Each

1. Choosing excess $\ce{H2}$ (Option A)

Mistake:

Thinking catalytic hydrogenation reduces amides to amines.

Why it’s wrong:

$\ce{H2}$ with a metal catalyst (like Pd, Pt, Ni) reduces alkenes, alkynes, nitro groups, and nitriles — but not amides. Amides are very stable to hydrogenation.

How to avoid:

Remember:

$\ce{H2}$/catalyst does not reduce the $\ce{C=O}$ of an amide. Only strong hydride donors like $\ce{LiAlH4}$ can do that.

2. Choosing $\ce{Br2}$ in aqueous NaOH (Option B)

Mistake:

Confusing this with the Hofmann rearrangement (which converts amides to amines with one fewer carbon).

Why it’s wrong:

$\ce{Br2}$/NaOH on an amide gives a primary amine with loss of one carbon (as $\ce{CO2}$).

Here, the product would be 1-phenylethanamine, not 2-phenylpropanamine.

How to avoid:

Check the carbon count:

• Starting amide: 9 carbons (phenyl + 3-carbon chain)
• Hofmann product: 8 carbons (loses the carbonyl carbon)
• Desired amine: 9 carbons (same chain length)

If the carbon skeleton must stay the same, do not use Hofmann rearrangement.

3. Choosing iodine + red phosphorus (Option C)

Mistake:

Thinking this reduces amides.

Why it’s wrong:

Iodine + red phosphorus is used to reduce alcohols to alkyl iodides (or sometimes carboxylic acids to alkyl iodides). It does not reduce amides to amines.

How to avoid:

Know the specific uses:

• $\ce{I2}$/red P → alcohols → alkyl halides …