Question

How many of the following reaction(s) product formed can also be prepared by Gabriel phthalimide synthesis:

• A. i, ii, iii
• B. ii, iii, iv
• C. ii, iii, v
• D. i, iii, v

Hint:

Gabriel phthalimide synthesis is used for preparing primary alkyl amines onlyIt does not work for aryl amines like anilineAlways first find the amine formed in Hofmann bromamide reaction, then check whether it is a primary aliphatic amine or not.

Answer 1

The Correct Option is C

Step 1: Identify the reaction taking place.
All the given reactions are examples of Hofmann bromamide degradation reaction.
In this reaction, an amide on treatment with \(Br_2 + KOH\) gives a primary amine having one carbon less than the parent amide.
General reaction:
\[ RCONH_2 \xrightarrow[\Delta]{Br_2 + KOH} RNH_2 \] Thus, we first find the amine formed in each case.

Step 2: Find the product formed in each reaction.
(i) \(\mathrm{PhCONH_2}\) is benzamide.
On Hofmann degradation:
\[ \mathrm{PhCONH_2 \xrightarrow[\Delta]{Br_2 + KOH} PhNH_2} \] So, the product is aniline.
(ii) \(\mathrm{PhCH_2CONH_2}\) is phenylacetamide.
On Hofmann degradation:
\[ \mathrm{PhCH_2CONH_2 \xrightarrow[\Delta]{Br_2 + KOH} PhCH_2NH_2} \] So, the product is benzylamine.
(iii) \(\mathrm{CH_3CONH_2}\) is acetamide.
On Hofmann degradation:
\[ \mathrm{CH_3CONH_2 \xrightarrow[\Delta]{Br_2 + KOH} CH_3NH_2} \] So, the product is methylamine.
(iv) \(\mathrm{CH_3CONHCH_3}\) is N-methyl acetamide.
Hofmann bromamide reaction is generally given by primary amides \(\mathrm{RCONH_2}\).
Since this is a substituted amide, it does not give a primary amine product suitable here.
(v) \(\mathrm{C_6H_{11}CONH_2}\) is cyclohexane carboxamide.
On Hofmann degradation:
\[ \mathrm{C_6H_{11}CONH_2 \xrightarrow[\Delta]{Br_2 + KOH} C_6H_{11}NH_2} \] So, the product is cyclohexylamine.

Step 3: Apply Gabriel phthalimide synthesis condition.
Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines only.
It does not prepare aryl amines like aniline efficiently.
Now check each product:
(i) Aniline \((\mathrm{PhNH_2})\):
This is a primary aromatic amine.
It cannot be prepared by Gabriel phthalimide synthesis.
(ii) Benzylamine \((\mathrm{PhCH_2NH_2})\):
This is a primary aliphatic amine because the amino group is attached to a side-chain carbon, not directly to the benzene ring.
It can be prepared by Gabriel phthalimide synthesis.
(iii) Methylamine \((\mathrm{CH_3NH_2})\):
This is a primary aliphatic amine.
It can be prepared by Gabriel phthalimide synthesis.
(iv) No suitable product for Gabriel synthesis.
This option is not applicable.
(v) Cyclohexylamine \((\mathrm{C_6H_{11}NH_2})\):
This is also a primary aliphatic amine.
It can be prepared by Gabriel phthalimide synthesis.

Step 4: Select the correct set.
The products which can also be prepared by Gabriel phthalimide synthesis are:
\[ \text{(ii), (iii), (v)} \] Hence, the correct option is:
\[ \boxed{(C)\ ii,\ iii,\ v} \] Final Answer:
\[ \boxed{(C)\ ii,\ iii,\ v} \]