Question

$A \xrightarrow{NH_2OH} B \xrightarrow{PCl_5} C \xrightarrow{H_2O} HOOC-(CH_2)_4NH_2$

$A + PhCHBrCOOEt \xrightarrow{base} D$, Compound D is?

• A. A cyclopentane ring is connected to a carbon atom. This carbon atom is connected to an epoxide ring (oxygen atom connected to the carbon atom of the cyclopentane ring) and a phenyl group. The same carbon atom is also connected to a COOEt group (ethyl ester).
• B. A cyclohexane ring is connected to a carbon atom. This carbon atom is connected to an epoxide ring (oxygen atom connected to the carbon atom of the cyclohexane ring) and a phenyl group. The same carbon atom is also connected to a COOEt group (ethyl ester).
• C. A cyclopentane ring is connected to a carbon atom. This carbon atom is connected to an oxygen atom and a phenyl group. The same carbon atom is also connected to a COOEt group (ethyl ester).
• D. A cyclopentane ring is connected to a carbon atom. This carbon atom is connected to an epoxide ring (oxygen atom connected to the carbon atom of the cyclopentane ring) and a phenyl group with a COOEt group (ethyl ester) at the ortho position.

Answer 1

Answer: (D)

Option (D)

Answer 2

1. The first reaction sequence converts A into HOOC–(CH₂)₄–NH₂ via oxime formation, rearrangement (using PCl₅) and hydrolysis. This shows that A is a cyclopentanone‐derived ester.

2. In the second reaction, A reacts with PhCHBrCOOEt in the presence of base. The resulting substitution and rearrangement is analogous to a benzilic acid rearrangement where the bromine‐substituted benzyl unit becomes attached to the aromatic ring in an “ortho” fashion.

3. Comparing the given options, only the structure labeled (D) shows a cyclopentane with an epoxide functionality at the alpha‐carbon and a phenyl group bearing a COOEt substituent at the ortho position—exactly as expected for compound D.