In the reaction given below
'B' is

Solution

Chemistry ma’am · 5 steps

Watch the five-membered ring next to the carbocation: it expands to a six-membered ring before H⁺ leaves.

1. Acid step. HCl protonates the basic -NH₂ to -NH₃⁺, so the nitrogen is no longer a nucleophile. It also protonates the -OH, and water leaves to give a tertiary, benzylic carbocation at C-1.
2. Ring expansion. C-2, next door, is the spiro carbon of a strained cyclopentane. One ring CH₂ shifts from C-2 to C-1 (a 1,2-alkyl shift). The five-membered ring becomes a six-membered ring, and the positive charge moves to the old spiro carbon, which is still tertiary.
3. Loss of a proton. The new carbocation loses an H⁺ from the neighbouring CH₂ of the middle ring. This gives a ring double bond in a hexahydrophenanthrene skeleton, with the -CH₂CH₂NH₃⁺ chain on the quaternary ring-fusion carbon.
4. Base step. KOH only neutralises the ammonium salt: -NH₃⁺ + OH⁻ → -NH₂ + H₂O. The carbon skeleton does not change.
5. So. B is the ring-expanded alkene that keeps a free -CH₂CH₂NH₂ chain, which is option (c).

The idea to keep. A carbocation next to a cyclobutane or cyclopentane carbon usually expands that ring by one. In acid, an amine sits as -NH₃⁺ and stays out of the reaction until base frees it.

Answer from the NTA final answer key. Solution by PrepWiser.