See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution **Identifying the smallest ketone and its next homologue:** - Smallest ketone: $CH_3COCH_3$ (acetone) - Next homologue: $CH_3COCH_2CH_3$ (butanone) **Reaction with $NH_2OH$:** Both ketones form oximes: $$CH_3COCH_3 + NH_2OH \rightarrow CH_3C(=NOH)CH_3 + H_2O$$ $$CH_3COCH_2CH_3 + NH_2OH \rightarrow CH_3C(=NOH)CH_2CH_3 + H_2O$$ **Counting distinct oximes:** 1. **Acetone oxime**: $CH_3C(=NOH)CH_3$ — **only 1 oxime** (symmetric ketone, no stereoisomerism) 2. **Butanone oxime**: $CH_3C(=NOH)CH_2CH_3$ — has a C=N double bond with two different groups attached - Forms **syn and anti isomers** (2 oximes due to restricted rotation around C=N) **Total oximes formed: 1 + 2 = 3 different oximes** **Optical activity:** - Acetone oxime: achiral (symmetric) - Butanone oximes: both syn and anti isomers are **achiral** (no chiral center) Therefore, **three different oximes are formed, but none are optically active**. **Answer: (B) Three different oximes are formed**