See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Ketones with MW = 100 that Give Racemic Products on $NaBH_4$ Reduction ## Step 1: Determine Molecular Formula MW = 100 with a ketone and $CH_3$ group (given in structure) $$C_nH_{2n}O = 100$$ $$n = 6 \Rightarrow C_6H_{12}O$$ ## Step 2: Identify All Ketone Isomers Possible ketones with formula $C_6H_{12}O$: - Pentan-2-one derivatives with one methyl branch - Hexan-2-one, hexan-3-one - Methylpentan-2-one isomers (with $CH_3$ branch at various positions) ## Step 3: Determine Which Give Racemic Products $NaBH_4$ reduction of a ketone produces racemic mixtures **only when**: - The ketone has a **chiral center created** during reduction - This occurs when the carbonyl carbon becomes chiral after reduction (has 4 different groups) **Ketones that give racemic alcohols:** 1. Hexan-3-one (symmetric ketone → achiral, but this doesn't apply) 2. **2-methylpentan-3-one** (asymmetric: $C=O$ carbon gets different groups) 3. **3-methylpentan-2-one** (creates new chiral center) 4. **4-methylpentan-2-one** (creates new chiral center) 5. **Pentan-2-one with methyl substitution** ## Step 4: Count Valid Isomers Ketones where reduction creates a new chiral center generating racemic mixtures: - 3-methylpentan-2-one - 4-methylpentan-2-one - 2-methylpentan-3-one - 5-methylpentan-2-one - Hexan-3-one (symmetric, gives meso compound, not racemic) **The answer is B (typically 3-4 ketones)** because only those ketones where the carbonyl carbon reduction produces a stereogenic center give true racemic products. Symmetric ketones give achiral products, not racemic mixtures.