Isomerism: optical — practice questions

First, draw the structure of 3-methylpentan-2-ol: CH₃CH(OH)CH(CH₃)CH₂CH₃.

Let's examine each carbon atom:

• C1 (in the end CH₃): Bonded to 3 H atoms. Not chiral.
• C2 (with the OH group): Bonded to H, OH, CH₃, and CH(CH₃)CH₂CH₃. These are four different groups. Therefore, C2 is a chiral centre.
• C3 (with the methyl group): Bonded to H, CH₃, CH(OH)CH₃, and CH₂CH₃. These are four different groups. Therefore, C3 is a chiral centre.
• C4 (in the ethyl group): Bonded to 2 H atoms. Not chiral.
• C5 (in the ethyl group): Bonded to 3 H atoms. Not chiral.
• The C in the methyl group at C3: Bonded to 3 H atoms. Not chiral.

This molecule has two chiral centres.

(i) The reaction is nucleophilic addition. Mechanism: Step 1: The cyanide ion (CN⁻) acts as a nucleophile and attacks the δ+ carbon of the planar carbonyl group. The π-bond in the C=O breaks. CH₃C(O)CH₂CH₃ + CN⁻ → CH₃C(O⁻)(CN)CH₂CH₃ Step 2: The intermediate anion is protonated by an HCN molecule (or H₂O) to form the product, 2-hydroxy-2-methylbutanenitrile. CH₃C(O⁻)(CN)CH₂CH₃ + HCN → CH₃C(OH)(CN)CH₂CH₃ + CN⁻

(ii) The carbonyl group in butanone is planar. The nucleophile (CN⁻) can attack the carbonyl carbon from above the plane or below the plane with equal probability. This leads to the formation of equal amounts of the two enantiomers of 2-hydroxy-2-methylbutanenitrile. This equimolar mixture is a racemic mixture. The optical rotations of the two enantiomers are equal in magnitude but opposite in direction, so they cancel each other out, resulting in the product mixture being optically inactive.