Question

The total number of optically active compounds formed in the following reaction is :

• A. Two
• B. Four
• C. Six
• D. Zero

Hint:

The substrate is a vinyl (enol) ether, CH₃CH₂C(CH₃)=CH-O-CH(CH₃)₂; HBr adds across C=C (H to the carbon away from oxygen, Br to the carbon bearing oxygen) and creates two stereocentres.

Answer 1

The Correct Option is B

The reactant (from the figure) is the enol ether \(\mathrm{CH_3CH_2-C(CH_3)=CH-O-CH(CH_3)_2}\).
H⁺ adds to the carbon not bearing oxygen, because the resulting cation \(\mathrm{CH_3CH_2-CH(CH_3)-CH=\overset{+}{O}-CH(CH_3)_2}\) is stabilised by the oxygen lone pair.
Br^- then attacks the cationic carbon: \(\mathrm{CH_3CH_2-C^*H(CH_3)-C^*H(Br)-O-CH(CH_3)_2}\).
The product has two different stereocentres (marked *), giving \(2^2 = 4\) stereoisomers (two pairs of enantiomers); none has a plane of symmetry, so all four are optically active.
Number of optically active compounds = 4.
Answer: (2) Four