Now, let's analyze each option:
(a) $(\mathrm{CH}_{3})_{2} \mathrm{CHCHO}$: There is no carbon atom bonded to four different groups, so this compound is not optically active.
(b) $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}$: The carbon
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