Answer:
D: CH2=CH-CH(CH3)-CH2-CH3 (R & S enantiomers)
E: CH3-CH2-(CH3)-CH2-CH3
(Please see the figures enclosed )
Explanation:
D is a racemic mixture (R & S) of 3-metyl-pent-1-ene, so it is optically inactive (although each of two enantiomers is optically active, the mixture is optically inactive. The reason is that two enantiomers are present in an equal amount).
E is optically inactive, so its structure has to be symmetric.