A compound D with the molecular formula C6H12 is optically inactive but can be resolved into enantiomers. On catalytic hydrogenation, D is converted to E (C6H14) and E is optically inactive. Propose structures for D and E. (Draw a three-dimensional formula for each using dashes and wedges around chiral centers.)

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.

because u need two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei to form isotopes.

so since they all have the same shape? i mean maybe because m& m and reeses are both chocolates