See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Catalytic hydrogenation (H2/Pt) proceeds via syn addition - both hydrogen atoms are delivered to the same face of the double bond simultaneously. Step 1 - Identify the starting material: The alkene is (E)- or (Z)-2,3-dideuterio-2-butene, i.e., (CH3)(D)C=C(D)(CH3). Each carbon of the double bond bears a methyl group and a deuterium. This makes both alkene carbons identical in substitution. Step 2 - Determine the geometry of the alkene: From the drawing, the CH3CH group and CH3 group are on opposite sides (or same side) of the double bond, with D groups placed accordingly. The key point is that both sp2 carbons bear the same two groups (methyl and deuterium), making this a symmetrically substituted alkene. Actually re-reading: left carbon has CH3CH- (ethyl-like, but it's just CH(CH3) attached) and D; right has CH3 and D. So the two carbons of the double bond are: C3 bearing (CH3CH-) and D, C2 bearing (CH3) and D — wait, this may be 3-methyl-2-deuterio... Let's treat it as: the alkene is such that hydrogenation produces two new stereocenters. Step 3 - Syn addition of H2: H2/Pt delivers two H atoms from the same face. If the alkene is (E)-configured (the two identical substituents on opposite sides), syn addition gives the (R,S) = meso compound predominantly AND the (R,R)/(S,S) pair depending on which face is attacked. For a symmetrically substituted (E)-alkene like (E)-(CH3)(D)C=C(D)(CH3): syn addition from one face gives (2R,3R) and from the other face gives (2S,3S) — these are enantiomers, forming a racemic mixture. For (Z)-alkene: syn addition gives the meso compound. Step 4 - From the drawing: the structure shown has CH3CH on one side and CH3 on the same side (both on top), with D on both carbons on the bottom — this corresponds to the (Z)-configuration... but the image shows CH3-CH on upper-left and CH3 on upper-right, D on lower-left and D on lower-right, suggesting (Z)-like arrangement where similar groups are cis. However, looking again: CH3-CH is on left-top, D on left-bottom; CH3 on right-top, D on right-bottom. This is (E)-configuration (the two methyl-bearing groups are trans if we consider CH3CH > D priority). For the (E)-alkene with syn H2 addition: attack from top face gives one enantiomer, attack from bottom face gives the other enantiomer with equal probability on an achiral catalyst. This produces equal amounts of (R,R) and (S,S) products = a racemic mixture. Step 5 - Why other options fail: (b) Diastereomers - the two products formed are enantiomers, not diastereomers of each other (they are mirror images). (c) Meso - meso would result from (Z)-alkene syn addition, not (E). (d) Constitutional isomers - impossible since same connectivity results. Therefore, the correct answer is A.