See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material. Me—C≡C—Et is 2-pentyne (pent-2-yne), CH3C≡CCH2CH3. Step 2: Reaction with Na/liquid NH3 is a Birch-type dissolving metal reduction of an alkyne, which gives trans (E) alkene via anti addition of two hydrogen atoms. So P is (E)-pent-2-ene (trans-2-pentyne), i.e., (E)-CH3CH=CHCH2CH3. Step 3: Reaction of (E)-pent-2-ene with Br2 in CCl4 is an electrophilic addition (anti addition of bromine across the double bond). The product Q is 2,3-dibromopentane. Step 4: Analyze the stereochemistry of Q. Anti addition of Br2 to (E)-pent-2-ene: The two carbons bearing bromine (C2 and C3) both become chiral centers. The substituents at C2 are: CH3, Br, H, and C3 chain; at C3: CH2CH3, Br, H, and C2 chain. These two chiral centers have different substituents (C2 has methyl, C3 has ethyl), so C2 and C3 are not identical — this molecule cannot be a meso compound. Step 5: Anti addition of Br2 to trans-2-pentene produces two enantiomers (2R,3R and 2S,3S dibromopentane) in equal amounts, forming a racemic mixture. This is because the trans alkene with anti addition gives the pair of enantiomers as a racemate. Step 6: A racemic mixture (equal amounts of two enantiomers) is a binary mixture that is optically inactive due to external compensation (the optical rotations cancel each other out in equal and opposite amounts). Step 7: Evaluate options: (a) Incorrect — internal compensation refers to a meso compound (single pure compound), which is not the case here since the two chiral centers bear different groups. (b) Correct — a binary mixture (racemate of 2R,3R and 2S,3S) that is optically inactive due to external compensation. (c) Incorrect — the mixture is not optically active; the racemate is inactive. (d) Incorrect — chiral centers are present. Therefore, the correct answer is B.