See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 — Formation of A: The starting material is pent-1-yne (CH3CH2CH2C≡CH), a terminal alkyne. Treatment with NaNH2 (a strong base) deprotonates the terminal alkyne to give the acetylide anion, which then reacts with CH3CH2Br (ethyl bromide) in an SN2 reaction. This extends the carbon chain by two carbons, giving hept-3-yne: CH3CH2CH2C≡CCH2CH3. Wait — let me re-examine the starting material. The structure drawn shows two lines before the triple bond and then ≡C-H, suggesting the starting material is pent-1-yne: CH3CH2CH2C≡CH. After alkylation with CH3CH2Br, A = hept-3-yne: CH3CH2CH2C≡CCH2CH3. Actually, looking more carefully at the structure, the starting material appears to be but-1-yne (CH3CH2C≡CH). Alkylation with CH3CH2Br gives hex-3-yne: CH3CH2C≡CCH2CH3. Step 2 — Formation of B (Birch-type reduction, dissolving metal): A (hex-3-yne or the internal alkyne) is treated with Na in liquid NH3. This is a dissolving metal reduction of an internal alkyne. This reaction selectively reduces an internal alkyne to a trans (E)-alkene via a radical anion mechanism. So B = (E)-hex-3-ene: CH3CH2CH=CHCH2CH3 (trans configuration). Step 3 — Formation of C: B (the trans-alkene) is treated with Br2 in CH2Cl2. Br2 adds to an alkene via anti addition (through a bromonium ion intermediate). Anti addition of Br2 to a trans-alkene gives the (R,S) or meso dibromide — specifically, anti addition to (E)-hex-3-ene gives the (3R,4S) and (3S,4R) enantiomers, i.e., the racemic anti-addition product: 3,4-dibromohexane with anti stereochemistry. Looking at the answer choices, option (a) shows a vicinal dibromide (two Br on adjacent carbons) consistent with anti addition of Br2 across a double bond of a symmetric internal alkene — this matches 3,4-dibromohexane (or the analogous structure shown). Why other options fail: - (b) 1,3-dibromopentane would require a different mechanism (not simple Br2 addition to an alkene). - (c) contains an NH2 group, which is not introduced in step 3 (Br2/CH2Cl2 adds no nitrogen). - (d) shows a dibromoalkene, which would result from addition to an alkyne, not an alkene; B is already an alkene. Therefore, the correct answer is A.