See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1: Understand the reaction mechanism. When SOCl2 reacts with an allylic alcohol via a 6-membered cyclic transition state (T.S.), it proceeds through an internal SN2-like (SNi') mechanism involving a cyclic sulfite intermediate. The 6-membered T.S. leads to allylic rearrangement with inversion/retention depending on geometry. Step 2: Analyze reaction A. Reactant A is CH3-CH=CH-CH2-OH (but-2-en-1-ol, i.e., the OH is at C1, double bond between C2 and C3). In a 6-membered cyclic T.S. with SOCl2, the chlorine is delivered to the gamma carbon (C1 = original OH carbon) or via allylic shift to C3. The 6-membered T.S. involves: O-S-O-Cl and the allylic system. The cyclic T.S. causes the Cl to be delivered to the allylic position away from the original OH, i.e., an SN2' (allylic rearrangement). The OH is at C1 (CH2-OH), so through the 6-membered T.S., Cl ends up at C3, giving CH3-CH(Cl)-CH=CH2. Wait, let me reconsider: the substrate is CH3-CH=CH-CH2-OH. The OH is at C4 (terminal CH2OH), double bond C2-C3. In the 6-membered cyclic T.S., the sulfur is bonded to O (at C4), and the Cl is delivered to C2 (gamma position via allylic shift). This gives: CH3-CH(Cl)-CH=CH2. So P = CH3-CH(Cl)-CH=CH2. Step 3: Analyze reaction B. Reactant B is CH3-CH(OH)-CH=CH2 (but-3-en-2-ol, OH at C2, double bond C3-C4). Through the 6-membered cyclic T.S., the OH is at C2, double bond at C3-C4. The Cl is delivered to C4 (gamma position), giving: CH3-CH=CH-CH2-Cl. So Q = CH3-CH=CH-CH2-Cl. Step 4: This matches option (c): P = CH3-CH(Cl)-CH=CH2 and Q = CH3-CH=CH-CH2-Cl. Step 5: Verify option (a) - P & Q are position isomers. P is 3-chlorobut-1-ene (Cl at C3, double bond C1-C2... rewritten: CH3-CH(Cl)-CH=CH2 means Cl at C2 counting from methyl end, i.e., 2-chlorobut-3-ene or but-3-en-2-yl chloride). Q is 1-chlorobut-2-ene (CH3-CH=CH-CH2Cl). Both have molecular formula C4H7Cl with one double bond - they are indeed position isomers (same molecular formula, same functional groups, different positions of Cl and double bond). So option (a) is TRUE. Step 6: Verify option (b) - P shows geometrical isomerism but Q does not. P = CH3-CH(Cl)-CH=CH2: the double bond carbons are CH= (bearing H and CH(Cl)CH3) and =CH2 (bearing two H's). Since =CH2 has two identical substituents (both H), P does NOT show geometrical isomerism. Q = CH3-CH=CH-CH2Cl: double bond carbons are CH= (bearing H and CH3) and =CH (bearing H and CH2Cl) - both carbons have different substituents, so Q DOES show geometrical isomerism. So statement (b) says P shows geometrical isomerism but Q does not - this is FALSE (it's the opposite). Step 7: Option (c) gives P = CH3-CH(Cl)-CH=CH2 and Q = CH3-CH=CH-CH2-Cl. This matches our analysis. TRUE. Step 8: Option (d) gives P = CH3-CH=CH-CH2-Cl and Q = CH3-CH(Cl)-CH=CH2. This is the reverse of what we derived. FALSE. Step 9: Therefore options (a) and (c) are correct, matching the given answer A, C. Therefore, the correct answer is A,C.