LIST-I contains reactions and LIST-II contains major products. | LIST-I | LIST-II | |--------|--------| | P. (see structure) | 1. (see structure) | | Q. (see structure) | 2. (see structure) | | R. (see structure) | 3. (see structure) | | S. (see structure) | 4. (see structure) | | | 5. (see structure) | Match each reaction in LIST-I with one or more products in LIST-II and choose the correct option.

(B) $\mathbf{P} \rightarrow 1,4; \mathbf{Q} \rightarrow 2; \mathbf{R} \rightarrow 4; \mathbf{S} \rightarrow 3$

(A) $\mathbf{P} \rightarrow 1,5; \mathbf{Q} \rightarrow 2; \mathbf{R} \rightarrow 3; \mathbf{S} \rightarrow 4$

(C) $\mathbf{P} \rightarrow 1,4; \mathbf{Q} \rightarrow 1,2; \mathbf{R} \rightarrow 3,4; \mathbf{S} \rightarrow 4$

(D) $\mathbf{P} \rightarrow 4,5; \mathbf{Q} \rightarrow 4; \mathbf{R} \rightarrow 4; \mathbf{S} \rightarrow 3,4$

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Chapter_pending

LIST-I contains reactions and LIST-II contains major products.

LIST-I	LIST-II
P. (see structure)	1. (see structure)
Q. (see structure)	2. (see structure)
R. (see structure)	3. (see structure)
S. (see structure)	4. (see structure)
	5. (see structure)

Match each reaction in LIST-I with one or more products in LIST-II and choose the correct option.

(A) $P \to 1, 5; Q \to 2; R \to 3; S \to 4$
(B) $P \to 1, 4; Q \to 2; R \to 4; S \to 3$ ✓ Correct
(C) $P \to 1, 4; Q \to 1, 2; R \to 3, 4; S \to 4$
(D) $P \to 4, 5; Q \to 4; R \to 4; S \to 3, 4$

Answer: (B)

Solution

Analyzing each reaction: (P) t-butyl bromide with water undergoes SN1 to give t-butanol (1) and elimination to give isobutylene (4). (Q) t-butyl bromide with methoxide undergoes E2 to give isobutylene (2). (R) t-butyl bromide with ammonia gives elimination to isobutylene (4). (S) t-butyl alcohol with HBr gives SN1 substitution to t-butyl ether (3). The correct matching is P→1,4; Q→2; R→4; S→3.

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Single correct option

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JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Chapter_pending

Solution

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