The compound A in the following reactions is: $$\begin{aligned} & \text { A } \xrightarrow[\text { (ii) Conc. } \mathrm{H}_{2} \mathrm{SO}_{4} / \Delta]{\text { (i) } \mathrm{CH}_{3} \mathrm{MgBr} / \mathrm{H}_{2} \mathrm{O}} \\ & \text { B } \xrightarrow[\text { (ii) } \mathrm{Zn} / \mathrm{H}_{2} \mathrm{O}]{\text { (i) } \mathrm{O}_{3}} \mathrm{C}+\mathrm{D} \end{aligned}$$

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Hydrocarbons

The compound A in the following reactions is:

A (i) CH_{3} MgBr / H_{2} O (ii) Conc. H_{2} SO_{4} /Δ B (i) O_{3} (ii) Zn / H_{2} O C + D

(A)(see structure)
(B)(see structure)
✓ Correct
(C)(see structure)
(D)(see structure)

Answer: (B)

Solution

Reading the sequence backwards, reductive ozonolysis of alkene B cleaves its double bond into two carbonyl fragments. B itself is the dehydration product of the alcohol formed when

C H_{3} M g B r

adds to the carbonyl of A. Taking A as phenylacetone,

C_{6} H_{5} C H_{2} C O C H_{3}

, the Grignard adds a methyl to give

C_{6} H_{5} C H_{2} C (C H_{3})_{2} O H

, which dehydrates toward the ring (Saytzeff) to

C_{6} H_{5} C H = C (C H_{3})_{2}

; ozonolysis of this alkene then yields benzaldehyde and acetone. Tracing the route back in this way identifies A as phenylacetone.

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Hydrocarbons

Solution

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