In the following reaction A is : (i) $\mathrm{Br}_{2}, \mathrm{~h} \nu$ (ii) KOH (alc.) (iii) $\mathrm{O}_{3}$ (iv) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S}$ (v) $\mathrm{NaOH}(\mathrm{aq})+\Delta$ (see product structure)

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Hydrocarbons

In the following reaction A is :

(i)

Br_{2}, h ν

(ii) KOH (alc.)
(iii)

O_{3}

(iv)

(CH_{3})_{2} S

(v)

NaOH (aq) + Δ

(see product structure)

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

Answer: (B)

Solution

Reading the sequence forward from the hydrocarbon: free-radical bromination with

B r_{2} /h ν

substitutes a bromine, alcoholic

KOH

eliminates

HBr

to give the cyclic alkene, ozonolysis with reductive

(C H_{3})_{2} S

workup cleaves the ring double bond to a dialdehyde, and warm aqueous

NaOH

then drives an intramolecular aldol condensation to the cyclic product shown. Tracing this chain backward, the starting compound A must be cyclohexane.

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option

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