4-Bromopentan-2-one is treated as shown: $\mathrm{CH_3\!-\!CO\!-\!CH_2\!-\!CH(CH_3)\!-\!Br} \;\xrightarrow[\;(ii)\ \mathrm{H_2O}\;]{\;(i)\ \mathrm{Mg}\;}\; \text{'A' (Major Product)}$ A is :

(B) 1-methylcyclopropan-1-ol (a cyclopropane ring bearing $-\mathrm{OH}$ and $-\mathrm{CH_3}$ on the same carbon)

(A) $\mathrm{CH_3\!-\!CO\!-\!CH_2\!-\!CH(CH_3)\!-\!MgBr}$ (the organomagnesium reagent itself, with the ketone intact)

(C) a gem-diol bearing a Grignard group, $(\mathrm{HO})_2\mathrm{C(CH_3)\!-\!CH_2\!-\!CH(CH_3)\!-\!MgBr}$ (one carbon drawn with two $-\mathrm{OH}$ groups and a terminal $-\mathrm{CH(CH_3)MgBr}$)

(D) $\mathrm{CH_3\!-\!CO\!-\!CH_2\!-\!CH(CH_3)\!-\!C(CH_3)(OH)\!-\!CH_2\!-\!CH(CH_3)\!-\!Br}$ (ketone at one end, a new tertiary $-\mathrm{OH}$ in the middle, $-\mathrm{Br}$ intact at the other end)

Chapter_pending — JEE Main · 2023 · 12 Apr 2023 · Shift 1 Question with Solution

4-Bromopentan-2-one is treated as shown:

C H_{3} - CO - C H_{2} - CH (C H_{3}) - Br (i) Mg (ii) H_{2} O ’A’ (Major Product)

A is :

(A) $C H_{3} - CO - C H_{2} - CH (C H_{3}) - MgBr$ (the organomagnesium reagent itself, with the ketone intact)
(B)1-methylcyclopropan-1-ol (a cyclopropane ring bearing $- OH$ and $- C H_{3}$ on the same carbon)✓ Correct
(C)a gem-diol bearing a Grignard group, $(HO)_{2} C (C H_{3}) - C H_{2} - CH (C H_{3}) - MgBr$ (one carbon drawn with two $- OH$ groups and a terminal $- CH (C H_{3}) MgBr$ )
(D) $C H_{3} - CO - C H_{2} - CH (C H_{3}) - C (C H_{3}) (OH) - C H_{2} - CH (C H_{3}) - Br$ (ketone at one end, a new tertiary $- OH$ in the middle, $- Br$ intact at the other end)

Answer: (B)

Solution

Mg generates a carbanion at the C–Br carbon (C4) of

C H_{3} COC H_{2} CHBrC H_{3}

. Because the carbonyl carbon (C2) lies two carbons away (a 1,3-relationship), the carbanion adds intramolecularly to the carbonyl carbon, closing a favourable three-membered ring; aqueous work-up protonates the alkoxide to an –OH. The major product is a cyclopropan-1-ol (1,2-dimethylcyclopropan-1-ol) — the cyclopropane ring bearing the –OH on the carbon that was the carbonyl carbon — option (B). Intramolecular cyclisation to the small ring dominates over intermolecular addition, so the dimeric adduct is not the major product.

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Single correct option

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Chapter_pending

Solution

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