NEET 2022 — Chemistry solution

NEET2022Chemistry

RMgX + CO $_{2}$ $dry ether$ Y $H_{3} O^{+}$ RCOOH

What is Y in the above reaction :

A
R $_{3}$ CO $^{-}$ Mg $^{+}$ X
B
RCOO $^{-}$ X $^{+}$
C
(RCOO) $_{2}$ Mg
D
RCOO $^{-}$ Mg $^{+}$ X

Solution

The Grignard reagent RMgX adds across one C=O bond of carbon dioxide. The nucleophilic carbon attacks the electrophilic carbon of CO $_{2}$ , transferring the alkyl group and placing the negative charge on oxygen, with Mg $^{+}$ X as the counter ion.

This gives the magnesium salt of the carboxylate, R-COO $^{-}$ Mg $^{+}$ X, which is intermediate Y.

Aqueous acidic workup (H $_{3}$ O $^{+}$ ) then protonates the carboxylate to deliver the free carboxylic acid RCOOH.

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