Ethylmagnesium bromide ($ ext{CH}_3 ext{CH}_2 ext{MgBr}$) is treated with dry ice (solid $ ext{CO}_2$) followed by aqueous acid workup. Identify the carboxylic acid formed and explain why this route can succeed even for a tertiary Grignard reagent, unlike the nitrile-hydrolys

Question

Ethylmagnesium bromide ($	ext{CH}_3	ext{CH}_2	ext{MgBr}$) is treated with dry ice (solid $	ext{CO}_2$) followed by aqueous acid workup. Identify the carboxylic acid formed and explain why this route can succeed even for a tertiary Grignard reagent, unlike the nitrile-hydrolysis route.

Accepted Answer

Ethylmagnesium bromide's ethyl group adds to the electrophilic carbon of $\text{CO}_2$ by nucleophilic addition, exactly as any carbanion-like nucleophile adds to a carbonyl-type carbon, giving the magnesium salt of propanoate; aqueous acid work-up then liberates the free acid, propanoic acid, $\text{CH}_3\text{CH}_2\text{COOH}$ . The nitrile-hydrolysis route to a carboxylic acid, by contrast, depends on first making the nitrile itself by an $\text{S}_\text{N}2$ displacement of a halide by cyanide -- and $\text{S}_\text{N}2$ requires backside attack on the halide-bearing carbon, which is sterically blocked when that carbon is tertiary (three bulky alkyl groups crowd the approach). Carbonation of a Grignard reagent has no such backside-attack requirement at all -- the alkyl group is already attached to magnesium and simply adds to the separate, small $\text{CO}_2$ molecule -- so it succeed …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

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$Q. How would you obtain the following? Butanoic acid from 1-chlorobutane$
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Since butanoic acid has the same number of carbons (4) as 1-chlorobutane, the conversion must NOT go through cyanide (which would add a carbon); instead, hydrolyse to the alcohol and then oxidise it fully to the acid.

1-Chlorobutane, $CH_3CH_2CH_2CH_2Cl$ , already has four carbons — exactly the number needed for butanoic acid, $CH_3CH_2CH_2COOH$ . So the terminal $-CH_2Cl$ carbon simply needs to become a $-COOH$ carbon, with no chain extension required (ruling out the usual $KCN$ chain-extension route used for acids with one extra carbon).

Step 1 — Hydrolysis (nucleophilic substitution by hydroxide):

$CH_3CH_2CH_2CH_2Cl + KOH \xrightarrow{aq.} CH_3CH_2CH_2CH_2OH\;(\text{butan-1-ol}) + KCl$

…

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$Q. On hydrolysis, benzamide forms:$
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CBSE 2021Set A1 markMCQ
$Q. The product of the reaction CH3CN --H3O+--> is$
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›Reveal solutionSolution

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Nitriles undergo hydrolysis. On complete hydrolysis with aqueous acid (H3O+), the –CN group is converted, via the amide, all the way to the carboxylic acid:

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…

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Q.Ethylmagnesium bromide ( $\text{CH}_3\text{CH}_2\text{MgBr}$ ) is treated with dry ice (solid $\text{CO}_2$ ) followed by aqueous acid workup. Identify the carboxylic acid formed and explain why this route can succeed even for a tertiary Grignard reagent, unlike the nitrile-hydrolysis route.

Concept understanding — Preparation of Carboxylic Acids

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