Wurtz reaction of bromoethane gives n-butane. | vedclass

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Question

(B) In Wurtz reaction, two molecules of bromoethane react with $2$ molecules of sodium to form $n$-butane: $2CH_3CH_2Br + 2Na \rightarrow CH_3CH_2CH_2

Vedclass · Accepted Answer

$CH_3-(CH_2)_3-COOH$

Vedclass · Answer

(B) In Wurtz reaction, two molecules of bromoethane react with $2$ molecules of sodium to form $n$-butane: $2CH_3CH_2Br + 2Na \rightarrow CH_3CH_2CH_2CH_3 + 2NaBr$ Decarboxylation of the sodium salt of a carboxylic acid with sodalime $(NaOH + CaO)$ removes one carbon atom as $Na_2CO_3$ to produce an alkane with one less carbon atom than the original acid. To obtain $n$-butane ($4$ carbons), the sodium salt must be derived from a pentanoic acid ($5$ carbons). The reaction is: $CH_3CH_2CH_2CH_2COONa + NaOH \xrightarrow{CaO, \Delta} CH_3CH_2CH_2CH_3 + Na_2CO_3$ Thus, $X$ is pentanoic acid, $CH_3(CH_2)_3COOH$.

Wurtz reaction of bromoethane gives $n$-butane. Sodium salt of $X$ on heating with sodalime also results in $n$-butane. Compound $X$ is

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