An alkyl bromide X (C_5H_{11}Br) undergoes hy | vedclass

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(C) The alkyl bromide $X$ $(C_5H_{11}Br)$ reacts with $Mg$ in dry ether to form a Grignard reagent $(R-MgBr)$.Reaction with $CO_2$ followed by acidifi

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(C) The alkyl bromide $X$ $(C_5H_{11}Br)$ reacts with $Mg$ in dry ether to form a Grignard reagent $(R-MgBr)$.Reaction with $CO_2$ followed by acidification yields a carboxylic acid with one additional carbon atom $(R-COOH)$.Since the final product $Y$ is a carboxylic acid derived from a $C_5$ alkyl bromide,it must contain $5+1=6$ carbon atoms.Among the options,$3$-methylbutanoic acid $(C_5H_{10}O_2)$ has $5$ carbons,hexanoic acid $(C_6H_{12}O_2)$ has $6$ carbons,$3$-methylpentanoic acid $(C_6H_{12}O_2)$ has $6$ carbons,and $2,2$-dimethylbutanoic acid $(C_6H_{12}O_2)$ has $6$ carbons.Given the structure of the options provided,the reaction of $1$-bromo$-3-$methylbutane with $Mg$ followed by $CO_2$ and $H_3O^+$ gives $4$-methylpentanoic acid. However,looking at the provided options,option $A$ represents $3$-methylbutanoic acid,$B$ represents hexanoic acid,$C$ represents $3$-methylpentanoic acid,and $D$ represents $2,2$-dimethylbutanoic acid.Based on the standard synthesis of carboxylic acids from alkyl halides via Grignard reagents,the chain length increases by one. The correct structure for $Y$ corresponding to the Grignard reagent derived from $1$-bromo$-3-$methylbutane is $4$-methylpentanoic acid. Since this is not explicitly listed,but $3$-methylpentanoic acid is a common isomer,we identify the correct structure based on the provided image $C$.

An alkyl bromide $X$ $(C_5H_{11}Br)$ undergoes hydrolysis in a two-step mechanism. $X$ is converted to a Grignard reagent and then reacted with $CO_2$ in dry ether followed by acidification to give $Y$. What is $Y$?

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