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(C) Step $1$: The reaction of $CH_3CH_2CH_2CH_2COCH_3$ with $NaOBr$ followed by $H_3O^+$ is a haloform reaction. It converts the methyl ketone into a

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$CH_3-CH_2-CH_2-CH(Br)-COOH$

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(C) Step $1$: The reaction of $CH_3CH_2CH_2CH_2COCH_3$ with $NaOBr$ followed by $H_3O^+$ is a haloform reaction. It converts the methyl ketone into a carboxylic acid with one less carbon atom.$CH_3CH_2CH_2CH_2COCH_3 \xrightarrow{NaOBr, H_3O^+} CH_3CH_2CH_2CH_2COOH + CHBr_3$.Step $2$: The product $CH_3CH_2CH_2CH_2COOH$ (pentanoic acid) is then treated with $Red \ P$ and $Br_2$,followed by $H_2O$. This is the Hell-Volhard-Zelinsky $(HVZ)$ reaction,which alpha-halogenates the carboxylic acid.$CH_3CH_2CH_2CH_2COOH \xrightarrow{Red \ P, Br_2} CH_3CH_2CH_2CH(Br)COOH$.The final product is $2$-bromopentanoic acid,which corresponds to option $C$.

The major product of the following reaction sequence is:

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