C_{4}H_{10}O xrightarrow{K_{2}Cr_{2}O_{7} / H | vedclass

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(B) The given sequence of chemical reactions is:$C_{4}H_{10}O$ $\xrightarrow{K_{2}Cr_{2}O_{7} / H_{2}SO_{4}} C_{4}H_{8}O$ $\xrightarrow{I_{2} / NaOH,

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Butan-$2$-ol

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(B) The given sequence of chemical reactions is:$C_{4}H_{10}O$ $\xrightarrow{K_{2}Cr_{2}O_{7} / H_{2}SO_{4}} C_{4}H_{8}O$ $\xrightarrow{I_{2} / NaOH, 	ext{Warm}} CHI_{3}$$(N)$Step $1$: The compound $N$ is oxidized by $K_{2}Cr_{2}O_{7} / H_{2}SO_{4}$ to a ketone $(C_{4}H_{8}O)$. This indicates that $N$ is a secondary alcohol.Step $2$: The ketone formed undergoes the iodoform reaction with $I_{2} / NaOH$ to produce iodoform $(CHI_{3})$. This confirms that the ketone must have a methyl group attached to the carbonyl carbon (i.e.,it is a methyl ketone).Butan-$2$-ol $(CH_{3}CH(OH)CH_{2}CH_{3})$ is a secondary alcohol. Upon oxidation, it yields butan-$2$-one $(CH_{3}COCH_{2}CH_{3})$, which is a methyl ketone and gives a positive iodoform test.Therefore, $N$ is butan-$2$-ol.

$C_{4}H_{10}O$ $\xrightarrow{K_{2}Cr_{2}O_{7} / H_{2}SO_{4}} C_{4}H_{8}O$ $\xrightarrow{I_{2} / NaOH, \text{Warm}} CHI_{3}$
Here, $N$ is

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$C_{4}H_{10}O$ $\xrightarrow{K_{2}Cr_{2}O_{7} / H_{2}SO_{4}} C_{4}H_{8}O$ $\xrightarrow{I_{2} / NaOH, \text{Warm}} CHI_{3}$Here, $N$ is