A compound with the molecular formula {C_4}{H | vedclass

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(D) $1$. The molecular formula ${C_4}{H_{10}}O$ corresponds to either an alcohol or an ether. $2$. The compound reacts with sodium,which indicates it

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sec-Butyl alcohol

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(D) $1$. The molecular formula ${C_4}{H_{10}}O$ corresponds to either an alcohol or an ether. $2$. The compound reacts with sodium,which indicates it is an alcohol $(R-OH + Na \rightarrow R-ONa + \frac{1}{2}H_2)$. $3$. Upon oxidation,it yields a carbonyl compound that does not reduce Tollen's reagent. This means the carbonyl compound is a ketone,not an aldehyde. $4$. For an alcohol to yield a ketone upon oxidation,it must be a secondary alcohol. $5$. The secondary alcohol with the formula ${C_4}{H_{10}}O$ is $sec$-butyl alcohol $(CH_3-CH(OH)-CH_2-CH_3)$,which oxidizes to butanone $(CH_3-CO-CH_2-CH_3)$. $6$. Butanone is a ketone and does not give a positive Tollen's test.

$A$ compound with the molecular formula ${C_4}{H_{10}}O$ reacts with sodium and gives a carbonyl compound upon oxidation,which does not reduce Tollen's reagent. What is the original compound?

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