An alcohol of formula C_9H_{12}O reacts with | vedclass

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(B) The reaction involves the oxidation of an alcohol to a carbonyl compound (aldehyde or ketone) using $Na_2Cr_2O_7$,which is a strong oxidizing agen

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$Ph-CH(OH)-CH_2-CH_3$

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(B) The reaction involves the oxidation of an alcohol to a carbonyl compound (aldehyde or ketone) using $Na_2Cr_2O_7$,which is a strong oxidizing agent.The change in molecular formula is from $C_9H_{12}O$ to $C_9H_{10}O$,which corresponds to the loss of two hydrogen atoms $(2H)$.This indicates that the alcohol is a secondary alcohol,which oxidizes to a ketone.Option $B$ $(Ph-CH(OH)-CH_2-CH_3)$ is a secondary alcohol. Upon oxidation,it forms $Ph-C(=O)-CH_2-CH_3$ (propiophenone),which has the formula $C_9H_{10}O$.

An alcohol of formula $C_9H_{12}O$ reacts with $Na_2Cr_2O_7$ to form a compound having formula $C_9H_{10}O$. The original alcohol might be

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