The compound shown below undergoes racemizati | vedclass

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(C) The compound is a chiral ketone. In the presence of an aqueous acid,the ketone undergoes keto-enol tautomerism. The acid protonates the carbonyl o

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(C) The compound is a chiral ketone. In the presence of an aqueous acid,the ketone undergoes keto-enol tautomerism. The acid protonates the carbonyl oxygen,followed by the removal of an $\alpha$-hydrogen to form an enol intermediate. The enol intermediate is planar at the $\alpha$-carbon,which removes the chirality at that center. When the enol tautomerizes back to the ketone,the proton can add from either side of the planar double bond,leading to the formation of both enantiomers,thus causing racemization. The structure representing the enol intermediate is $CH_3-C(OH)=C(OH)-CH_2CH_3$.

The compound shown below undergoes racemization on reaction with aqueous acid. Which of the following structures best represents the intermediate responsible for this process?

Similar Questions

Which of the following bridgehead carbocations is the most stable?

The correct stability order of the following carbocations is:
$I$. $H_2C^+-CH=CH-CH_3$
$II$. $H_2C^+-CH=CH-BMe_2$
$III$. $H_2C^+-CH=CH-NMe_2$
$IV$. $H_2C^+-CH=CH-OMe$

The bond dissociation energy of the $C-H$ bond for the compounds: $(I)\ CH_3-H, (II)\ CH_3-CH_2-H, (III)\ CH_2=CH-CH_2-H, (IV)\ C_6H_5-H$ follows the order:

Which one among the following carbocations is the most stable?

Intermediates formed in the reactions $P$ and $Q$ are:
$P: CH_3-CH=CH_2 \xrightarrow{HCl}$
$Q: CH_3-CH=CH_2 \xrightarrow[\text{Peroxide}]{HCl}$

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