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 $CH_3-C(OH)=C(OH)-CH_2CH_3$.

Column $-I$	Column $-II$
$A$. Free radical	$1$. Trigonal planar
$B$. Carbocation	$2$. Pyramidal
$C$. Carbanion	$3$. Linear

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

The compound $(X)$ will be:

Stability order of the following carbocations:
$(i) CH_3 - C^{+}(OCH_3) - CH_3$
$(ii) CH_3 - C^{+}(CH_3) - CH_3$
$(iii) CH_3 - NH - C^{+}H_2$
$(iv) CH_3 - C^{+}H_2$

Which of the following intermediates is not expected to be formed in the following reaction?

Which of the following represents the incorrect order of stability?

Match the intermediates given in Column $-I$ with their probable structure in Column $-II$.

Column $-I$ Column $-II$

$A$. Free radical $1$. Trigonal planar

$B$. Carbocation $2$. Pyramidal

$C$. Carbanion $3$. Linear

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