In the following species,the one which is lik | vedclass

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(C) The benzoin condensation involves the reaction of two molecules of benzaldehyde in the presence of a cyanide ion $(CN^-)$ catalyst to form benzoin

Vedclass · Accepted Answer

$Ph-C^-(OH)(CN)$

Vedclass · Answer

(C) The benzoin condensation involves the reaction of two molecules of benzaldehyde in the presence of a cyanide ion $(CN^-)$ catalyst to form benzoin. The mechanism begins with the nucleophilic attack of $CN^-$ on the carbonyl carbon of benzaldehyde. This is followed by a proton transfer (tautomerization) to form a carbanion intermediate,which is stabilized by the electron-withdrawing $CN$ group and the phenyl ring. The structure of this key intermediate is $Ph-C^-(OH)(CN)$. This carbanion then attacks another molecule of benzaldehyde to continue the reaction,eventually leading to the formation of benzoin after the elimination of the $CN^-$ catalyst.

In the following species,the one which is likely to be the intermediate during benzoin condensation of benzaldehyde is

Similar Questions

The correct order of decreasing stability of the following carbocations is:
$I. CH_3-CH^{+}-CH_3$
$II. CH_3-CH^{+}-OCH_3$
$III. CH_3-CH^{+}-CH_2-OCH_3$

For the following bond cleavages,use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify the reactive intermediate produced as free radical,carbocation,or carbanion.

What is the hybridization state of a carbocation?

Taking into account the stability of various carbocations and the rules governing mechanisms of carbocation rearrangements,which reaction is most likely to occur during the given reaction?

In which pair is the second ion more stable than the first?

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