The decreasing order of hydride affinity for | vedclass

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(C) Hydride affinity is inversely proportional to the stability of the carbocation.Stability order of the given carbocations:$B$ (Triphenylmethyl cati

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$C, A, D, B$

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(C) Hydride affinity is inversely proportional to the stability of the carbocation.Stability order of the given carbocations:$B$ (Triphenylmethyl cation) is the most stable due to extensive resonance with three phenyl rings.$D$ (Tricyclopropylmethyl cation) is highly stable due to cyclopropyl conjugation (bent bonds).$A$ (Allylic carbocation) is stabilized by resonance with the double bond.$C$ ($tert$-Butyl cation) is stabilized only by inductive effect and hyperconjugation.Stability order: $B > D > A > C$Since hydride affinity is inversely proportional to stability,the decreasing order of hydride affinity is: $C > A > D > B$.

The decreasing order of hydride affinity for the following carbocations is:
$A: CH_2=CH-C^+(CH_3)_2$
$B: (C_6H_5)_3C^+$
$C: (CH_3)_3C^+$
$D: (Cyclopropyl)_3C^+$
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The decreasing order of hydride affinity for the following carbocations is:$A: CH_2=CH-C^+(CH_3)_2$$B: (C_6H_5)_3C^+$$C: (CH_3)_3C^+$$D: (Cyclopropyl)_3C^+$Choose the correct answer from the options given below: