Determine the decreasing order of stability f | vedclass

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(D) To determine the stability of carbocations,we analyze the electronic effects of the substituents attached to the positively charged carbon atom.$(

Vedclass · Accepted Answer

$II > I > III$

Vedclass · Answer

(D) To determine the stability of carbocations,we analyze the electronic effects of the substituents attached to the positively charged carbon atom.$(I) \ CH_3-C^+(CH_3)-CH_3$ is a tertiary $(3^{\circ})$ carbocation. It is stabilized by the $+I$ effect of three methyl groups and hyperconjugation.$(II) \ CH_3-CH^+(OCH_3)$ is a secondary $(2^{\circ})$ carbocation. The oxygen atom in the $-OCH_3$ group has lone pairs,which provide strong $+M$ (mesomeric) stabilization to the carbocation,making it highly stable.$(III) \ CH_3-CH^+(COCH_3)$ is a secondary $(2^{\circ})$ carbocation. The $-COCH_3$ group exerts a strong $-I$ (inductive) effect and a $-M$ (mesomeric) effect,both of which destabilize the carbocation.Comparing these,$(II)$ is the most stable due to resonance stabilization ($+M$ effect). $(I)$ is more stable than $(III)$ because $(III)$ is destabilized by electron-withdrawing groups. Thus,the order of stability is $(II > I > III)$.

Determine the decreasing order of stability for the following carbocations: $(I) \ CH_3-C^+(CH_3)-CH_3$,$(II) \ CH_3-CH^+(OCH_3)$,$(III) \ CH_3-CH^+(COCH_3)$.

Similar Questions

Choose the most reasonable reaction intermediate for the following reaction.

Assertion : Carbanions like ammonia have pyramidal shape.
Reason : The carbon atom carrying negative charge has an octet of electrons.

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:

Which one of the following carbocations is the most stable?

The stability of carbanions follows the order:
$I: CH_3CH_2CH_2\bar{C}H_2$
$II: CH_3\bar{C}HCH_2CH_3$
$III: (CH_3)_3\bar{C}$
$IV: CH_3\bar{C}(Ph)CH_2CH_3$

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