The correct stability order for the following | vedclass

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Question

(D) The stability of carbocations is determined by resonance,hyperconjugation,and inductive effects.$(I)$ is a tertiary carbocation adjacent to an oxy

Vedclass · Accepted Answer

$(I) > (III) > (II) > (IV)$

Vedclass · Answer

(D) The stability of carbocations is determined by resonance,hyperconjugation,and inductive effects.$(I)$ is a tertiary carbocation adjacent to an oxygen atom. It is stabilized by resonance (lone pair donation from oxygen) and has $6$ $\alpha$-hydrogen atoms for hyperconjugation.$(III)$ is a secondary carbocation adjacent to an oxygen atom. It is stabilized by resonance,but has only $3$ $\alpha$-hydrogen atoms.$(II)$ is a secondary alkyl carbocation with $5$ $\alpha$-hydrogen atoms.$(IV)$ is a primary alkyl carbocation with $2$ $\alpha$-hydrogen atoms.Resonance stabilization by the oxygen atom is the most significant factor,making $(I)$ and $(III)$ more stable than $(II)$ and $(IV)$. Between $(I)$ and $(III)$,$(I)$ is more stable due to more hyperconjugation. Between $(II)$ and $(IV)$,$(II)$ is more stable due to more hyperconjugation.Therefore,the correct stability order is $(I) > (III) > (II) > (IV)$.

The correct stability order for the following species is:

Similar Questions

Which of the following represents the incorrect order of stability?

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

The stability of the following carbocations decreases in the order:

The shape of a carbonium ion (carbocation) is:

Arrange the following carbanions in the order of their stability:
$i$. $RC \equiv C^{\ominus}$
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$iii$. $R_2C = CH^{\ominus}$
$iv$. $R_3C - CH_2^{\ominus}$

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