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(C) Let us analyze each option:$A$: The stability of carbocations follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ} > 	ext{methyl}$. Thus,$(CH_3)

Vedclass · Accepted Answer

Order of heat of hydrogenation: $2	ext{-methylbut-2-ene} > 2	ext{-methylbut-1-ene} > 3	ext{-methylbut-1-ene}$

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(C) Let us analyze each option:$A$: The stability of carbocations follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ} > 	ext{methyl}$. Thus,$(CH_3)_3C^{+} > (CH_3)_2CH^{+} > CH_3CH_2^+ > CH_3^+$ is correct.$B$: The stability of carbanions is inversely proportional to the $s$-character of the carbon atom. $sp$ hybridized carbon $(C \equiv C^-)$ has $50\%$ $s$-character,$sp^2$ $(C=C^-)$ has $33.3\%$,and $sp^3$ $(C-C^-)$ has $25\%$. Higher $s$-character makes the carbon more electronegative,stabilizing the negative charge. Thus,the order is correct.$C$: Heat of hydrogenation is inversely proportional to the stability of the alkene. More substituted alkenes are more stable and have lower heat of hydrogenation. The stability order is $2	ext{-methylbut-2-ene} > 2	ext{-methylbut-1-ene} > 3	ext{-methylbut-1-ene}$. Therefore,the heat of hydrogenation order should be $3	ext{-methylbut-1-ene} > 2	ext{-methylbut-1-ene} > 2	ext{-methylbut-2-ene}$. This option is incorrect.$D$: Heat of combustion is proportional to the number of carbon atoms and inversely proportional to the stability of the isomer. For isomers,the more branched isomer is more stable and has a lower heat of combustion. $n	ext{-butane}$ is less stable than isobutane,so it has a higher heat of combustion. This is correct.

Which of the following options given below is incorrect?

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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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