The order of decreasing ease of abstraction o | vedclass

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Question

(B) The ease of abstraction of a hydrogen atom depends on the stability of the resulting radical or carbocation intermediate. Abstraction of $H_{a}$ l

Vedclass · Accepted Answer

$H_{a} > H_{c} > H_{b}$

Vedclass · Answer

(B) The ease of abstraction of a hydrogen atom depends on the stability of the resulting radical or carbocation intermediate. Abstraction of $H_{a}$ leads to a $3^{\circ}$ allylic radical/carbocation, which is highly stable due to resonance and hyperconjugation. Abstraction of $H_{c}$ leads to a $2^{\circ}$ allylic radical/carbocation, which is less stable than the $3^{\circ}$ allylic species. Abstraction of $H_{b}$ leads to a $2^{\circ}$ non-allylic radical/carbocation, which is the least stable among the three. Therefore, the order of decreasing ease of abstraction is $H_{a} > H_{c} > H_{b}$.

Column $-I$	Column $-II$
$A$. Free radical	$1$. Trigonal planar
$B$. Carbocation	$2$. Pyramidal
$C$. Carbanion	$3$. Linear

The order of decreasing ease of abstraction of hydrogen atoms in the following molecule is:

Similar Questions

Compare the stability of the following free radicals:
$(1)$ $CH_3 - \dot{CH} - CH_3$
$(2)$ $C_6H_5 - \dot{CH_2}$
$(3)$ $\dot{CH_2} - CH(CH_3)_2$
$(4)$ $\dot{CH_2} - CH_3$

Match the intermediates given in Column $-I$ with their probable structure in Column $-II$.

Column $-I$ Column $-II$

$A$. Free radical $1$. Trigonal planar

$B$. Carbocation $2$. Pyramidal

$C$. Carbanion $3$. Linear

Intermediates formed in the reactions $P$ and $Q$ are:
$P: CH_3-CH=CH_2 \xrightarrow{HCl}$
$Q: CH_3-CH=CH_2 \xrightarrow[\text{Peroxide}]{HCl}$

The stable carbocation formed in the reaction of benzene with $n$-propyl chloride in the presence of anhydrous $AlCl_3$ is:

Which of the following is the least stable carbocation?

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