Homolytic fission of the following alkanes forms free radicals CH 3- CH 3- CH 3- CH 2- CH 3- CH 32 CH - CH 3- CH 3- CH 2- CH CH 32The increasing order of stability of the radicals is-A- CH 33 C - CH 32 - - CH 2 CH 3- CH 3- C H - CH 3- CH 3- C H 2B- CH 32 C - CH 2 CH 3- CH 3- H - CH 3- CH 3- H 2- CH 33-C- CH 3- H 2- CH 3- CH 3- CH 33- CH 32- CH 2 CH 3D- CH 3- C H 2- CH 3- H - CH 3- CH 32- CH 2 CH 3- CH 33-

The correct option is D CH_3 .CH_2 lt;CH_3 .CH CH_3 lt;(CH_3 )_2.C CH_2 CH_3 lt;(CH_3)_3.CStability of radical follows similar order as carbocation follo ...

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Byju's Answer

Standard XII

Chemistry

Stability of Free Radical

Homolytic fis...

Question

Homolytic fission of the following alkanes forms free radicals
$C H_{3} - C H_{3}, C H_{3} - C H_{2} - C H_{3}, (C H_{3})_{2} C H - C H_{3}, C H_{3} - C H_{2} - C H (C H_{3})_{2} .$
The increasing order of stability of the radicals is:

(C H_{3})_{3} . C < (C H_{3})_{2} . C - C H_{2} C H_{3} < C H_{3} - . C H - C H_{3} < C H_{3} - . C H_{2}

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(C H_{3})_{2} . C - C H_{2} C H_{3} < C H_{3} - . C H - C H_{3} < C H_{3} - . C H_{2} < (C H_{3})_{3} . C

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C H_{3} - . C H_{2} < C H_{3} - . C H - C H_{3} < (C H_{3})_{2} . C - C H_{2} C H_{3} < (C H_{3})_{3} . C

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C H_{3} - . C H_{2} < C H_{3} - . C H - C H_{3} < (C H_{3})_{3} . C < (C H_{3})_{2} . C - C H_{2} C H_{3}

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Solution

The correct option is C $C H_{3} - . C H_{2} < C H_{3} - . C H - C H_{3} < (C H_{3})_{2} . C - C H_{2} C H_{3} < (C H_{3})_{3} . C$
Stability of radical follows similar order as carbocation follows.
( $methyl < 1^{o} < 2^{o} < 3^{o}$ ).
Moreover, $stability of free radical \propto Number of α - H$

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Similar questions

Q. Homolytic fission of the following alkanes forms free radicals

C H_{3} - C H_{3}, C H_{3} - C H_{2} - C H_{3}, (C H_{3})_{2} C H - C H_{3}, C H_{3} - C H_{2} - C H (C H_{3})_{2} .

The increasing order of stability of the radicals is:

Q. Rank the following cations in decreasing order of stability.

(C H_{3})_{2} C H - C H_{2} - C_{2}

C H_{3} - C^{+} H - C H (C H_{3})_{2}

C H_{3} - C^{+} (C H_{3}) - C H_{2} - C H_{3}

Q. Arrange the following alkanes in decreasing order of their heats of combustion.
(i)

C H_{3} - C H_{3} | C | C H_{3} - C H_{3} (N e o - p e n t a n e)

(ii)

C H_{3} - C H | C H_{3} - C H_{2} - C H_{3} (I s o - p e n t a n e)

(iii)

C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H_{3} (n - p e n t a n e)

Q. Decreasing order of stability of following alkenes is..
(i)

{C H}_{3} - C H = {C H}_{2}

(ii)

{C H}_{3} - C H = C H - {C H}_{3}

(iii)

{C H}_{3} - C_{3} = C H - {C H}_{3}

(iv)

{C H}_{3} - {C H}_{3} | C = {C H}_{3} | C - {C H}_{3}

C H_{3} - C H_{2} - \oplus C H_{2}

C H_{3} - C H | F - \oplus C H_{2}

C H_{3} - C H | B r - \oplus C H_{2}

C H_{3} - C H | C l - \oplus C H_{2}

Arrange the following carbocations in decreasing order of their stability:

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Homolytic fission of the following alkanes forms free radicals CH3−CH3,CH3−CH2−CH3,(CH3)2CH−CH3,CH3−CH2−CH(CH3)2. The increasing order of stability of the radicals is:

The correct option is C CH3−.CH2<CH3−.CH−CH3<(CH3)2.C−CH2CH3<(CH3)3.CStability of radical follows similar order as carbocation follows. (methyl<1o<2o<3o). Moreover, stability of free radical∝Number of α−H

Homolytic fission of the following alkanes forms free radicals
$C H_{3} - C H_{3}, C H_{3} - C H_{2} - C H_{3}, (C H_{3})_{2} C H - C H_{3}, C H_{3} - C H_{2} - C H (C H_{3})_{2} .$
The increasing order of stability of the radicals is:

The correct option is C $C H_{3} - . C H_{2} < C H_{3} - . C H - C H_{3} < (C H_{3})_{2} . C - C H_{2} C H_{3} < (C H_{3})_{3} . C$
Stability of radical follows similar order as carbocation follows.
( $methyl < 1^{o} < 2^{o} < 3^{o}$ ).
Moreover, $stability of free radical \propto Number of α - H$