Arrange the following free radicals in order of decreasing stability Methyl (I), Vinyl (II), Allyl (III), Benzyl (IV)

Question

Arrange the following free radicals in order of decreasing stability Methyl (I), Vinyl (II), Allyl (III), Benzyl (IV) (A)  I>II>III>IV  (B)  III>II>I>IV  (C)  II>I>IV>III  (D)  IV>III>I>II

Tardigrade Admin · Accepted Answer

On the basis of hyperconjugation the order of stability of given free radicals is as follows: underset textbenzyl F.r.C6H5 overset bulletCH2 > CH2 = underset textallyl F.r.CH - overset bulletCH2 > underset textmethyl F. r. overset bulletCH3 > underset textVinyl F.r.CH2 = overset bulletCH

Q. Arrange the following free radicals in order of decreasing stability Methyl (I), Vinyl (II), Allyl (III), Benzyl (IV)

$I > II > III > I V$

$III > II > I > I V$

$II > I > I V > III$

$I V > III > I > II$

On the basis of hyperconjugation the order of stability of given free radicals is as follows : $benzyl F.r. C_{6} H_{5} C ∙ H_{2} > C H_{2} = allyl F.r. C H - C ∙ H_{2} > methyl F. r. C ∙ H_{3} > Vinyl F.r. C H_{2} = C ∙ H$

Q. Arrange the following free radicals in order of decreasing stability Methyl (I), Vinyl (II), Allyl (III), Benzyl (IV)

I>II>III>IV

III>II>I>IV

II>I>IV>III

IV>III>I>II

On the basis of hyperconjugation the order of stability of given free radicals is as follows : benzyl F.r.C6​H5​C∙H2​​>CH2​=allyl F.r.CH−C∙H2​​>methyl F. r.C∙H3​​>Vinyl F.r.CH2​=C∙H​

$I > II > III > I V$

$III > II > I > I V$

$II > I > I V > III$

$I V > III > I > II$

On the basis of hyperconjugation the order of stability of given free radicals is as follows : $benzyl F.r. C_{6} H_{5} C ∙ H_{2} > C H_{2} = allyl F.r. C H - C ∙ H_{2} > methyl F. r. C ∙ H_{3} > Vinyl F.r. C H_{2} = C ∙ H$