Question

Give the decreasing order of the stabilities of the following free radicals:
a. $Ph\dot{C}HPh$
b. $Ph\dot{C}HCH=C{H}_2$
c. $Me\dot{C}HMe$
d. $Ph\dot{C}HMe$
e. $MeCH=CHC{H}_2\dot{C}{H}_2$
f. $Et-\dot{C}-M{e}_2$

• A. $a>b>f>c>d>e$
• B. $a>b>d>f>c>e$
• C. $b>d>f>a>c>e$
• D. $a>c>f>d>b>e$

Answer 1

The correct option is B $a>b>d>f>c>e$

The decreasing order of the stabilities of the free radicals is
$a\left(Ph\dot{C}HPh\right)>b(Ph\dot{C}HCH=C{H}_2)>d\left(Ph\dot{C}HMe\right)>f(Et-\dot{C}-M{e}_2)$

$>c\left(Me\dot{C}HMe\right)>e(MeCH=CHC{H}_2\dot{C}{H}_2)$
Diphenylmethane free radical is most stable due to the delocalization of electron due to resonance with two phenyl rings.
The stability decreases as the extent of the delocalization due to resonance or hyperconjugation decreases.
$MeCH=CHC{H}_2\dot{C}{H}_2$ is least stable as delocalization is not possible.
[$\left(a\right)Diphenylmethane>\left(b\right)BenzylorAllyl>Benzyl>3>2>1$ radicals].