You visited us 1 times! Enjoying our articles? Unlock Full Access!

Stability of Free Radical

Choose the co...

Question

Choose the correct stability order of the given free radicals.

I > I I > I I I > I V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I I > I > I V > I I I

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

I I = I > I V > I I I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I I I > I V > I I > I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $I I > I > I V > I I I$
In compounds (I), (III) and (IV) the -M effect and hyperconjugation effect are inoperative at meta positions. So we should consider the inductive effect of the groups. -I effect will decrease the stability of the free radical by withdrawing the electrons toward it itself. The order of -I effect is $N O_{2} > C N > O H$ . Therefore (III) is most destabilised.
In compound (II) the OH act as electron donating group through +M effect which is operative at para position. This increase the electron density at para position and stabilise the free radical. Hence compound (I) is most stabilised.
The order of stability of free radical is
$I I > I > I V > I I I$

Suggest Corrections

Similar questions

The correct order for decreasing stability of the following free radicals is

K_{2} C O_{3} (I), M g C O_{3} (I I), C a C O_{3} (I I I), B e C O_{3} (I V)

The stability of carbanions in the following

i) $R C \equiv ⊖ C$

ii)

iii) $R_{2} C = ⊖ C H$

iv) $R_{3} C - ⊖ C H_{2}$

is in the order of

+ I

- C H_{3}

- C D_{3}

- C T_{3}

- C H D_{2}

- I

- C N

- N O_{2}

- N H_{2}

- F

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program