Question

QUESTION 9 Arrange the following radicals in order of decreasing stability: CH3 a. (CH3)3C. b. H2CHC. C. CH3 d. (CH3)2CHCH2 e. (CH3)2CH. CH3 OOO A. a, b, c, d, e B. b, a, e, d, c OC. c, a, b, ed D.a, b, e, d, c E. b, a, c, d, e

Answer 1

1. Question a Radicals are like carbocation, that are electron deficient. So the stability of radicals will be as Carbocations : ay (CH3); e. a electron deficient HC-8-ah Carbon The presence of 3 CH₃ groups will exent tl effect and it will stabilise the electron 3° padical O deficient carbon. stabicity- 3.720710 Another effect is typen conjugation Here og ß controns are lydrogens are present So, a hyperconjugation structures are possible So it is very much stabilised. b) He? CHEC-et H CH2 – C11 = q --chitz cat Atso resonance stabiused 3² radical very much stable so it will be more stable than a I et alb. No stability faceton This is the most unstable among these.

dy (13)2 CH ciz et cz 136 sa 1. Ispadical, less stable. But there will be some tI effect by to S alkyl groups. is estable than ① but less stable than 6 But more So, it

© (CH3), étt - - * - "2"padical and more stable than On but less stable than So, the trend of decreasing stability is by ayenaye The answers is ®