AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Offline Centres

Following carbocation undergoes either methyl shift with rate constant $(K_{2})$ (or) hydride shift with rate constant $(K_{1})$ from same carbon at 300K. Both elementary reactions follow Arrhenius equation and have same pre-exponential factor

*labeled carbon represents radioactive carbon from which methyl or hydride shift occurs
If $K_{1} = \frac{1}{16} K_{2}$ , then find ( a – b) in units of K. cal/mol
where ‘a’ and ‘b’ represent energy of activation (k. cal/mol) for formation of
and from given cation respectively.
$R = 2 c a l m o l^{- 1} K^{- 1} \log_{e} 2 = 0.70$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is -1.68.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\frac{K_{1}}{K_{2}} = \frac{1}{16} E a_{1} - E a_{2} = 4 R T l n_{2} = 1 - 68 k . c a l / m o l$

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test