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Question: How to check the stability of any compound such as carbocations, etc Answer:stability of a organic compound such as carbocation is directly propotional to number of alpha hydrogen i.e the carbocation having more alpha hydrogen will more stable than the carbocation having less alpha hydrogen(answer was given by sir on askiitians)Sir,so if there are no alpha hydrogens in an option so how can we determine or if in the given question there are equal no of alpha hydrogens?

Question: How to check the stability of any compound such as carbocations, etc Answer:stability of a organic compound such as carbocation is directly propotional to number of alpha hydrogen i.e the carbocation having more alpha hydrogen will more stable than the carbocation having less alpha hydrogen(answer was given by sir on askiitians)Sir,so if there are no alpha hydrogens in an option so how can we determine or if in the given question there are equal no of alpha hydrogens?

Grade:12

1 Answers

Divyanshu Rawat
38 Points
6 years ago
For checking carbocation stability determining the no. of alpha hydrogen is the best method to work on and nearly all questions will get get solved by applying this trick.
But as you have mentiomed if the alpha hydrogen are not given kf the no. of alpha hydrogen are equal then we can check for stability by electronic effects ( Inductive effect, Mesomeric effect, electromeric effect ) due to the groups attached to the carbocation.
So to get this understand let us consider an example:
We have to decide stability b/w (CH3)3C-I and (CH3)3C-F
So if we talk about electronic effects here we know that alkyl group (CH3) shows +I effect and halogens show -I effects.
Also if we compare the -I effect of fluorine and Iodine, Iodine has less -I effect as compared to Fluorine.
 
So now we come on comparison:
Due to 3 alkyl groups the carbon at centre will have ∆- charge due +I effect and due to -I effect of halogen it will have ∆+ charge.
But as we compared above the -I effect of Iodine and fluorine.
So due to fluorine the ∆+ charge on central carbon will be greater than that of iodine so fluorine will more stabilise the -ve charge due to 3 alkyl groups.
So we can conclude that (CH3)3C-F will be more stable than (CH3)3C-I.
 
Hope it helps
Thankyou

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