**i tried using conjugate acid method to decide the order but I’m unable to decide between b’c’d.**

I tried using conjugate acid method to decide stability but I’m unable to differentiate between b,c,d based on inductive effect. There is no conjugation so no resonance.


In aprotic or very small polar solvents like chlorobenzene, solvated ions cannot form so there is no solvent effect. It follows a 'similar' order as that in gaseous phase.

Coming to the question, on protonation, option A has 2 Alpha Hydrogens (with respect to N+). B has 1 alpha Hydrogens, C has 6 alpha hydrogens and D has 4 alpha hydrogens.

Number of hyperconjugative structures = No. Of alpha hydrogens. So we would have C>D>A>B based on hyperconjugative stability.

But actually, B>A. This is because in B, there is a possibility of hydride shift which shifts the + charge on nitrogen to the carbon connected to it. On doing so, B has 4 alpha hydrogens.

Hence C>D>B>A is the answer. (although now D and B have same no of alpha hydrogens (same no of hyperconjugative structures), D is more stable than B as it would require some amount of energy to perform Hydride shift).


Not the answer you're looking for? Browse other questions tagged or ask your own question.