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**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.

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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).

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