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.
closed as off-topic by A.K., Mithoron, Todd Minehardt, Jon Custer, aventurin Nov 2 '18 at 10:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework questions must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: How do I ask homework questions on Chemistry Stack Exchange?" – A.K., Mithoron, Todd Minehardt, Jon Custer, aventurin
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).