As I was reading about physical properties of pyridines and substituted pyridines, I came across this:

2-methylpyridine boils at 129°C while 3-methylpyridine boils at 15° higher than 2-methylpyridine

And the reason that is given is:

This lowering of boiling point is explained in terms of hindrance to association offered by the groups in the vicinity of the nitrogen atom

Assuming that the book is talking about hydrogen bonding as "association", wouldn't 2-methylpyridine boil at higher temperature than 3-methylpyridine as the methyl group is closer to nitrogen atom in 2-methylpyridine and hence can form better hydrogen bonds than 3-methylpyridine ?

  • $\begingroup$ Pray, tell me why my question have been downvoted ? At least leave a reason for downvotes! $\endgroup$
    – Natasha J
    Commented Jun 19, 2023 at 17:15
  • 1
    $\begingroup$ Disregarding the question itself (which is fine IMO), there were a typo in the title ("32-methylpyridine"), so do not hesitate to proofread your own posts. Also, every quote needs to have a clearly stated source, i.e. here a reference is missing, which I would guess is the primary concern of downvoters. A brief guide how to cite a textbook is available, and make sure to include edition and page numbers. Otherwise it may be considered as plagiarism. $\endgroup$
    – andselisk
    Commented Jun 19, 2023 at 18:08

1 Answer 1


Two things:

  1. The nitrogen in pyridine doesn't hydrogen bond with saturated carbons. It's more likely that you'd get pi-pi stacking, or interactions between the lone pair and the aromatic ring of a second molecule. Having a methyl group near to the lone pair disrupts the latter interactions.

  2. If the nitrogen in pyridine did form interactions with the methyl group, 2-methylpyridine would still boil at a lower temperature. Internal interactions don't need to be broken for a compound to pass into the gas phase.


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