9
$\begingroup$

I know that in piperidine, the lone pair on the nitrogen can occupy either an axial or a equatorial position in the two chair forms. I also read that the axial position for the lone pair is preferred by a small margin in the gas phase (so intermolecular forces are minimized).

enter image description here

Now, this was presented as an argument against the idea that lone pairs are "sterically demanding" or "bulky" or repel other entities of negative charge more strongly than do bonding pairs, etc. The conclusion was that a lone pair must occupy less space than a hydrogen atom since it prefers the axial position in piperidine!

This argument seems correct at first, but could there be other factors affecting why the axial lone pair conformation might be more stable than the equatorial lone pair conformation? Also, are there any other examples in which the lone pair behaves "unexpectedly," at least with respect to elementary theory?

ETA: On the other hand one could make the argument that this isn't a apples to apples comparison; we are comparing a lone pair of electrons with a bonding pair of electrons plus a hydrogen atom. So it may be that this is a steric rather than an electronic (i.e. repulsive) effect.

Or could there be some sort of hyperconjugative effect going on? Carbon withdraws electron density from hydrogen; hydrogen becomes slightly partially positive, is attracted to lone pair?

$\endgroup$
  • $\begingroup$ There is an interesting article which looks at the idea of "bulky" lone pairs, saying it is a fix for VSEPR with no physical basis and proposing Bent´s rule as an easy alternative explanation for bond angles deviating from simple repulsion models. Source $\endgroup$ – Lucas Feb 14 '17 at 13:28
8
$\begingroup$

The energy difference between the two conformations in your example is extremely small. This makes it difficult to say that "this" is the reason why things are the way they are. Nonetheless, it is a very interesting question. Perhaps, like in the "how many angels can you draw on the head of a pin" question, the discussion around the question might be as significant as the answer.

What we need is more data. The phosphorus analogue of piperidine, phosphacyclohexane, has also been examined and the hydrogen resides in the axial position (no evidence for the equatorial isomer was observed).

enter image description here

There are important differences between these nitrogen and phosphorus analogues. To start with, phosphorus-carbon and phosphorus-hydrogen bond lengths (180-85 pm and 145 pm respectively) are typically longer than the analogous nitrogen-carbon and nitrogen-hydrogen bonds lengths (150 pm and 102 pm respectively). More important is the fact that the phosphorus is basically unhybridized, so the lone pair of electrons around phosphorous exists in a non-directional s-orbital. This makes conformational analysis of phosphacyclohexane easier because we don't have competing (hydrogen and lone pair) substituent effects to consider and dissect.

The observation that the hydrogen in phosphacyclohexane exists in the axial position suggests that there is an attractive 1-3 synaxial H-H interaction and/or a sterically repulsive 1-2 H-H gauche interaction. Given the long length of bonds to phosphorous, repulsive interactions seem unlikely and the attractive 1-3 synaxial H-H explanation was favored.

Another data point comes with N-methyl piperidine where the lone pair strongly prefers the axial position. This was attributed to a repulsive 1-3 synaxial Me-H interaction. The fact that the methyl group adopts the equatorial position also shows that the repulsive 1-2 Me-H gauche interaction must be of little consequence and any repulsion in the corresponding H-H case would be even smaller.

Piperidine itself appears to be somewhere in the middle, in between N-methyl piperidine and phosphacyclohexane. Certainly the repulsive 1-3 synaxial Me-H interaction observed in N-methyl piperidine has been reduced, but not eliminated.

Whether just the repulsive 1-3 synaxial H-H interaction has been reduced or whether there is now also a contribution from the attractive 1-3 synaxial H-H interaction seen in phosphacyclohexane cannot be determined.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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