0
$\begingroup$

The question asks what nitrogen in the ring of histidine has the shortest bond length.

I was thinking that since it has resonance, the initial thought of the nitrogen with the $\pi$ bond is incorrect. However, I imagine the resonance structure shown in the figure below makes the greatest contribution to the resonance hybrid. Therefore making my first guess correct. The question also asks for a single line of reasoning so I am confused how to present my thinking concisely.

enter image description here

$\endgroup$
0
1
$\begingroup$

It is correct that the $\ce{N}$ with the double bond has a localized lone pair, because of its $sp^2$ hybridization, due to which the lone pairs are in the same plane as ring, and hence not conjugated.

However, the other $\ce{N}$ is clearly in conjugation, hence, if you draw the $5$ possible resonance structures, you would notice that each bond in the ring is a double bond for exactly $2$ of them. However, since there is a positive charge on electronegative $\ce{N}$ in all of them except the one you have drawn, this structure would have maximum contribution. Thus, the shortest bond length indeed would be the one you've shown as a double bond in your figure, although in reality it would have bond order $<2$ due to resonance.

$\endgroup$

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.