1
$\begingroup$

Why is a C-Br bond considered polar? From electronegativity considerations, both carbon and bromine have very similar electronegativities - 2.5 and 2.8 - respectively.

Nonetheless, I am told that such a bond would be polar. We generally classify bond between atoms with EN differences < 0.5 as non-polar; why would C-Br be an exception?

$\endgroup$
  • $\begingroup$ What is your source for this view? Who thinks this bond is polar? $\endgroup$ – user467 Jun 10 '14 at 0:17
4
$\begingroup$

I poked around on the internet and found a lot of sites use the Pauling electronegativity scale (link 1, link 2). For carbon the value is 2.55 and for bromine 2.96. Not quite a 0.5 difference, but closer - and this "rule" is somewhat arbitrary to begin with. Also the dipole moment of bromobenzene is around 1.7 D (link 3), certainly not insignificant. I too would have said bromobenzene is polar. It is not strongly polar, whether it is "a little" polar, "mildly" polar, "somewhat" polar, etc. seems to be the question - but that's just semantics. Maybe the reason many people consider the $\ce{C-Br}$ bond to be polar is because it has a very observable effect on electrophilic aromatic substitution reaction rates. Specifically, why should the relative rate for electrophilic aromatic substitution at the meta position in bromobenzene be slower that the rate in benzene itself unless you invoke the inductive effects caused by a polar $\ce{C-Br}$ bond?

$\endgroup$
  • $\begingroup$ I think that's what the person who made this claim had in mind, and it makes sense, thank you! $\endgroup$ – Dissenter Jun 10 '14 at 22:56
  • $\begingroup$ Would a C-Br bond show up on IR spectroscopy? $\endgroup$ – Dissenter Aug 13 '14 at 16:19
  • 1
    $\begingroup$ Not sure, but I don't think so. A requirement for ir absorption is for a vibration to produce a net change in the molecular dipole moment. Couldn't stretching a non-polar bond in a polar molecule produce a change in the molecular dipole moment? $\endgroup$ – ron Aug 13 '14 at 16:40
  • 1
    $\begingroup$ That is possible, particularly with 13C-NMR. You'd need the right model compounds, but there is a substantial shift of around 180-200 ppm per unit charge in 13C-NMR spectroscopy. $\endgroup$ – ron Aug 17 '14 at 20:07
  • 2
    $\begingroup$ The only truly non-polar bonds are between the same elements in the same environment. Everything else is polar. In the $\ce{C-Br}$ bond already the different masses and the larger radii would induce polarity. Whoever proposed this rule made a very crude approximation. @Dissenter there is therefore no reason, why this bond should be IR inactive. It has a dipole moment to begin with, so every displacement of the atoms will induce a change of the dipole moment. The remaining question is only, how intense this signal will be compared to the other signals of a molecule. $\endgroup$ – Martin - マーチン Aug 18 '14 at 6:47

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.