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I've been trying to assign the peaks of 4-bromobenzophenone (I have the peaks, just need to assign them to the correct carbons), but my assignments and what it should be don't seem to agree. As part of this, I tried (and failed) to correctly assign the NMR of bromobenzene.

Bromobenzene's spectrum: https://www.chemicalbook.com/SpectrumEN_108-86-1_13CNMR.htm

Bromobenzene, my labelling of the carbon environments: https://imgur.com/a/NiVUd

Here's what I've thought so far for bromobenzene:

The halogen atom is a deactivating substituent for the cyclic ring. The two effects of the halogen are the inductive and conjugative effect.

1) The inductive effect falls away with distance. Therefore, carbon environment 2 experiences a stronger inductive effect than carbon environment 3, and likewise for carbon environment 3 compared to 4. Therefore, due to the inductive effect only, carbon 2 has a more positive, and so higher shift than carbon 3, which has a higher one than 4 i.e. 2>3>4.

2) The conjugative effect acts mainly on carbon 2 and 4 by resonance arrows (as bromobenzene is ortho/para directing). Therefore, carbon environment 3 has a more positive environment, and therefore higher shift, than 2 and 4 i.e. 3>2=4

However, when it comes to carbon 1, it is directly attached to the bromine. The inductive effect is strongest for the carbon 1, therefore, and the conjugative effect isn't the same strength here as for carbons 2/4 and so carbon 1 should be the highest shift?

According to my rationalisations, carbon 1 should have the highest shift, followed by 2/3 (order is hard to determine) and then 4. This is not seen in the spectra.

Where am I going wrong then? I don't see how you can (qualitatively) predict that:

1) carbon environment 1 actually has the lowest shift

2) carbon environment 2 has a higher shift than carbon environment 3 i.e. the increased inductive effect outweighs the increased conjugative effect for carbon environment 2.

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    $\begingroup$ Good question. For 13C the shifts are never as intuitive as that for 1H, because the factors dictating shifts can be different (see linked question in next comment). I find that this is especially so for aromatics. It is sensible to give a range in which you expect to find aromatic peaks, but trying to assign them based on 1H NMR intuition/heuristics will nearly always lead one down the wrong path. In general I find that a HSQC (plus HMBC if there's more than one substituent) is needed to confidently assign aromatic 13C peaks. $\endgroup$
    – orthocresol ♦
    Jan 10, 2018 at 20:11
  • $\begingroup$ Related: Nitrobenzene and other monosubstituted benzenes 1H vs 13C NMR assignments $\endgroup$
    – orthocresol ♦
    Jan 10, 2018 at 20:12
  • $\begingroup$ Thanks for the reply! I'm assigning this NMR as part of my course, however, so unfortunately I only have the 1H NMR and 13C NMR spectra seperately. With regards to the range, what range width would you be considering? For example, in my 13C spectrum of 4-bromobenzophenone, the aromatic peaks range from 132.7 ppm to 127.4 ppm. Would it be possible to separate the 132.7 ppm peak from the 127.4 ppm, or is this simply not feasible? $\endgroup$
    – chemistrystudent
    Jan 17, 2018 at 17:13

1 Answer 1

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Bromine is an example of what is sometimes described as the 'heavy atom effect', where chemical shift follows an opposite trend than would be expected due to electronegativity. These types of atoms (Br to a lesser extent than for I and Te) have a large electron cloud distribution. The electron cloud has a significant effect on shielding of the ipso carbon, effectively immersing the carbon within a cloud of electrons. This contributes to the diamagnetic shielding around the carbon nucleus and results in an upfield shift. As an aside, because of their size, Br and I often have significant γ effects also, which are usually steric in origin.

For calculating 13C shifts at C1, the trend between halogen-substituted benzene series follows what is expected based on electronegativities. So, qualitatively, one might expect shifts to be in the order:

F  >  Cl  >  Br  >  I  >  H

However, you just need to be aware of the heavy atom effect for Br and I, and that these will result in an upfield shift for the ispo. Hence, one can expect, qualitatively, that the progression of shifts would look like:

F  >  Cl  >  H  >  Br  >  I

This heavy atom effect is also well demonstrated in the substituted methane series:

    CH4     CH3X    CH2X2   CHX3    CX4 
Cl  -2.5    25.6    53.5    77.4    96.3
Br  -2.5    9.5     21.7    12.4    -29.4
I   -2.5    -24.0   -53.8   -139.7  -292.2

So, you can qualitatively predict shifts for C1, as long as you are aware of this contribution for Br. Predicting the order for C2/C3 is not so easy. Qualitatively, it is a reasonable assumption to say that the effect at C3 is negligible, that is, without any other knowledge, δ shift will be ~0. So you just need to look at the β carbon, or C2. Again, a reasonable guestimate might predict an upfield shift for those substituents with strong electronegativities. And, this holds for many cases. However, there are many exceptions to this, and the case becomes very complicated as soon as you start introducing multiple substituents. In this case, one should always assign peaks using 2D methods rather than just predictions.

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  • $\begingroup$ Thank you for your reply, it has been very helpful! However, regarding your comment that for C2, we may predict an upfield shift for substituents with strong electronegativities - shouldn't these substituents provide a downfield shift, conjugation of the halogen's electron pair notwithstanding? $\endgroup$
    – chemistrystudent
    Jan 17, 2018 at 17:20
  • $\begingroup$ Additionally, with regard to multiple substituents: seeing as my ultimate task is to assign the spectrum of 4-bromobenzophenone, would I be correct in the following assumptions: For the ring with both Br and the ketone group: I may be able to rank the relative shifts of the ipso and para carbons, but not the ortho/meta positions. Additionally, I cannot compare each ring's carbons ortho/meta carbons with each other, nor can I compare the carbons of each ring connected to the ketone group? $\endgroup$
    – chemistrystudent
    Jan 17, 2018 at 17:21

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