achiever #79 ochem Test 2

[TimeforDAT]

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ON achiever Test 2 Ochem #79

can someone help me with this H nmr problem? There is p-xylene and it shows two singlets. I'm not getting how that is?

I know there's going to be two peaks, one from the methyl group and one from the hydrogen bonded to the aromatic ring.

One peak from methyl Hs because they are in the same environment and won't be split by any hydrogens on adjacent carbons.

But what about the hydrogens on the aromatic ring? So looking at one carbon with the H, there is an adjacent carbon with n H, so shouldn't that split the signal into a doublet?

thanks

 

[AaBbZz]

When you have a benzene ring with two of the same substiuent groups attached at the para position the 4 hydrogens found on the ring are exactly equivalent and will give only 1 peak.

The other peak found is the one given by -CH3 substuient group and there is only 1 since both groups attached are the same, like you said.