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# Question on aromatic chemistry for A Level Please Help?

I'm talking about question c, the answer to both b and c parts is attached. I don't get how they identified A, couldn't it be any of the other structures? And how did they figure out the exact substitution for B. Please if you could explain fully that would be fantastic.

### 1 Answer

- Dr WLv 71 month agoFavorite Answer
look at the 4 compounds in answer "B". Assume 1 NO2 will be attached to the ring in the end.

Starting with the left compound (ethylbenzene) and numbering the ring clockwise with the ethyl group at the #1 position. If we add an NO2 group to position 2 or 6, we get the same compound (you can flip over the molecule and superimpose it so it's the same. Likewise if we add the NO2 to the "meta" position (3 or 5), we get a second compound. If we add in position 4 (para) we get a 3rd compound. For a total of 3 different possible products. Note that an alkyl group is an ortho para directing group. So we would expect nearly 100% of the product to be ortho or para (yes, we'll still have trace amounts of meta). Since there are 2 equal positions to make the 2-nitro-ethybenzene product (carbons 2 and 6) and only 1 position to make the 4-nitroethylbenzene product, if all else was equal we would expect a 2:1 ratio of 2-nitroethylbenzen to 4-nitroethylbenzene just by chance. However, the ethyl group adds a degree of "stearic hinderance" which reduces the chance of an ortho attachment and makes the product ratio about 50:50 ortho / para.

see this link. https://www.youtube.com/watch?v=K63p4fz4MEg

likewise, the 1,2-dimethylbenzene product directs the NO2 groups to all 4 open ring positions and we get a mixture of two products 1,2-dimethyl-3-nitrobenzene and 1,2-dimethyl-4-nitrobenzene because of symmetry. (Actually those numbers are incorrect, the nitro should be given priority and numbered "1" but you get the point right?)

in the 1,3-dimethylbenzene, the NO2 group directs to the 2, 4 and 6 positions making 2 distinct products (can you see that? positions 4 and 6 yield the same product)

the para-dimethylbenzene on the right directs the nitro group to ortho for either CH3 and all 4 open sites are now equivalent. So that nitrating the ring any 1 time gives only 1 product.