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Skyblitz
Oct11-04, 06:46 PM
This is a pretty basic polarity question but I'm having a hard time deciding.
I had to choose the two major products of methylbenzene with HNO3 and H2SO4.

I said that the two major products were O-methylnitrobenzene (1-methyl-2-nitrobenzene) and the other 1-methyl-4-nitrobenzene (p-methylnitrobenzene).

At anyrate, I have to chose which one would be more polar on TLC.

The more obvious choice is the 1-methyl-2-nitrobenzene which is what my friends were suggesting, however, if the methyl group is right beside the nitrobenzene, the methyl electron-donating would be negated by the nearby eletron-withdrawing nitrobenzene.

If it were the 1-methyl-4-nitrobenzene, the methyl group would make one side more electron dense, while the other side would have its electrons drawn in by the nitrobenzene group... this is where I'm confused since this means that both sides are electron-rich; therefore I'm a little stumped.

Any ideas?

Thank you. :grumpy:

chem_tr
Oct12-04, 01:00 AM
Please do not consider nitro group as you have methyl group, a strong mesomeric effect provider.

I think you nitrated toluene to obtain o- and p-nitrotoluene. I presume that the para-substituted product dominates due to steric relief, with somewhat less amount of ortho-product. There should be a seriously different dissolution behavior, I mean, one of the product must have a higher solubility in somewhere, so I don't think a TLC is really needed here. But it may be of great help, though.

In terms of polarity, we can't say that the nitro-side is more positive whereas the methyl group is a bit negative; there is a reduced activity compared to toluene is apparent only, due to nitro group's negative mesomeric effect.

Let's go on discussing this topic.