For 1,1,4,4-tetramethyl-1,2,3,4-tetrahydronaphthalene:


What would be the products following a standard nitration with $\ce{NO2+}$?

My reasoning is thus: enter image description here

Since the substituent group is alkyl then it is ortho- or para- directing. But surely there are steric hindrance factors to be considered due to the bulk of the ring which would mean para- is favoured instead of ortho. However, when I try to draw out the structure it seems the arrows are meta- instead!

  • $\begingroup$ I would think your product is the right one. What do you mean by "the arrows are meta"? $\endgroup$ – orthocresol Oct 18 '15 at 11:11
  • 1
    $\begingroup$ I think he is confused because at each para site is also meta site which is the other end of the alkyl substituent. And we were all told at the beginning that alkyl substituent are always para/ortho directing, but you can also see it as meta directing from which in my opinion stems the confusion (whoa, wait a minute! but i'm at the same time meta directing - something is rotten here.. hm..). $\endgroup$ – wuschi Oct 18 '15 at 11:21
  • 1
    $\begingroup$ Note that most of ortho/para directing substituent also increase electron density at meta positions, but only very slightly. So it is not that it would cancel somehow. $\endgroup$ – wuschi Oct 18 '15 at 11:24
  • 2
    $\begingroup$ The idea of "meta" or "para" is a bit different here since you have two alkyl groups and meta to one means ortho/para to the other, and vice versa. Look at this: i.imgur.com/TiLhXEH.png the carbons labelled a are o- to one alkyl group (stabilised a lot), but m- to another (stabilised a bit); the carbons labelled b are m- to one alkyl group (stabilised a bit), but p- to another (stabilised a lot). So, purely based on electronics, both positions are almost equally favoured. Now you turn to sterics to help you decide and as you said - b is favoured because less hindrance. $\endgroup$ – orthocresol Oct 18 '15 at 19:39
  • 1
    $\begingroup$ Essentially, since there are two substituents to start off with, you have to consider both their directing effects separately and add them up. The fact that a para site is also a meta site is not a contradiction because it is para to one substituent and meta to another substituent. The fact that the NO2 went to a site that is meta to one of the alkyl substituents does not mean that that alkyl substituent is meta-directing - as my previous comment said, the sum of the directing effects is almost the same for both the a and b carbons. $\endgroup$ – orthocresol Oct 18 '15 at 19:41

The molecule is symmetrical so the two positions on the far side of the alkyl groups are exactly the same. Similarly, the two positions closest to the alkyl groups are the same. For the far positions, each one is para with respect to one alkyl group and meta with respect to the other. This is not a contradiction because ortho, meta and para are purely relative terms.

I suspect that you are right that substitution will mostly occur at the far sites. The directing effect of the alkyl groups is essentially the same for every site because they are all ortho or para to one group and meta to another. Therefore, the steric effect of the bulky methyl groups adjacent to the near positions will disfavour substitution there.


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.