Which is the major and minor product formed from the reaction of benzyl methyl ether with dinitrogen pentoxide?

Nitration of benzyl methyl ether

I know that $\ce{NO2+}$ is behaving as an electrophile here and attacking the aromatic ring. Furthermore, the substituent directs electrophilic substitution to the ortho- and para-positions. I thought that since the ortho position is more sterically hindered, the para product should be the major product and the ortho product minor.

However, the answer I was given says the opposite, i.e. that the ortho product is major and the para product is minor. Where am I going wrong?

  • 1
    $\begingroup$ The ortho position isn't that hindered and there are two ortho positions compared to only one para so this probably accounts for the observed products. $\endgroup$
    – bon
    Nov 30, 2015 at 19:12
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    $\begingroup$ This isn't a literature reaction, and in general nitrations using N2O5 are not common. There is only one paper which mentions any kind of nitration of benzyl methyl ether, and the authors note that with a similar compound (PhCH2CH2OMe) the regioselectivity of nitration depends on the choice of reagent: pubs.acs.org/doi/10.1021/jo9606867. $\endgroup$ May 10, 2019 at 22:17

1 Answer 1


1) $\ce{NO+2}$ electrophile.


2) Benzene ring is attached to an activating group, so meta positions are enriched and electrophile has chances of attacking ortho/para positions.

The effect of the substituent in your example is probably similar to the effect of a methyl group (e.g. replacing a methyl hydrogen with a methoxy group won't have a big effect on the electron donating ability of the substituent). Look at the relative rates of nitration in the following figure.

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(image source)

We see that in toluene the ortho and para positions react 43 and 55 times (these two numbers are rather close) faster than in benzene, while reaction rate at the meta position changes little from benzene. This is consistent with the pi-donating ability of a methyl group adding electron density to the ortho and para positions making them more attractive to attack by an electrophile.

3) ortho position is more sterically hindered, hence para- major product and ortho minor product.

Look at the relative rate for the ortho position in t-butyl-benzene, here steric factors definitely play a role in slowing down attack at the ortho position. By comparison, the methyl group (and your substituent) is not very large and may slightly reduce the reaction rate at the ortho position in toluene (43 vs. 55 rather than both numbers being 55) or your example.

Another factor to consider is statistical in nature, there are two ortho positions, but only one para position. So, if everything else were equal, we would expect twice as much ortho product as para product.

In summary,

  • the electron donating effect of the substituent increases the rate at the ortho and para positions to a similar degree
  • steric factors might reduce the rate at the ortho position a small amount
  • statistical factors favor twice as much attack at the ortho position compared to the para position

This statistical effect is the largest of the three and is consistent with the ortho isomer being the major product.

One final cautionary note. When we discuss reaction rates we are really discussing kinetics - which product will form the fastest. However, electrophilic aromatic substitution is an equilibrium process. If you allow the reaction to run at high temperature, or for a long time, then thermodynamics can take over and we may wind up with a product composition based on relative product stability rather than kinetic factors.

  • $\begingroup$ Is there any example of nitration reaction on substituted benzene where there is a clear majority of para substitution? $\endgroup$
    – Aditya Dev
    Dec 1, 2015 at 3:15
  • $\begingroup$ sure, t-butyl-benzene $\endgroup$
    – ron
    Dec 1, 2015 at 3:43
  • $\begingroup$ If we take statistics, H-bonding would come into play right? Then if we take Intermolecular H-bonding > Intramolecular H-bonding.... ? $\endgroup$ Dec 1, 2015 at 12:25
  • $\begingroup$ @SujithSizon How would H-bonding come into play here? $\endgroup$
    – ron
    Dec 1, 2015 at 14:26
  • $\begingroup$ @ron intramolecular H-bonding between nitrogen attached oxygen(ortho case) and hydrogen attached to the starting/ending methyl group? $\endgroup$ Dec 1, 2015 at 14:36

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