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In a material that I use to study High School Organic Chemistry, the bromination of ethyl substituted benzaldehydes is given as follows:

Bromination of ethyl substituted benzaldehydes Source: FIITJEE Grand Masters Package - Chemistry (JEE Preparatory Material)

$\ce{-CHO}$ is a meta directing group, and $\ce{-C2H5}$ is an ortho, para directing group. So isn't some of the products given wrong?

For example, let's take the bromination of ortho ethyl benzaldehyde. Shouldn't the 2nd product's bromine be in a way that is in position 4 wrt $\ce{-C2H5}$ and position 3 wrt $\ce{-CHO}$?

In the meta ethyl benzaldehyde case, why is the $\ce{-CHO}$ group not affecting the position of $\ce{Br}$? Even though it is meta directing, $\ce{Br}$ doesn't seem to substitute the $\ce{H}$ in the meta position of $\ce{-CHO}$

I searched the internet for more information regarding such reactions, and could not find any helpful results for this case.

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Your first query is genuine and the 2nd product for the first case is wrong. It should be, as you noted as well, 5-bromo-2-ethyl-benzaldehyde.

Regarding your second query, you're wrong. Recall the general rule that:

  1. when more than one activators are present on the same benzene ring, the substitution is governed by the best activator alone
  2. when an activator and a deactivator is present on the same benzene ring, the substitution is governed by the activator alone
  3. when more than one deactivators are present on the same benzene ring, the substitution is governed by the weakest deactivator alone

So, here $\ce{Br}$ will go to the o/p position of ethyl group.


Note that in the high school context, I've assumed we're only talking about the major product, because otherwise, all types of crazy products are generally formed, but in smaller ratios

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