Question

Orgo II Question about reactivity of aromatics

Answer 1

@Michael @BRAINIAC sorry to be a pain but could you take a look at this whenever you get a chance? :S

Answer 2

The intermediate formed in these reactions would be meisenheimer complex which would have a negative charge at the site where the hydroxide ion attacks.

Answer 3

So if you have some electron withdrawing groups at the ortho/meta positions they'd stabilise the intermediate and increase the rate of rxn

Answer 4

The mechanism followed is \(S_NAr\)

Answer 5

not para? does that mean structures I and III take precedence over the others?

Answer 6

Oops that should be *para over there instead of meta

Answer 7

oh

then structure IV is the most reactive I guess?

Answer 8

Yep

Answer 9

between II and V does the meta group hinder the reaction in any way?

Answer 10

wait

Answer 11

oh, duh, since V has a para group then V takes priority over II

Answer 12

Yeah + I don't think that the meta group would interfere much.. because no2 isn't that big

Answer 13

my proposed solution: II, then I and III should be about the same, b/c they only have 1 o/p group each, then V, then IV?

Answer 14

wait

Answer 15

the problem says descending order so just flip that

Answer 16

I just talked to my professor and he says that V takes priority over II because of inductive effects of the nitro?

Answer 17

1 > 3
Because the size isn't a problem and the no2 on the ortho would also use the -I effect

Answer 18

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
I just talked to my professor and he says that V takes priority over II because of inductive effects of the nitro?
\(\color{#0cbb34}{\text{End of Quote}}\)
Thats correct
But the main reason why 5>2 would be the -m of the para group

Answer 19

-m > -i ->Priority order

Answer 20

alright, thank you

Answer 21

np :-)