Question

For the nucleophilic aromatic substitution (NAS) reaction of 1-chloro-2,4-dinitrobenzene with methoxide in methanol, what does the intermediate look like?

Answer 1

option (B)

Explanation:

First keep in mind that methanol is not donating the `"CH"_3"O"^-`ion.A strong base like sodium methoxide is donating it. It is OK to have `"CH"_3"O"^-` with methanol because they are in equilibrium(i.e sodium methoxide is not protonated by methanol).

This reaction is Nucleophilic aromatic substitution and a SNAr reaction. The 'Ar' stands for 'aromatic'.

First step) a very strong nucleophile (`"CH"_3"O"^-`) attacks the ring and bonds to the carbon which was already bonded to chlorine.This breaks one `pi` bond.This process is slow because this results in loss of aromacity which is not energy favorable.This pi bond is travels to the EWG group that is ortho or para to the leaving group and the double bond forms between the carbon and the EWG.In our reaction there.This complex formed is our intermediate.Because in our reaction their are two EWGs para and ortho to the chlorine so we can have two possible products and two possible complexes.But since only one complex is available in our options and that is option (B)

This photo is just an example.Not to be thought as the orginal reaction

2nd step) The reaction proceeds forward via the elimination of chlorine.Please keep in mind that the (`"CH"_3"O"^-`) cannot bond with the EWG because the EWG are already negative in charge

Answer 2

The intermediate is shown below in the diagram at the bottom.

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The reaction you have is:

The nitro (`-"NO"_2`) groups are highly-electron-withdrawing groups. Their dipole moment is something like over `"4 Debyes"` (whereas water is around `"1.85 Debyes"`).

Having two of them on the same benzene ring promotes nucleophilic aromatic substitution, where a nucleophile can displace a ligand.

Here, the nucleophile is `"CH"_3"O"^(-)`, and methanol is the solvent. In practice, sodium or potassium salts are used to improve yield.