Question

N,N-dimethylaniline is treated with `NaNO_2"/dil"HCL`.what will be the product/s?

Answer 1

N,N-dimethyl-p-nitrosoaniline will be the product.

Explanation:

This occurs as an electrophilic substitution at the para position of N,N-dimethylaniline.

The electrophile, Nitrosonium ion `"NO"^+` is formed in the reaction mixture as follows

`"NaNO"_2+ "HCl" → "NaCl" + "HNO"_2`

`"HO-N=O + H"^(+) → "H"_2stackrel(+)("O")"-N=O"`

`"H"_2stackrel(+)("O")"-N=O" → "H"_2"O" + stackrel(+)("N")"=O"`

`N,N"-dimethylaniline"+ stackrel(+)("N")"=O" → N,N"-dimethyl-"p"-nitrosoaniline" + "H"^+`

Here, the dimethylamino group activates the benzene ring at the ortho and para positions.

But the steric factor of this bulky group disfavors the ortho attack of the electrophile, nitrosonium ion `"NO"^+`

So, the attack at the para position occurs easily.