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

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Answer 1 · 2016-10-05T09:14:45

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

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.