2,6-dibromo Acetophenone Nitrogenation

Chemistry Level 3

$\ce{2,6-dibromo~acetophenone}$ is reacted in the following order:

$\ce{HNO3}$ in $\ce{H_2SO_4}$ .
$\ce{H_2}$ in excess ( $Xs$ ) with $Pt/C$ .

From among choices $A$ , $B$ , and $C$ ; choose the major product of this reaction scheme. If the major product is not among the choices, choose $\ce{No~Reaction}$ as your answer.

David's Organic Chemistry Set

David's Physical Chemistry Set

A

B

C

No Reaction

2 solutions

David Hontz
Jul 16, 2016

Both $Br$ s will act as ortho/para directors while the $acetyl$ group acts as a meta director. There are two carbons that all these groups will direct nitration to, but nitration of either will yield the same product
$H_2$ with a $Pt/C$ catalyst reduces the $acetyl$ group to a $2^{\circ}$ alcohol and $NO_2$ to $NH_2$ ; thus, the major product is choice $\boxed{A}$

$Answer \rightarrow \boxed{A} = \ce{2,4-dibromo-3- ( 1-hydroxy ~ ethyl) ~aniline}$

I think due to steric effect of Br no reduction of aldehyde is possible

Satyam Tripathi - 4 years, 6 months ago

That is a brilliant aspect to notice. Sterics of the bulky bromine substituents would have a noticeable effect if the aldehyde was instead a ketone connected directly to the ring. The aldehyde comes of the aromatic ring by a single carbon unit, which allows the reaction to proceed, while admittedly at a slow pace.

David Hontz - 4 years, 5 months ago

@David Hontz Platinum is not a strong catalyst that can reduce carbonyls.

Ankit Kumar Jain - 2 years, 10 months ago

Devansh Shringi
Jul 17, 2016

Basic nitration on para and ortho of bromine and meta of acetyl as nitration is electrophilic substitution reaction in first step . Reduction of keto to alkane and nitro to amine by catalytic hydrogenation in second step

2,6-dibromo Acetophenone Nitrogenation

David's Organic Chemistry Set

David's Physical Chemistry Set

2 solutions

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