1
$\begingroup$

enter image description here

The first reaction produces benzaldehyde, and the next one (perkin's condensation)produces Cinnamic acid.(X)

Now the treatment of X with $\ce{Br2/Na2CO3}$ is whats troubling me. $\ce{Na2CO3}$ being a base, abstracts the hydrogen from the $\ce{COOH}$ group. $\ce{Br2}$ reacts with the alkene portion to yield a cyclic intermediate. What next?

The solution claims that somehow The $\ce{CO2-}$ group leaves and a bromoalkene forms. The addition Of moist KOH in the next step results in $\ce{E2}$ elimination to yield option (c).

I cant quite digest the $\ce{CO2-}$ group leaving. I thought of a mechanism, which is similar to the syn-elimination in esters: but that doesnt work out well..

Any hints/ insights will be appreciated. The correct answer is (c).

$\endgroup$
2
$\begingroup$

The sodium carbonate is just there to mop up any $\ce{HBr}$. The $\ce{Br2}$ adds across the double bond to give cinnamic acid dibromide.

The treatment of cinnamic acid dibromide with $\ce{KOH}$ at elevated temperature eliminates 2 eq. of $\ce{HBr}$ to give Phenylpropiolic acid. This is known to decarboxylate through the intermediate acetylene anion.

$\endgroup$
4
  • 1
    $\begingroup$ I agree with answer (c) but, possibly, a different mechanism. The post says that a bromoalkene is formed first. Dibromocinnamic acid in base concertedly loses CO2 and bromide to give (Z)-(2-bromovinyl)benzene. This reaction is known and is stereospecific using amines as the base. [Note: Bromination of methacrylic acid followed by heating the dibromide in pyridine give 2-bromopropene]. KOH then effects E2 elimination of the vinyl bromide to give phenyl acetylene. I think this is a made-up problem and I think your mechanism, in the presence of KOH, would prevail in the flask! $\endgroup$ – user55119 Jul 30 '20 at 12:23
  • $\begingroup$ ...continued. Reference for the stereospecific decarboxylation: DOI:10.1016/j.tet.2005.02.043 $\endgroup$ – user55119 Jul 30 '20 at 18:03
  • $\begingroup$ Just a doubt, in our lectures the difference between alcoholic KOH and aqeuous KOH was taught. I guess it's aqueous KOH here with 200°c. It was taught to us that aqueous KOH can only do substitutions not E2. $\endgroup$ – vanshita rawat Sep 25 '20 at 3:32
  • $\begingroup$ @vanshitarawat note the temperature at which the reaction is - 473K - there is no KOH in solution as there no liquid water present. $\endgroup$ – Waylander Sep 25 '20 at 6:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.