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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).

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  • $\begingroup$ Br2 /sodacarbonate gives just anti addition to vicinal bromides. Just look carefully,the reaction is similar to treatment of vicinal bromide with soadamide(this rxn requires very strong base),KoH being less basic than sodamide ,thus additional heat has been provided. Thus 'C'would be formed. $\endgroup$
    – Ashwini patel 'A'
    Commented Dec 1, 2023 at 17:06

3 Answers 3

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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.

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    $\begingroup$ 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
    Commented Jul 30, 2020 at 12:23
  • $\begingroup$ ...continued. Reference for the stereospecific decarboxylation: DOI:10.1016/j.tet.2005.02.043 $\endgroup$
    – user55119
    Commented Jul 30, 2020 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
    Commented Sep 25, 2020 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
    Commented Sep 25, 2020 at 6:45
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An alternative to @Waylander 's explanation. First, we brominate cinnamic acid to form cinnamic acid dibromide and because of the base $\ce{Na2CO3}$ the cinnamic acid is deprotonated, and thereby both CO2 and Br are eliminated leading to the formation of bromostyrene and then KOH does its work.

Note: The answer may seem similar as of user55119 but here the reagent $\ce{Na2CO3}$is used and reaction conditions are similar to that of the author's question.

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Just trying to share my 2 cents here.

Firstly, brominate the cinnamic acid to furnish a dibromide(trans).

Then using Sodium carbonate, de-protonate the carboxylic acid.

The anion would conventionally prefer to form a ring but, on close observation we see that the ring ,if formed, would be a 4 membered one which is known to be an unstable form.(5 and 6 membered rings are preferred as products)

Hence the anion would cause CO2 and -Br to be eliminated from the molecule. Following a treatment by KOH, we would get the desired product.

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