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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).


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.

  • 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. $\endgroup$ 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

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.

  • 3
    $\begingroup$ Please use $\ce{...}$ for correct typesetting of chemical formulae; these should be upright and not italicised, see: chemistry.meta.stackexchange.com/q/86/16683. Compare: $\ce{Na2CO3}$ $\ce{Na2CO3}$ (correct) versus $Na_2CO_3$ $Na_2CO_3$ (wrong). Also, 'bromostyrene' is one word and should not be capitalised (unless occurring at the beginning of a sentence). There is also no reason to capitalise 'Author' and 'Question'. $\endgroup$
    – orthocresol
    Sep 8 '21 at 11:54
  • $\begingroup$ @orthocresol Now fixed $\endgroup$
    – RaMathuzen
    Sep 8 '21 at 14:05

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