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This question was asked in my book:
To my knowledge, side chain oxidation with KMnO4 occurs only when a benzylic hydrogen is present. However this doesn't contain a benzylic hydrogen and hence according to me it wouldn't give any reaction.
However the answer in my book states that Phenylacetylene will be oxidized to benzoic acid.
How does this take place then?
According to this source Chemistry Libre Texts
Alkynes, similar to alkenes, can be oxidized gently or strongly depending on the reaction environment. Since alkynes are less stable than alkenes, the reactions conditions can be gentler. For examples, alkynes form vicinal dicarbonyls in neutral permanganate solution. For the alkene reaction to vicinal dialcohols, the permanganate reaction requires a lightly basic environment for the reaction to occur. During strong oxidation with ozone or basic potassium permanganate, the alkyne is cleaved into two products.
Brief explanation
Hot alkaline solution of potassium permanganate oxidises a terminal alkyne according to the following process:
Triple bond between the first and second carbon atoms (they are marked in the figure above) converts into two carbonyl groups $\ce{C=O}$. The resulting molecule consists of a ketone part (blue) and an aldehyde part (red). The subsequent oxidative cleavage divides the molecule into a carboxylic acid with the $\ce{R1}$ radical and formic acid. But formic acid is oxidised in this conditions to carbon dioxide.
Reaction
Since ethenylbenzene is a terminal alkyne, it can be oxidised with a hot alkaline solution of potassium permanganate to form a carboxylic acid and carbon dioxide.
The oxidation of carbon atoms is described in half-reactions (all the oxidation states are marked in the figure below):
$\ce{C^0->C^{+3}}+3e^-$
$\ce{C^{-1}->C^{+4}}+5e^-$
Here is an equation of oxidation of ethenylbenzene with a hot alkaline solution of $\ce{KMnO4}$:
References
One possibility: the alkyne gets its $\alpha$ hydrogen through formation of a hydrate. The triple bonded carbon is electrophilic with its relatively high electronegativity, with the attack occurring on the remote carbon to form a stabilized (benzylic) carbanion:
$\ce{C6H5C#CH\overset{H_2O/OH^-}{<=>}C6H5\overset{-}{C}=CHOH<=>C6H5CH=CHOH}$
The form with the $\alpha$ hydrogen (or the tautomerized keto form which has two such hydrogens) then reacts in the usual way with alkaline permanganate.
