Question

The alkyne which will react with $KMn{O_4}$ to give pyruvic acid is
A. Ethyne
B. Propyne
C. Butyne
D. 2-pentyne

Answer 1

$KMn{O_4}$ is a Strong oxidizing agent it oxidizes alkyne to a carboxylic acid for the formation of pyruvic acid in the presence of $KMn{O_4}$ . We also know that pyruvic acid have 3 carbon atoms in its structural formula given by
$C{H_3} - \mathop C\limits_{\mathop {||}\limits_O } - \mathop C\limits_{\mathop {||}\limits_O } - OH$

Complete solution:
First of all we will check the number of carbons present in each of the options given in the question. After that we will find that Pyruvic acid contains three carbon atoms and it is prepared by oxidation of alkynes.
Only propyne has three carbon atoms while ethyne contains 2, butyne contains
4 and 2-pentyne contains 5 carbon atoms.
Pyruvic acid is the simplest alpha keto acid, it contains a carboxylic acid and a ketone functional group.
The molecular formula is $C{H_3} - \mathop C\limits_{\mathop {||}\limits_O } - \mathop C\limits_{\mathop {||}\limits_O } - OH$
During strong oxidation with potassium permanganate the alkyne is cleaved into two products.
When $KMn{O_4}$ act with propyne the ${C_2}$ and ${C_3}$ carbon get oxidised to co which can be represented by

{OH} \\\ | \\\ C \\\ | \\\ {OH} \end{array} - \begin{array}{*{20}{c}} {OH} \\\ | \\\ C \\\ | \\\ {OH} \end{array} - C{H_3}\xrightarrow{{ - 2{H_2}O}}\mathop {C{H_3} - \mathop C\limits_{\mathop {||}\limits_O } - \mathop C\limits_{\mathop {||}\limits_O } - OH}\limits_{Pyruvic{\text{ }}acid} $$ Propyne reacts with $$KMn{O_4}$$ in a neutral medium and is oxidized in pyruvic acid. So, that alkyne will be propyne. **Hence, option (B) is correct.** **Note:** At normal temperature pyruvic acid is a colorless liquid. It has a pungent smell. Its melting point is ${13.6^0}{\text{ C}}$ and the boiling point is ${165^0}{\text{ C}}$ . It is used as a starting material of biosynthesis of pharmaceuticals.