Question

$C{{H}_{3}}CH=CHCHO$ is oxidised to $C{{H}_{3}}CH=CHCOOH$ using
A. Alkaline potassium permanganate
B. Acidified potassium permanganate
C. Selenium dioxide
D. Osmium tetroxide

Answer 1

Hint: Oxidation of aldehyde in presence of a double bond is a very selective process. Any strong oxidising agent can not be used here as they can also oxidise the double bond. Therefore we have to use a mild oxidising agent for this selective oxidation.

Complete Step by Step Answer:
In the given question the activated aldehyde is named crotonaldehyde($C{{H}_{3}}CH=CHCHO$)which is oxidised to crotonic acid($C{{H}_{3}}CH=CHCOOH$) without oxidation of double bond in this compound. The given compound is activated because it is stabilised through the conjugation of a double bond with the carbonyl group present in this molecule.

Potassium permanganate $KMn{{O}_{4}}$is used as a strong oxidising agent. The rate of oxidation varies with the nature of the solvent whether it is acidic or basic. An acidic solution $KMn{{O}_{4}}$acts as a strong oxidising agent. AS in acidic solution it is reduced to a greater extent. Here is the oxidation number $(+VII)$ $MnO_{4}^{-}$ to $(+II)$in $M{{n}^{2+}}$.
$MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O$

But in the alkaline medium, the oxidising capacity of potassium permanganate is less efficient. Here $(+VII)$ $MnO_{4}^{-}$ is reduced to $(+VI)$ $MnO_{4}^{2-}$ .
$MnO_{4}^{-}+{{e}^{-}}\to MnO_{4}^{2-}$

Therefore we can use alkaline potassium permanganate to oxidise aldehyde to a carboxylic acid without breaking the double bond. But acidic potassium permanganate is a very strong oxidising agent and oxidises both the double bond and aldehyde group. That’s why here we can not use $KMn{{O}_{4}}$an acidic medium.
$C{{H}_{3}}CH=CHCHO\xrightarrow[KMn{{O}_{4}}]{Alkaline}C{{H}_{3}}CH=CHCOOH$

Selenium dioxide oxidises only allylic oxidation of alkenes to form allylic alcohol. Aldehydes are oxidised by selenium dioxide to form glyoxal, not carboxylic acid. Osmium tetraoxide oxidizes alkene compounds to give cis diol, not aldehyde. Therefore crotonaldehyde oxidised to crotonic acid using alkaline potassium permanganate.
Thus, correct option is (A)

Note: The oxidation by Neutral potassium permanganate is less efficient. Here manganese$(VII)$ in $MnO_{4}^{-}$ reduced to $Mn(IV)$ and thereby produces an insoluble $Mn{{O}_{2}}$as a byproduct. This insoluble compound needs to be removed from the reaction medium during the purification of carboxylic acid.