Question

Which alkene on heating with alkaline KMnO$_4$ solution gives acetone and a gas, which turns lime water milky:
(a) 2-Methyl-2-butene
(b) Isobutylene
(c) 1-Butene
(d) 2-Butene

Answer 1

In this reaction, alkaline potassium permanganate is used. As we know it is used for the displacement of carbon-carbon double bonds. It is mainly used to identify whether hydrocarbon is alkane, or alkene. So, we can identify the alkene in this reaction.

Complete step by step solution:

First, let us know about the alkenes. These are categorized under hydrocarbons having at least one carbon – carbon double bond.
Now, if we talk about the given options. The mentioned options belong to the category of alkenes.
As we know, when an alkene having two hydrogens at one end of the bond, reacts with alkaline potassium permanganate; the product obtained will be a ketone, carbon-dioxide, and water.
So, from all the given options, when 2-methyl-2-butene, 1-butene, and 2-butene does not show reaction with alkaline potassium permanganate.
They represent the dihydroxylation of alkenes with cold, dilute potassium permanganate to give diols.
Thus, we can say that isobutylene reacts with alkaline KMnO$_4$. The chemical reaction is
(CH$_3$)$_2$C=CH$_2$ $\overset{alkalineKMnO_4}{\rightarrow}$ (CH$_3$)$_2$C=O + CO$_2$
So, in the end we can conclude that isobutylene alkene reacts with a solution that gives acetone, and a gas carbon-dioxide which turns lime water milky.
Hence, the correct option is (B).

Note: Don’t get confused while writing the chemical reaction. We will not consider acidified potassium permanganate solution, as it reacts too rapidly to break the carbon-carbon double bond, it is rarely used in organic chemistry. So, we have used the alkaline potassium permanganate solution for the reaction with alkene.