A hydrocarbon of molecular formula ${\mathrm{C}}_6{\mathrm{H}}_{10}$ reacts with sodamide and the same on ozonolysis followed by hydrogen peroxide oxidation gives two molecules of carboxylic acids, one being optically active. Then, the hydrocarbon may be

In the compound, 

$\underset{\left(\mathrm{I}\right)}{{\mathrm{CH}}_2}=\underset{\left(\mathrm{II}\right)}{\mathrm{CH}}-\mathrm{C}\underset{\left(\mathrm{III}\right)}{\equiv \mathrm{CH}}$

The most acidic hydrogen atom is

The major product in the following reactions is

${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{C}\equiv \mathrm{CH}\underset{\left(\mathrm{ii}\right)\mathrm{HI}\left(1\mathrm{eq}\right)}{\overset{\left(\mathrm{i}\right)\mathrm{HCl}\left(1\mathrm{eq}\right)}{\to }}$

Which among the following depicts the correct order of acidity?

The correct sequence of reagents for the following conversion will be

The major product of the following reaction is

In the reaction


$\mathrm{x}$ and $\mathrm{y}$ , respectively, are

Match List$-\mathrm{I}$ with List$-\mathrm{II}:$

Choose the most appropriate match