JEE Main - 11th April 2015

Choose the correct set of reagents for the following conversion
trans$\left(\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_3\right)\to \text{ cis }\left(\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_3\right)$

Find out the major products from the following reactions.

$17\mathrm{mg}$ of a hydrocarbon (M.F. ${\mathrm{C}}_{10}{\mathrm{H}}_{16}$) takes up $8.40\mathrm{mL}$ of the ${\mathrm{H}}_2$ gas measured at $0°\mathrm{C}$ and $760\mathrm{mm}$ of $\mathrm{Hg}$. Ozonolysis of the same hydrocarbon yields

${\mathrm{CH}}_3-\underset{\underset{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-{\mathrm{CH}}_3,\mathrm{H}-\underset{\underset{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-\mathrm{H},\mathrm{H}-\underset{\underset{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\underset{\underset{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-\underset{\underset{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-\mathrm{H}$

The number of double bond/s present in the hydrocarbon is

The one giving maximum number of isomeric alkenes on dehydrohalogenation reaction is (excluding rearrangement)

The major products ‘$\mathrm{A}$’ and ‘$\mathrm{B}$’, respectively, are

A hydrocarbon 'X' with formula ${\mathrm{C}}_6{\mathrm{H}}_8$ uses two moles of ${\mathrm{H}}_2$ on catalystic hydrogenation of its one mole. On ozonolysis, 'X' yields two moles of methane dicarbaldehyde. The hydrocarbon 'X' is :

But-$2$-yne is reacted separately with one mole of Hydrogen as shown below:

$\underset{¯}{\mathrm{B}}\underset{\mathrm{liq} {\mathrm{NH}}_3}{\overset{\mathrm{Na}}{\leftarrow }}{\mathrm{CH}}_3-\underset{+{\mathrm{H}}_2}{\mathrm{C}\equiv }\mathrm{C}-{\mathrm{CH}}_3\underset{∆}{\overset{\mathrm{Pd}/\mathrm{C}}{\to }}\underset{¯}{\mathrm{A}}$

Identify the incorrect statements from the options given below:

A. A is more soluble than $\mathrm{B}$.
B. The boiling point & melting point of $\mathrm{A}$are higher and lower than $\mathrm{B}$ respectively.

C. A is more polar than $\mathrm{B}$ because dipole moment of $\mathrm{A}$ is zero.
D. ${\mathrm{Br}}_2$ adds easily to $\mathrm{B}$ than $\mathrm{A}$.

‘$\mathrm{X}$’ is: