Hydrocarbons

Addition of HCl to 3-methylbutene gives 2-chloro-2-methylbutane as major product.

The mechanism reaction is as follows

$ \text { 6-oxoheptanal heptane is } $

$ \text { It is formed by reductive ozonolysis of } $

$ \text { Hence, option (a) is the correct answer. } $

In the reaction(s),

$ \begin{aligned} & \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2 \xrightarrow{\mathrm{HBr}}(P) \\ & \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2 \xrightarrow[\mathrm{H}_2 \mathrm{O}_2 \mathrm{H}_2 \mathrm{O}_2 / \mathrm{OH}^{-}]{\left(\mathrm{BH}_3\right)_2}(Q) \end{aligned} $

The products are,

$ \text { because, } $

Br add according to, Markownikoff's rule, while in

$ \mathrm{C}_8 \mathrm{H}_8 \text { is } $

It is a derivative of benzene and decolorised the cold alkaline $\mathrm{KMnO}_4$ solution.

The reaction occurs as follows :

$ \mathrm{C}_8 \mathrm{H}_8+2 \mathrm{KMnO}_4 \longrightarrow \mathrm{KC}_8 \mathrm{H}_5 \mathrm{O}_2+2 \mathrm{MnO}_2+\mathrm{KOH}+\mathrm{H}_2 \mathrm{O} $

Addition of bromine can be used as the test for unsaturation with regard to colour change of reaction. e.g.

Unsaturated compound $+\mathrm{Br}_2 \rightarrow$ colour of $\mathrm{Br}_2$ rapidly disappears.

i.e. Bromine water is decolourised.

Lindlar catalyst is partially deactivated catalyst composed of $\mathrm{BaSO}_4$ coated with Pd poisoned with quinoline.

It reduce ( $\mathrm{C} \equiv \mathrm{C}$ ) to ( $\mathrm{C}=\mathrm{C}$ ) bond, occurs via syn-addition and gives cis-alkene.

Alkane with six or more C-atoms, when heated under pressure in the presence of a catalyst $\mathrm{Mo}_2 \mathrm{O}_3$ at 773 K at $10-20 \mathrm{~atm}$. First cyclise and then aromatise to give benzene.

The complete reaction is given as follows:

The reaction can be completed as follows:

The reaction take place as follows

HBr $\to$ H⁺ + Br^$-$

The benzyl carbanion is more stable than vinyl carbanion due to resonance. So, the carbanion become more stable in the transition state, and the alkyne is formed.