Hydrocarbons

The compound which is most reactive towards an electrophilic attack will be electron-rich in nature. So, out of benzene, toluene, chlorobenzene and phenol. Phenol is most electron-rich compound followed by toluene and then benzene, the chlorobenzene is least electron-rich compound since chlorine is electron-withdrawing group which makes benzene electron-deficient in nature.

Hence correct order is

(i) When glucose is heated with Hi, it completely reduces to yield $n$-hexane as HI is a reducing agent.

(ii) In next step, $n$-hexane reacts with $\mathrm{Mo}_2 \mathrm{O}_3$ at 773 K and $10-20 \mathrm{~atm}$ to form benzene. This reaction is known as aromatisation reaction.

Alkynes are reduced to cis-alkenes on reacting with $\mathrm{H}_2$ in presence of Lindlar's catalyst. So, $P$ is cis-pent-2-ene. This addition of $\mathrm{H}_2$ is syn-addition.

Alkynes are reduced to trans-alkenes on treating with Na in liquid $\mathrm{NH}_3$. Here, anti-addition of hydrogen takes place. So, Q is trans-2-pentene.

This reaction occurs in three steps

Mechanism

Step-I Alcoholic KOH eliminates HBr from the dihalide to form alkene.

Step-II In second step, $\mathrm{NaNH}_2$ is a strong base. So, it can further eliminate one more HBr molecule to form alkyne.

Step-III Alkynes on heating with red hot iron tube polymerise to form aromatic benzene ring.

Dry ether is the most sultable solvent for Wurtz reaction because

Attack of the alkene on bromine gives the bromonium ion, which is positive ion. Alkyl groups are electron donating and carbocation-stabilising groups.

$D$ has 2 alkyl groups, $C$ has one and $A$ has none, but the $B$ has a electron

withdrawing nitro group. So, higher the stability of bromonium ion formed higher will be the rate of bromination reaction.

So, the correct order of rate of reaction is $D>C>A>B$.

Greater the number of nitro ($-$NO₂) group in chlorobenzene, greater the rate of nucleophilic substitution reaction in aromatic halides.

When a nitro group is present at ortho and para position, it can withdraw electrons from the benzene ring.

Here for unknown formula given $\mathrm{C}_6 \mathrm{H}_{10}$, we can predict molecular structure of given formula by using double bond equivalent.

$\begin{aligned} & \mathrm{DBE}=\mathrm{C}-\frac{\mathrm{H}}{2}+\frac{\mathrm{N}}{2}-\frac{X}{2}+1 \\ & \Rightarrow \quad=6-\frac{10}{2}+0-0+1 \\ & \Rightarrow \quad=6-5+1=2 \end{aligned}$

Hence, the hydrocarbon X is 3-methylpent-1-yne.

When chlorobenzene is heated with 6% solution of NaOH at 623 K and under pressure 300 atm to form sodium phenoxide.

Further on acidified with dilute HCl, it will form phenol.

$ \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