Hydrocarbons

The complete reaction is as follows,

Let's go through the sequence of reactions step by step to identify the final product $(Z)$.

Step 1: $\mathrm{CH}_2=\mathrm{CH}_2 \xrightarrow{\mathrm{HBr}} X$

The addition of hydrogen bromide (HBr) to ethene ($\mathrm{CH}_2=\mathrm{CH}_2$) follows Markovnikov's rule, forming bromoethane ($\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}$) as the product.

$\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{HBr} \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}$

So, $X$ is $\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}$.

Step 2: $\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br} \xrightarrow{a q . \mathrm{KOH}} Y$

When bromoethane reacts with aqueous potassium hydroxide (KOH), nucleophilic substitution occurs, converting bromoethane to ethanol ($\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$).

$\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br} + \mathrm{KOH} \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} + \mathrm{KBr}$

So, $Y$ is $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$.

Step 3: $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \xrightarrow[\mathrm{I}_2 \text{ excess }]{\mathrm{Na}_2 \mathrm{CO}_3} Z$

Ethanol (ethyl alcohol) reacts with iodine ($\mathrm{I}_2$) in the presence of sodium carbonate ($\mathrm{Na}_2 \mathrm{CO}_3$) to form iodoform ($\mathrm{CHI}_3$). This reaction is known as the iodoform test, which is a common qualitative test for the presence of methyl ketones and ethanol.

$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} + 4\mathrm{I}_2 + \mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \mathrm{CHI}_3 + 5\mathrm{NaI} + \mathrm{HCOONa} + 2\mathrm{H}_2\mathrm{O}$

So, $Z$ is $\mathrm{CHI}_3$ (iodoform).

Therefore, the correct answer is:

Option C: $\mathrm{CHI}_3$

According to anti-Markownikoffs rule, in the presence of peroxide, addition of $\mathrm{HBr}$ takes place to unsymmeterical double bond in such a way that $\mathrm{Br}^{-}$ ions adds with carbon having more number of H-atoms.

$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{HBr} \xrightarrow{R \mathrm{OOR}} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$

The structure of acetylene is

In acetylene, both the C-atoms are sp-hybridised. Hence, in acetylene molecule, there are one sigma and two pi bonds are present between carbon atoms.

Electrophile addition reaction takes place via more stable carbocation.

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

The mechanism reaction is as follows

The complete reaction is given as follows:

The reaction can be completed as follows:

The reaction take place as follows

HBr $\to$ H⁺ + Br^$-$