In the following sequence of reactions, the products (A) to (G) are formed:
i. `2CH_4(g)overset(Delta)underset(1773K)rarr(A)(g)+B(g)`
ii. `4 mol of (A) overset(Delta)underset(Ni(CN)_4+THF)rarr(C)overset(O_3//o x id.)rarr(D) only`
iii. `(A)overset((1) 1 mol of NaNH_2)underset((2) C_2H_5I)rarr(E)underset(Red hot tube)overset(3 mol of E)rarrunderset((G) only)underset(darr O_3//Red.)F`
Compound (E) is

Options

A. Propyne
B. Butyne
C. But-2-yne
D. none of these

Explanation

To determine the product (E) in the given sequence of reactions, we will analyze each step systematically.

### Step 1: Identify Product A
The first reaction is:
\[ 2CH_4(g) \overset{\Delta}{\rightarrow} (A)(g) + B(g) \]

When two molecules of methane (\(CH_4\)) are heated at high temperature (1773 K), they undergo a process called pyrolysis or cracking, leading to the formation of ethyne (acetylene, \(C_2H_2\)) and hydrogen gas (\(H_2\)). Thus, we can conclude:
- Product (A) is \(C_2H_2\) (ethyne).

### Step 2: Analyze the Second Reaction
The second reaction is:
\[ 4 \text{ mol of } (A) \overset{\Delta}{\underset{Ni(CN)_4 + THF}{\rightarrow}} (C) \overset{O_3 // ox.}{\rightarrow} (D) \]

Here, 4 moles of ethyne (\(C_2H_2\)) are treated with a nickel complex in a tetrahydrofuran (THF) solvent. This typically leads to the formation of a compound that can undergo ozonolysis. However, since we are focused on finding product (E), we can skip the details of products (C) and (D) for now.

### Step 3: Analyze the Third Reaction
The third reaction is:
\[ (A) \overset{(1) 1 \text{ mol of } NaNH_2}{\underset{(2) C_2H_5I}{\rightarrow}} (E) \underset{\text{Red hot tube}}{\overset{3 \text{ mol of } E}{\rightarrow}} \underset{(G) \text{ only}}{\rightarrow} F \]

We start with product (A), which is \(C_2H_2\). When ethyne reacts with 1 mole of sodium amide (\(NaNH_2\)), the acidic hydrogen of the terminal alkyne is deprotonated, forming a sodium acetylide intermediate:
\[ HC \equiv C^- Na^+ \]

Next, this intermediate reacts with ethyl iodide (\(C_2H_5I\)) in an \(S_N2\) reaction, where the nucleophile (acetylide ion) attacks the ethyl iodide, leading to the formation of:
\[ HC \equiv C - C_2H_5 \]
This compound is 1-butyne (or but-1-yne), which is our product (E).

### Final Answer
Thus, the compound (E) is:
\[ \text{1-butyne (but-1-yne)} \]

### Summary
1. Product (A) is \(C_2H_2\) (ethyne).
2. Product (E) is \(C_4H_6\) (1-butyne).