Question

$\qquad CH_2-CH_2 \xrightarrow{xNaNH_2}$ $\qquad | \qquad |$ $\qquad Cl \qquad Cl$

Answer 1

Acetylene (Ethyne)

Answer 2

Given the molecule

$$\begin{array}{r} \text{Cl} \quad \quad \text{Cl} \\ | \quad \quad | \\ CH_2-CH_2 \end{array}$$

this is 1,2-dichloroethane. When treated with excess $NaNH_2$, a strong base, two successive E2 eliminations occur (first forming a vinyl chloride intermediate and then an alkyne), ultimately yielding acetylene (ethyne).

Reaction:

$$\text{ClCH}_2\text{-CH}_2\text{Cl} \xrightarrow{x \,\text{NaNH}_2} \text{HC}\equiv \text{CH}$$

Excess $NaNH_2$ deprotonates twice, eliminating two $HCl$ molecules from 1,2-dichloroethane to form acetylene.