In the reaction given below

An amino alcohol reacts with (i) HCl (ii) KOH to give 'B' Major product. 'B' is:

The given substrate is a β-amino alcohol. Denote it as $$HO\,-\,CH_2\,-\,CH_2\,-\,NH_2$$ (ethanolamine).

Step 1 : protonation with $$HCl$$
The basic $$NH_2$$ group gets protonated much faster than the -OH group, giving the ammonium salt

$$HO\,-\,CH_2\,-\,CH_2\,-\,NH_3^{+}\;Cl^{-} \qquad -(1)$$

In the ammonium form, nitrogen becomes a very good leaving group (it will depart as neutral $$NH_3$$).

Step 2 : treatment with strong base $$KOH$$
$$KOH$$ removes the proton from the alcoholic -OH to generate an alkoxide:

$$^{-}O\,-\,CH_2\,-\,CH_2\,-\,NH_3^{+} \qquad -(2)$$

Step 3 : intramolecular $$S_N2$$ displacement
The alkoxide oxygen (a strong nucleophile) attacks the adjacent carbon bearing $$NH_3^{+}$$, expelling neutral $$NH_3$$ and closing a three-membered ring.

$$^{-}O\,-\,CH_2\,-\,CH_2\,-\,NH_3^{+} \;\xrightarrow[\textit{intramolecular }S_N2]{}\; \begin{matrix} & O & \\ / & & \backslash \\ CH_2 &\; & CH_2 \end{matrix} \;+\;NH_3 \qquad -(3)$$

The three-membered ring containing oxygen is called oxirane (ethylene oxide).

Thus the major product ‘B’ is oxirane. Among the given choices this corresponds to Option A.

Answer: Option A which is: ethylene oxide (oxirane).