Question

Ethanol $\xrightarrow{{PB{r_3}}}$X $\xrightarrow{{alc.KOH}}$Y $\xrightarrow[{(ii){H_2}O,heat}]{{(i){H_2}S{O_4}roomTemperature}}$Z;
The product Z in the above reaction is:
A.$C{H_3}C{H_2} - OH$
B.$C{H_2} = C{H_2}$
C.$C{H_3}C{H_2} - O - C{H_2}C{H_3}$
D.$C{H_3}C{H_2} - O - S{O_3}H$

Answer 1

Sulphuric acid, ${H_2}S{O_4}$is known as the dehydrating agent. It removes water from the compound. It reduces carboxylic acids to alcohols and forms intermediate products ketones and aldehydes before reaching the final product.

Complete answer:
$PB{r_3}$converts alcohols into alkyl bromides. If the alcohol is primary (${1^\circ }$) and secondary$({2^\circ })$, $PB{r_3}$is preferred. When the alcohol has tertiary $({3^\circ })$ alkyl group, $HBr$ is used to get the desired alkyl halide. Since the primary reactant is (${1^\circ }$) alcohol, so $PB{r_3}$is used.
The reaction occurs as follows:
${C_2}{H_5}OH\xrightarrow{{PB{r_3}}}{C_2}{H_5}Br$
(Reactant)
Now, X is an alkyl halide, named as Ethyl bromide with chemical formula${C_2}{H_5}Br$.
KOH (alcoholic) supports elimination reactions. So alkyl halide gets reduced to alkene. The reaction proceeds as:
${C_2}{H_5}Br\xrightarrow{{Alc.KOH}}C{H_2} = C{H_2}$
And in the final step, though${H_2}S{O_4}$is a dehydrating agent, we add water in the next step, heating it simultaneously. The reaction occurs as:
$C{H_2} = C{H_2}\xrightarrow[{(ii){H_2}O,heat}]{{(i){H_2}S{O_4}roomTemperature}}{C_2}{H_5}OH$
The overall reaction is as follows:
${C_2}{H_5}OH\xrightarrow{{PB{r_3}}}{C_2}{H_5}Br\xrightarrow{{Alc.KOH}}C{H_2} = C{H_2}\xrightarrow[{(ii){H_2}O,heat}]{{(i){H_2}S{O_4}roomTemperature}}{C_2}{H_5}OH$

So option A. $C{H_3}C{H_2} OH$ is the correct answer to the given question.

Note:
KOH (alcoholic) dissociates to give$R{O^ - }$ and ${H^ + }$ions and this results in elimination reaction.Sulfuric acid is known as Oil of vitriol. With the variation in percentage concentration of this reagent, the reactants give different products.