Question

A compound X (having vinegar like smell) when treated with ethanol in presence of the acid Z, gives a compound Y which has a fruity smell. The reaction is:

${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+\mathrm{X}\stackrel{\mathrm{Z}}{\to }\mathrm{Y}+{\mathrm{H}}_2\mathrm{O}$

1. Identify Y and Z
2. Write the structural formula of X
3. Name the above reaction.

Answer 1

According to the question,

1. The product Z has a fruity smell which is indicative of ester.
2. We can next conclude that the reactant X must be Acetic acid$\left(C{H}_{3}COOH\right)$ since it smells like vinegar.

Using the above hints, we can conclude that the reaction given here is the Eterification of an acid.

The chemical reaction is as shown in the below.

$\underset{\mathrm{Ethyl}\mathrm{alcohol}}{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)}+\underset{\mathrm{Acetic}\mathrm{acid}}{{\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{aq}\right)}\stackrel{{\mathrm{H}}^{+}}{\to }\underset{\mathrm{Ethyl}\mathrm{acetate}}{{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5}\left(\mathrm{l}\right)+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

1. Y = Ethyl acetate$\left(C{H}_{3}COO{C}_2{H}_5\right)$, Z = concentrated Sulphuric acid$\left(H_{2}S{O}_4\right)$
2. X is Ethanoic acid$\left(C{H}_{3}COOH\right)$ whose structure can be drawn as follows:

iii. The reaction given here is called Esterification of an acid.