Question

Compound A reacts with $PC{{l}_{5}}$to give B which on treatment with KCN followed by hydrolysis gives propionic acid. A and B are respectively:
A.${{C}_{3}}{{H}_{8\,\,\,}}and\,\,{{C}_{3}}{{H}_{7}}Cl$
B.${{C}_{2}}{{H}_{8}}\,\,and\,\,{{C}_{2}}{{H}_{5}}Cl$
C.${{C}_{2}}{{H}_{5}}\,\,and\,\,{{C}_{2}}{{H}_{4}}C{{l}_{2}}$
D.${{C}_{2}}{{H}_{5}}OH\,\,and\,\,{{C}_{2}}{{H}_{5}}Cl$

Answer 1

Hint: Try to go backwards in the reaction and identify the reagents. For example, when a compound is hydrolysed and propionic acid is obtained, then the reagent will have one carbon less than propionic acid.


Complete step-by-step solution:
In order to answer our question, let us follow the steps of the reaction carefully. We have been given that compound A reacts with $PC{{l}_{5}}$ to give a compound B and B reacts with KCN which hydrolyses to form propionic acid. Now, propionic acid has the prefix ‘prop’ and that means it has 3 carbons in its chain. Propionic acid has the formula $C{{H}_{3}}C{{H}_{2}}COOH$, and this is the product after hydrolysis is done. Moreover, KCN is used for substitution in organic chemistry. KCN is used for substitution of halides. Moreover, the skeletal compound B has 2 carbons in it, as the third carbon is the acid group. So, we can assume that compound B is $C{{H}_{3}}C{{H}_{2}}Cl$, we have used Cl, because in the question, Cl is the only halide we have.
Verification: $C{{H}_{3}}C{{H}_{2}}Cl$ reacts with KCN to form $C{{H}_{3}}C{{H}_{2}}CN$ and when it hydrolyses, it gives $C{{H}_{3}}C{{H}_{2}}COOH$. Hence, B is ethyl chloride, and it is verified. Now, the compound A reacts with $PC{{l}_{5}}$ to give $C{{H}_{3}}C{{H}_{2}}Cl$. $PC{{l}_{5}}$ is the reagent that only reacts with alcohols, primary alcohols more preferably. So, A must be an alcohol on the form R-OH. Now, A will also have two carbons in it plus it is an alcohol. So, the possible structure of A is $C{{H}_{3}}C{{H}_{2}}OH$.
Verification: $C{{H}_{3}}C{{H}_{2}}OH$ reacts with $PC{{l}_{5}}$ to produce $C{{H}_{3}}C{{H}_{2}}Cl$ and that is what we have been given in the question. Hence, A is ethyl alcohol or ethanol, and it is verified. Following are the reactions:
$\begin{align}
& C{{H}_{3}}C{{H}_{2}}OH+PC{{l}_{5}}->C{{H}_{3}}C{{H}_{2}}Cl \\
& C{{H}_{3}}C{{H}_{2}}CL+KCN->C{{H}_{3}}C{{H}_{2}}CN \\
& C{{H}_{3}}C{{H}_{2}}CN+{{H}_{3}}{{O}^{+}}->C{{H}_{3}}C{{H}_{2}}COOH \\
\end{align}$

So, we obtain our correct answer as option D.

NOTE: It is to be noted that when an alcohol reacts with $PC{{l}_{5}}$, then along with the alkyl halide, chlorine gas is also liberated. However, alkyl halide is formed as the major product.