Question

An organic compound A with a molecular formula ${{C}_{4}}{{H}_{8}}{{O}_{2}}$ on alkaline hydrolysis gives two compounds B and C. C on acidification with dil. HCl gives D. oxidation of B with ${{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}_{2}}S{{O}_{4}}$ also gives D. identify A, B, C, and D, and explain the reactions involved.

Answer 1

Hint: When bases like sodium hydroxide or potassium hydroxide hydrolyze an ester is known as alkaline hydrolysis and carboxylate salt and alcohol forms as the product. This reaction is also called saponification because soaps are prepared by the alkaline hydrolysis of fats and oils.

Complete step by step solution:
Given a molecular formula${{C} _{4}}{{H}_{8}}{{O}_{2}}$. There are four isomeric esters with this molecular formula.
(1) Propyl formate - $HCOOC{{H}_{2}}C{{H}_{2}}C{{H}_{3}}$
(2) Isopropyl formate - $HCOOHCH {{(C{{H}_{3}})}_{2}}$
These two structures are esters with one carbon acid.
(3) Ethyl acetate is an ester with two carbon acid - $C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}$
(4) Methyl propionate is an ester with three carbon acid - $C{{H}_{3}}C{{H}_{2}}COOC{{H}_{3}}$
Step -1: compound A with the molecular formula ${{C}_{4}}{{H}_{8}}{{O}_{2}}$ on alkaline hydrolysis gives two compounds B and C.
Compound A will be ethyl acetate among four isomers of the given molecular formula${{C}_{4}}{{H}_{8}}{{O}_{2}}$because this ester involves alkaline hydrolysis which is known as saponification reaction. in this reaction, ethyl acetate upon alkaline hydrolysis gives carboxylic acid and alcohol
$C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}(A)+NaOH\to {{C}_{2}}{{H}_{5}}OH(B)+C{{H}_{3}}COONa(C)$
So, compound B = ethyl alcohol
The compound C = sodium acetate
Step-2: compound C, sodium acetate on acidification with dil. HCl gives D.
Sodium acetate reacts with dil. HCl gives carboxylic acid, which is compound D.
$C{{H}_{3}}COONa(C)+HCl\to C{{H}_{3}}COOH(D)+NaCl$
Step-3: oxidation of compound B, ethyl alcohol oxidation with ${{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}_{2}}S{{O}_{4}}$ also gives D.
Oxidation of ethyl alcohol (B) with acidified ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ gives acetic acid (D).
$C{{H}_{3}}C{{H}_{2}}OH(B)\overset{(O)}{\mathop \to }\,C{{H}_{3}}COOH(D)$

Hence, the compound A = ethyl acetate ( $C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}$ )
Compound B =ethyl alcohol (${{C}_{2}}{{H}_{5}}OH$ )
Compound C = sodium acetate ($C{{H}_{3}}COONa$ )
Compound D =acetic acid ($C{{H}_{3}}COOH$ )

Note: Unlike carbonic acids, esters are neutral compounds which are catalyzed with acid or a base is known as hydrolysis of esters. The reverse process of etherification is simply known as acidic hydrolysis, in which the ester is heated with a large excess of water containing a strong acid catalyst.