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Question: Which carboxylic acid (B) (Equivalent weight = 52 gm/eq) loses \[{\text{C}}{{\text{O}}_2}\] , which ...

Which carboxylic acid (B) (Equivalent weight = 52 gm/eq) loses CO2{\text{C}}{{\text{O}}_2} , which on heating gives an acid (C) (Equivalent weight=60 gm eq−1)\left( {\text{C}} \right){\text{ }}\left( {{\text{Equivalent weight}} = 60{\text{ gm e}}{{\text{q}}^{ - 1}}} \right) ?
A) R=CH3{\text{R}} = {\text{C}}{{\text{H}}_3}
C=CH3COOH{\text{C}} = {\text{C}}{{\text{H}}_3}{\text{COOH}}
N.E. (Ew)=52 gm eq−1{\text{N}}{\text{.E}}{\text{. }}\left( {{\text{Ew}}} \right) = 52{\text{ gm e}}{{\text{q}}^{ - 1}}
B) R=CH3{\text{R}} = {\text{C}}{{\text{H}}_3}
C=CH3−CH2−COOH{\text{C}} = {\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{COOH}}
N.E. (Ew)=52 gm eq−1{\text{N}}{\text{.E}}{\text{. }}\left( {{\text{Ew}}} \right) = 52{\text{ gm e}}{{\text{q}}^{ - 1}}
C) R=CH3{\text{R}} = {\text{C}}{{\text{H}}_3}
C=COOH{\text{C}} = {\text{COOH}}
N.E. (Ew)=52 gm eq−1{\text{N}}{\text{.E}}{\text{. }}\left( {{\text{Ew}}} \right) = 52{\text{ gm e}}{{\text{q}}^{ - 1}}
D) None of these

Explanation

Solution

The new carboxylic acid is a mono carboxylic acid. R represents an alkyl group. Using the equivalent weight of new carboxylic acid formed we can calculate the value of R.

Complete step by step solution:
When a carboxylic acid is heated then the removal of carbon dioxide occurs and the formation of an alkane occurs.
Since on heating the carboxylic acid B another carboxylic acid is formed that is C that means the carboxylic acid B must be a di carboxylic acid otherwise the product formed should be an alkane and not another carboxylic acid. Let us assume the general formula for the new carboxylic acid formed is:
R−COOH{\text{R}} - {\text{COOH}} and the equivalent weight is given to us. In this case the equivalent weight and molecular weight will be equal because it is mono carboxylic acid and the valence in case of mono carboxylic acid is 1. Hence we will find the molecular weight of R group as:
weight of R +12+16+16+1=60{\text{weight of R }} + 12 + 16 + 16 + 1 = 60
We have used the atomic masses of each element and the given weight of the acid C.
weight of R =15{\text{weight of R }} = 15. The only alkyl group with mass 15 is CH3{\text{C}}{{\text{H}}_3}.
This makes the formula for carboxylic acid C as CH3COOH{\text{C}}{{\text{H}}_3}{\text{COOH}}

Hence, the correct option is A.

Note:
The carboxylic acid B is malonic acid whose molecular formula is B=CH2(COOH)2{\text{B}} = {\text{C}}{{\text{H}}_2}{\left( {{\text{COOH}}} \right)_2}. Its valency is 2 because it donates 2 hydrogen. Its molecular weight is 104. We will get its equivalent weight by dividing the molecular weight with valency as 52. This is the equivalent weight provided to us and hence our answer stands correct.