Which carboxylic acid $(B)$ (Equivalent weight $= 52 g m e q^{- 1}$ ) loses $C O_{2}$ , which on heating gives an acid $(C)$ (Equivalent weight $= 60 g m e q^{- 1}$ )?

R = C H_{3} C = C H_{3} C O O H N . E . (E w) = 52 g m e q^{- 1}

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R = C H_{3} C = C H_{3} - C H_{2} - C O O H N . E . (E w) = 52 g m e q^{- 1}

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R = C H_{3} C = C O O H N . E . (E w) = 52 g m e q^{- 1}

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Solution

The correct option is A $R = C H_{3} C = C H_{3} C O O H N . E . (E w) = 52 g m e q^{- 1}$
Carboxylic acid produce another carboxylic acid after decaroxylation indicates it is dicarboxylic acid.
Therefore $(C)$ is monocarboxylic acid
$N . E . (E w) = 60$
$R - C O O H = R + 45 = 60$
$∴ R = 15 = C H_{3}$
$C = C H_{3} C O O H$

$(B)$ has one more $(- C O O H)$ group
Therefore, It is malonic acid $H O O C - C H_{2} - C O O H$ .
$N . E . (E w) = \frac{M w}{Ionisable H atoms} = \frac{104}{2} = 52 g m e q^{- 1} .$

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