The correct order of increasing acid strength of the compounds is:
$(A) CH_3CO_2H$
$(B) MeOCH_2CO_2H$
$(C) CF_3CO_2H$
$(D) (CH_3)_2CHCO_2H$

In the given reaction :
$CH_3-COOH$ $\mathop{\xrightarrow{(i) Br_2/P}}_{(ii) NaCN, (iii) H_2O/H^{+}/\Delta}$ $[X]$
$[X]$ will be :

$n$-Propanol on treatment with concentrated $HBr$ gives $P$. The product $P$ on reaction with $KCN$ gave the product $Q$. The product $Q$ on heating with aqueous acidic solution,furnished the product $R$. The product $R$ is

Give reactions of $(i)$ Acetic acid,$(ii)$ Benzoic acid,and $(iii)$ Phthalic acid with ammonia.

An optically active compound $'X'$ having molecular formula $C_4H_8O_3$ evolves $CO_2$ with $NaHCO_3$. $'X'$ on treatment with $LiAlH_4$ gives an achiral compound. Then $'X'$ is

The carboxylic acid which has maximum solubility in water is