Arrange the following labelled hydrogens in decreasing order of acidity.

Which of the following carboxylic acids could be resolved by reaction with an enantiomerically pure chiral amine?

The product $(A)$ will be:

What are $\underline{A}, \underline{B}, \underline{C}$ in the following reactions?
$(I)$ $(CH_3 CO_2)_2 Ca \stackrel{\Delta}{\longrightarrow} \underline{A}$
$(II)$ $CH_3 CO_2 H \underset{\text{Red P}}{\stackrel{HI}{\longrightarrow}} \underline{B}$
$(III)$ $2 CH_3 CO_2 H \stackrel{P_4 O_{10}}{\longrightarrow} \underline{C}$
$\underline{A} \quad \underline{B} \quad \underline{C}$

Given below are two statements:
Statement $I:$ Ethyl pent$-4-$yn$-$oate on reaction with $CH_{3}MgBr$ gives a $3^{\circ}$ alcohol.
Statement $II:$ In this reaction,one mole of ethyl pent$-4-$yn$-$oate utilizes two moles of $CH_{3}MgBr$.
In the light of the above statements,choose the most appropriate answer from the options given below:

$CH_3-C(=O)-OH$ $\xrightarrow{NaH^{14}CO_3} (A) (\text{gas})$ $\xrightarrow[(ii) H_3O^{+}]{(i) PhMgBr} (B)$
$CH_3-S(=O)_2-OH$ $\xrightarrow{NaH^{14}CO_3} (C) (\text{gas})$ $\xrightarrow[(ii) H_3O^{+}]{(i) PhMgBr} (D)$
Product $(B)$ and $(D)$ in the above reaction are: