Compound $(X)$ with molecular formula $C_4H_8O$ reacts with $2,4-DNP$ to form a derivative and gives a negative haloform test. The compound $(X)$ is:

The product $(A)$ is:

$A$ tertiary alcohol $H$ upon acid-catalysed dehydration gives a product $I$. Ozonolysis of $I$ leads to compounds $J$ and $K$. Compound $J$ upon reaction with $KOH$ gives benzyl alcohol and a compound $L$,whereas $K$ on reaction with $KOH$ gives only $M$.
$1.$ Compound $H$ is formed by the reaction of
$2.$ The structure of compound $I$ is
$3.$ The structures of compounds $J, K$ and $L$,respectively,are
$(A)$ $PhCOCH_3, PhCH_2 COCH_3$ and $PhCH_2 COO^{-} K^{+}$
$(B)$ $PhCHO, PhCH_2 CHO$ and $PhCOO^{-} K^{+}$
$(C)$ $PhCOCH_3, PhCH_2 CHO$ and $CH_3 COO^{-} K^{+}$
$(D)$ $PhCHO, PhCOCH_3$ and $PhCOO^{-} K^{+}$
Give the answer for questions $1, 2$ and $3.$

Provide the chemical conversion for the following:
$(i)$ $3-$phenylprop$-2-$enal from Benzaldehyde and Ethanal (Cross-Aldol Condensation).
$(ii)$ Benzyl alcohol from Benzene.

Explain the statement: "The $\alpha$-hydrogen of an aldehyde and ketone is acidic."

$C_{7}H_{10}O$ reacts with $CH_{3}MgBr$ to give a compound $C_{8}H_{12}O$ which gives a positive iodoform test. Identify the structure of $A$.