
- A$\begin{matrix}
C{{H}_{2}}=CH-CH-C{{H}_{3}} \\
\,\,\,\,\,\,\,\,\,| \\
\,\,\,\,\,\,\,\,\,\,\,\,\,OH \\
\end{matrix}$ - ✓

- C

- D







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$(A)$ Nucleotide form from nucleoside and phosphate group by ether linkage
$(B)$ In nucleoside, nitrogenous base link with hexose sugar of $DNA$ or $RNA$
$(C)$ $DNA$ and $RNA$ are differ in only nature of sugar
$(D)$ $A = T$ and $C \equiv G$ are hydrogen bond between nitrogenous base
$2 \mathrm{H}_{2}(\mathrm{g})+2 \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
the observed rate expression is, rate $=\mathrm{k}_{\mathrm{f}}[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right] .$ The rate expression of the reverse reaction is
$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{C{H_2} = CH - C{H_2} - C - H}
\end{array} \to C{H_3} - C{H_2} - C{H_2} - C{H_2}OH$
$A$, (molecular formula $\left.\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2}\right)$ with a straight chain structure gives a $C_{4}$ carboxylic acid. $A$ is :
$A \frac{{Li} {A} {H} {H}_{4}}{{H}_{3} {O}^{+}} \longrightarrow B \stackrel{\text { Oxidation }}{\longrightarrow} {C}_{4}-$ carboxylic acid