- A

- B

- C

- ✓Both $A\, \& \,C$



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$(1)\,\,\begin{array}{*{20}{c}}
{\,\,\,\,\,\,C{H_3}} \\
| \\
{{C_2}{H_5}C{H_2}C - OC{H_3}} \\
| \\
{\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(2)\,\,\begin{array}{*{20}{c}}
{{C_2}{H_5}C{H_2}C = C{H_2}} \\
{\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(3)\,\,\begin{array}{*{20}{c}}
{{C_2}{H_5}CH = C - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
($A$) carbon tetrachloride + methanol ($B$) carbon disulphide + acetone ($C$) benzene + toluene ($D$) phenol + aniline
$A$, (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