MCQ
Alkyl halides can be obtained by all methods except
- A$CH_3 CH_2OH + HCl/ZnCl_2 \to$
- B$C{{H}_{3}}-C{{H}_{2}}-C{{H}_{3}}-C{{H}_{2}}\xrightarrow{C{{l}_{2}}/UV\,light}$
- ✓$C_2H_5OH +NaCl \to$
- D$CH_3COOAg + Br_2/CCl_4 \to$

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$\begin{array}{*{20}{c}}
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{{C_6}{H_5} - CH - C{H_2} - C - C{H_3}}
\end{array}\mathop {\xrightarrow{{(i)\,NaOBr}}}\limits_{(ii)\,{H_2}O/{H^ + }\,(iii)\,\Delta } $ product
product will be :