MCQ
$CH_3Cl \to CH_4$
Above conversion can be achieved by
Above conversion can be achieved by
- A$Zn / H^+$
- B$LiAlH_4$
- C$Mg$ / (ether) then $H_2O$
- ✓all of these
✓
Answer
Correct option: D.
all of these
d
$C{H_3}Cl\xrightarrow[{\operatorname{Re} d.}]{{Zn/{H^ \oplus }}}C{H_4}$
$C{H_3}Cl\xrightarrow[{\operatorname{Re} d.}]{{Zn/{H^ \oplus }}}C{H_4}$
$C{H_3}Cl\xrightarrow[{\operatorname{Re} d.}]{{LiAl{H_4}}}C{H_4}$
$C{H_3}Cl\xrightarrow[\begin{subarray}{l}
Dry \\
ether
\end{subarray} ]{{Mg}}\mathop {\mathop C\limits^{.\,.} }\limits^\Theta {H_3}\mathop {Mg}\limits^ \oplus Cl\xrightarrow{{\mathop H\limits^ \oplus OR}}C{H_4}$
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