MCQ
$C{H_3}MgI$ will give methane with
- A${C_2}{H_5}OH$
- B$C{H_3} - C{H_2} - N{H_2}$
- C$C{H_3} - CO - C{H_3}$
- ✓Both $(a)$ and $(b)$
$C{H_3}MgI + {C_2}{H_5}OH \to C{H_4} + {C_2}{H_5}OMgI$
Alkyl group of Grignard’s reagent is involved in the formation of alkane.
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Given : $K_f (H_2O)$ = $2\ K\ kg\ mole^{-1}$, $K_{sp} (PbCl_2)$ = $4 × 10^{-6}$
(Assume molarity to be equal to molality) .....$^oC$
$C{H_3}C{H_2}OH\xrightarrow{{P + {I_2}}}A\xrightarrow[{Ether}]{{Mg}}B $ $\xrightarrow{{HCHO}}C\xrightarrow{{{H_2}O}}D,$
then compound $'D'$ is -