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प्रश्न
How will you bring about the following conversion?
Bromomethane to propanone
How will you bring about the following conversion in not more than two steps?
Bromomethane to propanone
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उत्तर १
\[\begin{array}{cc}
\ce{\underset{Bromomethane}{CH3 - Br} ->[Alc{.} KCN][C2H5OH, -H2O, \Delta] \underset{Acetonitrile}{CH3CN} ->[CH3MgBr/ether] [CH3 - C = NMgBr] ->[H^+/H2O][-NH3, -Mg(OH)Br] CH3 - C = O}\\
\phantom{....................................................................................}|\phantom{.........................................................}|\\
\phantom{.........................................................................................}\ce{CH3}\phantom{................................................}\ce{\underset{propanone}{CH3}}\\
\end{array}\]
उत्तर २
\[\begin{array}{cc}
\ce{\underset{Bromomethane}{CH3 - Br} ->[HC = C- Na+][Liq{.} NH3, -NaBr] \underset{Propyne}{CH3 - C ≡ CH} ->[Dil{.} H2SO4][HgSO4] CH3 - C - CH3}\\
\phantom{................................................................................}||\\
\phantom{................................................................................}\ce{\underset{propanone}{O}}\\
\end{array}\]
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संबंधित प्रश्न
Identify A, B, C, D, E, R and R1 in the following:

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A product (1) of above reaction is:-
The product formed in the first step of the reaction
\[\begin{array}{cc}
\ce{Br}\phantom{......}\\
|\phantom{.......}\\
\ce{CH3-CH2-CH-CH2-CH-CH3}\\
\phantom{...............}|\\
\phantom{................}\ce{Br}
\end{array}\]
with excess Mg/Et2O(Et = C2H5) is:
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Explain why Grignard reagents should be prepared under anhydrous conditions?
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