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Question: How will you bring Conversion of Methyl Bromide to Acetone?...

How will you bring Conversion of Methyl Bromide to Acetone?

Explanation

Solution

Alkyl halides is the compound containing one or more alkyl groups with one halogen atom. Alkyl halides can be prepared by reacting alcohols with thionyl chloride, phosphorus trihalide, etc. Acetone is also known as dimethyl ketone and it has ketone as a functional group.

Complete step by step answer:
Methyl chloride can be converted to acetone by the following steps.
Step:1 Formation of Grignard reagent
When methyl chloride reacts with magnesium and dry ether, it forms Methyl magnesium chloride(Grignard reagent). The reaction can be written as follows,
CH3CldryetherMgCH3MgClC{H_3}Cl\xrightarrow[{dryether}]{{Mg}}C{H_3}MgCl
Step:2 Addition of cyanide ions
When methyl nitrile reacts with Grignard reagent, it forms an adduct. The reaction can be written as follows,
CH3MgClCH3CNCH3C(CH3)NMgClC{H_3}MgCl\xrightarrow{{C{H_3}CN}}C{H_3} - C(C{H_3})N - MgCl
Step:3 Formation of acetone
When the formed adduct undergoes hydrolysis, it forms acetone. The reaction can be written as follows,
CH3C(CH3)NMgClH3O+CH3COCH3+H2NMgClC{H_3} - C(C{H_3})N - MgCl\xrightarrow{{{H_3}{O^ + }}}C{H_3}COC{H_3} + {H_2}N - MgCl
Thus, acetone can be prepared from methyl chloride by converting methyl chloride to Grignard reagent followed by the reaction with methyl cyanide and finally, through hydrolysis.

Note: When methyl chloride reacts with magnesium and dry ether, it forms Methyl magnesium chloride(Grignard reagent). The reaction can be written as follows,
CH3CldryetherMgCH3MgClC{H_3}Cl\xrightarrow[{dryether}]{{Mg}}C{H_3}MgCl
The formed Grignard reagent on reagent with acetaldehyde, it forms secondary alcohol. The reaction can be written as follows,
CH3MgClCH3CHOCH3CH(CH3)OHC{H_3}MgCl\xrightarrow{{C{H_3}CHO}}C{H_3} - CH(C{H_3}) - OH
It is known that when secondary alcohol undergoes oxidation, it leads to a ketone with the same number of carbon atoms. The reaction can be written as follows,
CH3CH(CH3)OHKMnO4CH3COCH3C{H_3} - CH(C{H_3}) - OH\xrightarrow{{KMn{O_4}}}C{H_3}COC{H_3}
If primary alcohol undergoes oxidation, it gives aldehyde with the same number of carbon atoms. This is the main distinction between the oxidation of primary and secondary alcohol.