
How are the following conversions carried out?
${\text{Methyl magnesium bromide }} \to {\text{2 - Methylpropan - 2 - ol}}$
Answer
581.4k+ views
Hint: Methyl magnesium bromide is a Grignard reagent which is mostly used to convert aldehyde and ketones into alcohol. Due to electronegativity differences they have some partial charges which force the reaction to proceed forward.
Complete step by step answer:
Methyl magnesium bromide: ${\text{C}}{{\text{H}}_3}{\text{MgBr}}$
Methyl magnesium bromide is a Grignard reagent. The Grignard reagents are basically represented as
${\text{R - Mg - X}}$ where ${\text{R}}$ is an alkyl group and ${\text{X}}$ is a halide.
Grignard reagents are mostly used in the formation of alcohols. Grignard reagent reacts with aldehyde and ketone to form the alcoholic species.
So to convert methyl magnesium bromide to ${\text{2 - Methylpropan - 2 - ol}}$ we can take help of a ketone, because with the help of ketone only we will be able to produce the secondary alcohol.
We will use propane or acetone in our reaction because it has three carbon atoms, when it is combined with methyl magnesium bromide one more carbon atom will be added to it and the ketone group will change into alcohol. This will result in our desired product ${\text{2 - Methylpropan - 2 - ol}}$.
So the reaction involved is:
\[{\text{C}}{{\text{H}}_{\text{3}}}{\text{MgBr + C}}{{\text{H}}_{\text{3}}}{\text{COC}}{{\text{H}}_{\text{3}}}\xrightarrow{{{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}}}{\text{C(C}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{3}}}{\text{OH + Mg(OH)Br}}\]
When methyl magnesium bromide attacks on acetone an adduct is formed which has partial charges on it due to the electronegativity differences and it is also termed as addition product because both the reactants combine here, to separate them and make our desired alcohol, we hydrolyse this adduct, on hydrolysis this adduct produces ${\text{2 - Methylpropan - 2 - ol}}$.
Note:
The preparation of Grignard reagent is also an important task. The Grignard reagents are prepared by reacting an organic halide with the magnesium metal. The magnesium ribbon is used to provide the magnesium metal in the formation of Grignard reagent.
Complete step by step answer:
Methyl magnesium bromide: ${\text{C}}{{\text{H}}_3}{\text{MgBr}}$
Methyl magnesium bromide is a Grignard reagent. The Grignard reagents are basically represented as
${\text{R - Mg - X}}$ where ${\text{R}}$ is an alkyl group and ${\text{X}}$ is a halide.
Grignard reagents are mostly used in the formation of alcohols. Grignard reagent reacts with aldehyde and ketone to form the alcoholic species.
So to convert methyl magnesium bromide to ${\text{2 - Methylpropan - 2 - ol}}$ we can take help of a ketone, because with the help of ketone only we will be able to produce the secondary alcohol.
We will use propane or acetone in our reaction because it has three carbon atoms, when it is combined with methyl magnesium bromide one more carbon atom will be added to it and the ketone group will change into alcohol. This will result in our desired product ${\text{2 - Methylpropan - 2 - ol}}$.
So the reaction involved is:
\[{\text{C}}{{\text{H}}_{\text{3}}}{\text{MgBr + C}}{{\text{H}}_{\text{3}}}{\text{COC}}{{\text{H}}_{\text{3}}}\xrightarrow{{{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}}}{\text{C(C}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{3}}}{\text{OH + Mg(OH)Br}}\]
When methyl magnesium bromide attacks on acetone an adduct is formed which has partial charges on it due to the electronegativity differences and it is also termed as addition product because both the reactants combine here, to separate them and make our desired alcohol, we hydrolyse this adduct, on hydrolysis this adduct produces ${\text{2 - Methylpropan - 2 - ol}}$.
Note:
The preparation of Grignard reagent is also an important task. The Grignard reagents are prepared by reacting an organic halide with the magnesium metal. The magnesium ribbon is used to provide the magnesium metal in the formation of Grignard reagent.
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