Question

CH3MgBr +CH3CHO=
(un the presence of H2O/H+)​

Answer 1

Answer:

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Answer 2

Answer:

To form primary alcohol by using the Grignard reagent then we have to use methanol as a reactant because only methanol can form primary alcohol, and with ketones, always tertiary alcohols are formed.

$\mathrm{CH}_{3} \mathrm{COCH}_{3}+\mathrm{CH}_{3} \mathrm{MgBr} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{3} \mathrm{CH}_{3}+\mathrm{COMgBr}$

Explanation:

• The reactants involved in the given reaction are acetone and Methyl magnesium bromide.
• When these two reactants interact double displacement of functional groups involved in the reactants occurs.
• Thus this reaction is an example of a double displacement reaction. Due to this CO in acetone is displaced by methyl from methyl magnesium bromide and methyl in methyl magnesium bromide is displaced by CO.
• Thus as per the conversion, the complete reaction is as follows:$\mathrm{CH}_{3} \mathrm{COCH}_{3}+\mathrm{CH}_{3} \mathrm{MgBr} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{3} \mathrm{CH}_{3}+\mathrm{COMgBr}$