
Acetone reacts with Grignard reagent to form:
A. $3^\circ $ alcohol
B. $2^\circ $ alcohol
C. Ether
D. No reaction
Answer
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Hint: The Grignard reagent is an organometallic compound having a chemical formula $R - Mg - X$. Here Mg represents a magnesium element.
R can be an aryl or alkyl group like $C{H_3}$ etc. and X could be a halogen atom-like $Br,Cl$ etc.
For example, $C{H_3} - Mg - Br$
Acetone belongs to the category of ketones and has the molecular formula ${({H_3}C)_2}C = O$
Complete step by step answer:
In the question, we have to find out the product from the reaction of acetone with the Grignard reagent.
We will discuss the reaction stepwise.
The reactants are acetone and Grignard reagent, so we can write it as follows:
${({H_3}C)_2}C = O + R - Mg - X \to ?$
Here, let us consider the Grignard reagent with the methyl group and halogen atom to be bromine.
${({H_3}C)_2}C = O + C{H_3} - Mg - Br \to ?$
The reaction of Grignard reagent proceeds in the presence of dry ether and the carbon (\[C = O\]) will be replaced. So, it will be represented as
${({H_3}C)_2}C = O + C{H_3} - Mg - Br\xrightarrow{{dryether}}{({H_3}C)_2}C - OC{H_3}MgBr$
In the third step, it will be further hydrolysed and tertiary alcohol will be formed. We can write it as
${({H_3}C)_2}C = O + C{H_3} - Mg - Br\xrightarrow{{dryether}}{({H_3}C)_2}C - OC{H_3}MgBr\xrightarrow{{Hydrolysis{{({H_3}O)}^ + }}}{({H_3}C)_2}C - (OH)C{H_3}$
The general reaction can be written as
${({H_3}C)_2}C = O + R - Mg - X \to {({H_3}C)_2}RC - {O^ - } \to {({H_3}C)_2}RCOH$
R can be an alkyl or aryl group.
Here, we can see that a tertiary ($3^\circ $) alcohol is formed.
So, the correct answer is Option A.
Note: When Grignard reagent reacts with an aldehyde group or compound it leads to the formation of $2^\circ $ alcohol. The formation of $2^\circ $ and $3^\circ $ alcohol is considered to be an example of $C - C$ bond formation.
Grignard reagents react in the presence of dry ether because
Ether has no acidic protons i.e. it would not donate any protons to the Grignard reagent. So, the Grignard reagent will be stable in presence of ether. As if any solvent used shows protonation (donate protons) then Grignard reagent will lead to the formation of hydrocarbons.
R can be an aryl or alkyl group like $C{H_3}$ etc. and X could be a halogen atom-like $Br,Cl$ etc.
For example, $C{H_3} - Mg - Br$
Acetone belongs to the category of ketones and has the molecular formula ${({H_3}C)_2}C = O$
Complete step by step answer:
In the question, we have to find out the product from the reaction of acetone with the Grignard reagent.
We will discuss the reaction stepwise.
The reactants are acetone and Grignard reagent, so we can write it as follows:
${({H_3}C)_2}C = O + R - Mg - X \to ?$
Here, let us consider the Grignard reagent with the methyl group and halogen atom to be bromine.
${({H_3}C)_2}C = O + C{H_3} - Mg - Br \to ?$
The reaction of Grignard reagent proceeds in the presence of dry ether and the carbon (\[C = O\]) will be replaced. So, it will be represented as
${({H_3}C)_2}C = O + C{H_3} - Mg - Br\xrightarrow{{dryether}}{({H_3}C)_2}C - OC{H_3}MgBr$
In the third step, it will be further hydrolysed and tertiary alcohol will be formed. We can write it as
${({H_3}C)_2}C = O + C{H_3} - Mg - Br\xrightarrow{{dryether}}{({H_3}C)_2}C - OC{H_3}MgBr\xrightarrow{{Hydrolysis{{({H_3}O)}^ + }}}{({H_3}C)_2}C - (OH)C{H_3}$
The general reaction can be written as
${({H_3}C)_2}C = O + R - Mg - X \to {({H_3}C)_2}RC - {O^ - } \to {({H_3}C)_2}RCOH$
R can be an alkyl or aryl group.
Here, we can see that a tertiary ($3^\circ $) alcohol is formed.
So, the correct answer is Option A.
Note: When Grignard reagent reacts with an aldehyde group or compound it leads to the formation of $2^\circ $ alcohol. The formation of $2^\circ $ and $3^\circ $ alcohol is considered to be an example of $C - C$ bond formation.
Grignard reagents react in the presence of dry ether because
Ether has no acidic protons i.e. it would not donate any protons to the Grignard reagent. So, the Grignard reagent will be stable in presence of ether. As if any solvent used shows protonation (donate protons) then Grignard reagent will lead to the formation of hydrocarbons.
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