
Which of the following reactions will produce isopropyl acetate in the best yield?
A.Acetyl chloride $ + $ 2- propanol$ \to $
B.Silver acetate $ + $ 2-chloropropane$ \to $
C.Acetic acid$ + $ 2- propanol/${H_2}S{O_4} \to $
D.All of these reactions work equally well
Answer
575.7k+ views
Hint:To answer this question, you should recall the concept of esterification. Esterification is the process of combining an organic acid \[\left( {{\text{RCOOH}}} \right)\] with an alcohol \[\left( {{\text{ROH}}} \right)\] to form an ester \[\left( {{\text{RCOOR}}} \right){\text{\;}}\] and water
Complete step by step answer:
This process of formation of ester requires five steps.
Step 1: Cation formation
Step 2: Delocalized carbocation where Carboxyl oxygen gets protonated to give delocalized carbocation making the carbocation a better electrophile.
Step 3: Transfer of proton where the proton is transferred to one of the hydroxyl groups to form a good leaving group
Step 4: Formation of the pi bond. The process involves the donation of electrons present over the hydroxyl group’s oxygen atom to a carbon atom which makes a pi bond by eliminating water.
Step 5: Ester formation
Among all the options acid chloride is the most reactive acid derivative hence undergoes the fastest esterification.
Therefore, we can conclude that the correct answer to this question is option A.
Note:
Hydrolysis is the most important reaction of esters. Acidic hydrolysis results in a carboxylic acid and an alcohol. Basic hydrolysis results in a carboxylate salt and alcohol.
You should remember in the following reaction: \[{{\text{R}}_{\text{1}}}{\text{COO}}{{\text{R}}_{\text{2}}} + {{\text{H}}_{\text{2}}}{\text{O}} \to {{\text{R}}_{\text{1}}}{\text{COOH}} + {{\text{R}}_{\text{2}}}{\text{OH}}\]
The forward reaction turns into pseudo-first-order in cases where a large concentration of water is involved. However, the rate constant depends on acidic catalytic factors and the net reaction rate is affected by the reverse reaction rate. The concentration of products formed i.e. organic acid and alcohol affects the net reaction rate by increased rate of the reverse reaction and by shifting of the reaction quotient.
Complete step by step answer:
This process of formation of ester requires five steps.
Step 1: Cation formation
Step 2: Delocalized carbocation where Carboxyl oxygen gets protonated to give delocalized carbocation making the carbocation a better electrophile.
Step 3: Transfer of proton where the proton is transferred to one of the hydroxyl groups to form a good leaving group
Step 4: Formation of the pi bond. The process involves the donation of electrons present over the hydroxyl group’s oxygen atom to a carbon atom which makes a pi bond by eliminating water.
Step 5: Ester formation
Among all the options acid chloride is the most reactive acid derivative hence undergoes the fastest esterification.
Therefore, we can conclude that the correct answer to this question is option A.
Note:
Hydrolysis is the most important reaction of esters. Acidic hydrolysis results in a carboxylic acid and an alcohol. Basic hydrolysis results in a carboxylate salt and alcohol.
You should remember in the following reaction: \[{{\text{R}}_{\text{1}}}{\text{COO}}{{\text{R}}_{\text{2}}} + {{\text{H}}_{\text{2}}}{\text{O}} \to {{\text{R}}_{\text{1}}}{\text{COOH}} + {{\text{R}}_{\text{2}}}{\text{OH}}\]
The forward reaction turns into pseudo-first-order in cases where a large concentration of water is involved. However, the rate constant depends on acidic catalytic factors and the net reaction rate is affected by the reverse reaction rate. The concentration of products formed i.e. organic acid and alcohol affects the net reaction rate by increased rate of the reverse reaction and by shifting of the reaction quotient.
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