Question

Select the ester that is formed when propanoic acid reacts with isopropyl alcohol (propan-2-ol) in the presence of heat and an acid catalyst.

Answer 1

Answer : The ester form in this reaction will be, Isopropyl propanoate.

Explanation :

When the carboxylic acid react with an alcohol in acidic medium result in the formation of an ester and this is known as Fischer's esterification.

Mechanism of this reaction :

Step 1 : Protonation of the propanoic acid.

In the presence of the an acid catalyst, protonation of carbonyl oxygen occurs which increases the electrophilicity of carbon.

Step 2 : Attack of isopropyl alcohol on the carbonyl carbon.

As the isopropyl alcohol is a weak nucleophile, it attacks on carbonyl carbon due to increases the electrophilicity.

Step 3 : Deprotonation and elimination of leaving group that means formation of ester.

Now loss of proton occurs from oxonium ion. This proton is captured by the oxygen of $OH^-$ group an hydroxyl group gets converted to a better leaving group named as $H_2O$ (water)

The mechanism are shown below.

Answer 2

The final product is an$\boxed{{\text{isopropyl propanoate}}}$.

Further explanation:

The chemical reaction between an acid and an alcohol molecule to form an ester is known as an esterification reaction. The reaction takes place in acidic medium, and a water molecule gets eliminated.

The reactants given in the question are propanoic acid (acid) and isopropyl alcohol (propan-2-ol) as alcohol. Therefore, the reaction in presence of catalyst is an esterification reaction, and the product is an ester after the removal of water molecule.

The steps involved in mechanism of esterification of propanoic acid and isopropy alcohol is as follows:

Step 1: Reaction is initiated from the protonation of propanoic acid.

Step 2: Then iso-propyl attack on the electrophilic carbonyl carbon.

Step 3: In the last step, initially the water molecule is removed, and then deprotonation occurs. Deprotonation occurs to produce the final product isopropyl propanoate.

The mechanism of esterification of propanoic acid and isopropyl alcohol is attached in the image.

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: brainly.com/question/3064603

2. Product formed when 2-propanol react with NaH: brainly.com/question/5045356

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Esterification

Keywords: Esterification, isopropyl alcohol, propanoic acid, fisher esterification, mechanism, step 1protonation of propanoic acid, step 2 attack on the electrophilic carbonyl carbon, step 3 deprotonation to produce isopropyl propanoate, product, isopropyl propanoate.