Questions: In an experiment, propan-1-ol is made to react with butanoic acid, in the presence of concentrated sulfuric acid as a catalyst. The product is: a. propyl propanoate b. ethyl propanoate c. propyl butanoate d. butyl propanoate

In an experiment, propan-1-ol is made to react with butanoic acid, in the presence of concentrated sulfuric acid as a catalyst. The product is: a. propyl propanoate b. ethyl propanoate c. propyl butanoate d. butyl propanoate
Transcript text: In an experiment, propan-1-ol is made to react with butanoic acid, in the presence of concentrated sulfuric acid as a catalyst. The product is: a. propyl propanoate b. ethyl propanoate c. propyl butanoate d. butyl propanoate
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Solution

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Solution Steps

Step 1: Identify the Reactants

In the given experiment, the reactants are propan-1-ol and butanoic acid.

Step 2: Understand the Reaction

The reaction between an alcohol (propan-1-ol) and a carboxylic acid (butanoic acid) in the presence of a concentrated sulfuric acid catalyst is an esterification reaction. This reaction produces an ester and water.

Step 3: Determine the Ester Formed

In an esterification reaction, the alcohol contributes the alkyl group (propyl from propan-1-ol) and the carboxylic acid contributes the acyl group (butanoate from butanoic acid). Therefore, the ester formed will be propyl butanoate.

Final Answer

\(\boxed{\text{c. propyl butanoate}}\)

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