Refer to the given passage and answer the following questions.
When alcohols react with carboxylic acids in the presence of concentrated sulphuric acid, compounds with fruity smell called esters are formed. Also, alcohols on oxidation in the presence of acidified `K_2 ,Cr_2,O_7` , form carboxylic acids.
If third member of alcohol family (homologous series) undergoes esterification reaction with second member of carboxylic acid family then, the name of ester formed and its formula will be respectively

Options

A. Ethyl propanoate `CH_3 CH_2 COOCH_2 CH_3`
B. Propyl ethanoate `CH_3 COOCH_2 CH_2 CH_3`
C. Ethyl butanoate `CH_3 CH_2 CH_2 COOCH_2 CH_3`
D. Ethyl ethanoate `CH_3 COOCH_2 CH_3`

Answer

To solve the problem, we need to identify the third member of the alcohol family and the second member of the carboxylic acid family, and then determine the ester formed from their reaction.

### Step 1: Identify the third member of the alcohol family
The alcohol family follows the general formula \( C_nH_{2n+1}OH \).
- The first member (n=1) is Methanol (CH₃OH).
- The second member (n=2) is Ethanol (C₂H₅OH).
- The third member (n=3) is Propanol (C₃H₇OH).

Thus, the third member of the alcohol family is **Propanol** (C₃H₇OH).

### Step 2: Identify the second member of the carboxylic acid family
The carboxylic acid family follows the general formula \( C_nH_{2n}O_2 \).
- The first member (n=1) is Formic acid (HCOOH).
- The second member (n=2) is Acetic acid (C₂H₄O₂).

Thus, the second member of the carboxylic acid family is **Acetic acid** (C₂H₄O₂).

### Step 3: Write the esterification reaction
Esterification is the reaction between an alcohol and a carboxylic acid to form an ester and water. The general reaction can be represented as:
\[ \text{Alcohol} + \text{Carboxylic Acid} \rightarrow \text{Ester} + \text{Water} \]

In our case:
- Alcohol: Propanol (C₃H₇OH)
- Carboxylic Acid: Acetic acid (C₂H₄O₂)

### Step 4: Determine the name and formula of the ester formed
The ester formed from Propanol and Acetic acid is called **Propyl acetate**.

To derive the formula:
- The alcohol contributes the propyl group (C₃H₇).
- The carboxylic acid contributes the acetate group (C₂H₃O).

Thus, the formula for Propyl acetate is:
\[ C₃H₇OOC₂H₃ \]
This simplifies to:
\[ C₅H₁₀O₂ \]

### Final Answer:
The name of the ester formed is **Propyl acetate** and its formula is **C₅H₁₀O₂**.

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