When alcohols react with carboxylix acids in the presence of concentrated sulphuric acid, compounds with fruity smell called edters 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 formed and its formula will be respectively

Options

A. ethyl propanoate, `CH_(3)CH_(2)COOCH_(2)CH_(3)`
B. propyl propanoate, `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)`

Explanation

To solve the problem step by step, 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 third member corresponds to \( n = 3 \):
- For \( n = 3 \):
\[ C_3H_{2(3)+1}OH = C_3H_7OH \]
Thus, the third member of the alcohol family is **Propan-1-ol** (or simply Propanol), with the formula \( C_3H_7OH \).

### 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 second member corresponds to \( n = 2 \):
- For \( n = 2 \):
\[ C_2H_{2(2)}O_2 = C_2H_4O_2 \]
Thus, the second member of the carboxylic acid family is **Ethanoic Acid** (or Acetic Acid), with the formula \( CH_3COOH \).

### Step 3: Write the Reaction for Esterification
Esterification is the reaction between an alcohol and a carboxylic acid in the presence of an acid catalyst (like concentrated sulfuric acid). The general reaction can be represented as:
\[ \text{Alcohol} + \text{Carboxylic Acid} \rightarrow \text{Ester} + \text{Water} \]
Substituting our identified compounds:
\[ C_3H_7OH + CH_3COOH \rightarrow \text{Ester} + H_2O \]

### Step 4: Determine the Ester Formed
The ester formed from Propanol and Ethanoic Acid is **Propyl Ethanoate**. The structure of the ester can be derived from the alcohol and the acid:
- The alcohol contributes the Propyl group (\( C_3H_7 \)).
- The carboxylic acid contributes the Ethanoate part (\( CH_3COO \)).

Thus, the formula of the ester formed is:
\[ C_3H_7COOCH_3 \]
This can also be written as:
\[ C_5H_{10}O_2 \]

### Step 5: Name of the Ester
The name of the ester formed is **Propyl Ethanoate**. This follows the naming convention where the alkyl part (from the alcohol) is named first followed by the acid part (with the suffix -oate).

### Final Answer
- **Name of the Ester**: Propyl Ethanoate
- **Formula of the Ester**: \( C_5H_{10}O_2 \)