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To determine the organic compound (A) with the molecular formula $C_6H_{12}O_2$, we need to analyze the information given in the question. The compound is hydrolyzed with dilute sulfuric acid ($H_2SO_4$) to produce a carboxylic acid (B) and an alcohol (C).
1. **Understanding the Hydrolysis Reaction**:
- Hydrolysis of an organic compound typically involves breaking down the compound into simpler components. In this case, compound (A) is likely an ester, which upon hydrolysis yields a carboxylic acid and an alcohol.
- The general reaction for the hydrolysis of an ester is:
\[
\text{Ester} + \text{Water} \xrightarrow{\text{H}_2\text{SO}_4} \text{Carboxylic Acid} + \text{Alcohol}
\]
2. **Identifying the Alcohol (C)**:
- The problem states that the alcohol (C) gives white turbidity immediately when treated with anhydrous zinc chloride ($ZnCl_2$) and concentrated hydrochloric acid ($HCl$). This is a characteristic reaction for primary alcohols, which form alkyl chlorides that can precipitate as white solids.
3. **Determining the Structure of Compound (A)**:
- Given that (A) is an ester and contains 6 carbon atoms, we can consider possible structures. The simplest ester that fits this description is ethyl acetate ($C_2H_5COOC_2H_5$), which has the formula $C_4H_8O_2$. However, we need a total of 6 carbons.
- A likely candidate is butyl acetate ($C_4H_8O_2$), which can be hydrolyzed to yield butanoic acid (B) and butanol (C).
4. **Final Verification**:
- The molecular formula of butyl acetate is $C_6H_{12}O_2$. Upon hydrolysis, it gives butanoic acid and butanol, both of which fit the criteria described in the question.
- The alcohol (butanol) will give a white turbidity with $ZnCl_2$ and $HCl$, confirming its identity.
Thus, the organic compound (A) is butyl acetate, which can be represented as:
\[
\text{Butyl Acetate: } C_4H_8O_2
\]
