An organic compound 'A' on treating with acidified potassium dichromated solution gives 'B' with molecular mass 60 g/mol. 'A' on heating with conc. `H_(2)SO_(4)` at 443 K produces a gas that decolourise bromine water. The compound 'A' is

Options

A. n-propyl alcohol
B. iso-propyl alcohol
C. ehtyl alcohol
D. acetaldehyde

Explanation

To solve the problem, we need to identify the organic compound 'A' based on the given reactions and molecular mass of the products. Let's break it down step by step:

### Step 1: Understanding the Reaction with Acidified Potassium Dichromate
The compound 'A' reacts with acidified potassium dichromate (K2Cr2O7), which is a strong oxidizing agent. This reaction typically oxidizes alcohols to aldehydes or carboxylic acids. The product 'B' has a molecular mass of 60 g/mol.

**Hint:** Identify common compounds with a molecular mass of 60 g/mol that are likely products of oxidation.

### Step 2: Identifying Compound 'B'
The only carboxylic acid with a molecular mass of 60 g/mol is acetic acid (ethanoic acid, CH3COOH). Therefore, we can conclude that 'B' is acetic acid.

**Hint:** Recall the molecular formula and molar mass of common organic compounds to identify 'B'.

### Step 3: Analyzing the Heating Reaction
Next, we consider the reaction of compound 'A' when heated with concentrated sulfuric acid (H2SO4) at 443 K. This reaction typically leads to dehydration of alcohols to form alkenes.

**Hint:** Remember that heating alcohols with concentrated sulfuric acid can lead to the formation of alkenes through elimination reactions.

### Step 4: Identifying the Gas Produced
The problem states that the reaction produces a gas that decolorizes bromine water. This indicates that the gas is likely an alkene, which is known to react with bromine water due to its unsaturation.

**Hint:** Alkenes are characterized by the presence of a double bond, which allows them to react with bromine.

### Step 5: Determining Compound 'A'
Since 'B' is acetic acid and it was formed from the oxidation of 'A', we need to determine what 'A' could be. The possible candidates are ethanol (C2H5OH) or acetaldehyde (CH3CHO). However, since the reaction with concentrated H2SO4 produces an alkene, we can conclude that 'A' must be ethanol.

**Hint:** Consider the structure of the compounds and their reactions to determine which one can produce the expected products.

### Step 6: Reactions Summary
1. Ethanol (C2H5OH) is oxidized by K2Cr2O7 to form acetic acid (CH3COOH).
2. Ethanol, when heated with concentrated H2SO4 at 443 K, undergoes dehydration to form ethene (C2H4), which can decolorize bromine water.

**Final Conclusion:**
The compound 'A' is ethanol (C2H5OH).

### Summary of Steps:
1. Identify the product 'B' from the oxidation reaction.
2. Determine the molecular mass of 'B' to find the correct compound.
3. Analyze the heating reaction to identify the gas produced.
4. Conclude the identity of compound 'A' based on the reactions.