Q. Isopropyl bromide on heating with a concentrated solution of alcoholic (ethanolic) KOH predominantly gives

Options

A. Propene
B. Propan-2-ol
C. Propan-1-ol
D. Isopropyl ethyl ether

Explanation

To solve the question regarding the reaction of isopropyl bromide with concentrated alcoholic KOH, let's break it down step by step:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is isopropyl bromide, which has the chemical structure \( \text{CH}_3\text{CHBrCH}_3 \).

2. **Understand the Reaction Conditions**:
- The reaction is conducted with concentrated alcoholic KOH. Alcoholic KOH is a strong base and promotes elimination reactions (dehydrohalogenation) rather than substitution reactions.

3. **Determine the Type of Reaction**:
- Given that we have a strong base (KOH) in an alcoholic solution, the reaction will favor an elimination reaction (E2 mechanism) leading to the formation of an alkene.

4. **Elimination of HBr**:
- In the elimination process, the bromine atom (Br) and a hydrogen atom (H) from adjacent carbon atoms are removed. This results in the formation of a double bond between the two carbon atoms.

5. **Formation of the Product**:
- The elimination of HBr from isopropyl bromide leads to the formation of propene, which has the structure \( \text{CH}_3\text{CH}=\text{CH}_2 \).

6. **Write the Balanced Reaction**:
- The overall reaction can be summarized as:
\[ \text{CH}_3\text{CHBrCH}_3 + \text{KOH (alc)} \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{HBr} \]

7. **Conclusion**:
- The predominant product of the reaction is propene.

### Final Answer:
Isopropyl bromide on heating with a concentrated solution of alcoholic KOH predominantly gives **propene**.

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