Question

How many isomers can a compound with molecular formula \( \text{C}_3\text{H}_5\text{Br} \) have?

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

When counting isomers in competitive exams like WBJEE, unless specifically stated as "structural isomers", always include geometrical and optical stereoisomers. Failing to consider the cis/trans isomers of 1-bromoprop-1-ene is a very common source of error.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for the total number of isomers (including both structural and stereoisomers) for the molecular formula \( \text{C}_3\text{H}_5\text{Br} \).

Step 2: Key Formula or Approach:
1. Calculate the Degree of Unsaturation (DU) to determine the structural features (double bonds or rings):
\[ \text{DU} = C + 1 - \frac{H + X}{2} \]
where \( C = 3 \), \( H = 5 \), and \( X = 1 \text{ (for Br)} \).
\[ \text{DU} = 3 + 1 - \frac{5 + 1}{2} = 4 - 3 = 1 \]
A DU of 1 indicates either one double bond (acyclic alkene) or one ring (cyclic cyclopropane).

Step 3: Detailed Explanation:
Let's systematically draw all possible isomers:
1. Acyclic structures (Alkenes with one double bond):
The carbon skeleton is a 3-carbon chain: \( \text{C=C-C} \).
- Case A: Bromine on C-1 (\( \text{CH(Br)=CH-CH}_3 \), 1-bromoprop-1-ene). This compound exhibits geometrical isomerism:
- Isomer 1: cis-1-bromoprop-1-ene
- Isomer 2: trans-1-bromoprop-1-ene
- Case B: Bromine on C-2 (\( \text{CH}_2\text{=C(Br)-CH}_3 \), 2-bromoprop-1-ene). This has no stereoisomers:
- Isomer 3: 2-bromoprop-1-ene
- Case C: Bromine on C-3 (\( \text{CH}_2\text{=CH-CH}_2\text{Br} \), 3-bromoprop-1-ene allyl bromide). This has no stereoisomers:
- Isomer 4: 3-bromoprop-1-ene
2. Cyclic structures (one 3-membered ring):
- The only ring with 3 carbons is a cyclopropane ring. Replacing one H with Br gives:
- Isomer 5: Bromocyclopropane
Thus, the compound can have a total of 5 isomers.

Step 4: Final Answer:
The total number of isomers is 5, which corresponds to option (D).