The hydrocarbon (molecular mass = 70) after reduction and chlorination gives a single monochloride . The hydrocarbon is

Options

A. pent - 1- ene
B. pent - 2- ene
C. 1,1 - dimethylcycloprop -1 - ene
D. cyclopentane

Answer

To solve the problem, we need to identify the hydrocarbon with a molecular mass of 70 that, after reduction and chlorination, yields a single monochloride. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the possible hydrocarbons
We know the molecular mass of the hydrocarbon is 70 g/mol. We need to check the options provided: pent-1-ene, pent-2-ene, 1,1-dimethylcyclopropene, and cyclopentane.

### Step 2: Calculate the molecular formula for each option
1. **Pent-1-ene (C5H10)**:
- Molecular mass = (5 × 12) + (10 × 1) = 60 + 10 = 70 g/mol
2. **Pent-2-ene (C5H10)**:
- Molecular mass = (5 × 12) + (10 × 1) = 60 + 10 = 70 g/mol
3. **1,1-Dimethylcyclopropene (C5H8)**:
- Molecular mass = (5 × 12) + (8 × 1) = 60 + 8 = 68 g/mol
4. **Cyclopentane (C5H10)**:
- Molecular mass = (5 × 12) + (10 × 1) = 60 + 10 = 70 g/mol

From this, we see that pent-1-ene, pent-2-ene, and cyclopentane all have a molecular mass of 70 g/mol.

### Step 3: Analyze the chlorination reaction
- **Pent-1-ene and Pent-2-ene**: These alkenes can produce multiple monochlorides due to the presence of different positions where chlorine can attach. Therefore, they do not meet the requirement of yielding a single monochloride.

- **Cyclopentane**: This cyclic hydrocarbon will yield only one type of monochloride upon chlorination. The chlorine can attach to any carbon in the ring, but since all carbons are equivalent in a symmetrical structure like cyclopentane, only one monochloride is formed.

### Step 4: Conclusion
Since cyclopentane is the only hydrocarbon that gives a single monochloride upon chlorination, we conclude that the hydrocarbon is:

**Cyclopentane (C5H10)**

### Final Answer:
The hydrocarbon is **Cyclopentane**.

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