Draw all structurally isomeric cyclic bromides with molecular formula `C_(4)H_(7)Br`.

Answer

To draw all structurally isomeric cyclic bromides with the molecular formula C₄H₇Br, we can follow these steps:

### Step 1: Determine the Double Bond Equivalent (DBE)
The formula for calculating the double bond equivalent (DBE) is:
\[ \text{DBE} = C - \frac{H}{2} + \frac{N}{2} + 1 \]
Where:
- C = number of carbon atoms
- H = number of hydrogen atoms
- N = number of nitrogen atoms (not applicable here)

For C₄H₇Br:
- C = 4
- H = 7
- Br is a halogen and can be counted as a nitrogen equivalent in the formula.

Substituting the values:
\[ \text{DBE} = 4 - \frac{7}{2} + 0 + 1 = 4 - 3.5 + 1 = 1.5 \]

Since we are dealing with cyclic compounds, we can interpret this as having one ring and one double bond or two rings.

### Step 2: Identify Possible Cyclic Structures
Given that we need to form cyclic structures, we can consider:
1. **Cyclobutane**: A four-membered ring.
2. **Cyclopropane derivatives**: A three-membered ring with substituents.

### Step 3: Draw the Isomeric Structures
1. **1-Bromocyclobutane**:
- Structure: A four-membered ring with bromine attached to one of the carbons.
- Formula: C₄H₇Br

2. **1-Bromo-1-methylcyclopropane**:
- Structure: A three-membered ring (cyclopropane) with a methyl group and bromine on the same carbon.
- Formula: C₄H₇Br

3. **1-Bromo-2-methylcyclopropane**:
- Structure: A three-membered ring (cyclopropane) with a methyl group on one carbon and bromine on a different carbon.
- Formula: C₄H₇Br

4. **1-Cyclopropyl-1-bromopropane**:
- Structure: A three-membered ring (cyclopropane) with a bromine on one carbon and a propyl group on the same carbon.
- Formula: C₄H₇Br

### Step 4: Verify the Structures
Each of these structures should have the molecular formula C₄H₇Br and should be cyclic bromides.

### Summary of Isomers
1. **1-Bromocyclobutane**
2. **1-Bromo-1-methylcyclopropane**
3. **1-Bromo-2-methylcyclopropane**
4. **1-Cyclopropyl-1-bromopropane**