The `IUPAC` name of `CH_(3)C(CH_(3))_(2)CH_(2)CH=CH_(2)` is

Options

A. (1) `2,2`-dimethylpent-`4`-ene
B. (2) `4,4`-dimethylpent-`1`-ene
C. (3) `1,1,1`-trimethylbut-`3`-ene
D. (4) `4,4,4`-trimethylbut-`1`-ene

Explanation

To determine the IUPAC name of the compound \( CH_3C(CH_3)_2CH_2CH=CH_2 \), we will follow a systematic approach.

### Step 1: Identify the Structure
The given compound can be expanded to show its structure:
- The compound consists of a carbon chain with a double bond and branching methyl groups.

### Step 2: Identify the Parent Chain
1. **Find the longest continuous carbon chain**:
- The longest chain in this compound has 5 carbon atoms, which is a pentane.

### Step 3: Number the Carbon Chain
2. **Number the chain**:
- To give the double bond the lowest possible number, we will number the chain from the end closest to the double bond.
- Numbering from right to left, we have:
- C1: CH2 (part of the double bond)
- C2: CH (part of the double bond)
- C3: CH2
- C4: C(CH3)2 (two methyl groups attached)
- C5: CH3

### Step 4: Identify Substituents
3. **Identify and name the substituents**:
- There are two methyl groups attached to the fourth carbon.
- Therefore, we denote this as "4,4-dimethyl".

### Step 5: Combine the Names
4. **Combine the names**:
- The parent chain is "pent" (for five carbons).
- The position of the double bond is at carbon 1, so we add "1" before the suffix.
- The suffix for alkenes is "ene".

### Final IUPAC Name
Putting it all together, the IUPAC name of the compound is:
**4,4-dimethylpent-1-ene**