Find out number of dimerize products obtain by following reaction.
`H_(3)C-Cl+H_(3)C-CH_(2)-Cl+H_(3)C-CH_(2)-CH_(2)-Clunderset("Dry ether")overset("Na")rarr`

Answer

To find the number of dimerized products from the given reaction involving alkyl halides in the presence of sodium and dry ether, we will analyze the possible combinations of the reactants. The reaction is a Wurtz reaction, which typically leads to the formation of alkanes through the coupling of alkyl halides.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants given are:
- CH₃Cl (Methyl chloride)
- CH₃CH₂Cl (Ethyl chloride)
- CH₃CH₂CH₂Cl (Propyl chloride)

2. **Understand the Wurtz Reaction**:
In the Wurtz reaction, two alkyl halides react with sodium to form a dimerized product (an alkane) and sodium halide as a byproduct. The general reaction can be represented as:
\[ 2R-X + 2Na \rightarrow R-R + 2NaX \]
where R is the alkyl group and X is the halide.

3. **Calculate Possible Dimerization Products**:
- **Self-dimerization of CH₃Cl**:
\[ 2 \text{CH}_3\text{Cl} + 2\text{Na} \rightarrow \text{C}_2\text{H}_6 + 2\text{NaCl} \quad (\text{Ethane}) \]
- **Self-dimerization of CH₃CH₂Cl**:
\[ 2 \text{CH}_3\text{CH}_2\text{Cl} + 2\text{Na} \rightarrow \text{C}_4\text{H}_{10} + 2\text{NaCl} \quad (\text{Butane}) \]
- **Self-dimerization of CH₃CH₂CH₂Cl**:
\[ 2 \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + 2\text{Na} \rightarrow \text{C}_6\text{H}_{14} + 2\text{NaCl} \quad (\text{Hexane}) \]

4. **Cross-dimerization Products**:
- **CH₃Cl with CH₃CH₂Cl**:
\[ \text{CH}_3\text{Cl} + \text{CH}_3\text{CH}_2\text{Cl} + 2\text{Na} \rightarrow \text{C}_3\text{H}_8 + 2\text{NaCl} \quad (\text{Propane}) \]
- **CH₃Cl with CH₃CH₂CH₂Cl**:
\[ \text{CH}_3\text{Cl} + \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + 2\text{Na} \rightarrow \text{C}_5\text{H}_{12} + 2\text{NaCl} \quad (\text{Pentane}) \]
- **CH₃CH₂Cl with CH₃CH₂CH₂Cl**:
\[ \text{CH}_3\text{CH}_2\text{Cl} + \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + 2\text{Na} \rightarrow \text{C}_5\text{H}_{12} + 2\text{NaCl} \quad (\text{Pentane, already counted}) \]

5. **List Unique Products**:
- Ethane (from self-dimerization of CH₃Cl)
- Butane (from self-dimerization of CH₃CH₂Cl)
- Hexane (from self-dimerization of CH₃CH₂CH₂Cl)
- Propane (from cross-dimerization of CH₃Cl and CH₃CH₂Cl)
- Pentane (from cross-dimerization of CH₃Cl and CH₃CH₂CH₂Cl)

6. **Count Unique Dimerized Products**:
The unique dimerized products are:
- Ethane
- Butane
- Hexane
- Propane
- Pentane

Thus, the total number of unique dimerized products is **5**.

### Final Answer:
The number of dimerized products obtained from the reaction is **5**.