The main product of the reaction would be 2- butene + chloroform `overset (NaOH) rarr`?

Options

A. butanoic acid
B. 2- methyl butanoic acid
C. 1,1,1 – trichloro – 2 – methyl butane
D. 1, 4 – butane diol

Answer

To solve the problem of the reaction between 2-butene and chloroform in the presence of sodium hydroxide (NaOH), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The main reactants are 2-butene (C4H8) and chloroform (CHCl3).

2. **Understand the Role of NaOH**:
- Sodium hydroxide (NaOH) acts as a base in this reaction. It can facilitate the deprotonation of chloroform and help in the formation of a nucleophile.

3. **Formation of the Nucleophile**:
- Chloroform (CHCl3) can react with NaOH to form a carbanion. The reaction can be represented as:
\[ CHCl_3 + NaOH \rightarrow CCl_3^- + H_2O + Na^+ \]
- Here, CCl3^- is a trichloromethyl anion, which is a good nucleophile.

4. **Nucleophilic Attack**:
- The trichloromethyl anion (CCl3^-) will then attack the double bond of 2-butene (C4H8). The reaction can be illustrated as:
\[ C_4H_8 + CCl_3^- \rightarrow C_4H_7CCl_3 \]
- This forms a new compound where the trichloromethyl group is added to the 2-butene.

5. **Elimination of Water**:
- The compound formed will have three chlorines and one hydrogen on the same carbon, which is unstable. Therefore, it will undergo elimination to form a double bond and release water:
\[ C_4H_7CCl_3 \rightarrow C_4H_6 + H_2O + HCl \]
- This results in the formation of an alkene.

6. **Final Product Formation**:
- The final product after elimination will be an alkene. In this case, the major product will be 2-methylbutanoic acid (C5H10O2) after the rearrangement and further reactions.

### Conclusion:
The main product of the reaction between 2-butene and chloroform in the presence of NaOH is 2-methylbutanoic acid.