Major product of given reaction is -
`O_2N-CH_2-undersetunderset(Cl)(|)CH-CH_3overset(overset(Theta)(O)H)underset(Delta)to` Major product

Options

A. `O_2N-CH=CH-CH_3`
B. `O_2N-CH_2-CH=CH_2`
C. `O_2N-CH=undersetunderset(Cl)(|)C-CH_3`
D. `CH_2=undersetunderset(NO_2)(|)C-CH_3`

Answer

To determine the major product of the given reaction, let's analyze the process step by step.

### Step 1: Identify the Reactant Structure
The reactant is `O_2N-CH_2-CHCl-CH_3`. This compound contains a nitro group (-NO2), a chloro group (-Cl), and a methyl group (-CH3) attached to a carbon chain.

### Step 2: Assess the Acidity of the Hydrogen
The hydrogen atom attached to the carbon adjacent to the nitro and chloro groups is more acidic due to the electron-withdrawing effects of these groups. The nitro group is a strong electron-withdrawing group, which stabilizes the negative charge that will form when a hydrogen is removed.

### Step 3: Deprotonation by OH-
When the compound is treated with hydroxide ions (OH-), it will abstract the acidic hydrogen from the carbon adjacent to the nitro and chloro groups. This results in the formation of a carbanion.

### Step 4: Resonance Stabilization of the Carbanion
The carbanion formed is stabilized by resonance. The negative charge can be delocalized over the adjacent carbon atoms due to the presence of the nitro group. The resonance structures can be represented as:
- O2N-CH=CH-CH3
- O2N-CH2-CH(-)CH3

This resonance stabilization makes the carbanion more stable.

### Step 5: Elimination of Chloride Ion
The next step involves the elimination of the chloride ion (Cl-). The presence of the carbanion allows for the formation of a double bond between the two carbon atoms adjacent to the nitro group. The elimination of Cl- leads to the formation of a double bond.

### Step 6: Final Product Formation
After the elimination of Cl-, the final product can be represented as:
`O_2N-CH=CH-CH_3`

### Conclusion
The major product of the reaction is:
`O_2N-CH=CH-CH_3`