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Predict the product of the reaction CH3-...

Predict the product of the reaction `CH_3-underset(H)underset(|)overset(CH_3)overset(|)C- underset(Cl)underset(|)overset(H)overset|C-CH_2-CH=CH_2overset("alc. KOH)rarr`

A

`(CH_3)_2C=CH-CH_2-CH=CH_2`

B

`(CH_3)_2CH-CH=CH-CH=CH_2`

C

`(CH_3)_2CH-CH=underset(CH_3)underset(|)C-CH=CH_2`

D

`(CH_3)_2C=CH-underset(CH_3)underset(|)C=CH_2`

Text Solution

AI Generated Solution

The correct Answer is:
To predict the product of the reaction given the starting material and the presence of alcoholic KOH, we can follow these steps: ### Step 1: Identify the Reactants The reactant is: - CH3C(CH3)(Cl)CH2CH=CH2 This compound has a chlorine atom (Cl) and is a branched alkyl halide. ### Step 2: Understand the Role of Alcoholic KOH Alcoholic KOH acts as a strong base. It can abstract acidic hydrogens from the reactant, leading to the formation of a carbon anion (carbanion) and facilitating elimination reactions. ### Step 3: Identify Possible Sites for Deprotonation In the given compound, there are two possible sites where the base can abstract a proton: 1. The hydrogen atom on the carbon adjacent to the chlorine (C adjacent to Cl). 2. The hydrogen atom on the carbon adjacent to the double bond (C adjacent to the double bond). ### Step 4: Formulate the Carbanions 1. **Possibility A**: If the base abstracts the hydrogen from the carbon adjacent to Cl, we form: - Carbanion: CH3C(CH3)(Cl)CH2^(-)CH=CH2 2. **Possibility B**: If the base abstracts the hydrogen from the carbon adjacent to the double bond, we form: - Carbanion: CH3C(CH3)(Cl)C^(-)H=CH2 ### Step 5: Analyze the Stability of the Carbanions - **Possibility A**: The carbanion formed is less stable due to the destabilizing effect of the chlorine atom. - **Possibility B**: The carbanion formed is more stable due to conjugation with the double bond. ### Step 6: Elimination Reaction Since the more stable carbanion is formed from **Possibility B**, we proceed with this pathway. The negative charge will shift to form a double bond, and the chlorine will leave as a chloride ion (Cl^-). ### Step 7: Write the Final Product After the elimination reaction, we will have: - Final Product: CH3C(CH3)=CH-CH=CH2 ### Summary of the Reaction The reaction of the compound with alcoholic KOH leads to the formation of an alkene through the elimination of HCl. ### Final Answer The product of the reaction is: - CH3C(CH3)=CH-CH=CH2
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