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CH(3)-CH(2)-underset(Br)underset(|)(CH)-...

`CH_(3)-CH_(2)-underset(Br)underset(|)(CH)-CH_(3)underset(Delta)overset("alc.KOH")rarrX ` (Major)

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D

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To solve the given question, we need to identify the major product formed when the specified alkyl halide reacts with alcoholic KOH. Let's break down the steps involved in this reaction: ### Step 1: Identify the Reactant The reactant is an alkyl halide represented as: \[ CH_3-CH_2-CH(Br)-CH_3 \] This structure indicates that we have a butyl bromide (specifically, 2-bromobutane). ### Step 2: Understand the Reaction Conditions The reaction is carried out with alcoholic KOH, which is a strong base. This suggests that the reaction will proceed via an elimination mechanism (E2 mechanism) to form an alkene. ### Step 3: Mechanism of Elimination In an elimination reaction, the halogen (Br) will be removed along with a hydrogen atom from a neighboring carbon. This process will lead to the formation of a double bond between the two carbons. ### Step 4: Identify Possible Elimination Pathways 1. **Elimination from Carbon 2 (where Br is attached)**: - Removing Br and a hydrogen from Carbon 3: \[ CH_3-CH=CH-CH_3 \] This product is 2-butene. 2. **Elimination from Carbon 3**: - Removing Br and a hydrogen from Carbon 2: \[ CH_3-CH_2-CH=CH_2 \] This product is 1-butene. ### Step 5: Determine Major Product Using Zaitsev's Rule According to Zaitsev's rule, the more substituted alkene is favored as the major product. In this case: - **2-butene (from the first elimination)** is more substituted (it has 3 alkyl groups attached to the double bond). - **1-butene (from the second elimination)** is less substituted (it has only 1 alkyl group attached to the double bond). ### Conclusion The major product (X) formed from the reaction of the given alkyl halide with alcoholic KOH is: \[ X = CH_3-CH=CH-CH_3 \] (2-butene)
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