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If a racemic mixture of 3 -methyl- 1-pentene is treated with HCl, how many different chloropentane (importantproducts only) would be formed?
    Correct answer is '5'. Can you explain this answer?
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    If a racemic mixture of 3 -methyl- 1-pentene is treated with HCl, how ...
    Introduction:
    When a racemic mixture of 3-methyl-1-pentene is treated with HCl, different chloropentane products can be formed due to the presence of a chiral center in the starting material. Chiral centers are carbon atoms bonded to four different substituents, causing the molecule to have non-superimposable mirror images. In this case, the chiral center is at the second carbon atom (numbered from the double bond end).

    Explanation:
    The reaction between 3-methyl-1-pentene and HCl proceeds through an electrophilic addition mechanism, where the electrophile is the proton (H+) from HCl. The proton is added to the carbon atom adjacent to the double bond, resulting in the formation of a carbocation intermediate. The chloride ion (Cl-) then acts as a nucleophile and attacks the carbocation, leading to the formation of different chloropentane products.

    Key points:
    The number of different chloropentane products formed can be determined by considering the possible orientations of the chloride ion attack on the carbocation intermediate. Here are the key points to understand:

    1. Chiral center: The chiral center in 3-methyl-1-pentene is the second carbon atom, which is bonded to a hydrogen atom (H), a methyl group (CH3), a propyl group (CH2CH2CH3), and an ethyl group (CH2CH3).

    2. Carbocation intermediate: The electrophilic addition of HCl to the double bond results in the formation of a carbocation intermediate. The chloride ion can attack the carbocation from either the front or the back face, leading to different products.

    3. Stereoselectivity: The attack of the chloride ion on the carbocation can occur with stereoselectivity. In this case, the attack can be either syn (on the same side as the methyl group) or anti (on the opposite side as the methyl group) with respect to the methyl group.

    Conclusion:
    Considering the chiral center, the carbocation intermediate, and the possible stereoselectivity of the chloride ion attack, it can be determined that five different chloropentane products can be formed. These products include:

    1. (R)-2-chloropentane (syn addition)
    2. (S)-2-chloropentane (syn addition)
    3. (R)-2-chloropentane (anti addition)
    4. (S)-2-chloropentane (anti addition)
    5. 3-chloropentane (resulting from rearrangement of the carbocation intermediate)

    These five products arise from the different combinations of the chiral center and the stereoselectivity of the chloride ion attack, resulting in a racemic mixture of the products.
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    If a racemic mixture of 3 -methyl- 1-pentene is treated with HCl, how many different chloropentane (importantproducts only) would be formed?Correct answer is '5'. Can you explain this answer?
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