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Preparation of alcohols from alkenes inv...

Preparation of alcohols from alkenes involves the electrophilic attack on alkene carbon atom. Explain its mechanism.

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To explain the mechanism of preparing alcohols from alkenes through electrophilic attack, we can break it down into several steps: ### Step 1: Formation of the Electrophile In this reaction, we start with an alkene, which contains a double bond (C=C). The double bond consists of a sigma (σ) bond and a pi (π) bond. The π bond is rich in electrons and can act as a nucleophile. When we introduce an electrophile, such as H⁺ (proton), it will attack the electron-rich π bond of the alkene. **Hint:** Identify the electrophile and the nucleophile in the reaction. ### Step 2: Electrophilic Attack ...
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Alkenes undergo electrophilic addition reactions. Electrophilic addition to unsymmetrical alkenes involve the formation of a more stable carbocation intermediate in the first step which is rate determining and carbocation is then attacked by a nuclephile to form product Which one of the following alkene reacts most readily with HCI ?

Alkenes undergo electrophilic addition reactions. Electrophilic addition to unsymmetrical alkenes involve the formation of a more stable carbocation intermediate in the first step which is rate determining and carbocation is then attacked by a nuclephile to form product +H_(2)O overset(H_(2)SO_(4))rarr Product

Knowledge Check

  • The isomeric alcohol which has a chiral carbon atom is:

    A
    n-butyl alcohol
    B
    iso-butyl alcohol
    C
    sec-butyl alcohol
    D
    tert-butyl alcohol

  • Which of the following alkenes is most reactive towards electrophilic addition reaction?

    A
    `H_(2)C=CH_(2)`
    B
    `CH_(3)-CH=CH_(2)`
    C
    D
    `H_(2)C=CH-Cl`

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    Alkenes undergo electrophilic addition reactions. Electrophilic addition to unsymmetrical alkenes involve the formation of a more stable carbocation intermediate in the first step which is rate determining and carbocation is then attacked by a nuclephile to form product HO-CH_(2)-CH_(2)-CH_(2)-CH=CH_(2)underset(NaHCO_(3))overset(I_(2))rarr major product formed is

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