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In principle, how many different alkyl chlorides would be formed on treatment of 3-methyl-2-pentanol with concentrated HCl in the presence of ZnCl2?
    Correct answer is '5'. Can you explain this answer?
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    Alkyl chlorides formation from 3-methyl-2-pentanol with concentrated HCl in the presence of ZnCl2

    When 3-methyl-2-pentanol is treated with concentrated HCl in the presence of ZnCl2, it undergoes an acid-catalyzed nucleophilic substitution reaction to form alkyl chlorides. The number of different alkyl chlorides formed depends on the number and positions of the available hydrogens in the starting alcohol.

    Step 1: Determining the number of available hydrogens
    To determine the number of available hydrogens, we need to identify the positions where the hydrogens can be replaced by chlorine atoms. In 3-methyl-2-pentanol, there are three different types of hydrogens:

    1. Primary hydrogens (attached to the carbon bonded to the hydroxyl group): There are no primary hydrogens in 3-methyl-2-pentanol.

    2. Secondary hydrogens (attached to the carbon adjacent to the one bonded to the hydroxyl group): There are two secondary hydrogens in 3-methyl-2-pentanol.

    3. Tertiary hydrogens (attached to the carbon two carbons away from the one bonded to the hydroxyl group): There are three tertiary hydrogens in 3-methyl-2-pentanol.

    Step 2: Substitution reactions
    Based on the number and positions of the available hydrogens, we can determine the possible substitution reactions that may occur. In this case, the following reactions can take place:

    1. Substitution of a secondary hydrogen: This leads to the formation of a secondary alkyl chloride. There are two secondary hydrogens, so two different secondary alkyl chlorides can be formed.

    2. Substitution of a tertiary hydrogen: This leads to the formation of a tertiary alkyl chloride. There are three tertiary hydrogens, so three different tertiary alkyl chlorides can be formed.

    Step 3: Counting the different alkyl chlorides
    Adding up the different alkyl chlorides formed from the substitution reactions gives us a total of 5 different alkyl chlorides:

    2 secondary alkyl chlorides + 3 tertiary alkyl chlorides = 5 different alkyl chlorides.

    Therefore, when 3-methyl-2-pentanol is treated with concentrated HCl in the presence of ZnCl2, a total of 5 different alkyl chlorides are formed.
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