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Question
\[\ce{CH3 - CH = CH2 ->[HI][Peroxide]}\]
The major product of the above reaction is,
Options
\[\ce{I - CH2 - CH = CH2}\]
\[\ce{CH3 - CH2 - CH2I}\]
\[\begin{array}{cc}
\ce{CH3-CH-CH3}\\
|\\
\ce{I}\end{array}\]\[\begin{array}{cc}
\ce{CH3-CH-CH2}\\
\phantom{.....}|\phantom{.....}|\\
\phantom{.......}\ce{I}\phantom{....}\ce{OH}
\end{array}\]
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Solution
\[\begin{array}{cc}
\ce{CH3-CH-CH3}\\
|\\
\ce{I}\end{array}\]
Explanation:
Identify the effect of peroxide
-
The peroxide (anti-Markovnikov) or Kharasch effect occurs only with HBr.
-
It does not occur with HCl or HI, because the free-radical mechanism is not energetically favorable for these acids.
With HI, the reaction follows the normal (Markovnikov) addition rule, not the peroxide effect.
Hydrogen (H) attaches to the carbon having more hydrogens, and iodine (I) attaches to the more substituted carbon of the double bond.
\[\ce{CH 3 - CH = CH 2 + HI -> CH 3 - CHI - CH 3}\]
This compound is 2-iodopropane.
Reason: Peroxide effect does not operate with HI, so normal (Markovnikov) addition occurs.
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\ce{CH3-CH-C2H5}\\
|\\
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(II)
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\[\begin{array}{cc}
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| benzyl halide | |
| aryl halide |
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\[\begin{array}{cc}
\ce{CH3}\phantom{.................}\\
|\phantom{...................}\\
\ce{CH3 - C - CH2 - Cl ->[Na/dry either] A}\\
|\phantom{...................}\\
\ce{CH3}\phantom{.................}
\end{array}\]
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