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Choose the Best Answer · Q13

Q.Major product of the below mentioned reaction is,
(CH3)2C=CH2 →ICl\xrightarrow{ICl} ?

(a) 2-chloro-1-iodo-2-methylpropane
(b) 1-chloro-2-iodo-2-methylpropane
(c) 1,2-dichloro-2-methylpropane
(d) 1,2-diiodo-2-methylpropane
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Step 1. In the interhalogen ICl, chlorine (electronegativity 3.16) is more electronegative than iodine (2.66), so the bond polarises as I(delta+)-Cl(delta-) -- iodine plays the electrophile's role here, analogous to H+ in HBr addition.

Step 2. Markovnikov's rule places the electrophile (I+) on the double-bond carbon with MORE hydrogens (the terminal =CH2 of (CH3)2C=CH2, which has 2 H's) and generates a carbocation at the OTHER, more substituted carbon.

Step 3. That carbocation forms at the carbon already bearing two methyl groups -- a TERTIARY carbocation, (CH3)2C+-CH2I -- clearly more stable than the alternative (a primary carbocation at the terminal carbon, which the reaction therefore avoids). …

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