Give the major product of the following reaction: $2 - m e t h y l p e n t - 2 - e n e + I C l \to M a j o r p r o d u c t$

4

- c h l o r o -

- m e t h y l p e n t -

- e n e

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4

- i o d o -

- m e t h y l p e n t -

- e n e

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3

- c h l o r o -

- i o d o -

- m e t h y l p e n t a n e

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2

- c h l o r o -

- i o d o -

- m e t h y l p e n t a n e

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Solution

The correct option is A $2$ $- c h l o r o -$ 3 $- i o d o -$ 2 $- m e t h y l p e n t a n e$

In compound $I C l$ , chlorine atom is more electronegative than iodine atom. So iodine acts as a positive charge and chlorine as a negative ion.
The first double bond is broken and the addition of iodine to double bond takes place to form the most stable carbocation. In this case, tertiary carbocation is most stable. Then chlorine ion is added to form $({C H_{3})}_{2} C C l - C H I - C H_{2} - C H_{3}$

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3 - E t h y l - 2 - m e t h y l p e n t - 2 - e n e

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