Which one of the following alkenes when treated with HCl yields majorly an anti Markovnikov product?

 

  • Option 1)

    CH_{3}O-CH=CH_{2}

  • Option 2)

    Cl-CH=CH_{2}

  • Option 3)

    H_{2}N-CH=CH_{2}

  • Option 4)

    F_{3}C-CH=CH_{2}

Answers (1)

(1) H_{3}CO-CH=CH_{2}

H_{3}CO-- e- releasing group

 

(2) Cl-CH=CH_{2}

Cl is a deactivating group but electrons are released by the +m effect

 

(3) H_{2}N-CH=CH_{2}

H_{2}N is e- releasing group

 

(4) F_{3}C-CH=CH_{2}

F_{3}C is e- withdrawing group

Hence :- F_{3}C-CH=CH_{2}  majorly gives antimarkonikov product.


Option 1)

CH_{3}O-CH=CH_{2}

Option 2)

Cl-CH=CH_{2}

Option 3)

H_{2}N-CH=CH_{2}

Option 4)

F_{3}C-CH=CH_{2}

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