Which one of the following alkenes when treated with HCl yields majorly an anti Markovnikov product?
Show Hint
- \(\mathrm{Cl{-}CH{=}CH_2}\)
- \(\mathrm{F_3C{-}CH{=}CH_2}\)
- \(\mathrm{CH_3O{-}CH{=}CH_2}\)
- \(\mathrm{H_2N{-}CH{=}CH_2}\)
The Correct Option is B
Solution and Explanation
\(\mathrm{Cl{-}CH{=}CH_2}\), \(\mathrm{CH_3O{-}CH{=}CH_2}\), \(\mathrm{H_2N{-}CH{=}CH_2}\): the heteroatom has a lone pair (+M effect) that stabilises the adjacent cation \(\mathrm{X{-}\overset{+}{C}H{-}CH_3}\), so addition follows Markovnikov rule.
\(\mathrm{F_3C{-}CH{=}CH_2}\): the \(\mathrm{CF_3}\) group has a very strong \(-I\) effect, so the cation \(\mathrm{F_3C{-}\overset{+}{C}H{-}CH_3}\) is badly destabilised. \(\mathrm{H^+}\) therefore adds to the carbon bearing \(\mathrm{CF_3}\), giving \(\mathrm{F_3C{-}CH_2{-}\overset{+}{C}H_2}\), and the product is \(\mathrm{F_3C{-}CH_2{-}CH_2Cl}\), the anti-Markovnikov product.
Answer: (2) \(\mathrm{F_3C{-}CH{=}CH_2}\)
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