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    Ans: c
    Why can't option d be the answer? There is no methyl group here. Please explain.

  • @anag first H+H^+H+ attacks on oxygen forming a +ve charge on it.
    Since there is no 3 degree or benzyllic carbon attached the mechanism will be SN2 (otherwise carbocation is formed and iodide is formed there). So I−I^-I attacks where hindrance is less. In dis case it attacks on ethyl side.

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