InterviewSolution
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InterviewSolution
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Consider the following reactions. Which staements(s) is/are wrong.A. The product by path `(I)` is `Me - CH = CH_(2) (I)`B. The product by path `(II)` is `Me - CH (OE) Me (II)`C. The products are mixture of `(I)` and `(II)` by both paths.D. Path `I` proceeds cia `E2` mechanism, while path `II` proceeds via `SN^(1)` mechanism. |
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Answer» Correct Answer - `c` Statement (c) is wrong. i. In path `(I), EtP^(o-)` is a strong base and with `2^(@) RX` groups. The `E2` product predominates over the `SN^(2)` product to give `(Me - CH = CH_(2))` ii. In path `(II), EtOH` is a weak base, but a better nuclophilc so `SN^(-1)` reaction is favored to give `MeCH(OEt)Me` |
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