InterviewSolution
Saved Bookmarks
| 1. |
Isotopic carbon-14 in (A) appears at new position (As in B) (A) reacts with `CH_(3)Ona`. Explain. |
| Answer» Correct Answer - `CH_(3)O^(-)` (nucleophile) attacks less submituted carbon (which is C-14 in this case) forming intermediate (C). | |