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Isotopic carbon-14 in (A) appears at new...

Isotopic carbon-14 in (A) appears at new position (As in B) (A) reacts with `CH_(3)Ona`. Explain.

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The correct Answer is:
`CH_(3)O^(-)` (nucleophile) attacks less submituted carbon (which is C-14 in this case) forming intermediate (C).
`(##ALN_CHM_C09(II)_S01_427_A01##)`
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