| I. Formation of A has proceeded with racemisation. |
| II. Formation of B has proceeded with inversion. |
A) I and It
B) Only I
C) Only III
D) None of these
Correct Answer: A
Solution :
\[{{S}_{N}}1\]reaction proceed through formation of carbocation which can form a pair of enantiomers. Thus, recemisation takes place (statement I).\[{{S}_{N}}2\]reaction takes place forming pantavalent intermediate which loses\[C{{l}^{-}}\]giving inverted alcohol. (8tatement II) Hence, statements I and II both are correct.
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