A) \[=9\]
B) \[{{K}_{b}}=C{{\alpha }^{2}}\]
C) \[\therefore \]
D) \[\alpha =\sqrt{\frac{2.5\times {{10}^{-5}}}{5\times {{10}^{-4}}}}=0.22\]
Correct Answer: C
Solution :
Product of reaction given in option (c) is, as \[(n+l)\] group present in toluene, activate the benzene nucleus towards electrophilic substitution due to its electron releasing nature.You need to login to perform this action.
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