A) \[\frac{d[B{{r}_{2}}]}{dt}=\frac{5}{3}\frac{d[B{{r}^{-}}]}{dt}\]
B) \[\frac{d[B{{r}_{2}}]}{dt}=\frac{3}{5}\frac{d[B{{r}^{-}}]}{dt}\]
C) \[B{{F}_{3}},NO_{2}^{-},NH_{2}^{-}\]
D) \[{{H}_{2}}O,\]
Correct Answer: D
Solution :
Key Idea \[=\frac{1000}{100}=10\,\text{m}{{\text{s}}^{-1}}\] in presence of P attacks on \[{{v}_{s}}=10+10=20m{{s}^{-}}^{1}\]carbon atom of acid and \[T=\frac{2\pi r}{v}=\frac{2\pi }{v}\times \frac{mv}{Bq}\] hydrogen is substituted by bromine. \[T=\frac{2\pi m}{Bq}\] This reaction is called Hell-Volhard Zelinsky (HVZ) reduction.You need to login to perform this action.
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