Arrange the following in incresing order of rate of Nitration :
(a)

Options

A. `b lt c lt a lt d lt e`
B. `d lt e lt a = c lt b`
C. `d lt a = c lt e lt b`
D. `a lt c lt b lt e lt d`

Answer

B

Explanation

On the basis that the rate determining step is the formation of the carbonium ion, then, with benzene as the reference compound, the rate is faster when the substituent activates the ring `(+I " and /or " + R, o//p)` and the rate is slower when the substituent activates `(-I " and /or " -R, m)`. Halogen is an slower when the substituent deactivates the ring `(-I, R, -I gt + R)`, but the orientation is `o//p`. Hence, the order of increasing rate of nitration is:
`underset(((-I, -R)))(PhNO_(2)) lt underset(-I gt +R)underset(((-I, -R)))(PhCl) lt PhH = C_(6)D_(6) lt underset((+I))(PhMe)`
Since proton removal from the carbonium ions is fast, no kinetic isotope effect is observed when H si replaced by D.Hence, the rate of nitration of PhH and `C_(6)D_(6)` are the same. Had a kinetic isotope effect been observed, this would mean that the second step had now become rate determining.