Question

Consider the reaction NaCH3COO (sodium acetate in acetic acid) with tert-butyl bromide (2-bromo-2-methylpropane). If the concentration of both the nucleophile/base and the substrate are doubled, what happens to the rate of reaction?

a. it doubles
b. it quadruples
c. new rate = 1/2 of original rate
d. it does not change
e. new rate = 1/4 of original rate

Answer 1

Answer: a. it doubles

Step-by-step explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

As the given halide is tertiary halide and the base is strong , it undergoes
`SN^1` reaction and thus rate depends only on the concentration of tert-butyl bromide.

`Rate=k[tertbutylbromide]^1`[base]^0

k= rate constant

If concentration of both the base and the substrate are doubled ,

`Rate'=k[2* tertbutylbromide]^1[2* base]^0}`

`Rate'=k2^1[tertbutylbromide]^1[base]^0` (2)

`Rate'=2* Rate`

Thus the rate of reaction doubles.