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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

Respuesta :

Answer:  a. it doubles

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 [tex]SN^1[/tex] reaction and thus rate depends only on the concentration of  tert-butyl bromide.

[tex]Rate=k[tertbutylbromide]^1[/tex][base]^0

k= rate constant

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

[tex]Rate'=k[2\times tertbutylbromide]^1[2\times base]^0}[/tex]

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

[tex]Rate'=2\times Rate[/tex]

Thus the rate of reaction doubles.

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