72,932 views
3 votes
3 votes
Consider the reaction of 2-chloro-2-methylpentane with sodium iodide.

Assuming no other changes, how would it affect the rate if one simultaneously doubled the concentration of 2-chloro-2-methylpentane and sodium iodide?
A) No effect.
B) It would double the rate.
C) It would triple the rate.
D) It would quadruple the rate.
E) It would increase the rate five times.

User SimonVT
by
2.8k points

2 Answers

26 votes
26 votes

Step-by-step explanation:

The given reaction represents the reaction between a tertiary alkyl halide that is 2-chloro-2-methylpentane and a nucleophile that is NaI.

This reaction favors SN1 mechanism which has order one.

So, the given reaction follows first-order kinetics.

For a first-order reaction, the rate law is:

rate =k [A]

That means the rate of the reaction is dependent on the concentration of reactants.

So, when the concentration of the reactant is doubled then, the rate of the reaction is also doubled.

Among the given options the correct answer is option B) It would double the rate.

Consider the reaction of 2-chloro-2-methylpentane with sodium iodide. Assuming no-example-1
14 votes
14 votes

Answer:

Step-by-step explanation:

The reaction between 2 chloro- 2 methyl pentane and sodium iodide takes place through SN2 mechanism . iodide ion is the nucleophile which attacks the substrate . The rate of such reaction depends upon concentration of both the nucleophile and the substrate .

Hence rate of reaction will be increased by 2 x 2 = 4 times.

option D ) is correct.

User Enezhadian
by
3.4k points