EK Chemistry 1001 #242

[mtravis2190]

Hi everyone,
The question I have refers to the following reaction mechanism
2-iodo-2-methylpropane---->2-methylpropane cation + iodide
2-methylpropane cation + water---->protonated 2-methylpropanol
protonated 2-methylpropanol+water---->2-methylpropanol+hydronium

Question 242 asks- If the reaction is first order in 2-iodo-2-methylpropane, this tell us that
a. The first step is most likely the slowest.
b. The first step is most likely the fastest.
c. The reaction would give a higher yield in the presence of a catalyst.
d. The reaction is not reversible.
Answer-a.

I don't understand how knowing the order of a reactant indicates whether the step in question is the fastest or slowest. I feel like I'm missing a very major point here.

Look forward to having this clarified! Thank you, all!

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[cjcarter]

A better way to approach this problem is to realize you are forming a tertiary carbocation. This is an Sn1 reaction, and formation of carbocations are usually the rate determining step/slowest step. I think organic chem knowledge will help you out with this question.

 

[DrknoSDN]

A better way to approach this problem is to realize you are forming a tertiary carbocation. This is an Sn1 reaction, and formation of carbocations are usually the rate determining step/slowest step. I think organic chem knowledge will help you out with this question.

I agree with cj, carbocation formation is going to be the slow step almost always. Knowing that would make this question much easier.

However, you don't have to know that to make a decent educated guess. If you were not given the complete reaction but instead given:
A > B
B > C
C > D

If you are told that (A > B) is first order then you should still assume it is the slow step when guessing. This is because it could only be the fast step if both other steps were also first (or zero) order and both other steps had a slower rate constant. In the answer choices the words "most likely" allow guessing for this question because of the other unknowns.

Also C and D can be eliminated because a catalyst does not effect yield, also all reactions are reversible to an extent. Some just much more unlikely due to the product having a much better (lower/more stable) energy. ...[quantum mechanics: very unlikely things still occur] =D

 

[mtravis2190]

Thanks so much! Make sense now!

 

[SweetBurger]

I'll try to explain it from a non organic chemistry standpoint.
It says that the reaction is first order in 2-iodo-2-methylpropane so we know that 2-iodo-2-methylpropane is part of the reaction rate and the reaction rate only includes reactants from the rate limiting step so you can know from that it is choice a.

 

[mtravis2190]

Wow! You made it so much easier than I was making it out to be! Thank you!