[theanonymous]

The reaction depends ONLY on the concentration of the alcohol, which is why we would call this reaction unimolecular (S_N1)

Is that the answer to this question?

Almost.  It is not quite an experiment yet.  Let me ask you a question.  If you double the concentration of bromoalkane, R-Br, does the rate double for the S_N1 case?  For the S_N2 case?

If you write the two rate equations side-by-side or on two adjacent lines of text, you will see one difference between them that suggests a second experiment.  Two final points.  One I interpreted the initial question to be, "How do you discriminate (decide) between the possibility that the reaction proceeds by the SN1 versus the SN2 mechanism"  A related question is "Does this particular set of conditions favor one mechanism over the other"  I would answer the two questions differently.  Two, I think it is better to say that the nucleophile is bromide ion, Br^1-, than to say it is bromine.

Omg this is so overwheliming...i've been working on this problem for 4 hours...
Can you help me formulate the answer?

Would I just say...

If you double the concentration of bromoalkane, R-Br, the rate DOES double for the S_N1 case.
However, for the S_N2 case, the rate wouldn't double?

I'm getting more confused because I feel like I'm going in circles..
And I have like 4 other problems to finish too.

[Babcock_Hall]

For SN2 Reactions, the rate law is: Rate = k[haloalkane][nucleophile]

The conversion of 2-Butyl alcohol to 2-butyl bromide upon treatment with aqueous HBr can't be an S_N2 reaction because the haloalkane and nucleophile is both Br (?)


For SN1 Reactions, the rate law is: Rate = k[R original molecule-X the leaving group]

Oops, I made a mistake in my earlier post, which I just now realized.  Sometimes halide ions (usually Br^-1 or I^1-) become the leaving group in these reactions and sometimes they are the nucleophile.  Sorry about the confusion.

For a general S_N2 reaction
rate = k[R-X][nucleophile] (X is the leaving group)
The starting material is the same in either the S_N1 or the S_N2 case.  The product is also the same, other than stereochemistry.  The product is the bromoalkane, and it can be represented as R-Br.

[theanonymous]

So the two rate laws side by side would be...

S_N2: Rate = k[R-X][nucleophile]
S_N2: Rate = k[R-H₂O][Br^-]

S_N1: Rate = k[R-H₂O]
The rate of this reaction depends on one thing; how quickly  the leaving group can leave so that the nucleophile Br^- can squeeze in.

?

[theanonymous]

For SN2 Reactions, the rate law is: Rate = k[haloalkane][nucleophile]

The conversion of 2-Butyl alcohol to 2-butyl bromide upon treatment with aqueous HBr can't be an S_N2 reaction because the haloalkane and nucleophile is both Br (?)


For SN1 Reactions, the rate law is: Rate = k[R original molecule-X the leaving group]

Oops, I made a mistake in my earlier post, which I just now realized.  Sometimes halide ions (usually Br^-1 or I^1-) become the leaving group in these reactions and sometimes they are the nucleophile.  Sorry about the confusion.

For a general S_N2 reaction
rate = k[R-X][nucleophile] (X is the leaving group)
The starting material is the same in either the S_N1 or the S_N2 case.  The product is also the same, other than stereochemistry.  The product is the bromoalkane, and it can be represented as R-Br.

Wait is that the answer? Like for describing two experiements that could allow me to determine whether the reaction proceeeds though an Sn1 or Sn2 reaction??

[Babcock_Hall]

Omg this is so overwheliming...i've been working on this problem for 4 hours...
Can you help me formulate the answer?

Would I just say...

If you double the concentration of bromoalkane, R-Br, the rate DOES double for the S_N1 case.
However, for the S_N2 case, the rate wouldn't double?

I'm getting more confused because I feel like I'm going in circles..
And I have like 4 other problems to finish too.

I made a mistake in two of my previous comments, and I was able to edit only one of them just now (I wrote the wrong rate law).  You starting material is an alcohol, and your product is a bromoalkane, and I had it backwards.  I am multitasking, and my focus was not all that it should have been.

If I double the concentration of the alcohol, R-X, I will double the rate in either the S_N2 case or the S_N1 case.  Make sure you understand this in terms of the two rate equations before going forward.  Now what happens to the rate if I double the concentration of the nucleophile instead?

[Babcock_Hall]

So the two rate laws side by side would be...

S_N2: Rate = k[R-X][nucleophile]
S_N2: Rate = k[R-H₂O][Br^-]

S_N1: Rate = k[R-H₂O]
The rate of this reaction depends on one thing; how quickly  the leaving group can leave so that the nucleophile Br^- can squeeze in.

From the mathematical point of view, what does the S_N2 rate equation have that the S_N1 rate equation does not have?

[theanonymous]

EDIT:

Would this be a better way to formulate my answer?

Experiment 1)
If you double the concentration of the of the alcohol, R-X, you would double the rate of the reaction in either the S_N2 reaction OR the S_N1 reaction.

Experiment 2)
Reminder: The rate law for S_N2 has the concentration of the nucleophile in it: k[R-X][Br^-]
Whereas the rate law for the S_N1 just has the alcohol (original molecule with the leaving group) in it: k[R-X].

In contrast to Experiment 1, if you double the concentration of the nucleophile, the rate of the Sn2 reaction would be dependent on BOTH the alcohol AND the nucleophile.

[Babcock_Hall]

EDIT:

Would this be a better way to formulate my answer?

Experiment 1)
If you double the concentration of the of the alcohol, R-X, you would double the rate of the reaction in either the S_N2 reaction OR the S_N1 reaction.

Experiment 2)
Reminder: The rate law for S_N2 has the concentration of the nucleophile in it: k[R-X][Br^-]
Whereas the rate law for the S_N1 just has the alcohol (original molecule with the leaving group) in it: k[R-X].

In contrast to Experiment 1, if you double the concentration of the nucleophile, the rate of the Sn2 reaction would be dependent on BOTH the alcohol AND the nucleophile.

I think you have the basic idea down, and you just need to tweak the wording a little bit.  For experiment 2), you correctly say that when the concentration of nucleophile is doubled, the rate of the S_N2 reaction would double.  You should say that the rate of the S_N1 reaction would stay the same.

[theanonymous]

Sweeeeet.
Ok!

Thanks a bunch!!!

I never thought I'd be able to understand this in time. It's starting to make more sense, thanks to you...Now i have to bookmark this page

[orgopete]

Hint, if you repeated the reaction with twice the HBr present, how would that affect the rate? If the reaction were SN1, how would that affect the reaction? SN2?