[discodermolide]

No, although that is the ultimate reaction.
The OH of the alcohol accepts a proton from the HBr. Then it leaves as H₂O to give a secondary carbocation.
What next?

Read this page

http://www.organic-chemistry.org/namedreactions/nucleophilic-substitution-sn1-sn2.shtm


If you wish me to draw the mechanism I will.

[theanonymous]

Hint: What factors affect the rate of S_N1 and S_N2 reactions? Think about the differences.

Alcohol structure has an effect on reaction rate

S_N1 - a unimolecular reaction (only one thing undergoes a chemical change); tertiary and secondary carbocations
S_N2 - a bimolecular reaction (two things undergoe a chemical change; methyl and primary carbocations

?

Right according to your definition above the conversion of butane-2-ol to 2-bromobutane must be a unimolecular process. That is S_N1.
So what do you think is the mechanism for this reaction, what happens first?
Read this page carefully:
http://en.wikipedia.org/wiki/SN1_reaction

Google SN1 reaction conditions and see what you get

I did and there's a lot of stuff...

hate my life.

[theanonymous]

No, although that is the ultimate reaction.
The OH of the alcohol accepts a proton from the HBr. Then it leaves as H₂O to give a secondary carbocation.
What next?

Read this page

http://www.organic-chemistry.org/namedreactions/nucleophilic-substitution-sn1-sn2.shtm


If you wish me to draw the mechanism I will.

I would greatly appreciate that!!

[discodermolide]

Don't panic, it's really simple.
Here is my mechanism. Others may not like it but it is simple enough.

So what about the experimental conditions to distinguish between S_N1 and S_N2?

[theanonymous]

Don't panic, it's really simple.
Here is my mechanism. Others may not like it but it is simple enough.

So what about the experimental conditions to distinguish between S_N1 and S_N2?

Sn1
- the rate-determining step is unimolecular
- Sn1 reaction takes place in  three steps

That's it?

[discodermolide]

No it is not three steps. The intermediate in brackets is only there to show you it's structure. It probably does not exist in the true sense of the word.
But I asked you how you would distinguish between S_N1 and S_N2 reactions with respect to the reaction conditions. This is also a part of your question. The info is in the links I sent you.

[Babcock_Hall]

For S_N2 reactions of haloalkanes, the rate law is:
rate = k[haloalkane][nucleophile]
The "2" means bimolecular (first order in the haloalkane and first order in the nucleophile)

I suggest you try this:
1.  Identify the haloalkane and the nucleophile in your example.
2.  Write the rate law for the S_N1 reaction.
rate = ?
Hint:  what does the "1" tell you?
EDT
I see that you have made a good start on S_N1 reactions in that you know that they are unimolecular.  Now you just have to translate the difference between unimolecular and biomolecular reactions into an experiment.

[theanonymous]

No it is not three steps. The intermediate in brackets is only there to show you it's structure. It probably does not exist in the true sense of the word.
But I asked you how you would distinguish between S_N1 and S_N2 reactions with respect to the reaction conditions. This is also a part of your question. The info is in the links I sent you.

Sn2 reactions always proceed by inversion of configuration; two different molecules in a reaction; subsitition of 2 nucleophiles.

Is that it?

Omg at this point I have no idea.
I just keep getting the wrong answer.

Can you please tell me what the answer is?

[Babcock_Hall]

Sn2 reactions always proceed by inversion of configuration; two different molecules in a reaction; subsitition of 2 nucleophiles.

So far, so good.  Do S_N1 reactions proceed with inversion?  If not, what do you observe?

[theanonymous]

Ohhhhhh!!!!!!
No SN1 reactions don't proceed with inversion!! Racemization occurs right (?)

S_N1 Reactions
1) The leaving group attached to the central carbon (C) leaves - a carboncation (C⁺) is formed
2) The nucleophile comes in and attacks the carbocation (C⁺). Nucleophile is free to attack from EITHER side; racemization occurs - one enantiomer is converted into two different versions of the molecule (R and S); half the product has retained the original configuration, and about half is inverted.




S_N2 Reactions
1) Two molecules are involved in the actual transition state
2) 1 leaving group departs while 1 nucleophile comes in and attacks the backside of the molecule
3) Result - inversion of configuration at the stereocenter

[theanonymous]

Sn2 reactions always proceed by inversion of configuration; two different molecules in a reaction; subsitition of 2 nucleophiles.

So far, so good.  Do S_N1 reactions proceed with inversion?  If not, what do you observe?

So how would I go about "describing two experiments/measurements you could make to determine whether the reaction proceeds via an S_N1 or an S_N2 reaction"?

Or is the answer just the conditions I posted a few seconds ago?

[Babcock_Hall]

One way is to determine the stereochemistry of the product to see whether the product has inverted stereochemistry or is a racemic mixture.  I think that doing an experiment with plane-polarized light would be the easiest way to obtain an answer.  One down, one to go.

I wrote the rate equation for an S_N2 reaction in an earlier message, and the rate equation for S_N1 reactions looks similar but not identical.  It might be helpful to remember that rate constants (symbolized by the letter k) are just the proportionality constant between the rate of the reaction and the concentration(s) of reactant(s).  For S_N1 reactions, the rate of the reaction is
rate = ?

[theanonymous]

One way is to determine the stereochemistry of the product to see whether the product has inverted stereochemistry or is a racemic mixture.  I think that doing an experiment with plane-polarized light would be the easiest way to obtain an answer.  One down, one to go.

I wrote the rate equation for an S_N2 reaction in an earlier message, and the rate equation for S_N1 reactions looks similar but not identical.  It might be helpful to remember that rate constants (symbolized by the letter k) are just the proportionality constant between the rate of the reaction and the concentration(s) of reactant(s).  For S_N1 reactions, the rate of the reaction is
rate = ?

For SN2 reactions of haloalkanes, the rate law is:
rate = k[haloalkane][nucleophile]
The "2" means bimolecular (first order in the haloalkane and first order in the nucleophile)

I suggest you try this:
1.  Identify the haloalkane and the nucleophile in your example. Haloalkane and nucleophile is Br (?)
2.  Write the rate law for the SN1 reaction.
rate = k[original molecule with the leaving group attached to it]
Hint:  what does the "1" tell you?
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?

[theanonymous]

So my answer to this whole question would be.

1) We can determine the stereochemistry of the product to see whether the product formed has its configuration inverted (S_N2 reaction) or if it is a racemic mixture (S_N1 reaction).

2) We can also determine whether or not the reaction is unimolecular or bimolecular by writing out the rate law.

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]

The reaction depends ONLY on the concentration of the alcohol, which is why we would call this reaction unimolecular (SN1). Therefore, the conversion of 2-Butyl alcohol to 2-butyl bromide upon treatment with aqueous HBr is an S_N1 reaction.

[Babcock_Hall]

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 R-X, 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 S_N1 versus the S_N2 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.