[Violagirl]

Hi, I had a few questions from a lab we did last week with analysing the peaks of 2-methyl-1-butene and 2-methyl-2-butene. I know that 2-methyl-2-butene is the favored product and that 2-methyl-1-butene is more volatile. My one question, since 2-methyl-1-butene is the volatile substance, would it be considered kinetic? And vice versa for 2-methyl-2-butene being thermodynamic? Or do I have that backwards?

2. For a reaction of 2-methyl-2-butanol and sulfuric acid, a student instead uses hydrochloric acid instead while following the same procedure. During the final distillation of product, the student finds the boiling point of the liquid to be at 85 degrees C and the GC analysis shows a large peak at 2.5 min. What would be the explanation for these observations?

3. Using NMR to determine the ratio of 2-methyl-1-butene to 2-methyl-2-butene, one alkene has a singlet appear at 4.69 pp and the other contains a quartet like peak at 5.2 ppm. Assign these peaks to the correct structrue and determine how many Hs are represented by each peak?

[Violagirl]

I thinik for number two, your main diffrence really is that you'll end up with different retention times using a different catalyst for the reaction as the reaction is taking place under different condiitions then it would be if HCl had been used. Right?

Also would 2-methyl-1-butene be considered the kinetic product as it has the higher boiling boint and most likely will be the favored product at higher temperatures? Vice versa for 2-methyl-2-butene too then.

Number 3, I have still not had any luck on it...

[Nosterius]

Hi Violagirl,

Although this seems ilke an analytical chemistry lab, you should look at it from an organic chemistry point of view. The reaction you are doing is a dehydration of an alcohol, which has to go through a carbocation intermediate (see http://en.wikipedia.org/wiki/Alcohol#Dehydration ). From this carbocation, you can form the alkene by eliminating different hydrogen atom.

If you are using hydrochloric acid instead of sulfuric acid, one of the principal difference is the identity of the conjugate base. Chloride anions are more nucleophilic than sulfate anion (or hydrogenosulfate), which means that it can react with your carbocation in a different way.

In your case, the thermodynamic product is directly related to the stability of the alkene form. Do you have any clue of which one it is, and why? I don't think it has a direct link with the boiling point though (correct me if I'm wrong).

[Violagirl]

Hi thanks so much for your input! The more stable product is 2-methyl-2-butene as it's more substituted and follows Zaitsev's rule. Hence, it'll be thermodynamic then. And that clears up my confusion with the HCl over H2SO4 as the solvent instead.

For last problem:

Using NMR to determine the ratio of 2-methyl-1-butene to 2-methyl-2-butene, one alkene has a singlet appear at 4.69 pp and the other contains a quartet like peak at 5.2 ppm. Assign these peaks to the correct structrue and determine how many Hs are represented by each peak?

I have had a lot of difficulty with assigning peaks related to structures. I know you have to look at your neighboring hydrogens to determine the number of peaks that will occur. For this one though, it seems like the number of peaks should be similar...

[fledarmus]

Kinetic and thermodynamic products don't have anything to do with volatility - they have to do with stability. The kinetic product is the one which goes through the lower energy transition state, which means it takes less energy to get the reaction to go. Consequently, the reaction forming the kinetically favored product would go faster. The thermodynamic product is the one which which has the lower-energy product, which means it is more stable.

If one product is more volatile than the other, you can drive an equilibrium reaction by boiling off the more volatile product. By LeChatelier's principle, if you remove one of the products of a reaction equilibrium, you drive the reaction to form more of that product, and eventually your entire reaction would go to the product that you are removing.

For question 2, think about the products that you are forming from your reaction. What reactions can they undergo?

For question 3, how many protons can couple with the alkene protons in 2-methyl-1-butene? In 2-methyl-2-butene? What multiplicities would that give you?

[Violagirl]

So I tried applying the N+1 rule to both products but still seem to be having trouble...For the 2-methyl-1-butene, I got what I think was 2 neighbors but that would result in a triplet I think...And for the 2-methyl-2-butene, I also got 3 so I'm not sure what I'm not doing right.

[Violagirl]

Oh! I read where you wrote what the neighbors for alkene are! So for 2-methyl-2-butene, that would be quartet right as there are 3 surronding methyl groups and then applying N+1, you get a quartet. Is this right??

[fledarmus]

No, you are looking at the number of hydrogens attached to a carbon adjacent to a proton attached to your alkene. Try this:

1) Draw your structures carefully, explicitly adding each hydrogen

2) in 2-methyl-1-butene, how many hydrogens are attached to the carbons which form part of the double bond?

3) in 2-methyl-2-butene, how many hydrogens are attached to the carbons which form part of the double bond?

4) for the hydrogens attached to a carbon forming part of the double bond in 2-methyl-1-butene, how many carbons are attached directly to that carbon? How many hydrogens are attached to them?

5) for the hydrogens attached to a carbon forming part of the double bond in 2-methyl-2-butene, how many carbons are attached directly to that carbon? How many hydrogens are attached to them?

If you drew your structures carefully in step 1 and work your way step by step through the molecule, I think you will find the answer  

[Violagirl]

So would the 2-methyl-2-butene be my quartet peak then? It has 2 C's and 1 H to the carbons attached to the double bonds.

[fledarmus]

Remember, it is a proton NMR. All you are concerned about is protons. As you said, 2-methyl-2-butene has one proton attached to a carbon with a double bond. Look at the two carbons attached to that carbon. How many protons are attached to them? And are those protons identical?

[Violagirl]

I'm looking at it again but must not be seeing it. It sounds like I'd have it the other way around then?

[Violagirl]

So I turned in the assignment for this problem but wanted to see if I could get another explanation for it. I was wondering if you look at the neighbors of the carbons containing the double bond to get an idea of the number of peaks that would appear? I'm not used trying to determine peaks at all.

[fledarmus]

Since you have already turned in the assignment 

Here is 2-methyl-2-butene.


If you look at the double bond, only one of those carbons (carbon 3, following the IUPAC numbering, the one on the left in the picture) has a proton. The NMR signal from that proton would be split by protons on the neighboring carbons, carbon 2 (which is the other carbon in the double bond, to the right) and carbon 4. Carbon 2 doesn't have any protons, which means the protons on carbon 3 would be split only by those on carbon 4. Carbon 4 has three identical (by rotation) protons, so the proton on carbon 3 would be split into a quartet (N+1). This would be the "quartet like peak at 5.2 ppm" and one H would be represented by the peak.

I couldn't find a picture of 2-methyl-1-butene that had all the protons, but for the purposes of this example, this picture of 2-methylpropene should be sufficient:

You just have to imagine an extra methyl group on one of the two arms on the right.

If you look at the double bond on this molecule, you will see that carbon 1 (the one on the left) has two protons. They are not quite in identical magnetic environments, because one is cis to an ethyl group and the other trans, but they are so close together that instead of splitting each other, they fall on top of each other and appear to be a singlet (unless you have a really powerful magnet!). They would be split by any protons on adjacent carbons, but there is only one adjacent carbon, carbon 2 which is at the other end of the double bond, and it doesn't have any protons. The protons further out, on the two methyl groups (actually a methyl group and an ethyl group in our vision), are too far away to have an effect. Consequently, all you see is a singlet, but the peak represents two H's. So this would be the "singlet appear[ing] at 4.69 ppm" and would represent two protons.

[Violagirl]

I feel like I understand it a lot more now. Thanks for your *delete me*