[w451208]

How do you calculate the Number of NMR singles you will get. I was under the assumption that similar proton environments will count as one but this is not true for number 12. Why is that?

[saden99]

What did you think the answer was for #12 and I could explain from there...

However, to answer the question--that has what I assume is an ether (I'm not sure why it's drawn that way but...meh). The ether is going to effect the shielding of the protons depending on their proximity to the oxygen.

[w451208]

I thought the answer would be 5 for number 12. This is because I counted the ch2 bonds on the right as one.

[saden99]

I assume I'm looking at the ones you described and those would fit into the explanation. One of the groups is farther than the other and would be more downfield on the NMR.

[w451208]

Like this one. Why is it five when I counted 4?



Also why is this 4?

[saden99]

One way you can look at 16 based on your assumption of 4 is counting how many carbons away those "1's" are. This isn't a terribly good method...but an explanation at least. The 1's on the left are three carbons away while the "1" on the right is 2 carbons away. Again, proximity to that oxygen. On an actual NMR, the peaks won't be extremely far apart, but there will be noticeably  different peak locations between the 2 proton sets.

For the second...I'm not much of an NMR expert by an means (in fact I don't really like NMR's too much because I'm weird), but if you draw the molecule with the hydrogens and label the different hydrogens a, b and c then work out their integration you'll see that the proton on the carbon connected to the bromine interacts with both proton sets at each end.

I could be wrong on the second explanation, but again I'm sure the first has to do with proximity to the oxygen in the ether group.

[sjb]

For 16 - ethyl 3-pentyl ether, I think there will be actually two signals where you have indicated 1 for environment 2, consider diastereotopic protons. Can you draw a Newman projection or similar where these two protons have the same environment?

[orgopete]

If the problem is identifying chemically equivalent hydrogens, then the following can be a useful alternate. For hydrogens that you suspect to be equivalent, replace them with another atom, e.g., Cl or F. If the name remains the same, the hydrogens are equivalent. If the name is different, then the hydrogens are not chemically equivalent. If you replace the hydrogens of the CH3-group of ethanol with a fluorine, you will get 2-fluoroethanol for all replacements. Therefore, the hydrogens are all equivalent.

For #16, replacing just one of the methyl hydrogens with a fluorine would give,
1-fluoro-3-ethoxypentane
1-fluoro-3-ethoxypentane
3-(2-fluoroethoxy)pentane
Therefore, two of the three methyl group hydrogen atoms are equivalent and ethoxy one is not.

For #20, the CH with the Br attached is a chirality center. Therefore, replacing the hydrogens of the CH2Cl group gives another chirality center, so the hydrogens are diastereogenic and are not chemically equivalent. One is (1S) and the other (1R). I have used the same naming strategy and arbitrarily started with only one of the possibilities for the bromine. There would be an additional pair of mirror images.

[sjb]

Hmm, that's a good spot. I didn't notice the triplication of a "1" label. Do you feel that the protons at 2 are equivalent then? Seems to me you could have diastereomeric pairs of compounds?

[w451208]

Thanks for your answers. They have helped a lot. Also could you look at this one and tell me why there are 4 signals when I perform the substitution by orgopete I only get 3

[sjb]

Can you draw out the 3 compounds you've found (replace the H by, say F) ?

[w451208]

No problem.

[sjb]

Is there free rotation around all the bonds in this structure?

[w451208]

I believe so. Everyone but the double bonds.

[sjb]

I believe so. Everyone but the double bonds.

So is the structure the same if you replace the hydrogen that points down, compared to that at an angle, on the end of the double bond?