do the protons of alcohols show up in nmr spectra?

[thebillsfan]

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see subj

 

[crazybob]

in the proton NMR spectra, they should. http://www.cem.msu.edu/~reusch/VirtualText/Spectrpy/nmr/nmr1.htm

 

[thebillsfan]

in the proton NMR spectra, they should. http://www.cem.msu.edu/~reusch/VirtualText/Spectrpy/nmr/nmr1.htm

yeah i shouldve known that.

if you know anything about gas chromatography: why do substances with a higher boiling point have a longer retention time in the apparatus?

yes, i was actually asked this question! who knew?

 

[crazybob]

http://orgchem.colorado.edu/hndbksupport/GC/GC.html

Retention Time (RT)

The retention time, RT, is the time it takes for a compound to travel from the injection port to the detector; it is reported in minutes on our GCs. The retention time is measured by the recorder as the time between the moment you press start and the time the detector sees a peak. If you do not press start at the same time you inject your sample, the RT values will not be consistent from run to run.

Factors which affect GC separations

Efficient separation of compounds in GC is dependent on the compounds traveling through the column at different rates. The rate at which a compound travels through a particular GC system depends on the factors listed below:

• Volatility of compound: Low boiling (volatile) components will travel faster through the column than will high boiling components
• Polarity of compounds: Polar compounds will move more slowly, especially if the column is polar.
• Column temperature: Raising the column temperature speeds up all the compounds in a mixture.
• Column packing polarity: Usually, all compounds will move slower on polar columns, but polar compounds will show a larger effect.
• Flow rate of the gas through the column: Speeding up the carrier gas flow increases the speed with which all compounds move through the column.
• Length of the column: The longer the column, the longer it will take all compounds to elute. Longer columns are employed to obtain better separation.

Generally the number one factor to consider in separation of compounds on the GCs in the teaching labs is the boiling points of the different components. Differences in polarity of the compounds is only important if you are separating a mixture of compounds which have widely different polarities. Column temperature, the polarity of the column, flow rate, and length of a column are constant in GC runs in the Organic Chemistry Teaching Labs. For each planned GC experiment, these factors have been optimized to separate your compounds and the instrument set up by the staff.

 

[MCATBear]

As far as GC question goes, I'm pretty sure it's just that the apparatus heats a certain # of degrees per second, and the more volatile substances vaporize at a lower temperature, so they do it sooner. Higher boiling points don't vaporize until higher temperature (when more time has gone by).

 

[loveoforganic]

Sorry, that's not right at all. In GC, you want to vaporize your entire sample instantly. The gassified sample then passes through a column (basically a long tube) with the help of an inert gas such as helium. The column is packed with a (typically polar) substance. Different compounds in the sample have different affinities for the column packing and thus elute at different rates.

 

[SuperSaiyan3]

Does anybody know any specific NMR proton values that we should memorize for MCAT day?

I memorized that:

vinyl: 5-6
aromatic: 6-8
aldehyde: 9-10
carboxylic acid: 10-12

any others?

 

[boaz]

see subj

Yes, it will show up on the NMR, BUT keep in mind that it will not be involved in splitting.

 

[boaz]

yeah i shouldve known that.

if you know anything about gas chromatography: why do substances with a higher boiling point have a longer retention time in the apparatus?

yes, i was actually asked this question! who knew?

The "higher boiling point" is probably meant to mean that it is a polar molecule, for example ethanol vs. propane. The more polar the substance the less it is attracted to the stationary phase, the less time it spends in that phase, the earlier it leaves the column.

Asking the question that way is bery unusual; "boiling point", by definition, has nothing to do with retention time.

 

[thebillsfan]

the link that crazy bob gave makes it seem like the higher bp is important for retention time in GC for a different reason than polarity. it doesnt really explain why....why would low boiling compounds move faster than high boiling compounds? if both compounds are vaporized, it shouldnt matter what their BP is.

 

[boaz]

the link that crazy bob gave makes it seem like the higher bp is important for retention time in GC for a different reason than polarity. it doesnt really explain why....why would low boiling compounds move faster than high boiling compounds? if both compounds are vaporized, it shouldnt matter what their BP is.

I stand corrected. I can't believe I didn't learn about this in my analytical chemistry class. Polarity was constantly emphasized with no mention of b.p.

I think the reason is that the larger dispersion forces in the larger molecules result in electronic interaction with the stationary phase.

 

[wanderer]

see subj

Yes, unless they're deuterated.

 

[thebillsfan]

ishchayll...are you assuming, then, that compounds with a higher BP have larger dispersion forces? I mean, the polarity thing would overshadow dispersion forces anyway, so i dont see how BP could really be that important if its only contribution is more VDW forces

 

[boaz]

ishchayll...are you assuming, then, that compounds with a higher BP have larger dispersion forces? I mean, the polarity thing would overshadow dispersion forces anyway, so i dont see how BP could really be that important if its only contribution is more VDW forces

If there is a significant difference in polarity of the two compounds, then polarity will also determine retention time. If there is only a difference in molecular size and little/no difference in polarity, you will still see a difference in retention time--because of size. Decane and propane have the same polarity but they will have very different retention times because of size.

 

[sleepy425]

as far as the original question is concerned (about alcohol protons showing up in NMR spectra), it depends on the NMR solvent used. The alcohol proton will usually show up if a non-exchanging deuterated solvent is used. All of the organic solvents (DMSO-d6, CDCl3, etc) are non-exchanging, so in these solvents you'll see the alcohol protons. D2O (water), on the other hand, is an exchanging solvent, which means that the alcohol protons will exchange with the deuterons very rapidly so you won't see it in the NMR spectrum. Deuterated alcohol solvents also exchange with alcohol analytes.

amines and amides also exchange rapidly in exchanging solvents. the alpha protons of ketones and aldehydes exchange as well, but much more slowly, and it is easily possible to get a nearly perfect NMR spectrum in D2O of most of these. Also, if you keep taking the NMR spectrum every hour for a little while, you'll see the alpha proton peak start to disappear as exchange occurs. Finally, doubly activated alpha protons exchange rapidly. For example, look at oxaloacetic acid. If you take a spectrum of oxaloacetic acid in D2O, you get 0 peaks. Why? Because the two carboxylic acid protons exchange rapidly, and the two alpha protons are relatively acidic and also exchange rapidly, leaving the molecule fully deuterated. However, if you take the spectrum of oxaloacetic acid in DMSO-d6, you get peaks.

as ishchayill mentioned, somtimes alcohol and amine protons do not participate in splitting (spin-spin coupling). There is nothing fundamentally different between a heteroatomic proton and a neighboring methylene proton except exchange. Rapid exchange can sometimes lead to loss of the splitting. However, this does not always happen and I have definitely seen split amine protons.

 

[sleepy425]

whoaaaa, ok, for GC, everyone is kind of making the polarity thing out to be more than it is. Column polarity is generally a very minor portion of the GC separation. GC primarily separates based on relative vapor pressure of compounds in the column. The higher the compound's normal boiling point, the lower its vapor pressure will be. The idea is that your sample is vaporized and enters the column. The sample will partition between the gaseous mobile phase and the column material (stationary phase) depending on the compound's vapor pressure under the column conditions used. If a compound has a low vapor pressure, it will partition more with the stationary phase, so it will move more slowly. If the compound has a high vapor pressure, it will partition more with the mobile phase, and will elute quickly. That's why compounds with high boiling points (low vapor pressure) will elute more slowly since they are partitioned more greatly in the stationary phase, while compounds with low boiling points (high vapor pressure) will elute more quickly, since they are partitioned more greatly in the mobile phase.

 

[loveoforganic]

Hm, so are the transitions to the stationary phase essentially transitions back to the liquid phase?

 

[sleepy425]

Hm, so are the transitions to the stationary phase essentially transitions back to the liquid phase?

Yeah, the stationary phase of the GC column is a liquid. So yes, the transitions are to the liquid phase. The transitions are rapid, so all of the molecules of a given analyte are constantly exchanging. However, at any given time, a certain fraction of the molecules are in the mobile (gas) phase and a certain fraction are in the stationary (liquid) phase. This fraction depends primarily on the boiling point of the analyte. Molecules with a higher fraction in the gaseous mobile phase will travel through the column faster.

 

[Hemichordate]

So how exactly is carbon NMR different from H-NMR? Ek doesn't really explain it.

 

[wanderer]

So how exactly is carbon NMR different from H-NMR? Ek doesn't really explain it.

Carbon NMR tells you how many different kinds of carbon there are in a compound. It is especially useful in a case where a carbon does not have a hydrogen directly attached to it (like if it's quaternary), since there will be no signal for that carbon on an H-NMR. It is generally not as useful as H-NMR because there is no signal splitting in C-NMR.

 

[sleepy425]

So how exactly is carbon NMR different from H-NMR? Ek doesn't really explain it.

Actually, they're not very different. They use the same basic principles, except in 1H NMR, the nucleus under study is a proton, and in 13C NMR, a C-13 nucleus is under study. In 13C spectra, each unique carbon will show up as a peak. Just like proton spectra, 13C (and all other types of NMR) exhibit chemical shift which is affected by electron density.

A very important difference between 1H and 13C spectra is that the proton spectrum uses the major isotope of hydrogen as the nucleus, while the 13C spectrum uses a minor isotope of carbon, C-13, as the isotope. So the 13C NMR signal is much much weaker than the 1H NMR signal. The consequence is that you either have to have a much more concentrated sample for the 13C measurement, or you have to take more scans for a 13C spectrum (when you take a 1H NMR spectrum, you typically take 8 or 16 repeated scans and the computer averages them. This usually takes a minute or two. when you take a 13C NMR, you might take 1000 repeated scans. This takes a lot longer, which is why many people run 13C NMRs overnight and pick up the results in the morning).

Most 13C spectra are obtained as "decoupled" spectra. Here's what that means: Just as 1H spectra are split by spin-spin coupling with neighboring protons, 13C spectra can be split by spin-spin coupling with both 1H and 13C nuclei. Since splitting of C13 spectra due to neighboring protons can clutter the spectrum too much, a special pulse of RF radiation is used that excites all of the protons in the molecule so they can't cause spin-spin coupling.

As far as 13C-13C splitting, this is not observed because the abundance of the 13C nucleus is so low that the likelihood of a single molecule possessing two C-13 nuclei is negligible.

The result of this is that you don't see splitting in typical 13C spectra like you do in 1H spectra.

You can also integrate 13C peaks to see how many identical carbons of each type you have, just like 1H spectra.

 

[thebillsfan]

nice, sleepy. also note that while the chance of having two c13 in one molecule is low, the chances of having two c13 ADJACENT to each other in the same molecule is even lower!