Lecture Outline Alcohols

Lecture outline Alcohols — R—O—H shows up at δ 2 - 5 depending on concentration — usually appears as a singlet, often broad (i.e. broader than the peaks of Hs attached to Cs); – O–H protons normally do not couple to neighboring Hs because exchange of Hs among ROH molecules (and traces of water) is fast (≈ 10–5 sec). H H C H O H singlet (usually); often broad (br) no coupling (usually) In general, the NMR signal of a proton arises from a time-average of all the environments the proton "feels" in the time required for the instrument to measure the frequency of NMR transition (≈ 10–2 to 10–3 sec) Recall that splitting arises from the small perturbations on the magnetic field caused by the various combinations of neighboring α and β spins. Each neighboring spin is either α or β during the timescale of the NMR transition, so each combination gives a separate, sharp peak of a multiplet. If this were the case in methanol, the CH3 protons would appear as a doublet. But under ordinary conditions the H on oxygen rapidly exchanges with other such Hs, so lots (100 or 1000) of different Hs will have spent a little time on a particular O during the timescale of the CH3 protons' NMR transition. So those methyl protons "feel" a magnetic field perturbed, not by either one α or one β spin sitting on the O, but by many neighboring spins, about half of which are α and half β. The result is no net effect on the methyl Hs' NMR signal. The same is true for the signal of the hydroxyl H — each one of these Hs visits hundreds of methanol molecules in the time required for the NMR transition, so each one feels the cumulative effects of many different methyl H spin combos — again, the splitting is "averaged out", and the –OH appears as a singlet. In dilute solutions of very pure alcohols or at low temperature, exchange is slowed, and splitting (i.e. coupling between the H on the O and the Hs on the neighboring C) can be observed, therefore, the signals of methanol would be a methyl doublet and an –OH quartet. Thus, the splitting (or lack thereof) is a function of the rate of –OH proton exchange — this is one of many examples of dynamic effects in NMR (more to follow...) Now for a brief (though important) digression concerning deuterium and NMR solvents... The NMR experiment is usually done be dissolving a few mg of compound in about 300 µl of solvent in a 5-mm o.d. thin-walled (fragile) glass tube. If one used a "normal" (protio) solvent, like CHCl3, benzene, acetone, etc, the solvent peak would be enormous, and could potentially obscure all kinds of interesting stuff. Thus, we usually use deuterated solvents for proton-NMR spectra, e.g., CDCl3, benzene-d6, acetone-d6. The D (2H) nucleus resonates at a very different frequency from the 1H nucleus, so the solvent is virtually invisible. The solvent is not quite invisible, however. Most deuterated solvents are made from their "normal" H-containing counterparts by exchanging H for D chemically (via an acid-base reaction, for example). This is a good way to replace 99% of the H with D, or even 99.9%, or perhaps more, but a little H always remains. So all one sees of the solvent is a tiny peak from a bit of residual protio impurity. Because their chemical shifts are known accurately, those little residual H peaks from solvent can often be used as a reference to set the δ scale instead of adding TMS. For example, with CDCl3 the tiny CHCl3 peak shows up at δ7.26. (A handy compilation of exact chemical shifts of residual solvent peaks in NMR spectra is HE Gotlieb, V Kotlyar, A Nudelman JOrgChem 1997, 62, 7512-15. The data tables from this paper are taped up on the wall of most NMR instrument rooms.) –O–H protons can be identified by D2O-exchange — Keeping in mind that Hs on O exchange rapidly (under most conditions), what effect would adding a drop (i.e. lots) of D2O have on the NMR spectrum of an alcohol? We saw that the potential coupling between –OH protons and their neighbors is often wiped out by rapid swapping of Hs among alcohol molecules. This is an example of chemical exchange. Dynamic (rate-dependent) effects also arise from conformational changes — e.g., consider ring-flip of d11-cyclohexane (C6D11H) all CD2 groups D eq – 80 °C + 25 °C eq H D H ax H H ax at –80°C, the lifetime of each chair ≈ 0.5 sec, so the ring-flip is "slow on the NMR timescale" — NMR measures separate transitions for Hs in the equatorial and axial positions. at +25°C, the lifetime of each chair ≈ 5 µs (5 x 10–6 sec), so the ring-flip is "fast on the NMR timescale" — i.e. the chairs interconvert faster than the time required to measure the frequency of the NMR transitions. NMR measures a frequency for the spin-flip that corresponds to a timeaverage of the axial and equatorial environments. A more correct view of coupling... So far we've described coupling as a through-space effect, i.e., the protons "feel" each others' spins "through space", in the same way that magnets "sense" each others' fields. How would we then expect vicinal (i.e. between Hs on adjacent Cs) coupling constants to change with the dihedral angle between the Hs? The actual dependence takes the form of the function below, called the Karplus curve — ! 13 ! 10 J (Hz) 0° ! 90° 180° What does this tell you about the mechanism of interaction between the nuclear spins? Is it through-space like the interaction between two little bar magnets? We'll spend a few minutes on how this works. In most acyclic compounds, rapid rotation around σ-bonds leads to a time-averaging of Jvalues for the 3 staggered conformations, and we observe Js of about 7 Hz — another dynamic effect. When rotation is slow (low temp) or restricted (as in rings), we see different J-values for different dihedral angles. This is why NMR spectra of cyclic compounds are often complex. "Complex", of course, means lots of good information for the thoughtful, brave, and mathematically inclined; scary blobs to be swept under the bed for the rest of us.