Tertiary protons

In -isosparteine (47) both tertiary protons at C-6 and C-11 are arranged in the CIS position to the free electron pair on nitrogen as indicated by the NMR spectrum (80). Much earlier X-ray analysis showed that all rings in crystalline a-isospa teine (46) are present in stable chair conformations (SJ). A comparative rate (extrapolated) at 65 for a-isosparteine (5.0) and sparteine (1.0) has been calculated (82). Also it has been reported that (8-isosparteine gave the dehydro derivative under mild conditions and the didehydro under more drastic conditions (times, temperatures not given) (60). 

Solid-state NMR spectroscopy has not found as wide an application in soils as it has in other fields. The great advantage of NMR is that it is specific for specific elements that is, it is tuned to a specific element and other elements are not detected. NMR spectroscopy shows the environment or multiple environments in which a particular element exists. For example, in a proton NMR spectrum, primary, secondary, and tertiary protons can be differentiated, as can protons attached to oxygen, nitrogen, and other atoms. 

Chemical shifts of magnetically active nuclei, such as [79MI171], Vicinal and geminal scalar H,H and H,C coupling constants Line widths of proton resonances of tertiary protons... 

Let us take the case of adjacent carbon atoms carrying, respectively, a pair of secondary protons and a tertiary proton, and consider first the absorption by one of the secondary protons ... 

The magnetic field that a secondary proton feels at a particular instant is slightly increased or slightly decreased by the spin of the neighboring tertiary proton increased if the tertiary proton happens at that instant to be aligned with the applied... 

Next, what can we say about the absorption by the tertiary proton ... 

It is, in its turn, affected by the spin of the neighboring secondary protons. But now there are two protons whose alignments in the applied field we must consider. There are four equally probable combinations of spin alignments for these two protons, of which two are equivalent. At any instant, therefore, the tertiary proton feels any one of three fields, and its signal is split into three equally spaced peaks a triplet, with relative peak intensities 1 2 1, reflecting the combined (double) probability of the two equivalent combinations (Fig. 13.10),... 

Jy Sec. 13.11) in the doublet is exactly the same as the separation of peaks in the triplet. (Spin-spin coupling is a reciprocal affair, and the effect of the secondary protons on the tertiary proton must be identical with the effect of the tertiary proton on the secondary protons.) Even if they were to appear in a complicated spectrum of many absorption peaks, the identical peak separations would tell us that this doublet and triplet were related that the (two) protons giving the doublet and the (one) proton giving the triplet are coupled, and hence are attached to adjacent carbon atoms. 

If the base that is used is large, and so has a large steric demand, would it be more likely to remove a primary or tertiary proton, assuming that they each had the same pK... 

The sterically hindered base would be more likely to remove the primary proton, because it would not be able to reach the crowded tertiary proton. Bearing these factors in mind, which proton is most likely to be removed when CH3C0CHC02Et is treated with a general base ... 

The OEt- anion may still gain access to the tertiary proton, which is the most acidic, and so remove it. However, LDA has a much larger steric demand and so cannot approach a tertiary proton without incurring some steric hindrance. Thus, it attacks the next most acidic proton that is readily available, which in this case is one of the terminal methyl protons. Draw the two anions that result from the removal of the two different protons. 

Step 1 This is the rate-determining step and is bimolecular. Protonation of the double bond occurs in the direction that gives the more stable of two possible carbocations. In this case the carbocation is tertiary. Protonation of C-2 would have given a less stable secondary carbocation. 

The largest peak, at high field represents pendant methyls in propylene units. It is characteristically split hy the tertiary hydrogen. By area integration, about 20% of the nominal methyl proton peak is due to overlap of absorption from chain methylenes. This overlap is consistent with a reported syndiotactic triplet [6]. The absence of a strong singlet peak in the methylene range indicates the virtual absence of amorphous polymer in the atactic PP, which could possibly be due to the head-to-head and tail-to-tail units. The low field peak represents the partial resolution of tertiary protons which are opposite the methyls on the hydrocarbon chain. 

Polyethylene has also been modified by extrusion at 200°C in the presence of isoprene-styrene rubbers or butadiene-styrene rubbers. The formation of graft copolymers has been shown by solvent extraction tests and IR spectroscopy. Grafts of the first composition are more easily induced by the tertiary proton reactions in the isoprene units. The crack resistance of polyethylene reportedly increases exponentially with the amount of graft content [234]. 

GaCls gave 2-naphthylated frans-perhydronaphthalene The C-C bond formation predominantly took place at the 2-position of naphthalene and the 3-position of perhydronaphthalene. Notably, cis-perhydronaphthalene reacted much more effectively than the trans isomer, which indicated that the equatorial tertiary proton, and not the axial proton of the cycloalkane, was activated selectively. This is an interesting example of the stereoselective C-H activation of alkanes. Chatani reported the carbonylations of adamantane and methyIcyclohexane with CO using GaCb, which also involved the formation of carbocations (Scheme 7.57) [91]. 

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