Between four protons

There are always two arrows. One is drawn coming from the base and grabbing the proton. The second arrow is drawn coming from the bond (between the proton and whatever atom is connected to the proton) and going to the atom currently connected to the proton. That s it. There are always two arrows. Each arrow has a head and a tail, so there are four possible mistakes you can make. You might accidentally draw either of the heads incorrectly, or you might draw either of the tails incorrectly. With a little bit of practice you will see just how easy it is, and you will realize that acid-base reactions always follow the same mechanism. 

Runanine (17) was isolated from the roots of Stephania sinica, a species found in the Chinese provinces of Heibei, Gueizhou, and Yunnan (35). The H-NMR spectrum of runanine (17) (Table II) revealed the presence of two aromatic protons, C-5 methylene protons, one N-methyl, and four methoxyl groups. An NOE effect (10% enhancement) was observed between the protons of two methoxyl groups (53.79 and 3.80) and the aromatic protons (56.47 and 6.64), but the same phenomenon was not observed for the other methoxyl protons (53.61 and 4.05). Therefore, the former methoxyls should be situated on ring A. From the further observation of an NOE (22.6% enhancement) between the aromatic C-4 proton (56.64) and one (53.00) of the C-5 methylene protons, it was assumed that the two methoxyl groups (53.79 and 3.80) should be located at C-2 and C-3, respectively. The absence of signals for olefinic... 

Figure 2. Plots of the chemical shift differences between the enantiomers of 1.2SM isopropyl methylsulfoxide in the presence of (-)-TFPE at 28° in CCI4. Each of the four proton sets shows nonequivalence o, SCHs , CCH3 0, CCH3 , CH. Reprinted with permission from Tetrahedron Lett. 1974, 2295-2298.

The emission of a helium nucleus in the final stage regenerates the initial carbon-12. The latter thus plays the role of a catalyst. The overall result is the fusion of four protons into a helium nucleus. At high temperatures, this cycle dominates over the proton-proton chain. Indeed thermal agitation facilitates penetration of the relatively high electrical barrier between proton and carbon nucleus. Whatever hydrogen fusion mechanism is prevalent, the star s mass determines the rate at which it consumes its nuclear fuel, and hence also its lifetime. The higher its mass, the more quickly it bums. 

To reduce an O2 molecule to two molecules of H2O, a total of four electrons are needed, which are supplied by cytochrome c (pink, top left) and initially given off to Cua- From there, they are passed on via heme a and heme as to the enzyme s reaction center, which is located between heme as and Cub. The reduction of the oxygen takes place in several steps, without any intermediate being released. The four protons needed to produce water and the ions pumped into the intermembrane space... 

The Q cycle accommodates the switch between the two-electron carrier ubiquinone and the one-electron carriers—cytochromes b562, b566, clt and c—and explains the measured stoichiometry of four protons translocated per pair of electrons passing through the Complex III to cytochrome c. Although the path of electrons through this segment of the respiratory chain is complicated, the net effect of the transfer is simple QH2 is oxidized to Q and two molecules of cytochrome c are reduced. 

Like Complex III of mitochondria, cytochrome b6f conveys electrons from a reduced quinone—a mobile, lipid-soluble carrier of two electrons (Q in mitochondria, PQb in chloroplasts)—to a water-soluble protein that carries one electron (cytochrome c in mitochondria, plastocyanin in chloroplasts). As in mitochondria, the function of this complex involves a Q cycle (Fig. 19-12) in which electrons pass, one at a time, from PQBH2 to cytochrome bs. This cycle results in the pumping of protons across the membrane in chloroplasts, the direction of proton movement is from the stromal compartment to the thylakoid lumen, up to four protons moving for each pair of electrons. The result is production of a proton gradient across the thylakoid membrane as electrons pass from PSII to PSI. Because the volume of the flattened thylakoid lumen is small, the influx of a small number of protons has a relatively large effect on lumenal pH. The measured difference in pH between the stroma (pH 8) and the thylakoid lumen (pH 5) represents a 1,000-fold difference in proton concentration—a powerful driving force for ATP synthesis. 

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