2-Methylpropane proton

Positive values of A[8( H) —8( H)] are obtained for the eight P-diketones given in Table 10. These were discussed earlier (Section 1, p. 271) and they confirm that the hydrogen bonding is strong. A negative value for this quantity has also been observed (Clark et al., 1989) for the protonated form of l,3-diphenyl-2-methylpropane-l,3-dione. The keto form of 1,3-diphenyl-2-methylpropane-l,3-dione is the preferred species in the solid phase. Protonation of one of the carbonyl oxygens clearly provides the impetus for the... 

We can do the same with a CH group, and in the left-hand side of Table 15.4 we take a series of isopropyl compounds, comparing the measured shifts with those for the central proton (CHMe3) or carbon (CHMej) of 2-methylpropane. We set two of the substituents as methyl groups and just vary the third. Yet again die shifts for die same substituent are broadly the same. 

In addition to catalyzing hydroformylation, the platinum SPO complexes are excellent hydrogenation catalysts for aldehydes (as already demonstrated by the side products of hydroformylation), in particular, in the absence of carbon monoxide. Further, in ibis process, the facile heterolytic splitting of dihydrogen may play a role. The hydrogenation of aldehydes requires the presence of carboxylic acids, and perhaps the release of alkoxides from platinum requires a more reactive proton donor than that available on the nearby SPO. For example, 4 hydrogenates 2-methylpropanal at 95 °C and 40 bar of H2 to give the alcohol, with a TOF of 9000 mol moN h (71). 

Where the template has been an aromatic amine of low basidty, interaction with carboxylic acid groups has proved too weak to allow selective imprinting. However, Sherrington et al. [112] have shown that the sulphonic add residue in 2-acrylamido-2-methylpropane sulphonic acid is very useful under these circumstances the much higher acidity of this add allows protonation of the weak base template, and hence sufficiently strong electrostatic interaction for imprinting to be achieved. 

In conventional hydration of alkenes by water in the presence of a strong acid, there is little direct stereochemical consequence, but we mention it as a useful contrast to the content of the next section. A proton initially adds to the alkene, and in the case of 2-methylpropene (44) gives the intermediate carbocation 45 (Scheme 4.9), which is then captured by the nucleophile, water, and yields the product 2-methylpropan-2-ol (46). 

NMR Aldehydes are readily identified by the presence of a signal for the hydrogen of CH=0 at 8 9-10 ppm. This is a region where very few other protons ever appear. Figure 17.14 shows the NMR spectrum of 2-methylpropanal [(CH3)2CHCH=0)],... 

The pH-rate profile for the hydrolysis of the A-methylimine of 2-methylpropanal is shown in Figure 7.6. The curve is similar to that for aromatic ketones with EWG substituents. The rate increases in the pH range 0. 5, where decomposition of the zwitterionic intermediate is rate controlling. In the pH range 4.5-8, the rate decreases and then levels off This corresponds to the transformation of the protonated imine to the less reactive neutral form. Above pH 8, the rate is again constant, as the increase in [ OH] is compensated by the decrease in the amount of protonated imine. 

The spectra can be predicted for the alkanes butane and isobutane (or 2-methylpropane). The peaks should appear in the 1-1.5 ppm chemical shift region according to Table 3.3. Butane, CH3CH2CH2CH3, has two types of protons as noted in Fig. 3.15(a). Isobutane also has two types of protons, shown in Fig. 3.15(b). Therefore both spectra should have two absorption peaks. In butane, the methyl protons should be split by the adjacent methylene protons into a triplet the methylene protons would be split by the methyl protons into a quartet. We would predict that the proton NMR spectmm of butane would look like the schematic spectrum in Fig. 3.15(a), with the relative peak areas shown. Isobutane would show a very different splitting pattern. There are nine chemically equivalent protons (marked b on the structure) on the three methyl groups the peak for these nine protons will be split into a doublet by the single a type proton on the middle carbon. The peak for the single proton will be split into (9 + 1) = 10 peak multiplet by the b type protons, with the relative peak areas as shown schematically in Fig. 3.15(b). 

So, how can we choose between these The solution is in the H NMR spectrum, which is shown below. There are only two peaks visible one at 3.3 and one at 1.1 ppm. It s quite common in NMR spectra not to see signals for protons attached to O or N (you will see why in Chapter 13) so we can again rule out all structures with more than two different types of H attached to C. Again, we are left with A and B, confirming our earlier deductions. But the chemical shift of the signal at 5 3.3 teUs us more it has to be due to H atoms next to an oxygen atom because it is deshielded. The industrial emulsifier must therefore be A 2-amino-2-methylpropan-l-ol. 

The reaction ofHBr with 2-methylpropene produces only 2-bromo-2-methylpropane, for the same reason regarding carbocations stability. Here, in the first step (i.e., the attachment of the proton) the choice is even more pronounced—between a tertiary carbocation and a primary carbocation. Thus, l-bromo-2-methylpropane is not obtained as a product of the reaction because its formation would require the formation of a primary... 

Ionic addition reactions of alkenes are quite regioselective. For instance, adding concentrated HCl to 2-methylpropene produces largely 2-chloro-2-methylpropane and a much smaller amount of l-chloro-2-methylpropane. This can be explained by examining the energies of the two carbocation intermediates that can be formed by adding a proton in the first step of the reaction ... 

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