2.2- Dimethylbutane spectrum

Equivalent atoms appear at the same chemical shift value. Figure 4.10 shows the proton-decoupled carbon spectrum for 2,2-dimethylbutane. The three methyl groups at the left side of the molecule are equivalent by symmetry. 

CH3COCH2CI CICH2CHO CH2O CH3COCH3 CH3CHO Sketch the expected proton NMR spectrum of 3,3-dimethylbutanaL... 

Very low temperature studies on 2,3-dimethylbutane 27 revealed that the concept of the gauche conformations 27b and 27c as the energy minima is correct " . In Figure 7, the well-resolved low-temperature spectrum is reproduced and the assignments given. The barriers of rotation were estimated to be of the order of 4.5kcalmol" ... 

Carbon atom relaxation times influence the intensity of peaks in a spectrum. When more protons are attached to a carbon atom, relaxation times become shorter, resulting in more intense peaks. We expect methyl and methylene groups to be relatively more intense than the intensity observed for quaternary carbon atoms where there are no attached protons. Thus, a weak-intensity peak is observed for the quaternary carbon atom at 30 ppm in 2,2-dimethylbutane (see Figure 27.6). 

Chiral 2-phenyl-3,3-dimethylbutane (IV) may be assumed as a suitable conformational model . Indeed this molecule has a restricted conformational mobility as supported by the invariability of the UV and CD spectra by lowering temperature, and the dominant conformation appears to have the phenyl ring and H bond coplanar [10] (Figure 1) as found also for polystyrene and poly-a-vinylnaphtalene in the crystalline state [11]. The UV spectrum of (IV) is very similar to that of the phenylalkanes (III) and of the styrene copolymer (la). 

In 1,2,2-trichloropropane, each of the two sets of hydrogen atoms produces a single peak, called a singfet. This is by no means always the case in fact, the NMR spectra of most organic compounds contain many sets of peaks. The spectrum of l,l,2-tribromo-3,3-dimethylbutane provides an example (Figure 14.9). 

Why a binomial expansion The spin of a hydrogen atom can be either aligned or opposed to the applied field. Therefore, if there is one hydrogen on each of two bonded atoms, as in 1,1,2-tri-bromo-3,3-dimethylbutane (Figure 14.9), each can see the other as either spin-up or spin-down with equal probability. Therefore, each appears in the spectrum as a doublet. 

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