Carbon atom contributions

These absorptions are ascribed to n-n transitions, that is, transitions of an electron from the highest occupied n molecular orbital (HOMO) to the lowest unoccupied n molecular orbital (LUMO). One can decide which orbitals are the HOMO and LUMO by filling electrons into the molecular energy level diagram from the bottom up, two electrons to each molecular orbital. The number of electrons is the number of sp carbon atoms contributing to the n system of a neuhal polyalkene, two for each double bond. In ethylene, there is only one occupied MO and one unoccupied MO. The occupied orbital in ethylene is p below the energy level represented by ot, and the unoccupied orbital is p above it. The separation between the only possibilities for the HOMO and LUMO is 2.00p. 

For the minute, imagine an HF-LCAO treatment of just the jr-electrons in ethene where each carbon atom contributes just one electron and one atomic orbital of the correct symmetry to the conjugated system. Without any particular justification except chemical intuition, we make the following assumptions. 

These equations lead to the energy values E = a + ft and E = a — ft. Both a and ft represent negative quantities, and each carbon atom contributes one electron to the n bond, so the energy level diagram can be shown as illustrated in Figure 5.15. 

In general, C NMR spectra cire simpler than proton NMR spectra. Instead of integration, the height of each absorption peak is approximately proportional to the number of carbon atoms contributing to the peak. It s only approximately proportional because the more hydrogen attached to the carbon atom, the greater the peak height. 

First, add up the total valence electrons. Each carbon atom contributes 4 electrons, and each hydrogen contributes 1, for a total of 18 valence electrons. Next, pick a central atom. The only choice available is carbon, because hydrogen can only have one bond and thus can never be a central atom in a Lewis structure. You have three carbons, so just connect the three into a carbon chain. In formulas containing just carbon and hydrogen, you ll always find the carbons bonded together in a chain with the hydrogens bonded around the outside. 

It is assumed that each carbon atom contributes one 7r-electron and one 2pr atomic orbital to the system, so that... 

We will consider the application of the Hiickel molecular orbital method to the benzene molecule and we will first see what happens when we do not make use of symmetry. The benzene molecule has a framework of six carbon atoms at the comers of a hexagon and each carbon atom contributes one 7r-electron. The tt-electron MOs will be constructed from six 2pc atomic orbitals, each located at one of the carbon atoms, thus, c... 

Bonds to other carbon atoms contribute 0 to the oxidation level. 

In conjugated polyenes, each carbon atom contributes one p orbital and one electron to the jt bonding of the system, as illustrated in Fig. 3.5.2. So the jt molecular orbitals have the general form... 

Nylon 66. The word nylon was established as a generic name for polyamides, one class of the new high molecular weight linear polymers. The first of these, and the one still produced in the largest volume, was nylon 66 or polyhexamethylene adipamide. Numbers are used following the word nylon to indicate the number of carbon atoms contributed by the diamine and dicarboxylic acid constituents, in this case hexamethylenediamine and adipic acid, respectively. 

Now in these carbocyanines each carbon atom contributes one, the two nitrogen atoms, one at both ends, together three n electrons. If there are Z atoms in the chain with Z — 2W + 9 f°r the above chemical formula, then there are N = Z + 1 electrons and the length L has to be put equal to / (Z + 1) where l is the average separation and 1/2 / is also added for the two ends. Thus L = / N or... 

Alternation of the CC bond lengths along the chain and the existence of a large energy gap are well-established facts in PA (see Chapter 12, Section II.C.2). However, since each carbon atom contributes one tt electron, there is at first sight no obvious reason why CC bonds should not be equivalent. If they were, and taking into account the electron spin, the tt electrons should generate a half-filled band such a material is a metal. If there is bond alternation, the one-dimensional unit cell is doubled and a gap opens at the Brillouin zone boundary the material is a semiconductor. 

Each carbon atom contributes one electron to the 7r-orbitals four carbon atoms means four 7r-electrons. Since electrons are indistinguishable, we may assume that these four electrons as distributed over the entire four carbon chain. Or, if we are attempting to apply a simple model, these four electrons are in a one-dimensional box with a length equal to the length of the carbon chain. And you probably thought that this simple model was one of these cases of use only in a lecture, not in the real world ... 

Suppose you have a mole of geraniol. Its molar mass is 154 g/mol. Of this mass, how many grams do the carbon atoms contribute The formula shows that one molecule of geraniol includes ten atoms of carbon. Therefore, 1 mol of geraniol contains 10 mol of carbon. Multiply the mass of 1 mol of carbon by 10 to get the mass of carbon in 1 mol of geraniol. 

Every atom of the ring is sp hybridized and has three bonding orbitals. These orbitals lie in a common plane and are 120° apart. Each carbon atom contributes one electron to each of its sp orbitals, thereby forming a bonds with three other atoms, and has one electron left over, which occupies a 2p orbital. 

Since all carbon atoms contribute to the tt system of polyimide, and the vast majority does in polystyrene, knowledge of the (preferential) orientation of the TT orbitals is sufficient to derive the (preferential) orientation of the pol3nmer chains, d vice versa. Therefore, the preferential orientation of the polyimide chains along the rubbing direction results in a preferential orientation of the 7T orbitals perpendicular to the rubbing direction. For polystyrene, on the other hand, the preferential orientation of the tt orbitals is parallel to the rubbing direction. 

A final average of EC values over all rings then provides a global aromaticity index per benzenoid ring. We recall that all catafusenes have the molecular formula C4r+2H2r+ and that each of the n = 4R -h 2 carbon atoms contributes with one... 

The PPP-model of r-conjugated molecule assumes that each carbon atom contributes a 2p orbital and a single r-electron to the r-system. The PPP parametriza-tion of the Hamiltonian matrix elements includes resonance (electron hopping) integrals for the connected carbon atoms n and v ... 

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