Region for Alkenes

As mentioned above, the C=C stretching frequencies appear in this region. The exact position in each case is determined by the structural groups attached to the C=C unit. The intensity of the 

We can summarize some of the factors which influence the position of the C=C stretching frequency as follows  

The substitution of hydrogens around the double bond determines the position of the stretching frequency of the double bond. 

When halogens are substituted for the hydrogens next to the double bond, the stretching frequency shifts to higher values. For example, one fluorine substituted in ethylene will shift the band for ethylene from 1628 to 1650 cm while if a second fluorine atom is substituted on the same atom, the band will appear at 1730 cm The presence of the halogen also enhances the intensity of the band. 

When the double bond is conjugated with an aromatic ring, the stretching frequency is shifted to lower values compared to a nonconjugated alkene of similar structure. The shift is usually not greater than 30 cm  

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The five-proton signal at 7.28 ppm is typical of a phenyl group, C6H5, and the one-proton signals at 5.35 and 5,11 ppm are in the region for alkenic pro-... 

The proton chemical shifts of the protons directly attached to the basic three carbon skeleton are found between 5.0 and 6.8 ppm. The J(H,H) between these protons is about -5 Hz. The shift region is similar to the region for similarly substituted alkenes, although the spread in shifts is smaller and the allene proton resonances are slightly upfield from the alkene resonances. We could not establish a reliable additivity rule for the allene proton shifts as we could for the shifts (vide infra) and therefore we found the proton shifts much less valuable for the structural analysis of the allene moiety than the NMR data on the basic three-carbon system. 

Electrostatic potential map for alkene A shows negatively-charged regions (in red) and positively-charged regions (in blue). 

For more detail about the structure of the alkene, concentrate on the 800-1000 cm" region. As mentioned, this area is indicative of out-of-plane bending for alkenes. 

The most valuable information for alkenes is obtained from analysis of the C—H out-of-plane region of the spectrum, which extends from 1000 to 650 cm These bands are frequently the strongest peaks in the spectrum. The number of absorptions and their positions in the spectrum can be used to indicate the substitution pattern on the double bond. 

Aromatic hydrocarbons have many IR active vibrations, resulting in complex spectra. The C H aromatic absorption occurs above 3000 cm , but in the same region as that for alkenes. The aromatic... 

The hexane spectrum (Figure 14.15) shows that the C-H stretch at 2850-2960 cm-i and the signals at 1350-1470 cm-i correlate with the absorptions in Table 14.3. Benzene derivatives and other aromatic compormds usually show absorption for the C-H units at 3000-3100 cm" but also at 675-870 cm i (in the fingerprint region). Note the subtle shift of the C-H absorption to lower energy for the aromatic compounds. Other compounds that have a C-H absorption are those for alkenes and alkynes. The CsC-H absorption is at lower energy than the C=C-H absorption, which is lower in energy than the C-C-H absorption for alkanes. 

Many nonbenzenoid compounds can have strong bands in the 1600-1450 cm" region. For example, the alkene CH2=CHCH(CH3)CH2CH3... 

Alkenes have out-of-plane bending modes in the 1000—800 cm" region. The absorptions are suffi-ciendy intense to be useful in assigning structures (Table 2.3). Terminal alkenes produce the two most rehable absorption patterns. The two C—H bonds of terminal alkenes— R2C=CH2 —bend in concert, and the absorption occurs in the 895—885 cm region. For compounds of the type RCH=CH2, the two methylene C—H bonds absorb in the 910—905 cm" region, and the other C—H bond absorbs in the 995—985 cm region. These two types of terminal alkenes can therefore be distinguished. 

The region of high electron density between the doubly bonded carbon atoms gives alkenes an additional reactivity and in addition to burning and reacting with halogens, alkenes will add on other molecules for example ... 

The structure of ethylene and the orbital hybridization model for its double bond were presented m Section 2 20 and are briefly reviewed m Figure 5 1 Ethylene is planar each carbon is sp hybridized and the double bond is considered to have a a component and a TT component The ct component arises from overlap of sp hybrid orbitals along a line connecting the two carbons the tt component via a side by side overlap of two p orbitals Regions of high electron density attributed to the tt electrons appear above and below the plane of the molecule and are clearly evident m the electrostatic potential map Most of the reactions of ethylene and other alkenes involve these electrons... 

FIGURE 18.11 As a bromine molecule approaches a double bond in an alkene, the atom closer to the ethene molecule acquires a partial positive charge (the blue region). The computation that produced this image was carried out for the point at which the bromine molecule is so close to the double bond that a carbon-bromine bond is starting to form. 

It is possible to take advantage of the differing characteristics of the periphery and the interior to promote chemical reactions. For example, a dendrimer having a non-polar aliphatic periphery with highly polar inner branches can be used to catalyse unimolecular elimination reactions in tertiary alkyl halides in a non-polar aliphatic solvent. This works because the alkyl halide has some polarity, so become relatively concentrated within the polar branches of the dendrimer. This polar medium favours the formation of polar transition states and intermediates, and allows some free alkene to be formed. This, being nonpolar, is expelled from the polar region, and moves out of the dendrimer and into the non-polar solvent. This is a highly efficient process, and the elimination reaction can be driven to completion with only 0.01 % by mass of a dendrimer in the reaction mixture in the presence of an auxiliary base such as potassium carbonate. 

It follows that aromaticity can be determined from an NMR spectrum. If the protons attached to the ring are shifted downfield from the normal alkene region, we can conclude that the molecule is diatropic, and hence aromatic. In addition, if the compound has protons above or within the ring (we shall see an example of the latter on p. 69), then if the compound is diatropic, these will be shifted upfield. One drawback to this method is that it cannot be applied to compounds that have no protons in either category, for example, the dianion of squaiic acid (p. 69). Unfortunately, NMR is of no help here, since these spectra do not show ring currents. ... 

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