Chemistry, 5E

6.3 Structure of Substituted Alkanes

6.3a Alkyl Halides Simple alkyl halides are nearly tetrahedral, and therefore the carbon to which the halogen is attached is approximately sp³ hybridized. The bond to halogen involves an approximately sp³ carbon orbital overlapping with an orbital on the halogen for which the principal quantum number varies from n = 2 for fluorine to n = 5 for iodine. The C―X bond lengths increase and the bond strengths decrease as we read down the periodic table. Some typical values for methyl halides are shown in Figure 6.16, which gives both C―X bond lengths (in angstrom units) and C―X bond strengths (in kilocalories per mole and kilojoules per mole). Remember: These values refer to homolytic bond breaking (formation of two neutral radicals; see Fig. 1.44).

FIGURE 6.16 The C―X bond of an alkyl halide is formed by the overlap of a singly occupied carbon sp³ orbital with a singly occupied halogen orbital.

6.3b Alcohols Alcohols are derivatives of water, and it is no surprise to see that they rather closely resemble water structurally. The bond angle is expanded a bit from 104.5° in H―O―H to approximately 109° in simple alcohols R―O―H.

The oxygen–hydrogen bond length is very little changed from that of water, but the carbon–oxygen bond is a bit shorter and stronger than the carbon–carbon bond in ethane (Fig. 6.17).

FIGURE 6.17 Bond lengths and angles for water, methyl alcohol, and ethane.

PROBLEM 6.5 What is the approximate hybridization of the oxygen atom in methyl alcohol (H₃C―O―H angle = 108.9°). How do you know?

The orbital interaction diagram of Figure 6.18 shows the construction of the carbon–oxygen bond, as well as typical bond energies. Notice that an electron in an orbital near the very electronegative oxygen is more stable than it would be in an orbital near carbon.

FIGURE 6.18 A graphical description of the formation of the carbon–oxygen σ bond.

FIGURE 6.19 The carbon–nitrogen bond in amines is shorter than the carbon–carbon bond in alkanes.

6.3c Amines In simple amines, the carbon–nitrogen bond distance is a little shorter than the corresponding carbon–carbon bond distance in alkanes. This phenomenon is similar to that found in the alcohols and ethers we just discussed (Fig. 6.19).

Simple amines are hybridized approximately sp³ at nitrogen and thus are pyramidal. The bond angles are close to the tetrahedral angle of 109.5° but cannot be exactly the tetrahedral angle.

PROBLEM 6.6 Draw N,N-dimethylethanamine clearly showing the three-dimensional perspective at the nitrogen. Use the convention of solid and dashed wedges and include nitrogen’s lone pair.

PROBLEM 6.7 Why is the carbon–nitrogen bond distance in aniline slightly shorter than the carbon–nitrogen bond in methylamine?

FIGURE 6.20 Some bond angles in simple amines.

For amines in which all three R groups are identical, there will be only a single R―N―R angle, but for less symmetric amines, the different R―N―R angles cannot be the same (Fig. 6.20).

The most important structural feature of amines comes from the presence of the pair of nonbonding electrons acting as the fourth substituent on the pyramid. In alkanes, where there are four substituents, there is nothing more to consider once we describe the pyramid. In amines there is, because the pyramid can undergo an “umbrella flip,” amine inversion, forming a new, mirror-image pyramid (Fig. 6.21). Because the nitrogen can be a stereogenic center, this inversion has an interesting outcome.

FIGURE 6.21 Amine inversion interconverts enantiomeric amines: (a) initial amine, (b) higher energy halfway point, (c) final inverted amine.

Draw, Visualize, Understand

Inversion of an Amine

The amine inversion process and the resulting racemization is sometimes difficult to visualize when first encountered, so let’s work through it step-by-step.

1. Draw a three-dimensional representation of a chiral amine. For example, you could use the convention of solid and dashed wedges to draw (R)-N-methyl-1-propanamine. The lone pair is taken as having the lowest atomic number. Let’s draw the molecule with the lone pair pointing up (Fig. 6.21a). At the beginning of the inversion process, the pyramid begins to flatten, and the R―N―R bond angles increase.

2. At the halfway point along the path to the inverted pyramid, the transition state is reached. Here the molecule is planar, with approximately 120° R―N―R bond angles. At this point, the hybridization of the central nitrogen is sp² (Fig. 6.21b).

3. The inversion continues until the pyramidal shape is re-obtained (Fig. 6.21c). This new molecule is the mirror image of the initial amine.

PROBLEM 6.8 Why don’t we see the inversion of a stereogenic carbon?

An important concern is the rate of this inversion process. If it is slow at room temperature, then we should be able to isolate appropriately substituted optically active amines. If inversion is fast, however, racemization will also be fast as the inversion process interconverts enantiomers. In practice, this is what happens. Barriers to inversion in simple amines are very low, about 5–6 kcal/mol (21–25 kJ/mol), and isolation of one enantiomer, even at very low temperature, is difficult. Inversion is simply too facile, and the rapid inversion interconverts enantiomeric amines (Fig. 6.22).

FIGURE 6.22 Rapid inversion racemizes optically active amines.

PROBLEM 6.9 In contrast to simple amines, aziridines (three-membered rings containing a nitrogen) can often be separated into enantiomers. For example, the activation energy (ΔG ^‡) for the inversion of 1,2,2-trimethylaziridine is about 18.5 kcal/mol (77.5 kJ/mol), much higher than that for simple amines. Explain.

PROBLEM 6.10 If a phenyl group (Ph) is attached to the nitrogen atom of the aziridine, the barrier to inversion decreases. Explain.

6.3d Ethers The carbon–oxygen bond length in ethers is similar to that in alcohols, and the opening of the angle as we go from H―O―H (104.5°) to R―O―H (~109°) to R―O―R (~112°) continues. Because the angle in ethers is approximately 112°, the oxygen is hybridized approximately sp³.