LECTURE 3 STRUCTURE AND STEREOCHEMISTRY OF ALKANES

Transcription

1 LECTURE 3 STRUCTURE AND STEREOCEMISTRY OF ALKANES 1. Molecular Formulas. Alkanes are hydrocarbons, which have only sp 3 -hybridized carbon atoms, i.e. carbon atoms that form only σ-bonds. Such hydrocarbons, lacking completely any double or triple bonds are known as saturated hydrocarbons. If one writes the structural formulas of the first few members of the alkane family, one sees a trend, namely that each next member is obtained from the previous by the formal insertion of a C 2 -group, called a methylene group. It reflects on the molecular formulas as well, as higher members of the class are obtained by summing up the current number with one more C 2. Thus, C 4, C 2 6, C 3 8, etc. This phenomenon, of a series of compounds, derived by the regular addition of C 2 -groups, is often encountered in chemistry. Such series of compounds are usually known as homologous series, and the members are called homologs. Overall, an alkane molecule could be represented as a sequence of methylene groups, C 2, capped on both sides by two more hydrogen atoms. This amounts to a general formula: C n 2n Nomenclature. With the exception of the first four members (methane, ethane, propane, butane) all the rest are named based on the root of the Greek word for the particular number of carbon atoms, with the suffix ane. Also, if necessary,

2 additional prefixes are attached to denote a particular isomer. Thus, pentane, isopentane, neopentane. The systematic approach to naming compounds and the set of rules, which governs it, is usually called nomenclature. Several have been suggested and practiced in the past, but the contemporary and accepted is the IUPAC nomenclature. Names, given in accordance with IUPAC are known as IUPAC names or systematic names, unlike the pre-iupac names, which are known as common or trivial. There is an established set of rules as to how to name alkanes: 1) Form the base for the compounds name. This is the name of the longest continuous carbon chain (main chain) in the molecule. Other groups attached to the main chain are considered as substituents. If two chains of equal length can be found in a particular molecule, use the one that gives a greater number of substituents. 2) Number the main chain, starting with the end, which is nearest a substituent. 3) The substituents names are derived in the same fashion as the parent hydrocarbons, but instead of suffix ane, put a suffix yl. Thus, methane methyl, pentane pentyl, hectane hectyl, etc. The substituents also take prefixes, whenever necessary, to differentiate between different isomeric substituents. Several possibilities: a) if the substituent is a straight chain, then nothing or n- (for normal); b) iso-, if it contains the iso-group at the end

3 of the chain; With the exception of the isopropyl, all other isogroups, as well as all n-groups have a carbon atom attached to the main chain, which is connected to only one other carbon atom. Such carbon atoms are known as primary (1 o ). Examples. If this carbon is connected to two other carbon atoms, then the substituent receives the prefix sec-. Such carbon atoms are known as secondary (2 o ). Examples. If this carbon atom is bound to three more carbons, then prefix tert-. Such carbon atoms are known as tertiary (3 o ). 4) When two or more substituents are present, list the substituents in alphabetical order. If some of them are identical, list them together with a prefix identifying their total number: di- (two), tri- (three), tetra- (four), penta- (five), etc. Example. 5) When a more complex substituent is encountered, one may need, for the naming of the substituent, to apply the entire procedure outlined above, i.e. select a main chain, etc. But as usual, the substituent as whole should always bear the suffix yl, instead of ane. 3. Physical Properties of Alkanes: Solubilities, densities, boiling points and melting points. Many of these properties can be explained on the basis of the already discussed London forces, acting between such molecules. This is the explanation for their insolubility in water and other polar solvents. They are less dense than water (~ 0.7), and so they float on top. Oil spills.

4 Boiling points steadily increase with the increase of carbon atoms, since bigger molecules have larger surfaces and thus attract each other better. Branched alkanes (i.e. not with straight chains) boil at slightly lower temperatures, also fitting within the London forces model. Melting points actually depend on the ability of molecules to pack tightly, and because of that some symmetrically branched alkanes actually pack better and melt at higher temperatures. Otherwise, the straight chain materials have higher melting points than branched ones. Also, even-number- C-atom chains pack better than odd-number-c-atom chains, and this causes slightly higher melting points for the former. There are two smooth lines one for the even-numbered and one for the odd-numbered. 4. Uses and Sources of Alkanes. To be read by the students. 5. Reactions of Alkanes. Alkanes are the least reactive among the many classes of organic compounds. The reason is in the fact that they contain only single C C and C bonds, which do not easily enter into reactions. There are a few general processes, pertinent to all of them: a) Combustion alkanes can burn, in presence of air or pure oxygen, giving carbon dioxide and water as final products. This process is the foundation of contemporary energy production, in power plants, vehicle engines, etc. C n 2n+2 + (3n+1)/2 O 2 nco 2 + (n+1) 2 O Examples!

5 b) Cracking and hydrocracking in presence of certain catalysts, longerchain alkanes can break into shorter ones, one of the chains always containing a double bond. If the process is done in presence of hydrogen, both chains are saturated. Thus dodecane cleaves to give pentene and heptane, but in presence of hydrogen the products are pentane and heptane. c) alogentation under proper conditions halogen molecules (F 2, Cl 2, Br 2 and I 2 ) can react with alkanes. The reaction is called a substitution, because hydrogen atoms in the starting alkane molecule are replaced by the same number of halogen atoms. Depending on the amount of halogen, type of halogen and reaction conditions, different degrees of substitution can be achieved. Thus methane can give with chlorine four products: chloromethane, dichloromethane, trichloromethane (chloroform) and tetrachloromethane (carbon tetrachloride). Give example! The reaction usually does not start before heating the mixture to certain temperature or shining light upon it. It is said in such cases that heat or light serve as initiators and the corresponding reaction needs to be initiated. Fluorine is the most active of halogens and reactions with it are usually fast and difficult to control. Iodine, on the other hand is too unreactive and reactions are either very slow or do not occur at all. Chlorine and bromine exhibit moderate activity and rate of reaction. More details in Chapter 4.

6 6. Structure and conformational analysis of alkanes. We did mention in the previous chapter, that materials such as alkanes, consisting entirely of single, σ-bonds, will have all of their carbon atoms in the sp 3 -hybridization. This in its turn pre-determines a tetrahedral arrangement around each carbon, and such is indeed the structure of methane. We will now go ahead and explore in details the structures of other simple alkanes: A. Ethane we saw in Chapter 2 that rotation and twist around a σ-bond is free and does not lead to any change of the properties of the bond. Because of this, one can have an infinite number of conformations, i.e. different arrangements due to rotation around the C C bond. Before we go any further, it is a good idea to familiarize ourselves with some techniques, used to represent conformations. One is the Newman projection: One has to imagine looking straight down the C C bond of ethane. The back (more remote) carbon atom is represented as a circle, with three bonds, at a 120 o angle, pointing out from it. The front carbon is represented as a point in the center of the circle, with three bonds, at a 120 o angle, converging at this point: view along this direction

7 (ydrogens attached to front carbon are in black, those at back carbon in blue). Whenever we would like to represent rotation around the C C bond, we could just rotate the front carbon and keep the arrangement in the back intact. ere are some of the conformations: q = 0 o q = 60 o Notice that in the Newman projection, one could define an angle between any two bonds, one each from the front and the back carbon. This angle is of great importance in discussing conformations. It is known as a dihedral angle (θ) or torsional angle. In the projection above left, the torsional angle between the two C bonds is 0 o. At 0 o dihedral (torsional) angle, the C bonds exactly overlap (eclipse) each other, and this conformation is known as the eclipsed conformation of ethane. At a dihedral (torsional) angle of 60 o (above right), the C bonds are as far away from each other as possible. This conformation is known as the staggered conformation of ethane. At any other angle, the relative positioning of the C bonds is an intermediate of these two.

8 All these other conformations are collectively known as skew conformations. Another technique, used for expressing conformations are the sawhorse structures, generated by looking down at an angle at the C C bond. Can be misleading, because of the way bonds are represented, through simple straight lines. Requires more imagination. We did emphasize on the fact that any degree of rotation around the σ-bond is equally favored, as far as the properties of the σ-bonds go. owever, other factors now enter into play, which indeed make the ethane conformations to be actually non-equivalent. If you take a closer look at the Newman projections above, you will notice that in the eclipsed conformation, the three C bonds of the front carbon atom exactly overlap with the three C bonds of the back carbon. Such overlap places the corresponding electron clouds of these bonds into a close contact with each other, causing a certain repulsive interaction to take place. This interaction makes the eclipsed conformation to be about 3 kcal/mol more energetic than the staggered conformation, which means that each overlap of a pair of C bonds (one from each C- atom) contributes about 1 kcal/mol in energy increase. As soon as we increase the dihedral angle, the unfavorable interaction, due to the overlap of the electron clouds, starts decreasing. When we reach 60 o of dihedral angle, the bonds are at maximum distance from each other, and therefore in this conformation, which is exactly the staggered one, the energy should be at minimum. Than as

9 the angle increases, the bonds get closer again maximum energy at 120 o, then minimum at 180, then maximum at 240, and so on. In other words, in steps of 60 o of dihedral angle change, the energy alternates between a maximum and minimum value. Show the process! The fact that the energy of the eclipsed conformation is higher by 3 kcal/mol in comparison with the staggered means that the system will experience certain resistance to rotation from staggered to eclipsed, i.e. there will be certain strain, associated with this process. It is known as torsional strain, and the energy required to overcome it torsional energy. The above outlined procedure of detailed analysis of the energy changes, associated with the transitions from one conformation to another, is known as conformational analysis. B. Propane there are three carbon atoms and regardless which C C bond we pick, there will be one carbon bearing three -atoms and the other carbon with two -atoms and one methyl group (C 3 ). Same analysis and results as with ethane, but energy of eclipsed conformation is slightly higher, due to the fact that one of the interactions is a C bond with a C C 3 bond (instead of C ) bond. The overall energy difference between eclipsed and staggered conformation is 3.3 kcal/mol, which means that the eclipsing interaction of a C and a C C 3 bond is 1.3 kcal/mol. Show the energy diagram and major Newman projections!

10 C. Butane the line of reasoning is still the same. But there are now three C C bonds. If we select any of the two terminal ones, the analysis should be analogous to the one offered for propane. If we select the C 2 C 3 bond, then we have more distinct conformations. One way to distinguish them is by using the dihedral angle between the two C C 3 bonds (see the scheme below). At 0 o angle the conformation is called totally eclipsed. At 60 o it is known as gauche conformation, at 120 o as eclipsed conformation, and at 180 o as anti conformation (in which the two C C 3 bonds are exactly opposite to each other. Upon further increase of the angle in 60 o steps, the picture is a mirror image of the one so far described. q = 0 o q = 60 o q = 120 o q = 180 o C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 As expected, the eclipsed conformations have higher energy, both the totally eclipsed and the eclipsed. It can also be reasoned that the totally eclipsed, in which the two C C 3 bonds exactly face each other, should be of higher energy, compared to the eclipsed, in which each C C 3 bond overlaps with a C bond. The reason is that the C 3 groups are bigger and therefore the electron clouds would be expected to unfavorably interact to an even greater extent. One can also say that since in the anti conformation the two C C 3 bonds are exact opposite,

11 it should have the lowest energy. It is indeed true, with the eclipsed conformation being 3.6 kcal/mol higher in energy than the anti (1 interaction of two C bonds + two interactions of C and C C 3 bond = 1kcal/mol + 2 x 1.3 kcal/mol = 3.6 kcal/mol). The totally eclipsed conformation is 5 kcal/mol higher in energy in comparison with the anti. What is interesting to notice is that the gauche conformation is 0.9 kcal/higher than the anti, i.e. even at 60 o dihedral angle, the two C 3 groups feel each other, and this increases the energy of the molecule. This kind of unfavorable interaction between two bulky groups is known as steric strain or steric hindrance. In fact, the 5 kcal/mol energy increase of the totally eclipsed vs. the anti conformation is a sum of torsional and steric strain. One must not forget that values of the magnitude of 3 5 kcal/mol are actually not very high and at room temperature the molecules possess enough energy to be able to overcome these kind of barriers, generated by torsional and/or steric strain. The result is that rotations continue to occur, but certain conformations are more preferred than others. The above analysis can be used to analyze larger chains. In every case the all-anti conformation would be preferred, but in reality the chains are mixed gauche and anti conformations, since energy barriers of the magnitude of 5 kcal/mol are easy to overcome at room temperature and rotations do occur continuously.

12 7. Structure and conformational analysis of cycloalkanes. A. Structure and nomenclature these are saturated materials, which contain rings of carbon atoms. Examples. Their physical properties are similar to those of alkanes. Names are formed by adding the prefix cyclo- to the base, the latter again being derived from the Greek word designating the number of carbon atoms. For substituted cycloalkanes the cycle is frequently chosen as the base of the name. Examples. In rare instances the cycle can be considered as a substituent, as in the case of two cycloalkane units being joined. Examples. B. Cis-Trans Isomerism in a simplified picture, the cycle can be considered as a plain, with an up and down face. If there are two substituents, and they point towards the same face, then cis-isomer; if they point towards opposite faces, trans-isomer. Examples. C. Stabilities of cycloalkanes ring strain. All cycloalkanes consist of sp 3 - hybridized carbons ONLY, which implies bond angles of o. But a closer inspection of cycloalkanes leads to the conclusion that the actual angles, especially in the small cycles (rings) are very far from this value 60 o in cyclopropane, 90 o in cyclobutane. This forced contraction of the angles imposes a different type of strain on such systems angle strain. Draw cyclopropane and then draw the hybrid orbitals responsible for the C C bonds. On top of the angle strain, these systems will have significant torsional strain from the eclipse of C bonds (Draw a Newman projection along any of the C C bonds in cyclopropane or cyclobutane to show the eclipse). Together, the angle + torsional strain

13 compound the so-called ring strain. ow do we measure ring strain through the heats of combustion. Think of it this way: If you have more energy stored in a particular material, and you burn it, then more heat would be evolved. The ring strain is exactly the extra energy that the molecule might possess, which would be released upon burning. The only thing that should not be forgotten: The evolved energy has to be divided by the number of C C bonds in the ring, or as people call it, the evolved energy has to be normalized per bond. Then it needs to be compared to the value per bond in a system that has zero ring strain, for which purpose the most convenient would be an open-chain system, since it has, by its very nature, zero ring strain. This performed, it comes out that cyclopropane has the highest ring strain per bond, followed by cyclobutane, cyclopentane and cyclohexane, the latter having zero ring strain. With further increase of the ring size the values tend to vary somewhat irregularly. D. Cyclopropane largest ring strain, due to very unfavorable angle strain + large torsional strain. Because of this high ring strain, cyclopropane is chemically different from the other cycloalkanes; it is considerably more reactive, tending to undergo ring-opening reactions. C

14 E. Cyclobutane also unfavorable angle strain, but avoids partially the torsional strain by a slight folding, further contracting the bond angles to 88 o. Check the Newman projection. F. Cyclopentane very small angle strain (slight deviation from the ideal o ) and avoids torsional strain by assuming an envelope conformation, which undulates, i.e. the ring carbons undergo successive up and down motions so that the ring flap moves around the ring. Check the Newman projection. G. Cyclohexane zero ring strain, by assuming a puckered conformation, in which all angles are and there are no eclipsing bonds (i.e. torsional strain). It is the so-called chair conformation. The boat conformation also has angles, but is higher in energy because of torsional strain, also interference of the flagpole hydrogens. ow does the boat avoid this strain by assuming a twist-boat conformation. The highest in energy is the half chair conformation. Draw the conformational energy diagram of cyclohexane. Make sure you see the advantage of the chair vs. boat conformation, or the twist-boat vs. boat conformation. The complete energy diagram describes the transition from chair, through half-chair, to twist boat, which, through a boat, is converted to the other twist boat, the latter, via the other half-chair, eventually goes to the other chair. This process is continuous, known as the chair-chair interconversion.

15 If you take a closer look at the chair you will see that there are two types of hydrogens. Through the center of the chair passes a vertical axis of symmetry. Six of the hydrogen atoms, one per carbon, have their bonds parallel to this axis axial hydrogens. The other six (one per carbon) are called equatorial. An important issue eventually becomes how to draw these bonds (and hydrogens) up and down. ere is an easy way: It is not difficult to see that the axial C bonds point either up or down, and they alter, i.e. up, down, up, etc. It is more difficult with the equatorial. But remember this: If the axial bond at a particular carbon is UP, then the equatorial bond at the same carbon has to be DOWN. Remember and practice this, as it will be very important when it comes to writing various disubstituted cis- or trans-isomers. Learn to properly draw chairs and boats; you will find it very useful in the future. axis of symmetry Monosubstituted cyclohexanes the substituent could be axial or equatorial. Which is more stable and why? The equatorial is more stable; its energy is 1.7 kcal/mol lower. This can actually be predicted:

16 The axial has two gauche relationships, with the methylene group carbons C3 and C5, i.e. 2 x 0.9 kcal/mol (we know the latter value from analysis of butane!!) = 1.8 kcal/mol, which is very close to the actual. Another thing to notice here: The axial group is close to the axial hydrogens at C3 and C5, causing a slight unfavorable overlap of electron clouds. This additional interference, valid only for the axial group, is known as 1,3-diaxial interaction. It actually plays a very significant role for bulkier groups. Disubstituted cyclohexanes there are several possibilities: a) Cis-1,2-disubstituted one axial and one equatorial group; b) Trans-1,2-disubstituted two axial OR two equatorial. Which is more stable and why? c) Cis-1,3-disubstituted two axial OR two equatorial groups. Which is more stable and why? d) Trans-1,3-disubstituted one axial and one equatorial; e) Cis-1,4-disubstituted one axial and one equatorial; f) Trans-1,4-disubstituted two axial or two equatorial groups. Which is more stable and why? Consider the case of differently-sized substituents. The bulkier always is preferably equatorial!! Two additional points of attention here: 1) Learn to draw and recognize the cis- and trans-isomers in a flat (i.e. two-dimensional) drawing, in which they are represented by solid or dashed wedges. 2) Consider the case of very bulky groups in axial positions how can this

17 change the entire conformation of the system, in order for the bulky group to avoid the unfavorable interactions.. Bicyclic systems they can be fused, bridged and spirocyclic. Nomenclature. Cis- and trans-decalin.