bond is lower than the energy of the C(sp 3 ) C(sp 3 ) bond of

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1 3. Alkenes and Alkynes. 3.1 ybridization in Alkenes. When C is bound to four groups, as we saw with methane, it is sp 3 -hybridized, which means mathematically that the s orbital and all three p orbitals are added up and divided by four to make four new orbitals. We can hybridize orbitals in other ways, too. If we combine the s orbital with two p orbitals and leave the third one unaltered, we have sp2 hybridization. An sp 2 -hybridized atom has three equivalent sp 2 orbitals, each with 1/3 s and 2/3 p character, and one unadulterated p orbital. The sp 2 orbitals are 120 apart. If we combine the s orbital with one p orbital and leave the other two unaltered, we have sp hybridization. An sp-hybridized atom has two equivalent sp orbitals, each with 1/2 s and 1/2 p character, and two unadulterated p orbitals. The sp orbitals are 180 apart. The two p orbitals are perpendicular to each other and to the line containing the sp orbitals. Under what circumstances do we find C(sp 3 ), C(sp 2 ), and C(sp) hybridization, and how does C "choose" a particular hybridization? Suppose C 4 were sp 2 -hybridized. Three atoms would be coplanar, and the fourth, whose s orbital overlapped with the p orbital, would have a 90 angle with respect to the other three. Moreover, half the p orbital's bonding density would be wasted on the side of the C atom opposite the unique atom. So we can see that sp3 hybridization is best for a C atom that is bound to four different atoms. owever, when a C atom is bound to three different atoms, the situation changes. Consider ethane (ethylene, 2 C=C 2 ). If the C atoms were sp 3 -hybridized, we would need to use two hybrid

2 orbitals to construct the double bond. They would need to be canted with respect to the C C axis, thus putting only a small amount of electron density between the two C nuclei. Not a good situation! To put maximum electron density between the two C nuclei, the C atom must place an orbital in the same plane as the C bonds. This arrangement is achieved by sp 2 hybridization. The three C(sp 2 ) orbitals on one C atom are used to form two σ bonds to atoms and one σ bond to the other C. The p orbital is used to form the second bond to the other C atom. The C(p) orbitals overlap with each other lengthwise, giving a bond that does not look like the σ bonds we have seen already. This bond is a π bond. It has a node, i.e., a plane in which no electron density resides. The C(p) orbital has a higher energy than the C(sp 2 ) orbital used to make the σ bond, and the C(p) C(p) overlap is poorer than the C(sp 2 ) C(sp 2 ) overlap, so the energy of the C C π bond is higher than the energy of the C C σ bond. Whenever you have a double bond between two atoms, there is one σ MO and one π MO. The energy of the C(sp 2 ) C(sp 2 ) σ bond is lower than the energy of the C(sp 3 ) C(sp 3 ) bond of

3 ethane, because the C(sp 2 ) orbitals are lower in energy than a C(sp 3 ) orbital. Just like there is a σ* orbital associated with a σ orbital in the bond, so there are anti-bonding orbitals associated with the C(sp 2 ) C(sp 2 ) σ bond and the C(p) C(p) π bond. A C atom that makes one double bond and two single bonds is sp2- hybridized. 3.2 (No ) Ro tation in Alkenes. We saw that the two ends of ethane can rotate with respect to one another with only a small barrier to rotation due to the loss of hyperconjugative stabilization in the eclipsed isomer. Is the same true in ethylene? If you look down the C C axis in ethylene, you will see that each C bond on one C has a 180 dihedral angle with a C bond on the other C. A 90 rotation about the C C bond gives a new conformational diastereomer of ethylene, in which no C bond has a 180 dihedral angle with another C bond. As a result, you might expect that ethylene would be predominantly planar, with a small barrier to rotation. In fact, the barrier to rotation in ethylene is very high, about 66 kcal/mol. Why? In the 90 isomer of ethylene, the two p orbitals are perpendicular to each other, so they don't overlap. No overlap, no bond. In other words, rotation about the C=C bond in ethylene breaks the π bond, increasing the energy of the electrons associated with that bond, so it is a very high energy process. Since room t emperature supplies about 15 kcal/mol of energy, there is no rotation about C=C π bonds at room

4 t emperature! 3.3 Subs tituted and igher Alkenes. Configurational Dias tereomers. The four C bonds in alkenes are identical. Replacing an with another group results in a vinyl compound. E.g., vinyl chloride is C2=CCl. The "vinyl" you are familiar with is polyvinyl chloride (PVC), a polymer that is made by linking multiple vinyl chloride units together. (The electrons linking the units together come from the π bond of vinyl chloride, so PVC has no π bonds.) Another name for vinyl chloride is chloroethylene. In ethene, "eth" means "two carbons", and "ene" means that there's a double bond. If we replace an in ethene with C 3 to get a three-carbon compound with a double bond, we call it propene. Propene has three different kinds of C atoms. Two are sp 2 - hybridized, and the third is sp 3 -hybridized. The energy of the C(sp 3 ) bond is higher than the energy of the C(sp 2 ) bond because the energy of the C(sp 3 ) orbital is higher than the energy of the C(sp 2 ) orbital. We number the C atoms in propene from one end to another. The ends of propene are different, so we can number it in two different ways. We do it so that the atoms of the π bond have the lowest numbers possible. We can replace one of the atoms of propene with Cl, to get the skeletal isomers 1-chloropropene, 2-chloropropene, or 3- chloropropene. Actually, there are two kinds of 1- chloropropene. There is the kind where the Cl atom is near the C 3 group attached to C(2), and there is the kind where the Cl atom is near the atom attached to C(2). The relationship between these two compounds is that they have the same atom-to-atom connections, but different shapes, so they are

5 stereoisomers. They are stereoisomers that have different internal dimensions, so they are diastereomers. Because rotation to interconvert one diastereomer into the other is a high-energy process, requiring cleavage of a π bond, they are configurational diastereomers. (In the past, isomers such as these have been called double-bond isomers or geometric isomers.) Any alkene in which one C of the C=C bond is attached to two different groups, and the other C is also attached to two different groups, will have two diastereomeric forms. For alkenes in which each C is attached to one and one other group, we call these two forms cis and trans. So cis-1- chloropropene is the diastereomer in which the Cl and the C 3 group are on the same side of the π bond, and trans-1- chloropropene is the diastereomer in which the Cl and the C 3 group are on opposite sides of the π bond. Remember "cis same side". In a moment we will see another, more general way of naming the two diastereomers of alkenes. If we replace a terminal in propene with a C 3 group, we generate either 1-butene or 2-butene. (If one replaces the internal in propene with C 3, we generate 2-methylpropene, also known as isobutylene.) The number designates the position of the C=C bond in the chain. Not only do 1-butene and 2-butene have their π bond in different locations, but, by necessity, the C bonds have also been moved around, so 1- butene and 2-butene are skeletal isomers. Just as there are in 1-chloropropene, there are two diastereomers of 2-butene: cis- 2-butene and trans-2-butene. They cannot interconvert by rotation, so they are configurational diastereomers. The interconversion of cis and trans alkenes requires cleavage of the C=C π bond. This can happen when light of the proper wavelength is absorbed by an alkene, which promotes one electron from the π to the π* orbital. Because 1/2 bond + 1/2 anti-bond = no bond, there is now no π bond and the molecule is free to rotate about what used to be the π bond.

6 Nomenclature igher alkenes are named as you would expect. The root designates the number of C atoms in the longest chain containing the double bond, the suffix "ene" designates an alkene, the number preceding the root designates the position of the double bond in the chain, and any prefixes designate substituents. An alkene with two double bonds is called a diene, one with three is called a triene, etc., and more than one number is used before the root to designate the position of the double bonds: e.g., 1,3-butadiene or 1,4-cyclohexadiene. A cycloalkene is named so that the C's of the π bond are 1 and 2 and so that the first substituent has as low a number as possible. All cycloalkenes up to 8-membered rings must be cis, so we don't need to indicate whether they are cis or trans in their name. (Try to make a model of transcyclohexene!) Larger cycloalkenes, however, must be so designated: e.g., trans- or cis-cyclodecene. A compound XC=CY may be classified as cis or trans. But what does one do for a compound like, say, 1-iodo-1- bromo-2-chloro-1-propene? This compound can also exist as two diastereomers, but it's not clear which one is cis and which one is trans. In these cases we use the E/Z nomenclature. First we need to assign priorities to the four groups attached to the double bond. Look at the atomic numbers of the two atoms attached to each C of the double bond. One C has I and Br attached: I has higher priority. The other C has C and Cl attached: Cl has higher priority. (It is helpful to circle the higher-priority group attached to each C.) If the two highpriority groups are on the same side of the double bond, the compound is "zusammen" (German for together), or Z; if they are on opposite sides, the compound is "entgegen" (German for against), or E. In the name of the compound, the E or Z follows immediately after the number indicating the position of the double bond.

7 If the two atoms directly attached to a C are the same, one looks for the first difference between the groups attached to each of those atoms to determine which is higher priority. For example, in the compound below, C2 is attached to C1 and C6: no difference. C1 is attached to O, O, and O (the double bond to O is counted as two O atoms); C6 is attached to S,, and. The heaviest atom attached to C6 is heavier than the heaviest atom attached to C1, so C6 has higher priority. Similarly, C3 is attached to C4 and C7: no difference. C4 is attached to C5, C9, and ; C7 is attached to C8,, and. The heaviest atom attached to C4 is the same as the heaviest atom attached to C7, but the second-heaviest atom attached to C4 is heavier than the second-heaviest atom attached to C7, so C4 has higher priority. The two high-priority groups, C4 and C6, are against one another, so the isomer is E. We assign priority to groups in compounds so that we can name the compounds. It is purely a matter of language. It has no chemical reality whatsoever. 3.4 S t abilit y of Alkenes. There are two rules involved in evaluating the relative stability of two alkenes: 1. Trans alkenes are lower in energy than cis alkenes (except when the alkene is in a ring with 10 or fewer members). 2. More substituted alkenes (alkenes with more non- groups attached to the C=C bond) are lower in energy than less substituted alkenes. The reason that cis alkenes are higher in energy than trans alkenes is simple. In a cis alkene, the two large groups are both on the same side of the π bond, so they bump into one another, creating a steric interaction. This interaction is not present in the trans isomer. The steric interaction can cost as little as 0.9 kcal/mol for 2-butene, but it can be larger if larger groups are attached to the C=C bond.

8 The reason that more substituted alkenes are lower in energy than less ubstituted ones is more subtle, and it is still a matter of controversy. Jones gives one explanation; I will give another one, again involving hyperconjugation. The C=C π bond has both a bonding and antibonding orbital associated with it. The π* orbital is not very high in energy, and it can interact with C σ bonds of the type C=C C. (In other words, the C bonds must involve a C that is adjacent to but not part of the C=C bond.) The more C σ bonds that can hyperconjugate with the π bond, the lower in energy the molecule is. Other bonds, e.g. C C bonds, can also interact hyperconjugatively with the C=C π* orbital, as long as they involve a C that is adjacent to but not part of the C=C bond. We mentioned earlier that all cycloalkenes up to 8-membered rings must be cis. The trans isomers are much too high in energy to exist at room temperature. In a trans alkene, the two allylic C atoms are so far apart that one needs at least four C atoms to bridge them. In bridged bicyclic compounds, the C atoms at the bridging points (those C atoms attached to three C atoms of the ring system) cannot participate in a π bond. The reason is that the p orbital of the bridgehead C atom cannot overlap with the p orbital on a neighboring C atom. This phenomenon is called Bred t 's rule. Bredt's rule applies only to smaller bicyclic compounds. In large rings, Bredt's rule can be violated (and is violated regularly).

9 3.5 ybridization in Alkynes. The same logic that told us that the C atoms in ethylene cannot be sp 3 -hybridized tells us that the C atoms in ethyne (acetylene, C C) cannot be sp 3 - or sp 2 -hybridized. Instead, each C atom is sp-hybridized. The s orbital is averaged with just one p orbital to make two new hybrid sp orbitals that are pointing 180 from one another. The energy of the sp orbital is halfway between that of an s orbital and that of a p orbital. One of the sp orbitals is used to make a σ bond with the other C, and the other is used to make a σ bond with. The two remaining, unhybridized p orbitals are used to overlap with two p orbitals on the other C atoms to make two π bonds. Whenever you have a triple bond between two atoms, there are one filled σ MO and two filled π MOs (and the corresponding empty anti-bonding MOs). Again, there are anti-bonding orbitals associated with the C(sp) C(sp) σ bond and the C(p) C(p) π bonds. A C atom that makes one triple bond and one single bond is sp-hybridized. Even though it's hard to tell, the π electrons around the C C bond form a hollow cylinder of electron density. 3.6 ybridization in eteroatoms. eteroatoms also hybridize. We treat most lone pairs as if they were groups to which the heteroatom was bound; usually they reside in hybrid orbitals. If a heteroatom has only σ bonds to its neighbors, it is sp 3 -hybridized. (Examples: dimethyl ether, ammonia.) If a heteroatom has one π bond to one neighbor, it is sp 2 -hybridized. (Examples: pyridine, acetone, protonated acetone.) If a heteroatom has two π bonds to its neighbors, it is sp-hybridized. (Acetonitrile.) These hybridizations have consequences for both structure (bond angles and lengths) and reactivity (lone pairs with certain energies). To summarize: ybrid orbitals are used to form σ bonds and to contain lone pairs. Unhybridized p orbitals are used to form π bonds or are empty. To determine the hybridization of an

10 atom, count the number of σ bonds and lone pairs not used in resonance; you need that many hybrid orbitals. Problems. For each of the following compounds: (1) Fill in any non-bonding pairs of electrons. (2) Indicate the kinds of orbitals (π and σ) that are made. (3) Indicate the hybridization of each non- atom. (4) Indicate the approximate bond angles. (a) Formaldehyde (methanal), 2 C=O. (b) 2-butynal N- methylimine, C 3 C C C=N C 3. (c) Methyl acetate, C 3 CO 2 C 3. (d) Nitrilamine, 2 N CN. (e) Trimethyl borate, (C 3 O) 3 B. We said that lone pairs are put in hybrid orbitals. If the lone pair can be used in resonance, however, it must be in a p orbital for maximum overlap to occur. Therefore: ybrid orbitals are used to form σ bonds and to contain lone pairs not used in resonance. Unhybridized p orbitals are used to form π bonds, hold lone pairs used in resonance, or are empty. Be careful when determining hybridization of heteroatoms in cyclic compounds. Pyridine, pyrrole, furan all have sp 2 - hybridized heteroatoms. In pyrrole and furan one lone pair is involved in resonance; in pyridine the lone pair is not involved in resonance. 3.7 igher alkynes Replacement of an atom in acetylene with a C 3 group gives propyne, "prop" because there are three carbons, and "yne" because there is a triple bond. Propyne has two kinds of atoms, one attached to C(sp) and one attached to C(sp3). Replacement of the latter with another C 3 group gives 1- butyne, and replacement of the former gives 2-butyne, where the number is associated with the position of the triple bond. igher alkynes are called pentyne, hexyne, heptyne, octyne, etc., with associated numbers. Isopropylacetylene is also called 3-methyl-1-butyne. A compound with both a double and a triple bond is called an "enyne", and the positions of the bonds are given numbers this way: 2-octen-6-yne has the double bond in position 2 and the triple bond in position 6. (Not e the double bond is given the smaller number.)

11 The linear nature of the alkyne makes it very difficult to make cycloalkynes. Cyclooctyne is the smallest cyclic alkyne isolable at room temperature. 3.8 Alkyne acidit y The C(sp) orbital is much lower in energy than the C(sp 2 ) or the C(sp 3 ) orbital. As a result, a pair of electrons in such an orbital is quite strongly stabilized, and a carbanion in which the lone pair is in a C(sp) orbital is quite low in energy. This means that RC C alkynes are much more acidic at the terminal C(sp) than most hydrocarbons. On the scale we use to measure acidity, the pka scale, ethyne (acetylene) falls at 25. It is less acidic than 2 O (15) but much more acidic than ethene (37) or ethane (~45). Even N 3, an acid with a central atom more electronegative than C, has a pka of 37, much less acidic than acetylene. 3.9 Degrees of unsaturation Alkanes have the general formula C n 2n+2. Alkenes have two fewer atoms than alkanes, so they have the formula C n 2n, as do cycloalkanes. Compounds with two π bonds (alkynes, alkadienes), one π bond and one ring (cycloalkenes), and two rings (spiro, fused, or bridged alkanes) have the formula C n 2n 2. In fact, for each π bond or ring in a compound, two atoms are removed from the formula. Conversely, if we want to draw a hydrocarbon with a particular formula, C x y, we can calculate how many rings or π bonds it has by the following formula: # rings or π bonds = [ 2x + 2 y ] / 2 The number of rings or π bonds in a compound is called its degrees of unsaturation (a bit of a misnomer, actually, because an unsaturated compound refers to one with a π bond, not a ring, but there you go). Calculating the number of degrees of unsaturation is a quick way of making sure that you have drawn a compound with the right number of atoms. For example, if you were asked to draw one example of C 30 54, the easy way to do it would be to calculate the number of degrees of unsaturation as follows: #

12 rings or π bonds = [ 2(30) ] / 2 = 4 It has four degrees of unsaturation, so any example of C would have to have 30 C atoms and four rings, three rings and one π bond, two rings and two π bonds, one ring and three π bonds, or four π bonds. Suppose your compound has halogen atoms, C x y X z? When you replace a atom with a halogen atom, you don't change the number of rings or π bonds, so halogens can be treated like atoms. Thus: # rings or π bonds = [ 2x + 2 (y + z) ] / 2 Since we're on the subject, let's learn how to deal with chalcogens (O, S) and pnictogens (N, P). If we look at EtO and acetone, C 3 C(=O)C 3, we can see that the number of degrees of unsaturation of each compound C x y O w is the same as C x y (none in the case of EtO, one in the case of acetone). So, in calculating the number of degrees of unsaturation in a compound with O or S, C x y O w X z, we ignore the O atoms. # rings or π bonds = [ 2x + 2 (y + z) ] / 2 What about N atoms? Since nitrogen is trivalent, an organonitrogen compound has one more hydrogen than an equivalent hydrocarbon has, and we therefore subtract the number of nitrogens from the number of hydrogens to arrive at the equivalent hydrocarbon number. Thus, the N-containing compound, C x y N v, has the same number of degrees of unsaturation as C x ( y v ). N C 5 9 N = C 5 8 N removed two unsaturations: one ring and one double bond For example, Trimethylamine, NMe 3 (C 3 9 N), has no degrees of unsaturation, like propane with the formula C 3 8.

13 Acetone hydrazone, C 3 C(=NN 2 )C 3 (C 3 8 N 2 ), has one degree of unsaturation like an hydrocarbon with the formula C 3 6 (e.g. C 3 C=C 2 ) Acetonitrile, C 3 C N, (C 2 3 N) has two degrees of unsaturation like the hydrocarbon with the formula C 2 2 (C C) So, for C x y N v O w X z, # rings or π bonds = [ 2x + 2 (y + z v) ] / 2 Not e that Si, P, and S are treated like C, N, and O, respec tively.