1 Lecture 15

Alkanes represent one of four families (having only carbon and ). These families include , , , and aromatics. The family will provide our main examples for a discussion of the variations of shape in chains and rings resulting from rotation around single bonds. Without a pi bond, the alkanes are less reactive than the other hydrocarbon families and they undergo a more limited number of reactions. The generic term aliphatic is often used to indicate that are not aromatic. If rings are present, this is often modified to alicyclic.

Hydrocarbon Families (carbon and hydrogen only)

Aliphatics and Alicyclics Aromatics H

H C H Alkanes Alkenes Alkynes C C H H H H C C H C H H C C H C C H3C C C CH3

H H H CH3 H ethane propene 2-butyne

cyclopentane cyclobutene cyclooctyne

Cumulenes H H CCC H H 1,2- (allene)

Conformations

Single bonds allow a wide range of motions depending on such factors as bond angles, bond lengths, polarities and whether an open chain or a ring is present. The different possible shapes that result from rotations about single bonds are called conformational changes and the different structures are referred to as conformational . The energy necessary to cause these changes in conformation is usually in the range of 1-15 kcal/mole. This is lower than ambient thermal background energy (approximately 20 kcal/mole at room temperature), and conformational changes tend to occur very rapidly (typically, many thousands of times per second). Organic molecules therefore exist in a dynamic world of constant change. 2 Lecture 15 In general, chains have a wider range of rotational motions possible than do rings. Chains have the potential to rotate 360o about their single bonds, while rings are limited in rotation by attachments to other ring atoms. However, as chains get very long they are less able to rotate freely, due to entanglements with nearby molecules. As rings get very large, they are more able to rotate, even as far as 360o.

chain ring both chain and ring

Newman Projections

A Newman projection focuses on two, attached atoms. The viewing perspective is straight on, looking down the linear bond connecting the two atoms. Sighting down the C-C bond in ethane is a simple, first example. The front carbon (or atom) is represented as a dot while the rear carbon (or atom) is represented as a large circle.

H H front carbon (atom) H C C H rear carbon (atom) H H ethane

The front carbon in ethane has three other attached groups; in this case they are all hydrogen atoms. Additionally, with the front-on view we lose our three dimensional perspective and instead of seeing the 109o bond angles from the hydrogen atoms leaning forward, we see a flat 360o, divided three ways. The angles between the hydrogen atoms are drawn as though they are 120o apart. The angles are not really 120o, but we can’t see this from the flat perspective of the page. Since the dot is the front carbon and nothing is blocking its view, we draw in the bond connections all the way down to the dot.

H H H C C H H H ethane

The rear carbon of ethane also has three additional attached groups, all hydrogen atoms. All three hydrogen atoms are leaning away. We don't see this in a flat Newman projection, and once again the angles between the rear hydrogen atoms appear to be 120o. Since these hydrogen atoms are behind the rear carbon, the lines showing these bonds only go down as far as the circle and then stop, disappearing behind the rear carbon.

3 Lecture 15

H H H C C H H H ethane

Finally, we fill in whatever the three groups are on the front carbon atom and whatever the three groups are on the rear carbon atom. In ethane, there are six hydrogen atoms at these positions.

H

H H In addition to the bond connecting H H the two atoms of a Newman projection H (behind the dot), there are three C C additional bonds on the front carbon H atom and three additional bonds on the H H H H rear carbon atom. ethane H

staggered conformation

A special perspective was chosen in this example. The position of the groups on the front carbon bisect the positions of the groups on the rear carbon in our straight-on view. In ethane, this is the most stable conformation. It goes by the name of staggered conformation. However, it is by no means the only conformation possible. In drawing Newman projections, what is usually done is to select only certain conformations of extreme potential energy (both high energy and low energy) and emphasize these. We have already viewed the low energy staggered conformation in ethane. To view the high energy conformation we must rotate one of the carbon atoms by 60o in our flat two dimensional perspective (I’ll choose the front one).

H H H H H C C H C C H H H H H H ethane ethane The rear carbon atom is 60o H (arbitrarily) fixed, while H H the front carbon atom is The front C-H bonds are H H rotated by 60o about the slightly misdrawn so that o C-C bond. the atoms on both the 60 front and rear atoms can be seen. H H H H H H 60o H staggered conformation eclipsed conformation

4 Lecture 15 If we mark one of the rear hydrogen atoms and one of the front hydrogen atoms (with a square below) to distinguish them from the others, we can see that energetically equivalent, but technically different structures are possible by rotating one hydrogen through 360o (front atom in this case). In doing this we have arbitrarily fixed the rear carbon atom and its attached groups (all hydrogen atoms here).

H HH H HH H HH H H H H rotate rotate H rotate H H rotate H H o rotate rotate 60 o o o 60 60 60o 60 60o H H H H H H H H H H H H HH H H H H H H H H H H o o 0o 60 120o 180o 240o 300 360o staggered eclipsed staggered eclipsed staggered eclipsed staggered conformation conformation conformation conformation conformation conformation conformation

Rotating the three hydrogen atoms (in front) past the three hydrogen atoms (in back) raises the potential energy by 2.9 kcal/mole over the staggered arrangements. Room temperature thermal background energy is approximately 20 kcal/mole, which allows the 2.9 kcal/mole energy fluctuations to occur many tens of thousands of times per second. The two CH3 groups can be viewed as spinning propeller blades (only the bonds are spinning faster). The increase in energy due to eclipsing interactions is referred to as torsional strain. Since there are three such H-H interactions in ethane, the increase in energy for each must be (2.9kcal/mole)/(3) ≈ 1.0 kcal/mole.

= symbol for a hydrogen atom to keep track of rotations

H H H H H H H H

H H H H H H H H H H HH H H HH H H H H Torsional strain per H/H interaction = (2.9 kcal/mole)/3 ≈ 1.0 kcal/mole. 4.0

3.0 2.9 kcal potential mole energy kcal 2.0 mole repeats 1.0

0.0

rotational 0o 60o o o o 300o o angle 120 180 240 360

dihedral o o o o o o o angle 180 120 60 0 60 120 180

Torsional strain specifically refers to an increase in potential energy due to eclipsing interactions of groups on adjacent atoms. In a more general sense, electron-electron repulsions through space are sometimes referred to as steric effects. Steric effects are more severe when groups are larger, sometimes having many atoms and pairs of electrons. The equilibrium percentages can be roughly calculated if the relative energy differences of each conformational are known (we are assuming ∆Go ≈ ∆Ho). In the case of ethane, a 2.9 kcal/mole potential energy difference predicts a preponderance of staggered conformations. Over 99% of the molecules will be staggered in ethane.

5 Lecture 15 - ∆G - ∆H - (2900 cal/mol) 2.3RT 2.3(2 cal/mol-K)(298K) eclipsed 2.3RT -2.12 1 = K = 10 ≈ 10 = 10 = 10 = 130 staggered

HH

H H HH H kcal DH = 2.9 mole eclipsed = 1 H H 131 = 0.008 (0.8%)

H H H 130 staggered = = 0.992 (99.2%) 131

Substitution of Hydrogen with other groups

Switching a hydrogen atom with a methyl group barely changes our analysis of the conformations possible. You might expect the difference in energy between the staggered and eclipsed conformations would be greater because of the larger size of a methyl group over a hydrogen atom, and you would be correct. The eclipsed conformation is 3.4 kcal/mole higher in potential energy than the staggered conformation. Using 1.0 kcal/mol as the energy cost of H/H eclipsing interactions from the ethane calculation (there are two of them), we can calculate a value of 1.4 kcal/mol for the energy cost of a CH3/H eclipsing interaction.

Torsional strain of eclipsed propane = 3.4 kcal/mole = [(H-H) x 2] + (CH3-H)

Torsional strain per CH3/H interaction = (3.4 kcal/mole) - [2 x (1.0 kcal/mole)] = 1.4 kcal/mole

H3C CH CH3 H 3 H H H H

H H H H H 4.0 H HH H H kcal 3.4 3.0 mole potential energy kcal 2.0 mole repeats 1.0

0.0

rotation by o o o 0 60 120 As with ethane, the equilibrium percentages can be roughly calculated from the value of ∆Ho (assuming ∆Go ≈ ∆Ho). For propane, a 3.4 kcal/mole potential energy difference predicts an even greater preponderance of staggered conformations (99.7%) than was found in ethane (99.2%). 6 Lecture 15

- ∆H - (3400 cal/mol) 2.3(2 cal/mol-K)(298K) eclipsed 2.3RT -2.48 1 = K ≈ 10 = 10 = 10 = 302 staggered

staggered = 302 303 = 0.997 (99.7%) 1 eclipsed = = 0.003 (0.3%) 303

Butane is our next example. In butane we have four attached carbon atoms and we also have even more choices in how to draw the Newman projections. We could use the C1-C2 bond or the C2-C3 bond. (Why not include the C3-C4 bond?) In addition we could view our projection from either perspective C1ÆC2 or C2ÆC1 and C2ÆC3 or C3ÆC2. In butane, the C1ÆC2 and C2ÆC1 perspectives produce slightly differing views, while the views of C2ÆC3 and C3ÆC2 are equivalent because they are at the center of a symmetrical molecule.

C1 C2 H H H H C C 1 2 34 1 2 H C C C C H

C2 C3 H H H H

C2 C3

CH3CH2 CH3CH2 CH3 CH3

H H H H H H H H

H H H H H H H H

H H CH3 CH3

C C C1 C2 C1 C2 C2 C3 2 3

These two perspectives are energetically These two perspectives are energetically equivalent, but differ in where the ethyl equivalent, and similar in appearance. group, CH3CH2, is drawn (front or back).

Using the C2ÆC3 bond, we can generate six different Newman projections before we begin to repeat ourselves. We need to add the three additional groups at each of the carbon atoms. On the C2 carbon atom of butane, two hydrogen atoms and a methyl group are attached. On C3 there are also two hydrogen atoms and a methyl group attached. If we arrange these groups in a staggered conformation with one methyl pointing up (the rear carbon) and the other methyl pointing down (the front carbon), we obtain the lowest energy staggered conformation. If you build molecular models, this will likely be intuitively obvious to you. The larger methyl groups are as far apart in space as they can possibly get. To simplify our analysis we will fix the rear carbon atom, while we rotate the front carbon atom in 60o increments to obtain several additional conformational isomers. 7 Lecture 15

= symbol for a hydrogen atom to keep track of rotations "anti" no special name "gauche" "syn" "gauche" no special name "anti" (antiperiplanar) (anticlinal) (synclinal) (synperiplanar) (synclinal) (anticlinal) (antiperiplanar) CH CH CH3 CH3 CH3 CH3 3 3H CH3 H CH3 H H H H3C H H CH3 H

H C H H 3 H H H H H H H H H H H H H CH3 H H H H CH3 CH3 H 1 23 4 56repeat ≈ 66% < 1% ≈ 17% << 1% ≈ 17% < 1%

5.0

4.0

3.0 potential energy 4.5 kcal 2.0 3.7 3.7 mole repeats 1.0 0.8 0.8 0.0

rotational 0o 60o o o o 300o o angle 120 180 240 360

dihedral angles of o 180o 120o 60o 0o 60o 120o 180 CH3 groups

8 Lecture 15

3n Energy (CH ) + O2 n CO + n H O + 2 n 2 2 2 (heat or work)

cycloalkanes

Cycloalkanes have a variable starting energy based on the 3n cycloalkanes = n (CH2) + 2 O2 (gas) number of (CH2)s in the ring. The all look different here.

o ∆H reaction

Identical final products per (CH2). n CO2 + n H2O They all look the same here.

Total o ∆Ho ∆H com b com b Difference from Total CH2's ring n = per mole per (CH2) n-alkane CH2 in ring strain

C3 -499.8 -166.6 9.2 3 27.6 C4 -656.0 -164.0 6.6 4 26.4 C5 -793.4 -158.7 1.2 5 6.2 C6 -944.8 -157.4 0.0 6 0.0 C7 -1108.3 -158.3 0.9 7 6.2 C8 -1269.2 -158.6 1.2 8 9.7 C9 -1429.6 -158.7 1.4 9 12.6 C10 -1586.8 -158.6 1.3 10 12.6 C11 -1743.1 -158.4 1.0 11 11.3 C12 -1893.4 -157.6 0.2 12 2.4 C13 -2051.9 -157.8 0.4 13 5.2 C14 -2206.5 -157.4 0.0 14 0.0 C15 -2363.5 -157.5 0.1 15 1.9 C16 -2521.0 -157.5 0.1 16 2.0 *C17 -2673.2 -157.2 -0.2 17 -3.3* n-alkane -157.4 - - -

All energies in kcal/mole * It is possible that C has fewer gauche interactions than a straight chain. 17

1. angle strain (overlap) = classical strain = Baeyer strain 2. torsional strain = eclipsing strain = Pitzer strain 3. van der Waals strain = steric strain = Prelog strain

9 Lecture 15 Rings are classified by size as small (n = 3, 4), normal (5, 6, 7), medium (8, 9, 10, 11) and large (12 and larger). The small and normal size ring systems are commonly encountered in nature and laboratory work. The most common rings, by far, are six atom and five atom rings. A few larger rings and polycyclic ring systems are included at the end of this topic for reference purposes. We will not emphasize their special complexities in our course.

Potential Range of Ring Strain per CH (possible reason) Energy 2

kcal small ring systems = 3-4 atoms 9.2 and 6.6 (mainly angle strain and torsional strain) mole normal ring systems = 5-7 atoms 1.2, 0.0 and 0.9 (C5 is mainly torsional strain and 10 C7 is torsional and van der Waals strain)

8 medium ring systems = 8-11 atoms 1.2, 1.4, 1.2 and 1.0 (mainly van der Waals strain)

large ring systems = 12 and larger -0.2 to + 0.4 (mainly van der Waals strain) 6

4

2

0

3 4 5 6 7 8 9101112 13 14 15 16 17

small normal medium large

Ring size (CH2)n Cyclohexane

Cyclohexane rings (six atom rings in general) are the most well studied of all ring systems. They have a limited number of, almost strain free, conformations. Because of their well defined conformational shapes, they are frequently used to study effects of orientation or steric effects when studying chemical reactions. Additionally, six atom rings are the most commonly encountered rings in nature. As in the previous ring examples, cyclohexane does not choose to be flat. Slight twists at each carbon atom allow cyclohexane rings to assume much more comfortable conformations, which we call chair conformations. (Chairs even sound comfortable.) The chair has an up and down shape all around the ring, sort of like the zig-zag shape seen in straight chains (...time for models!).

10 Lecture 15

C C C C C C

lounge chair - Used to kick back and relax while you study your organic chemistry. chair conformation

Cyclohexane rings are flexible and easily allow partial rotations (twists) about the C-C single bonds. There is minimal angle strain since each carbon can approximately accommodate the 109o of the tetrahedral shape. Torsional strain energy is minimized since all groups are staggered relative to one another. This is easily seen in a Newman projection perspective. An added new twist to our Newman projections is a side-by-side view of parallel single bonds. If you look carefully at the structure above or use your model, you should be able to see that a parallel relationship exists for all C-C bonds across the ring from one another.

There are three sets of parallel C-C bonds in cyclohexane rings.

Two simultaneous Newman projection views are now possible as shown below. Remember, that any two bonds on opposite sides of the ring can be used and they can viewed from front or rear directions. H H

H H H H H2 C C H C H H C H H Newman C H projection C C H H H C H H2 H H H H Chair conformations can be viewed down any two parallel All groups are staggered in chair bonds (from either side) as two parallel Newman projections. conformatios of cyclohexane rings. In this example, the view is from the left side.

Once the first bond is drawn in a Newman projection of a chair conformation of cyclohexane ring, all of the other bonds (axial and equatorial) are fixed by the staggered arrangements and the cyclic connections to one another. Once you have drawn all of the bonds in a Newman projection, you merely fill in the blanks at the end of each line, based on the substituents that are present. Boat Conformation A Newman projection viewing down the two parallel bonds of the boat conformation clearly shows the increased torsional strain from eclipsing interactions. 11 Lecture 15 H2 C H2 H C H H H Newman projection

H H Boat conformation H H of cyclohexane All groups are eclipsed in a boat conformation of cyclohexane rings, increased torisional strain.

A chair does not immediately become a boat, and then the other chair. There is an even higher transitory conformation, which is call a half chair. Due to the torsional strain and bond angle strain this conformation is even higher in potential energy than the boat. We can consider the half chair as a transition state that exists only on the way to a chair or boat conformation. The potential energy of the half chair is estimated to be about 11 kcal/mole. A potential energy diagram for interconverting the two possible chair conformations would look something like the following.

chair 1 half chair 1 twist boat 1 boattwist boat 2 half chair 2 chair 2 12

10 Background thermal energy is about 20 kcal/mole, more than enough to make all of 8 these conformations accesible.

PE 6 ≈ 11 ≈ 11

4 ≈ 5.0 ≈ 6.4 ≈ 5.0 2

0 Energy changes of cyclohexane conformations.

We will view the interconversion of the two cyclohexane conformations, simplistically, as chair 1 in equilibrium with chair 2 via the boat conformation.

12 Lecture 15 van der Waals strain, (flag pole interactions)

Boat Chair 1 Groups along parallel bonds in the Chair 2 ring are eclipsed, which increases torsional strain. Groups pointing All groups are staggered, which All groups are staggered, which into the center of the ring on the minimizes torsional strain energy. allows maximum stabilization. carbon atoms pointing up crowd one another, which increases van der Waals strain.

Equilibration, back and forth, between the two chair conformations is rapid at room temperature and occurs on the order of 80,000 times per second. At lower temperatures interconversion is much slower (there is less thermal background energy). At -40oC the interconversion occurs about 40 times per second, at -120oC interconversion occurs about once every 23 minutes and at -160oC it is estimated to occur once every 23 years. The equilibration can occur in either of two directions as shown below.

a boat 1

a b

chair 1 chair 2 b

boat 2

Notice that all axial positions (top and bottom) become equatorial positions with the flip-flop of two chairs. Of course, all equatorial positions become axial positions at the same time.

= axial = equatorial = equatorial axial on top becomes equatorial on top = axial

transition via "boat"

equatorial on top becomes axial on top

chair 1 axial on bottom becomes chair 2 equatorial on bottom

13 Lecture 15 Since the two chair conformations are so much more stable than the boat or half chair over 99.99% of the molecules are in one of the two chair conformations. If all of the ring substituents are hydrogen atoms, there should nearly be a 50/50 mixture of two indistinguishable chair structures.

−∆G - (6400 cal/mole) boat 2.3RT K = = 10 = 10 (2.3)(2 cal/mol-K)(298 K) = 10-4.60 = 2.1x10-5 = 0.002% chair 99.998%

−∆G - (0 cal/mole) chair 1 2.3RT (2.3)(2 cal/mol-K)(298 K) 0 1 50% K = chair 2 = 10 = 10 = 10 = = 1 50%

Real molecules actually exist that illustrate most of the cyclohexane conformations mentioned, above, as substructures (chair, boat and twist boat). has four chair cyclohexane rings in its complicated tricyclic arrangement. Twistane has a good example of a twist boat conformation and is isomeric with adamantane. Norbornane has a boat conformation locked into its rigid bicyclic framework with a bridging CH2 holding the boat shape in place. Each of these is highlighted below. This is another excellent opportunity to let your models help you see what we have been discussing, above.

chair (from back) boat twist boat chair chair

chair side view front view Adamantane Twistane - has a twist boat conformation Every face is a Norbornane - has a chair conformation. boat conformation

Mono Substituted Cyclohexanes A single substitution of a hydrogen atom with another group complicates the analysis of cyclohexane conformations. There are still two rapidly equalibrating chair conformations, but they are no longer equal in potential energy. H

R H

R "R" is on the bottom in both conformations. Of course, if you turn the ring over, then "R" is on the top in both conformations. R is equatorial in the left conformation and axial in the right conformation.

A Newman perspective using the C1ÆC2 and C5ÆC4 bonds can help to evaluate which of these two conformations is more stable. Both chair conformations have an all staggered orientation about the ring, but chair 2 has gauche interactions of the substituent and the ring, which are not present in chair 1. 14 Lecture 15 H H H H "R" is on the bottom in both H conformations. Of course, if 3 H 2 H 2 4 3 6 H 5 1 H you turn the ring over, then 5 1 R H 4 "R" is on the top in both conformations. R is equatorial H 6 H H R in the left conformation and H axial in the right conformation. Newman projections:

C5 C4 and C1 C2

H H H H H H H H

H H H R H H H R

Equatorial "R" is anti to the ring on two sides. Axial "R" is gauche to the ring on two sides. Only one shows in this Newman projection. Only one shows in this Newman projection.

If we view down the C1ÆC6 bond and the C1ÆC2 bond, it is clear that an axial R substituent is gauche with both sides of the ring. Any axial substituent in a cyclohexane ring really has two gauche interactions. H H

4 2 3 6 4 2 H 5 1 H 3 6 5 1 H

H R R viewed from the rear viewed from C1 C6 C1 C2 the front

Newman projection: Newman projection:

H H

H ring C2 ring C6 H

ring C5 H ring C3 H R R is also gauche on the R other side of the ring R is also gauche on the other side of the ring

We have previously seen that a CH3/CH3 gauche interaction raises the potential energy about 0.8 kcal/mole. 15 Lecture 15 CH3 CH 3 gauche CH3

CH3

anti CH3/CH3 = 0 kcal/mole gauche CH3/CH3 = 0.8 kcal/mole

Two such relationships should raise the energy by approximately (2)x(0.8 kcal/mole) = 1.6 kcal/mole. The actual value observed for an axial CH3 in cyclohexane relative to equatorial is 1.7 kcal/mole, very close to what we expect.

− ∆G - (1700 cal/mole) chair 1 2.3RT 1 = 10 = 10 (2.3)(2 cal/mol-K)(298 K) = 10-1.2 = 6% K = chair 2 = 0.058 = 17 94%

12 half chair 1 half chair 2

10

8 boat 6% PE 6 94% ≈11 ≈11 4 ≈ 6.4 chair 2 2 chair 1 kcal ∆G = 1.7 mole 0 Energy changes of methylcyclohexane conformations.

The two gauche interactions of axial R substituents and the ring carbon atoms in cyclohexane structures, are called 1,3-diaxial interactions. In the axial position, the substituent R, is forced close to the other two axial groups on the same side of the ring. Since these are both three atoms away from the ring carbon atom with the R substituent, the 1,3-diaxial descriptor is appropriate.

R H H 1,3-diaxial interaction 1 3' (≈ two gauche interactions) 2' 3 2

Problem 10 - Which boat conformation would be a more likely transition state in interconverting the two chair conformations of methylcyclohexane...or are they equivalent? Explain your answer.

16 Lecture 15 flip "R" side down first R boat 1

first boat 2 R R

flip non-"R" side down

When the CH3 group is equatorial in the chair conformation, the equatorial CH3/ring relationships are anti and highlighted below. There is one such relationship on each side of the methyl. Just as in butane, the anti configuration is preferred since it minimizes torsional repulsion of the larger groups. anti H H ring A A ring CH3 H H B CH3 is anti to both sides of the ring when equatorial.

CH3 anti H

ring H B ring CH3 H

A wide range of substituents has been studied and almost all show a preference for the equatorial position. Several examples are listed in the table below. The actual energy difference between equatorial substituents and axial substituents is often called the A value (axial strain). Larger A values indicate a greater equatorial preference for the substituent due to larger destabilizing 1,3-diaxial interactions (or double gauche interactions). axial

equatorial S

S axial PE ∆Go = equatorial S = substituent group (table of energy values below)

17 Lecture 15 Substituent ∆Go (A value) Substituent ∆Go (A value) -H 0.0 -CH2OH 1.8 -CH3 1.7 -CH2Br 1.8 -CH2CH3 1.8 -CF3 2.4 -CH(CH3)2 2.1 -O2CCH2CH3 1.1 -C(CH3)3 > 5.0 -OH 0.9 -F 0.3 -OCH3 0.6 -Cl 0.5 -SH 1.2 -Br 0.5 -SCH3 1.0 -I 0.5 -SC5H6 1.1 -CH=CH2 1.7 -SOCH3 1.2 -CH=C=CH2 1.5 -SO2CH3 2.5 -CCH 0.5 -SeC5H6 1.0 -CN 0.2 -TeC5H6 0.9 -C5H6 (phenyl) 2.9 -NH2 1.2(C6H5CH3), 1.7(H2O) -CH2C5H6 (benzyl) 1.7 -N(CH3)2 1.5 (C6H5CH3), 2.1(H2O) -CO2H 0.6 -NO2 1.1 -CO2 2.0 -HgBr 0.0 -CHO 0.7 -HgCl -0.2 -MgBr 0.8

The energy differences in the table between equatorial and axial positions for substituent groups from methyl through ethyl, isopropyl and t-butyl appear puzzling. Ethyl definitely is larger than methyl and isopropyl is larger than either of those two, but they all have similar preferences for the equatorial position. The value for the t- butyl group, on the other hand, suddenly jumps to a much higher value. Examination of the three dimensional axial conformation provides insight into this observation. 5 PE 4 equatorial axial difference in ∆G kcal 3 mole 2 1

methyl ethyl isopropyl t-butyl 18 Lecture 15 Compare axial interactions on the top face of the two conformations. Rotation is possible around R this bond, which allows H H R R different "R" groups to face H R H C in towards the center of the ring. H C C R H

R chair 1H chair 2

R R R substituent ∆G Keq axial PE H H H methyl +1.7 5% / 95% ∆G = equatorial H H CH3 ethyl +1.8 5% / 95% H CH3 CH3 isopropyl +2.1 3% / 97% CH3 CH3 CH3 t-butyl >5.0 0.02% / 99.98%

H H Severe crowding R C when a nonhydrogen R H C atoms faces into the H middle of the ring. H H

19 Lecture 15 Disubstituted Cyclohexanes Here is an excellent example of why memorization does not work in organic chemistry. Not only is it easier to learn a limited number of basic principles and logically use them in newly encountered situations, it is down right impossible to memorize every possible situation you might encounter. When we add a second substituent to monosubstituted cyclohexane rings, the possibilities increase tremendously. Seven flat ring structures are drawn below, which just emphasize the top and bottom positions on the ring. Flat structures can be drawn as time average approximations between two interconverting chairs, although in actuality none of the structures are flat. These structures are only used as a short hand to show that substituents are on the same side or are on opposite sides. In biochemistry they are sometimes called Haworth projections and commonly used with cyclic sugar molecules. We will examine stereochemical subtleties of these structures in our next chapter.

R R R R R R R R

R R

R R

R R 1,1-di "R" cis-1,2-di "R" trans-1,2-di "R" cis-1,3-di "R" trans-1,3-di "R" cis-1,4-di "R" trans-1,4-di "R"

Even though we have gone from a single monosubstituted cyclohexane to seven disubstituted cyclohexane rings, the situation is even more complicated yet! Just as our monosubstituted cyclohexane had two chair conformations, there are two possible chair conformations for each of the flat structures shown above. In some cases the two conformations are equivalent in energy, but in other cases one conformation is preferred. We need a systematic method of analysis or we will quickly become hopelessly lost in the wilderness of flip-flopping cyclohexane rings. You should adopt the following strategy for every cyclohexane analysis.

Method to analyze cyclohexane conformers with a variable number of substituents.

1. Draw a cyclohexane chair using a bond-line formula.

2. Add in axial and equatorial positions (the ring points to the axial positions).

3. Add substituents to positions according to the name of the structure. It is generally easier to visualize substituents drawn on the extreme left or extreme right carbon atoms, because those bonds will be in the plane of the paper, so these are good places to draw in your first substituent. This approach is a 4. Draw the other chair conformation (flip up and flip down). lot more reasonable than brute force 5. Use the table of A values to account for the energy expense of an axial substituent. memorization. 6. In addition to 1,3-diaxial interactions, look for gauche interactions whereever substituents are substituted 1,2 (vicinal substitution).

7. Evaluate which conformation is preferred from the available energy values and determine what is the approximate equilibrium distribution.

-∆G K = 10 2.3RT

20 Lecture 15 Example 1 cis 1,4-dimethylcyclohexane

1.Draw the bond line formula of a chair.

2. Add axial and equatorial positions (ring points axial).

You can simplify this step somewhat by just adding the axial and equatorial positions of the substituted 4 carbon atoms. Include both axial and equatorial positions at substituted positions, even if a hydrogen 1 2 3 atom occupies one of them. If you can draw the first structure, you should be able to draw any possible structure. ...or just this, if 1,4-disubstituted.

3. Add the indicated substituents. Generally the easiest positions to fill are the positions on the far left and far right carbon atoms (1 and 4, just above), since their representation is drawn in the plane of the page. At least the first substituent should be drawn at one of these positions and then fill in any other substituent(s) as required.

CH3 cis = same side, so both CH3 groups should be on the same side (top side H H3C 4 as written in this example). If you H3C 4 1 2 3 turned the structure over, the two 1 2 3 methyl groups would still be on the same H side, but that side would be the bottom. H 1,4 substitution tells One CH3 is axial and one us that the second CH is equatorial, but both substituent goes here. 3 are on the same side.

4. Draw the other conformation (flip up, flip down). Axial/equatorial interchange occurs but the structure is still cis.

CH3 flip down CH3 flip H up H3C 4 2 H 1 1 3 CH3 3 2 4 H H

5. There is one axial CH3 (+1.7 kcal/mole) and one equatorial CH3 in each conformation.

6. 1,2-gauche interactions are not applicable in this example, since there are no vicinal substituents.

7. Since the two conformations are equivalent in energy we would expect a 50/50 mixture.

21 Lecture 15 − ∆G - ( 0 cal/mole) chair 2 2.3RT (2.3)(2 cal/mol-K)(298 K) 1 K = = 10 = 10 = 100 = = 50% chair 1 1 50%

12 half chair 1 half chair 2 10

8 50% boat PE 50% 6 chair 1 chair 2 4

2 ∆Goverall = 0.0 kcal kcal ∆G = 1.7 ∆G = 1.7 mole mole 0 Energy changes of cis-1,4-dimethylcyclohexane conformations. Zero is a strain free cyclohexane reference point.

Example 2 trans-1,2-dimethylcyclohexane 1. See above. 2. Add axial and equatorial positions wherever substituents are present (the ring points to axial).

1 2

3. Add the indicated substituents.

trans = opposite side, so the second CH3 should be on the bottom side H H3C since we drew the first CH3 on the 1 2 top side. H3C 1 2 H CH 1,2 substitution tells 3 H us that the second Both CH3 groups are substituent goes here. equatorial, but are on the opposite sides. 4 & 5. Draw the other conformation and estimate the energy expense of each conformation. CH flip H flip 3 up down H3C H 1 2 H CH3 H CH3 kcal 0 axial substituents kcal 2 axial substituents (opposite sides) = 2(1.7) = 3.4 1 gauche interaction = 0.8 mole mole 0 gauche interaction

22 Lecture 15 6. Chair 1 has a gauche relationship between the two CH3 substituents, which increases energy of that conformation by +0.8 kcal/mole.

H3C

One gauche CH3 interaction.

7. The potential energy difference between the two conformations is (3.4 – 0.8) = 2.6 kcal/mole. Use this value to calculate Kequilibrium. − ∆G - ( 2,600 cal/mole) chair 2 2.3RT (2.3)(2 cal/mol-K)(298 K) 1 1.3% K = = 10 = 10 = 10-1.9 = = chair 1 79 98.7% 12 half chair 1 half chair 2 10

≈1% 8 chair 2 boat PE 6 ≈99% chair 1 4 kcal ∆Gchange = (3.4 - 0.8) = 2.6 mole

2 kcal ∆G = 3.4 kcal mole ∆G = 0.8 mole 0 Energy changes of trans-1,2-dimethylcyclohexane conformations.

Example 3 trans- 1-t-butyl-3-methylcyclohexane 1. See above.

2. Add axial and equatorial positions wherever substituents are present (the ring points to axial).

3. Add the indicated substituents.

CH3 CH trans = opposite side, so the CH3 3 C H3C should be on the bottom side since 1 we drew the t-butyl on the top side. C H3C 3 H3C H 1 H H3C 3

H CH3 1,3 substitution tells us that the second The t-butyl group is equatorial substituent goes here. and the CH3 group is axial, and they are on opposite sides.

23 Lecture 15 4 & 5. Draw the other conformation and estimate the energy expense of each conformation.

CH3 H3C CH flip H C 3 3 C up C flip H H3C down H H C 2 3 H H CH3

CH3

kcal kcal 1 axial CH3 substituent = 1.7 1 axial t-butyl substituent > 5.0 0 gauche interaction = 0 mole mole 0 gauche interaction = 0.0

6. Gauche relationships are not applicable because substituents are not 1,2 (vicinal).

7. The potential energy difference between the two conformations is (5.0 – 1.7) = 3.3 kcal/mole. Use this value to calculate Kequilibrium.

− ∆G - ( 3,300 cal/mole) chair 2 2.3RT = 10 = 10 (2.3)(2 cal/mol-K)(298 K) -0.004 1 = 0.3% K = chair 1 = 10 = 255 99.7% 12 half chair 1 half chair 2 10

8 0.3% 99.7% boat chair 2 PE 6

kcal 4 G = (5.0 - 1.7) = 3.3 chair 1 ∆ change mole kcal 2 ∆G > 5.0 mole kcal ∆G = 1.7 mole 0 Energy changes of trans-1-t-butyl-3-methylcyclohexane conformations.

24 Lecture 15 Problem 14 – Draw all isomers of dimethylcyclohexane and evaluate their relative energies and estimate an equilibrium distribution for the two chair conformations.

Use the following approximate energy values for substituted cyclohexane rings.

One axial methyl group increases the potential energy by 1.7 kcal/mole, Two axial methyl groups, on the same side (cis), increase the potential energy by 5.5 kcal/mole, Three axial methyl groups, on the same side, increase the potential energy by 12.9 kcal/mole and 1,2 gauche methyl groups increase the potential energy by 0.8 kcal/mole. a. Potential Energy kcal mole 6.0

5.0

4.0

3.0

2.0 ∆G = 2x(1.7) = 3.4

1.0

0.0

AB CD EF GH IJ KL MN

trans-1,4 cis-1,4 trans-1,3 cis-1,3 trans-1,2 cis-1,2 1,1 25 Lecture 15 Problem 15 – Estimate a value for the strain energy of the trimethylcyclohexane structures provided. Use the same energy values from the previous problem.

One axial methyl group increases the potential energy by 1.7 kcal/mole, Two axial methyl groups, on the same side (cis), increase the potential energy by 5.5 kcal/mole, Three axial methyl groups, on the same side, increase the potential energy by 12.9 kcal/mole and 1,2 gauche methyl groups increase the potential energy by 0.8 kcal/mole. 1. Reference compound 2 34 (no features with strain)

5 6 7 8

9 10 11 12

13 14 15 16

17 18 19 20

21 22 23 24

26 Lecture 15 Problem 16 - Both cis and trans 1-bromo-3-methylcyclohexane can exist in two chair conformations. Evaluate the relative energies of the two conformations in each isomer (use the energy values from the table presented earlier). Estimate the relative percents of each conformation at equilibrium using the difference in energy of the two conformations from you calculations. Draw each chair conformation in 3D bond line notation and as a Newman projection using the C1ÆC6 and C3ÆC4 bonds to sight down.

2 Br CH3 1 3

6 4 5 1-bromo-3-methylcyclohexane

Problem 17 – a. Our assumption is that it is mainly electron-electron repulsion that contributes to the increased torsional strain of eclipsed or gauche conformations. Using this reasoning, predict whether a carboxylic acid or carboxylate anion would have a larger A value as an axial substituent on cyclohexane. Check your prediction in the table of A values, p 46. Make your prediction before you look.

O OH O O C C

H H

carboxylic acid carboxylate substituent substituent b. Both -CH3 and -CF3 are approximately the same size, but their A values are 1.7 and 2.4 kcal/mole, respectively. Does this seem consistent with your conclusions from part a?

CF CH3 3 methylcyclohexane trifluoromethylcyclohexane

Problem 18 - Draw the other chair conformation for each part below. If an axial X is assumed to be less energetically expensive than an axial methyl, which conformation of each part is more stable? a. X b. X

H C H 3 H H

H CH3

27 Lecture 15 c. H d. CH3

H C X X 3 CH3

H H f. e. X CH3 H

X H H

H3C CH3

Problem 20 - Oxidation of alcohols (ROH) to carbonyls (C=O) is a reaction we will study later. It is thought to occur in two steps. The first step is formation of an inorganic ester group with an oxide of chromium. The second step involves a base removing a proton, formation of a π bond and elimination of the large chromium complex.

O H O Base CrO3 O O Cr O O Cr O

H step 2 step 1 O O cyclohexanol chromium ester of cyclohexanone cyclohexanol

Assuming we don’t know whether step 1 or step 2 is the slow step of the reaction (a pretty safe assumption at this point in the book), make a prediction about whether an axial alcohol or an equatorial alcohol (shown below) would oxidize faster to the cyclohexanone product. First assume that step 1 is the slow step of the reaction and then assume that step 2 is the slow step of the reaction. The large t-butyl group is used to conformationally lock the cyclohexane ring into one preferred conformation. Explain your answers. H OH Which alcohol reacts slower...... if step 1 is the slow step of the mechanism...? OH H ...if step 2 is the slow step of the mechanism...? equatorial "OH" axial"OH"

Problem 22 - Make an estimate of the increase in potential energy for two axial methyl groups on the same side of a cyclohexane using the following data (A value for methyl = 1.7 kcal/mole). Explain your reasoning.

equilibration over a catalyst These two structures are not conformational isomers. kcal They are stereoisomers. ∆G = -3.7 mole