Pi Bonding

Chem 106E. Kwan

-Bonding

Scope of Lecture

Eugene E. Kwan

Helpful References

-bonding

1. Weinhold, F.; Landis, C. Valency and Bonding: A NaturalBond Orbital Donor-Acceptor Perspective Cambridge,Cambridge University Press, 2005.

2. Clayden, J.; Moran, W.J. "The Twisted Amide 2-Quinuclid-one: 60 Years in the Making." ACIE 2006, 45, 7118-7120.

3. Carroll, F.A. Perspectives on Structure and Mechanism inOrganic Chemistry. Pacific Grove, California, Brookes/ColePublishing Company, 1998.

Key Questions(1) Which is more stabilized by conjugation?

-Bonding

s-cis s-transbisected

(2) Are there significant differences in the -bonding ofbenzene and borazine?

NB

NBN

B

(3) Is this an amide? How do -lactams work?

N O

N

SHROCN

MeMe

COOHO

(4) Is this aromatic?

to * hyperconjugation cooperative andanticooperative

arrays

the Hückelapproximation

strained olefinstwisted amides

Hückel vs.Möbius topologyand aromaticity

0 20 40 60 80 100 120 140 160 180

0

1

2

3

4

5

6

7

8

rela

tive

ener

gy (k

cal/m

ol)

C=C-C=C dihedral angle (degrees)

slightlytwisteds-cis

s-trans

bisected

s-trans:

S = 0.22

estimated energy/interaction: 14.9 kcal/mol

s-cis:

S = 0.20

estimated energy/interaction: 12.8 kcal/mol

Chem 106E. Kwan

Conjugated OlefinsLecture 6: -Bonding

Consider butadiene. Please rank its conformers in terms oftheir stability:

s-cis s-transbisected

Of course, the bisected form is "not conjugated," and thereforequite unstable. In donor-acceptor terms, it does not possesscomplementary, bi-directional to * interactions.

Here is butadiene's rotational energy profile (B3LYP/6-31g(d)).The C=C-C=C dihedral angle is incrementally frozen, and allthe other geometric parameters are allowed to relax. This issometimes called a "relaxed scan."

The /* overlap in the s-trans conformation is a bit better thanin the s-cis conformation, a general trend in dienes. Here arethe PNBOs:

Thus, the ranking is s-trans (0 kcal/mol), slightly twisted s-cis(3.5 kcal/mol), bisected (7.6 kcal/mol). The twisting is due tothe steric interaction between the in-plane C-Hs.

We can consider the perturbative estimates of the to *donor-acceptor interactions a measure of the "conjugationenergy." (Many complicated definitions of this exist, but weshall not trouble ourselves with such details.) Note that(14.9 - 12.8) x 2 = 4.2, which is about the s-cis/s-trans difference.

Do you think the stabilization in acrolein is bigger or smallerthan in butadiene?

butadiene

O

acrolein

Chem 106E. Kwan


In acrolein, there are two donor-acceptor interactions (shownbelow as PNBOs):

(1) (C=C) to *(C=O) (S=0.26, interaction energy: 20.4 kcal)

(2) (C=O) to *(C=C) (S=0.17, interaction energy: 6.5 kcal)

So, the electronegativity imbalance in the carbonyl groupimproves interaction (1) but makes interaction (2) worse.Overall, the stabilization is worse. This is a generalfeature of donor-acceptor interactions: equal, bidirectionalinteractions are better than unequal ones.

Given your knowledge, what can you predict about glyoxal?

The overall estimate of resonance energy, 27.1 kcal/mol, is lessthan in butadiene, where it was 29.8 kcal/mol. (All of these calculations were done at B3LYP/6-31g(d).)

OO

H

H

The directionality of the donor-acceptor interactions is onceagain matched (look at the higher isosurfaces on the inside).However, not all double bonds are created equal; these(C=O) to *(C=O) interactions are only worth 7.7 kcal/mol.

S=0.19

-0.25-0.41 -0.22

-0.23

+0.21 +0.23

What do you think happens with all s-trans hexatriene? Willthe resonance energy here just be _____ times the to *interaction in butadiene?

Chem 106E. Kwan


(1) (inside) to *(outside): S=0.22

inside

outside

(2) (outside) to *(inside): S=0.22

Evidently, the inside olefin is a bit different from the ones on theoutside. Let's add up all the resonance stabilizations:

2 x 15.7

2 x 14.4total resonance energy

is 60.2 kcal/mol

In butadiene, the resonance energy/olefin is 14.9 kcal/mol; hereis 20.1 kcal/mol. So there is a cooperative benefit to havingthree olefins together.

In octatetraene, the effect is even stronger:

inside to outside: 2 x 15.9inside to inside: 2 x 16.4outside to inside: 2 x 14.6

average: 15.6 kcal/mol

As the chain gets longer, there are more and more olefins onthe inside and fewer terminal ones (by proportion). So theresonance stabilization/olefin levels out eventually.

Why are the hyperconjugations on the interior stronger?

The polyolefins get slightly positively charged at the ends, asshown by the NBO charges below. Note the alternation:

As noted above, this asymmetry reduces stabilization.

-0.10

-0.24

-0.20

-0.40

Sometimes, the olefins are not in a linear array, but are "cross-conjugated" like in the example below. Will this arrangementbe more or less stable than the linear array in the hexatrieneexample given above?

Chem 106E. Kwan

Cross-ConjugationLecture 6: -Bonding

Now that you know about these "anticooperative effects,"can you predict which of these will have more resonancestabilization?

The resonance energy is less. Here are the second-orderperturbative estimates from the NBO program:

12.0(S = -0.20)

11.7(S = -0.20)

Certainly, overlap is reduced in this s-cis array, but the chargedistribution shows an additional effect (in s-cis-butadiene, theestimate was 12.8 kcal/mol):

The chain is somewhat postively charged in its interior. Thishas an obvious polarizing disadvantage. Another way to thinkabout it is that the central olefin is "busy" donating to twoacceptors, and is not as good of a donor.

Note that typical resonance structures predict some sort ofpolarization, although they don't indicate its direction:

In general, cross-conjugated systems ("open charge transfer")are less stable than linearly-conjugated systems ("closedCT"). What can you predict about this system?

B vs.H

BH

H

Same thing: the average donations are 24.2 kcal/mol (left)and 26.2 (right).

NB

NBN

B

The compound on the right is "borazine," the "inorganicbenzene." See Pietro, Inorg Chem 1989 28 544.

Benzene is much more stable. Here's a simple donor/acceptorresonance explanation:

(1) The Pauling electronegativities of boron and nitrogen are 2.04 and 3.04, so even though boron is "electron deficient," it still isn't more electronegative than nitrogen.

(2) Therefore, the B-N bond is a great acceptor from the B direction.

(3) And the B-N bond is a great donor from the N direction.

If we represent the donor-acceptor interactions with resonancearrows, we get:

Chem 106E. Kwan

Cooperative and AnticooperativityLecture 6: -Bonding

N

BN

B

NB

The donations go more or less clockwise through the ring.NBO estimates place the energy of each interaction at about36.8 kcal/mol (S=0.36), which is a very large overlap and acorrespondingly large interaction.

In benzene, there are two sets of resonance arrows:

clockwise counterclockwiseEach interaction isn't as good (20.4 kcal/mol, S=0.20), butthere are twice as many!

Benzene is the best example of closed CT, a series ofcomplementary bi-directional donor-acceptor interactions.This is "cooperative behavior." In other words, when a bunchof donor-acceptor interactions are lined up correctly, each onegets stronger than it would on its own. This is a generalphenomenon; for example, cyclic arrays of correctly arrangedhydrogen bonds are also stronger together than separately.

What about quinone?O

O

Cross-conjugation is present here, and we already know thatis an anticooperative pattern. The NBO analysis shows:

O

O

Another way to think about it is that in cooperative arrays,resonance arrows can be drawn in either direction, all the wayaround the array. You can't do that here:

17.8 kcal/mol

6.6 kcal/mol

O

O

O

O

O

O

O

O

betterthan

Chem 106E. Kwan Lecture 6: -BondingAromaticity: The Hückel PictureBenzene is often referred to as "aromatic." Many measures ofaromaticity, many of them controversial, have been proposed.For a good discussion, see Chapter 2 in Computational OrganicChemistry, by Stephen Bachrach. Fortunately, we need not betoo concerned with such details here.

So, by now, you're probably asking: what about the 4n+2 rulethat I learned before?

To address this important question, we need to go throughsome background. In 1930, Erich Hückel proposed a verysimple method for understanding the -bonding in a semiquantitative way.

For an array of -bonds, one assumes:

(1) a minimal basis of hydrogen-like p-orbitals;

(2) a "-effective Hamiltonian" Heff; and

(3) a simplified diagonal/adjacent-to-diagonal form for Heff in the minimal basis of p-orbitals.

(I will show you exactly what this means on the next page.)While crude, this theory gives good predictions and intuitionabout the origin of conjugative/aromatic stabilization in arraysof -bonds.

So what about cyclobutadiene? Clearly, I can draw somecyclic resonance arrows for that:

However, these resonance arrows are meaningless, becausethey don't represent any stabilizing donor-acceptor interactions.Here is why:

So the /* overlap is zero, which means the Hamiltonianmatrix element H(,*)=0, and therefore there is no interaction.The bond lengths are therefore of (more or less) single anddouble bonds. Cyclobutadiene has been isolated in very coldmatrices as a rectangular structure: Michl JACS 1988 110 2648.

B3LYP/6-31g(d)

Chem 106E. Kwan Lecture 6: -BondingAromaticity: The Hückel Picture In Hückel theory, the secular determinant has a very simpleform. One assumes that:

ij ij iji H j

where

1 if 1 if is bonded to and

0 if 0 if is not bonded to ij ij

i j i ji j i j

For a linear polyene, this means the secular determinant is:

0 00

00 0

0 0 0

EE

E

E

For a system with n p-orbitals, this is an n-th order polynomialin E whose roots are the energies of the system. One findsthat the k-th energy level is given by:

2 cos1k

kEn

22 coskkEn

(linear system)

(cyclic system)

One can picture the cyclic energy levels with the Frost-Muslincircle (J Chem Phys 1953 21 572):

One pictures the vertices of the cyclic system as lying on theedge of a circle. (The ring is oriented with one of its pointsdownward.) The intersections are the energies. Here it isfor benzene:

energy

Benzene has a total of six -electrons, which is n=1 in the4n+2 rule (don't confuse this with the n p-orbitals describedearlier.) For cyclobutadiene, we get:

energy

Here, n=1 in the antiaromatic 4n rule. Note that if we fillup the electrons here, we see that it could be a tripletdiradical. However, in real life, it behave as two isolateddouble bonds, which means that it should be treated as twodisconnected ethylene molecules, not one cyclobutadiene.

Chem 106E. Kwan Lecture 6: -BondingAromaticity: The Hückel Picture So why is benzene "extra stable," and cyclobutadiene"extra unstable"? In a system with n -electrons, there is anestimated variational energy of na before variational mixing.This means that an isolated p orbital in the system underconsideration would have an energy of . What happens aftervariational mixing (i.e., conjugation)?

benzene

number of-electrons

6

totalenergy

averagestabilization

6 + 8 8/6 = 1.33

butadiene 4 4 + 4.48 4.48/4 = 1.12

cyclobutadiene 4 4 + 2 2/4 = 0.5

Note that and are pretty big negative energies. So benzeneis extra stable and cyclobutadiene is extra unstable relative tobutadiene. From the geometry of the Frost-Muslin circle, it iseasy to see that 4n+2 geometries should be stabilized relativeto 4n ones, and that the stabilization gets relatively smaller asthe ring gets bigger.

How Good is the Hückel Approximation?It's qualitatively good, but quantitatively bad.

(1) The HOMO-LUMO gaps give the right trend in predicting the frequency (color) of aromatic compounds (although the usual anti Koopmans'-type arguments must apply)

(2) The barrier to rotation in butadiene is 4 kcal/mol. Using values for from benzene, the predicted barrier is 18 kcal/ mol.

(3) The patterns of charge and electron distribution in substituted aromatics is qualitatively reproduced.

In many textbooks, Hückel calculations are used to calculatea variety of parameters such as bond order, charge, etc.Although this may give reasonable qualitative results, the factthat these parameters are being estimated from a non-orthogonal guess basis means that the results are notnecessarily meaningful. In plainer language, the p-orbitals youput into the calculation are not the real p-orbitals in the system,so it's not sensible to talk about how many electrons are inthem.

However, we can compare the Hückel secular determinantwith its ab initio-derived counterpart. In this comparison, theKohn-Sham operator in NAO basis for the ab initio methodB3LYP/6-311++G** is used.

(1) Using NBO-derived values for just the "minimal p-type NAO basis," and the full secular determinant gives energies for the occupied orbitals that are within 1-2 kcal/mol of experimental values.

Thinking of aromatic systems in terms of only their p-orbitals is quantitatively correct.

(2) If the Hückel approximation is used with the same values, the errors jump to 5-11 kcal/mol.

Neglecting everything but neighbor-neighbor interactions is a very coarse approximation.

(3) The ab initio values for and are -69 kcal/mol and -87 kcal/mol. Experimental estimates range from 17-20 kcal/ mol to 100 kcal/mol, depending on whether thermodynamic or spectral data are used to estimate the parameter.

full discussion: Valency and Bonding pg. 208-214.

Quantitative Comparison

Chem 106E. Kwan Lecture 6: -BondingStrained Olefins

The stereochemical requirement that the component p-orbitals in an olefin be correctly aligned is fairly stringent. Insome cases, however, it is possible. Trans-cyclooctene, forexample, is the smallest stable cyclic alkene known:

Ph2(O)P OH

OO

NaH

DMF

O O

White Tet: Asy 2004 15 3123

91%

(Many other methods are available, and trans-cycloocteneshave been known since the 1950s: Cope JACS 1953 75 3212;Ziegler Ann 1950 1 567)

Clearly, the olefin in trans-cyclooctene is a "stereocenter."In this crystal structure, with NBOs calculated at B3LYP/6-31G*,the olefin is very twisted (the H-C-C-H dihedral angle is 139°,which is 41° out of plane):

The twisted * can be seen quite clearly. Apparently, themolecule is stabilized in this form by (C-C) to * donation (twodonations of about 6.6 kcal/mol each).

Fox JACS 2008 130 3760Twisted Amides During World War II, there was an intense synthetic efforttowards penicillin. Although it was known that penicillin wasvery active towards hydrolysis, and this might be the key to itsbiological activity, the core structure was not known (NMRspectroscopy and X-ray crystrallography were not yet ready).Woodward proposed that the structure was a -lactam, whileRobinson disagreed, arguing that the strong resonance inamides should preclude such behavior.

In modern language, Woodward's argument was that thestrong n(N) to *(C=O) donation that ordinarly makes amidesso stable is disrupted by ring strain:

Stoltz Nature 2006 441 731(also see Wasserman commentaryand Clayden ACIE 2006 45 7118)

N

SHROCN

MeMe

COOHO

activated carbonyl group

Chem 106E. Kwan Lecture 6: -BondingTwisted Amides Eventually, Dorothy Hodgkin published the X-ray structureof the potassium salt of penicillin, removing all doubt. This isthe benzyl derivative:

Dexter JCS PT1 1978 185

Clearly, the amide is quite bent. The dihedral angles indicatedwith the dotted lines are 11° (red) and 39° (blue). Thepotassium ion lies above the structure and is not shown.Interestingly, the evolution of beta-lactamases is a key pathway for antibiotic resistance.

Many attempts have been made to synthesize other twistedamides. Kirby (ACIE 1998 37 785) has prepared an aza-adamantanone:

N CO2HAc

1. HCl

2. heat

N O

N N OMe

BF4

NO

O

with PPh3=CH2 with Me3OBF4 with propyleneglycol

This anomalous reactivity is more like that of an amine anda ketone, rather than an amide. In formamide, NBO placesthe n(N) to *(C=O) donation at 59.8 kcal/mol. Here (B3LYP/6-31g(d), it is <0.5 kcal/mol due to improper symmetry. Thereis still a substantial n(O, p-type) to *(C-N) donation present(28.4 kcal/mol):

Chem 106E. Kwan Lecture 6: -BondingTwisted Amides In 2006, Stoltz finally prepared 2-quinuclidonium tetra-fluoroborate, a very strained amide (Nature 2006 441 731):

N3

OHBF4

N N

OH

BF4

H OBF4

The key to this strategy is that it avoids traditional amide-formation reactions which require aqueous workup orchromatographic steps. How does it work? Why is therea mixture of products? (This is the Schmidt-Aubé reaction.)

38% 12%

O NN

NH O N

NN

HN

OH

tautomerize

There are two products because the extrusion of nitrogencan be accompanied by bond migration. Calculations showthat the O-protonated form is much higher in energy than theN-protonated form.

O NN

NH

tautomerizeN

H O

mixture oftetrafluoroborate

salts

1. MeOH2. Boc2O N

Boc

CO2Me

NH

CO2Me

56% 34%

Obviously, this is a very reactive "amide." The IR frequency and13C NMR carbonyl shift of 175.9 ppm indicate that this is morelike an anhydride or acid chloride. In fact, it is so unstable thatthere is no known way to free-base it. Here is its X-ray crystalstructure (counterion removed):

1822 cm-1

Evidently, even partially twisted amides (-lactams) or evenprotonated twisted amides (this example) are highly unstableentities which are very susceptible to nucleophile-inducedcleavage. Even though the delocalization, namely, the n(O)to *(C-N) interaction, is still present, this uni-directionalanomeric-type effect is not enough to make this moleculestable.

Hückel vs. Möbius Topology

If it were possible to free-base it, the amide would be fullytwisted (B3LYP/6-31g(d)):

Chem 106E. Kwan Lecture 6: -BondingTwisted Amides

In larger ring systems, it is possible to twist arrays of olefinswithout incurring a large penalty in strain. One idea is thata large cyclic array of olefins in a Möbius topology should be"aromatic":

extranode

In 1964, Heilbronner (TL 1964 29 1923) used Hückel theory topredict that cyclic arrays of olefins with Möbius topology and4n electrons would be aromatic:

ij ij iji H j cosij ij ij iji H j

Hückel topology Möbius topology

2 12 cosk

kE

n

The energy levels of Hückel-benzene, hexatriene, andMöbius-benzene are compared below:

Möbius topologyHückel topologyMöbiusHückel

22 coskkEn

Evidently, Möbius-benzene is anti-aromatic. In general,what is Hückel aromatic is Möbius anti-aromatic and viceversa.

energy

conjugated

a [16]-annulene

a [12]-annulene

Hückel vs. Möbius Topology

Chem 106E. Kwan Lecture 6: -Bonding

The idea of aromaticity is controversial when larger rings areconsidered. (Schleyer OL 2002 4 3431)

(1) Annulenes are cyclic arrays of double bonds. [10]- annulene is the next 4n+2 aromatic hydrocarbon after benzene. It has many conformers, is not planar, and may or may not be aromatic. However, [14]-annulene and [18]-annulene seem to be somewhat aromatic.

(2) The degree of "bond alternation," the difference in adjacent bond lengths, should be relatively small in an aromatic compound.

(2) Here is a picture (B3LYP/6-31G*) of the lowest energy conformer of [12]-annulene. It has Hückel topology, but a conformer with Möbius topology is calculated to lie only 4.4 kcal/mol higher in energy, and seems to be aromatic. Similar results are found with [16]-annulene, one of whose isomers is calculated to have a very small bond alternation of (CC)=0.019. However, it is not the ground state by far.

In 2003 Herges and coworkers synthesized the [16]-annulene,shown above, which they suggest is Möbius-aromatic.(Nature 2003 426 819)

(1) As Schleyer points out, bond alternation should be considered along the whole 16-carbon path, not just the bridge. This raises the estimate from CC=0.049 Å to a much larger CC=0.157 Å. (JACS 2005 127 2425)

(2) Its NICS(0) value (see Chem Rev 2005 105 3842) is -3.4 ppm. Typically, aromatic and antiaromatic compounds have a NICS(0) near of -9.7 (benzene) -27.6 ppm (cyclobutadiene).

(3) The aromatic stabilization energy (ASE) of this compound is almost the same as benxene itself

Thus, no one has umambiguously produced a Möbius-aromatic structure.

The Cyclopentadienyl Cation Controversy


Although it is unlikely you will run into Möbius-aromaticity inany of "ordinary" ground states, the aromaticity/antiaromaticitycriteria are valuable ways to think about transition states. Forexamples, reactions which would proceed via an antiaromaticcarbanion are likely to be very difficult:

Br H

77 K surface,vacuum

SbF5SbF5Br

triplet,very unstable

Wasserman ACR 1977 10 27

Me HMe

MeMe

Me

ACIE 2002 41 1429

Similarly, the cyclopentadientyl cation is expected to beantiaromatic. Indeed, both theoretical calculations and ESRmeasurements show that the cyclopentadienyl cation is veryunstable, and exists as a ground state triplet:

NH3 (l)

NaNH2

Naantiaromatic;not observed

Na

In contrast, the aromatic cyclopropenyl cation has beenisolated and is stable in acid:

1 M HClstable

aromatic

Therefore, it was surprise when Lambert and coworkersreported this reaction:

+ Ph3C+TPFPB-

Me

Me

MeMe

Me

+ Ph3CH+ TPFPB-(TPFPB = tetrakis(pentafluorophenyl)borate,

a very non-nucleophilic counterion)

Although many scientists had tried to prepare this compoundbefore, none had succeeded. After two rounds of review, bya panel of three "experts," the paper was accepted in April of 2002. However, by May, it was shown by Bertrand and coworkers that the putative cyclopentadienyl cation was, in fact, merely cyclopentenyl cation. The paper was retracted immediately. (ACIE 2002 41 2211 and refs. therein). Whatwent wrong? Here is its X-ray crystal structure:

Cowley Chem Commun 2002 1520

(1) very weak 1H-13C NMR correlation

(2) very broad, featureless peak (instead of expected quartet)

predicted: 1.36 Åactual: 1.51 Å

marked pyramidalizationexplained as “crystalpacking effects”

13C: 60 ppm(!)

The Cyclopentadienyl Cation Controversy


Me HMe

MeMe

Me

Me

Me

MeMe

Me

Confirmation

also synthesizedindependently

by Cowley

From The Editor"...this antiaromatic cation should, in fact, be highly unstable,and for decades attempts at its preparation and definitecharacterization have failed--however, perhaps this time thecounterion...had worked wonders... As the Editor I regretthat we did not examine the original manuscript morecritically, and that its publication has resulted in unnecessarywork to reproduce the results experimentally andtheoretically."

From Lambert"Because of the evidence presented in ref. [1], I am retractingthe conclusions of ref. [2], which were entirely my own andimply no reflection on the part of my co-workers (whoseexperimental and theoretical work is valid.)

From Thomas Müller (an independent observer)"Quantum-mechanical calculations show that the distortionswhich are apparent in [cyclopentadienyl cation] stronglydestabilize [it]. Therefore, it is unlikely that relatively weakcrystal packing forces are responsible for the elongatedtrans pyramidalized formal C=C bond...on the other handcalpculations...provide strong evidence that the speciesobtained by Lambert and coworkers is indeed thepentamethylcyclopentene cation..."

Cowley Chem Commun 2002 1520

+ CF3SO3H

Pi Bonding

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