Structure Determination

How can the structure of an unknown compound be determined?

Before modern analytical tools were developed (for organic chemists this means the development of NMR (~1950s) or IR (first used ~1900, not readily obtained until mid 1900s), the only experimental evidence for the identity of an was elemental analysis

A chemist could thus determine the % of various elements in an unknown sample

For example, if a sample had 1 part carbon and 4 parts (CH4) what is the constitution?


How to test the possible structures?

1 Structure Determination

In order to distinguish the possible constitutional , experimentalists would react the compound in a substitution reaction and detect the number of isomeric products

H H Cl H C H H C Cl H C H Obtain 1 H H H

Same compound

H Cl H H H C H H C Cl H C Obtain isomers H H H

Different compounds

By reacting methane in a variety of reactions, it was determined that a monosubstitution always produced only 1 isomer, therefore all 4 must be equivalent

2 Structure Determination

In addition to the constitution, however, what was the configuration?

How are the 4 equivalent hydrogens arranged in space? H H H C H H C H C H H H H H H Planar Pyramidal Tetrahedral

H H H C Cl H C H C H H Cl Cl H H Each possible configuration would generate only 1 isomer with monosubstitution (thus the 4 hydrogens are at equivalent positions) React twice By logic, eventually a Cl H tetrahedral geometry Cl was proposed H C Cl Cl C Cl H C H Cl C H Cl Cl H Cl C H H H Cl H

Two products Two products One product! 3 Structure Determination

This was the state of affairs until the beginning of the 20th century, for any new compound the elemental analysis was obtained and then the structure was inferred by systematically running reactions or by comparing to analogous compounds

Could a model be developed to predict the structure of organic compounds?

In 1916 G.N. Lewis published a paper that hypothesized that bonding occurred by sharing pairs between (in contrast to ionic bonds)

We now call these structures “Lewis dot structures” by explicitly designating all as a dot

Only designate H B C N O F valence electrons

The constitution could thus be predicted by filling the outer shell of electrons for each (second row atoms thus need 8 electrons and thus an )

H O H Two electrons shared “” Filled outer shell for all atoms 4 Structure Determination

Original course notes from Lewis (1902) The cube indicated does not indicate easily how multiple bonds are formed, or what is the angle between various bonds


C C ???

The Lewis dot structures work well to determine the constitution of a (atoms are bonded in a way that allows a filled octet rule to form),

but does not indicate configuration 5 Structure Determination

The intuitive arguments of Lewis to determine constitution by sharing electrons to form a filled outer shell, however, were given mathematical basis by Heitler and London (1927)

Studied binding energy of H2 molecule

H H Figure from Heitler and If electron 1 was only London* associated with hydrogen 1 and electron 2 was only associated with hydrogen 2

H1 e1 H2 e2 If electron constraint was Calculated bond energy a removed and each electron fraction of actual can interact equally with both nuclei

e1 H1 e H2 2 Energetic basis for a “bond” Calculated bond energy much -sharing of electrons between two atoms closer to actual *W. Heitler, F.Z. London, Z. Physik, 1927, (44), 455-472 6 Valence Bond

The intuitive description proposed by Lewis where bonds are formed between two atoms by sharing valence electrons in order to obtain a filled octet for the outer shell is basis of “” This theory is still the underlying principle most organic chemists use to rationalize chemical reactions and predict chemical properties

Two major results from valence bond theory:

1) Concept of Resonance

Resonance structures result from electrons being associated with different nuclei

For H-H molecule

e1 e1 H1 e1 H2 e2 H1 e2 H2 e1 H1 H2 H1 H2 e2 e2

ΨMOL = c1 Ψ1 + c2 Ψ2 + c3 Ψ3 + c4 Ψ4

Actual structure is a combination of all contributing structures with appropriate weighting factors

Called “principle of linear combination” 7 Resonance in Organic Compounds

What is resonance? (also called ‘delocalization’) Look at a nitro group


The negative charge on the oxygen could be placed on either oxygen using Lewis structures


Which structure is correct?

It turns out neither structure is correct, but the charge is delocalized onto both oxygens -This process of being able to delocalize the charge onto more than one atom is called resonance (Resonance is a result of not being able to draw an accurate structure using one ) 8 Resonance in Organic Compounds

“Rules” of Resonance

1. All resonance structures must be valid Lewis structures (e.g. cannot have 10 electrons on one carbon in one structure)

2. Only electrons move (cannot move nuclei, only electrons – usually double bonds or lone pairs connected through an extended p orbital system)

3. Number of unpaired electrons must be constant

9 Resonance in Organic Compounds

Concept of resonance allows explanation of a number of chemical properties

How does resonance explain acidity?

Consider pKa of organic




H3C H3C H 4.8 OH O Resonates similar

to nitro group

Both structures place a negative charge on oxygen after loss of proton,

but the pKa difference is greater than 11

10 Resonance in Organic Compounds

A is a common resonance source

O -H O O


Once the acid is deprotonated, the negative charge is located on one oxygen

The charge can be delocalized (resonated) onto the other oxygen

Neither structure is correct, but rather the negative charge can be delocalized over both oxygen atoms

Resonance structures are simply a result of one Lewis structure being incomplete in describing the location of electrons A double headed arrow always means two (or more) resonance forms

Two arrows mean two chemically distinct structures 11 Resonance in Organic Compounds

Comparison of for Ethoxide versus anion


The excess negative charge is more stable on the acetate anion that can resonate, thus the is more acidic

12 Resonance in Organic Compounds

Resonance also allows explanation for the concept of

Electrons in a need not be shared equally between two nuclei

For H-Cl molecule

e1 e1 H e1 Cl e2 H e2 Cl e1 H Cl H Cl e2 e2

ΨMOL = c1 Ψ1 + c2 Ψ2 + c3 Ψ3 + c4 Ψ4

Unlike with H2 where the Ψ1 and Ψ2 wavefunctions were more stable than the charged wave functions, with HCl the Ψ4 is the most stable, therefore highest coefficient

The electrons are more stable closer to Cl than H, therefore the chlorine atom is said to be more “electronegative” than the hydrogen atom

These resonance considerations therefore also cause bond dipoles in unsymmetrical bonds

13 Electronegativity Tables

Linus Pauling first established values to associate with each element (there have been many different values computed, but the trend is the same)

Elements toward the upper right hand of the periodic table are more electronegative

Also can predict the relative electronegativity of two atoms by their relative placement in the periodic table

H (2.3) Li (0.9) Be (1.6) B (2.1) C (2.5) N (3.1) O (3.6) F (4.2) Cl (2.9)

H3C OH Br (2.7) 2.5 3.6 I (2.4) The numbers are a relative indication of how much the electrons are ‘attracted’ to a certain atom

As the number becomes larger, the electrons are attracted more by that atom 14 Resonance in Organic Compounds

The bond dipoles resulting from unsymmetrical bonds also can affect the acidity

O O pKa H 4.8 H3C OH H3C O


H 0.7 Cl3C OH Cl3C O

Anion with trichloroacetic acetate is stabilized by inductive effects from polarized C-Cl bonds

O Cl C O Cl Cl

15 Induction

Induction refers to electron movement “through bonds” -All bonds between different atoms are polar and the electrons are closer to the more electronegative atom “on time average”

As the electronegative atom is further removed, the inductive effect is less (inductive stabilization is through bonds, therefore if there are more bonds to transverse the effect is less)


X pKa The electronegative H 4.8 fluorine pulls electron I 3.2 density away from F 2.6 carboxylate 16 Field Effect

Can have similar effect merely through space (field) rather than through bonds (inductive)

H Cl Cl H H Cl Cl H CO2H CO2H

Which is more acidic?

H Cl Cl H H Cl Cl H CO2 CO2

17 Resonance in Organic Compounds

Important to Remember: Not all resonance structures need to contribute equally

If two resonance structures are not of equal energy, then they will not contribute equally to the actual structure


Octet rule obeyed Positive charge on on all atoms less electronegative atom

This leads to major and minor contributors

18 Resonance Forms


Deeper blue color on carbon (less charge on carbon)

Actual “hybrid” structure

19 Resonance Forms


H C CH H3C CH3 H3C CH3 H3C CH3 3 3

major minor inconsequential Only a resonance form if spins are paired

Not all resonance forms need to contribute equally, but rather a weighting factor is given to each resonance form depending upon its importance to the actual form

Typically forms with more bonds are more important (often due to forms with less bonds typically have atoms without an octet)

Placing charge on the more electronegative atoms is important

The number of paired electrons must be constant 20 Resonance Forms

Factors affecting stability of resonance structures:

- Placement of charge



minor major

When the only difference is the location of , structure is more stable when anion is placed on more electronegative atom

21 Resonance Forms

- Amount of Charge

Also related to number of bonds in a structure


While structure with four formal charges shown is a “valid” resonance form, if structure is dramatically higher in energy then it is practically an irrelevant resonance form

22 Resonance Forms

- Octet rule is important


Octet rule not obeyed major

Having all atoms with a filled octet rule is more stable than a resonance form that only has 6 electrons in one outer shell

Even if this requires a positive formal charge to be placed on a more electronegative atom

Second row atoms are always more stable with a filled outer shell

23 Resonance Forms

Curved arrows represent movement of electrons

As already observed in acid/base reactions, a curved arrow indicates movement of electrons


Arrows always show where electrons are moving

Formal charges on atoms are a result of electrons moving

24 Resonance Forms

Drawing resonance structures properly is an aid to predict location of electrons

Remember actual structure is a hybrid of all relevant resonance forms

These resonance forms allow a chemist to predict where excess electron density is located in a molecule

Excess negative charge is located on three carbon atoms, not on all five equally 25 Empirical Evidence for Resonance

Chemical properties of molecule are not like one resonance form

Have already observed this with acidity difference between ethanol and

pKa O O 4.8 H3C H3C H OH O


26 Empirical Evidence for Resonance

Observe also with dipole values for compounds with resonance


4.47 D

Interatomic distances do not correspond to single, double, or triple bonds


Each C-C bond is part single and part

27 Valence Bond Theory

2) Second conceptual advancement with valence bond theory is idea of “hybridization”

To understand why the concept of hybridization was introduced we need to become reacquainted about the structure of an atom and how bonds between different atoms form

An atom consists of three types of particles: Proton (positively charged) Neutron (neutral) Electron (negatively charged)

Ironically the word “atom” comes from the Greek language meaning “indivisible”, as it was thought to be the smallest particle that did not have smaller constituents

The number of protons determines the element (also is the atomic number) Carbon therefore has 6 protons located in the nucleus

Usually the nucleus also contains an identical number of neutrons as protons If the number is different it is called an isotope

These two particles have similar mass

(~1830 times greater than an electron) 28 Consider a Carbon Atom

Nucleus – means “kernel of a nut”

1 fm Nucleus of atom 6 protons (red), 6 neutrons (blue) Nucleus of atom -held together by nuclear forces (region where protons and neutrons reside)

Shell of atom (region where electrons reside)

1 Å

Nucleus size is ~2 fm (1 fm = 10-15 m), atom size is ~1 Å (1 Å = 10-10 m)

For an uncharged 12C atom, there are 6 protons, 6 neutrons and 6 electrons

Therefore the nucleus, which is responsible for ~3600/1 parts of the mass, only encompasses ~1 x 10-15 part of the volume (remember V = 4/3 πr3)

29 Electrons

Unlike the protons and neutrons which are in the nucleus (a relatively fixed point) we cannot say with certainty where an electron is located at a certain time (Heisenberg uncertainty principle)

What we can say is that ‘on time average’ the electrons are located in orbitals (regions of space) *much bigger region than the nucleus

The electrons are not randomly placed, nor do they reside in simple circular orbits

Schrödinger developed a formula that could describe the properties of the electrons (called Ψ)

This wave function, Ψ, mathematically describes the shape of an orbital where the electrons resides and the square of the wave function, Ψ2, is proportional to the probability of finding an electron in a given volume

30 Electrons

Each electron is described by a set of four quantum numbers (Pauli principle: no two electrons may have the same values of all four quantum numbers)

n l

principal quantum number Related to shape of orbital Integral values, n = 1, 2, 3, etc. Integral values, l = 0, 1, 2, …(n-1) Indicative of “shell” the electron resides Therefore if n=1, l must be 0 If n=2, l can be either 0 or 1 Atoms where n=1 (H and He) are first shell If n=3, l can be 0, 1 or 2 Atoms where n=2 (Li, Be, B, C, N, O, F, Ne) Each value of l represents a different orbital shape are second shell, and so on l=0 is s orbital The higher the value of n, the greater the l=1 is p orbital average distance of the electron from l=2 is d orbital nucleus and thus the greater the electron’s l=3 is f orbital energy (naming or orbitals derives from description from early spectroscopists of metal alkali lines as sharp, principal, diffuse and fundamental) 31 Electrons

Each electron is described by a set of four quantum numbers (Pauli principle: no two electrons may have the same values of all four quantum numbers)

m1 s Related to orientation of the orbital in space Spin quantum number (x, y, z coordinates in three dimensions) Can only have two values (-1/2 or +1/2) Has integral values –l,…., 0, …., +l By convention, the two possible spin states Thus if l=0 then m1 must be 0 are typically indicated by either having the If l=1, then m1 can be -1, 0, or +1 spins pointing up or down If l=2, then m1 can be -2, -1, 0, +1, +2

Orbitals of the same shell (n) and shape (l) With two electrons, if the spins are opposite have the same energy regardless of m1 they are stated to have paired spins, if the (thus all 3 p orbitals in the same shell have spins are pointing in the same direction then the same energy, 2px=2py=2pz) they are stated to have unpaired spins

paired unpaired 32 Electronic Configuration

If the four quantum numbers are known for each electron, then the electrons can be placed into their respective orbitals by knowing the relative energy difference

Since the first row atoms (H and He) have only one orbital (1s) it is fairly easy to write the electronic configuration for these atoms (either have one electron or two paired electrons)

Consider the second row atoms

The second row has a total of 5 orbitals (1s, 2s, 2px, 2py, 2pz)

2p The first shell orbitals are lower in Energy 2s energy than second shell and p orbitals in the 2nd shell are higher in energy 1s than s orbitals in the second shell

2 2 2 With Carbon (1s 2s 2p ), however, the question arises as to where the second 2p electron goes Hund’s rule: for a given electronic configuration, the state with the greatest number of unpaired spins has the lowest energy 2 2 1 1 Therefore carbon is 1s 2s 2px 2py where the spins are parallel 33 Shape of Orbitals

Solutions of the Schrödinger equation (EΨ = ĤΨ) will describe the wave equation (Ψ), since Ψ2 = probability of finding electron it also describes the shape of orbitals

1s orbital radial density plot radial probability plot

!2 never goes to zero 4!r2"2

r r (distance from nucleus) (distance from nucleus)

three-dimensional shape

34 Shape of Orbitals

As the number of shells increases, the presence of nodes increases (nodes are regions of space where there is zero probability of finding an electron)

The number of nodes is equal to one less than the principal quantum number (therefore for n=1 there are 0 nodes, for n=2 there is 1 , for n=3 there are 2 nodes)

All second shell orbitals thus have 1 node

2s 2px 2py 2pz Actual shape is p orbitals have node at the nucleus different than typically stylized “dumbbell” shaped p orbitals 35 Structure of Methane

Molecules are made by combining atomic orbitals to form bonding regions for the electrons

Using the outer shell orbitals of methane, the compound results from combining

the 2s, 2px, 2py and 2pz orbitals of carbon with the four 1s orbitals of hydrogen

The electronic configuration for atomic carbon thus has a lower energy 2s orbital and three degenerate 2p orbitals that are each orthogonal to the others

The valence electrons will thus have 2 in the lower energy 2s orbital and 1 in each of the

degenerate 2px and 2py orbitals

To form the methane molecule, therefore the 1s orbital of each hydrogen must form bonds with the orbitals that have electrons This bonding model has many problems: 1) implies different energy and of C-H bonds 2) Two of the C-H bonds must have a 90˚ bond angle 3) Have too many electrons in an orbital 36 Hybridization Model for Bonding

Instead of using atomic orbitals for bonding, a different model considers first hybridizing the atomic orbitals to form “hybridized” orbitals

Same rules apply for combining atomic orbitals to form hybrid orbitals

1) Get same number of hybridized orbitals as starting atomic orbitals used to form hybrid

2) Shape of hybridized orbitals is obtained by the mathematical addition of the wave functions for the atomic orbitals

The name (designation) of hybridized orbitals merely refers to the number and type of atomic orbitals used in the formation

37 sp Orbital

Combine one s orbital with one p orbital

Notice the relative shape difference between bonding and antibonding lobes -Allows more overlap!

If the orbitals are subtracted then an identical hybridized orbital is obtained directed 180˚ from the first

Bonds formed from the two sp hybridized orbitals will thus have a 180˚ bond angle

38 sp2 Hybridization

-Can also hybridize by combining one s orbital with two p orbitals (would allow formation of three covalent bonds – one from each sp2 hybridized)

Look in the x-y plane, three sp2 hybridized orbitals pz is coming in and out of the plane

All three sp2 orbitals are in the same plane (120˚ apart from one another)

39 sp3 Hybridization

To form four equivalent bonds carbon can hybridize all of its valence orbitals (three p and one s to form four sp3 hybrids)

The four sp3 hybridized orbitals have a bond angle of 109.5˚

Forms a tetrahedral geometry

All bonds can thus have the same bond length and angle, unlike the model using atomic orbitals

40 Hybridization Model for Bonding

When a hybridized orbital is used to form a bond with an atom, a new bonding and antibonding are formed

These bonds have the electron density cylindrically symmetric about the internuclear axis

Bonds that are symmetric about the internuclear axis are called sigma (σ) bonds

Sigma bonds and of electrons (if they are not involved in resonance) use hybridized orbitals for the electrons

When 2nd row atoms have the same , they use sp hybridization for two bonding orbitals, sp2 hybridization for 3 bonding orbitals, and sp3 hybridization for 4 bonding orbitals

Knowing the structure thus allows chemists to predict the hybridization and also the geometry for the compound

41 Bonding in Unsymmetrical Compounds

In methane there are 4 identical bonds between carbon and each of the four hydrogens

The carbon atom thus adopts a sp3 hybridization and each H-C-H bond angle is 109.5˚ for a perfect tetrahedron geometry

When one of the C-H bonds is replaced with a different atom, however, the perfect tetrahedron geometry is no longer present (The C-Br bond length is obviously longer than the C-H bonds, thus not a tetrahedron)

We still approximate the carbon as being sp3 hybridized, it is very close as seen by geometry, but we realize this is an approximation

42 Variable Hybridization

As seen, the hybridization affects the geometry of a compound

Atomic orbitals need not be “hybridized” in integer numbers, need not add exactly one s orbital with 2 p orbitals to yield exactly a sp2 hybridized orbital

As the amount of s and p orbital ratios are changed, the geometry changes

Pure s sp sp2 sp3 Pure p %s 100 50 33 25 0 %p 0 50 67 75 100 Bond < ~ 180˚ 120˚ 109.5˚ 90˚

As %p increases in a hybridized bond, the bond angle decreases

As %s increases in a hybridized bond, the electrons are held closer to the nucleus (since s orbitals are closer to the nucleus on time average than p orbitals)

The geometry is thus intimately related to the hybridization of the atom 43 Variable Hybridization

To determine the amount of hybridization for each bond in a nonsymmetrical structure, first consider how to determine the hybridization for a symmetrical methane Place carbon at center of box and consider placement of four hydrogens at corners of a box for an ideal tetrahedron geometry z

C y

a x

Distance for the C-a bond (one of the four hydrogens in methane)

2 2 2 1/2 Pa = (x + y + z )

44 Variable Hybridization

Distance for the C-a bond (one of the four hydrogens in methane)

2 2 2 1/2 Pa = (x + y + z ) C y By symmetry x = y = z

2 1/2 Therefore, Pa = (3x ) = √3 x a x x = Pa / √3

Each composite orbital = (s + x + y + z) = (s + [3/√3]Pa)

Typical hybrid orbital therefore = s + λPa λ = mixing factor sp3 = sp(λ2)

2 Total s character is Σi 1/(1 + λ i) = 1

Geometry is also determined as 1 + λµcosθij = 0 λ = mixing coefficient of bond 1 µ = mixing coefficient of bond 2 θ = angle between bonds 1 and 2 45 Variable Hybridization

Use formula for symmetrical molecule like methane H H C H H

λ of all four bonds is thus √3

Therefore the H-C-H bond angle can be determined

1 + λµcosθij = 0

1 + (√3)(√3)cosθij = 0

cosθij = -1/3

θij = 109˚28’

Pure s sp sp2 sp3 Pure p %s 100 50 33 25 0 λ 0 1 √2 √3

θ 180˚ 120˚ 109.5˚ 90˚ 46 Variable Hybridization

What does formula suggest about nonideal tetracoordinate carbon compounds?



(when all four substituents are not the same on a carbon, the bond angle is not a perfect tetrahedron anymore!)

1 + λ2cos(112˚) = 0

λ2 = 2.67

The C-C bonds in propane use a sp2.67 hybridized orbital from the central carbon

A molecule can place substituents further apart by using a hybrid orbital with less p character (greater p character results in smaller angles for bonds)

47 Variable Hybridization

How to determine the H-C-H bond angle in propane?

The total s character in all four bonds must equal to 1 2 Σi 1/(1 + λ i) = 1

2 2 Therefore 2/(1 + λ C-C) + 2/(1 + λ C-H) = 1

λ2 = 2.67 for both C-C bonds

2 2/(1 + 2.67) + 2/(1 + λ C-H) = 1

2 2 if solve for λ C-H, λ = 3.38 for both C-H bonds

Therefore the H-C-H bond angle in propane can be solved by 1 + (√3.38)(√3.38)cosθH-C-H = 0

θH-C-H = 107.4˚

Factors affecting the ratio of s and p orbitals used to form bond: 1) Electrons in bonds from s orbitals are closer to the nucleus than p orbitals 2) As the percent s character increases in a bond, the bond angle is larger 48 Variable Hybridization

If the p character is increased in a bond, allows electrons to be closer to other nucleus


1) When carbon is bonded to a more electronegative atom, the other atom prefers more p character so electrons can be closer

2) Sterically the greater the p character, the smaller the angle, therefore larger atoms prefer larger angles and more s character

Consider the haloform series


HCF3 108.8˚ The length of the C-X bond also HCCl3 110.4˚ matters as a shorter bond length the HCBr3 110.8˚ greater the potential steric interactions

HCI3 113.0˚

49 Variable Hybridization

What does this variable hybridization model predict for highly strained systems?

Consider cyclopropane



The line angle drawing implies a

Using atomic orbitals, this angle is too small (the smallest angle possible is 90˚ if pure p orbitals were used for the σ bonds)

More importantly an organic chemist wants to know where the electrons are located for the bond between the carbons (Valence Bond Theory assumption is that bonds are a result of sharing electrons between two bonded atoms)

If we know where the electron density is located, we can understand the properties of these highly strained systems 50 Variable Hybridization

It has been determined that the

2 1 + λ C-H cos(118˚) = 0

2 λ C-H = 2.13

2 2/(1 + 2.13) + 2/(1 + λ C-C) = 1 C-C bond uses more λ2 = 4.53 C-C p character in bond

1 + 4.53 cos(θC-C) = 0

Model predicts that electron density is not located directly between two carbons

This is what is referred to as “bent” bonds 51 Bent Bond Theory for Bonding

Some argue that this idea of “bent” bonds should be used instead of the concept of hybridization in bonding in all organic compounds

Instead of using the idea of σ and π bonds in organic compounds, the bonding is merely a result of overlap of tetrahedral bonding for a carbon atom

Also instead of using the sp3 hybridization model for a saturated carbon, rationalize the tetrahedral geometry is a result of placing the electrons in the four bonds to carbon as far apart as possible (thus at the corners of a box in a tetrahedral geometry)

This is easy to rationalize for saturated carbons, as the model relative to hybridization bonding is nearly identical

Bonding occurs by overlapping The bond angles in ethane orbitals to form new bond would thus closely match 109.5˚ in methane

Tetrahedral carbon with 109.5˚ bond angles 52 Bent Bond Theory for Bonding

How would this “bent” bond theory account for multiple bonds? (hybridization assumes the formation of π bonds to account for multiple bonds)

Instead of sp3 in methane, 2 H C C H H C C H ethene has a sp hybridization H H H H with p orbitals overlapping for the second π bond

Instead of using sp2 hybridization, bent bond theory would merely align two of the tetrahedral orbitals toward each other to form the two bonds

Bonds would bend toward This model would thus also predict each other to allow 2 bonds that all hydrogens in ethylene between both atoms would be in the same plane


53 Bent Bond Theory for Bonding

In therefore three bonds would form between the two carbons by aligning the three orbitals toward each other

This bent bond theory also allows for the prediction that the C-C bond length is longest in ethane but decreases in ethene and is shortest in acetylene

Concept of hybridization is still prevalent in models (and we will use this description also), but it is only a model to explain experimental observations

and not the only model

54 Bonding Theory in Organic Compounds

Valence Bond Theory: Electrons are located in discrete pairs between specific atoms

Molecular Orbital Theory: Electrons are located in the molecule, not held in discrete regions between two bonded atoms

Most organic chemists think intuitively about compounds using valence bond theory, we consider reactions with certain functional groups by considering how the discrete bond reacts

In a particular reaction, a discrete bond breaks and a new bond forms between two atoms

It would be more useful, however, if we could determine where exactly the electrons are located in a molecule

Are bonds truly a result of sharing of electrons between two atoms (valence bond theory), or are the electrons shared with the entire molecule ()

How can we determine where the electrons are located? 55