Physical Organic Chemistry and Conformational Analysis

Question 1

What 5 tools can be used to provide indirect evidence for a mechanism?

Answer 1

Regio and stereochemistry of products and intermediates.

Structures of by-products.

Isotope labelling (common isotopes are ¹³C, ²D and even ¹⁸O).

Observing intermediates by: NMR, IR, UV-Vis, GC/HPLC, MS and EPR.

Question 2

What 2 ways can short-lived intermediates be observed?

Answer 2

Use another selective mechanism only possible with the intermediate such as the Diels-Alder reaction.

Spin traps such as nitroxides can stablise radicals.

Question 3

Give a way an S_n1 and S_n2 reactions can be told apart only comparing the reactant and product.

Answer 3

The reaction centre will invert in an S_n2 reaction.

Question 4

What do mechanisms provide and hence why are they important to study?

Answer 4

The provide detailed descriptions of all elementary steps in a given reaction.

Question 5

How can the possible conformations of a molecule change its reactivity? For example a long alkyl chain.

Answer 5

Conformations depending on solvent and temperatures can form that ‘wrap up’ and be blocked from reacting.

Question 6

Which ring system size is the least strained in theory? How can this be demostrated? What is the take home message from this?

Answer 6

5-membered are the least strained in theory based on angle of the shapes.

Plotting the ring size against the heat of combustion per CH2 group shows that 6 membered rings are the most stable.

Cyclohexane rings and larger are considered strain free.

Question 7

What is the actual shape taken by 3, 4, 5 and 6 membered rings? Comment on their stability and draw each of them.

Answer 7

3 - has to be planar, eclipsed C-H bonds are very strained

4 - a V shaped stucture forms with two atoms at the base, the twisting reduces eclipsing

5 - ‘open envolope’ shape, C-H bonds are almost ax/eq so are called psuedo axial and equitorial

6 - most stable form is the chair conformation but the boat and twist boat can also occur, the chair makes all C-H bonds staggered

Question 8

Draw the ring flip mechanism.

Answer 8

Question 9

Sketch, with values, the energy graph for a ring flip.

Answer 9

Question 10

Why is axial ^tBu so much less favoured than axial ⁱPr?

Answer 10

ⁱPr only has 2 methyl groups so the hydrogen can be oriented to face the other axial groups to minimise the steric clash. With ^tBu, no possible rotation is possible to avoid the steric clash.

Question 11

When looking at flat cyclohexane ring, how can you tell if groups will be equitorial or axial?

Answer 11

Define one direction for one atom as eq, the atoms adjacent will have eq positions in the other direction to the originial. The equitorial position will alternate in and out of the page around the ring.

Question 12

How does a ^tBu group affect cyclohexane rings?

Answer 12

It acts as a locking group, where the conformation of the ^tBu group is in the axial position is heavily disfavoured.

Question 13

How does the position of a leaving group on a cyclohexane ring affect the rate of substitution with a nucleophile?

Answer 13

The rate of nucleophilic attack is approximately 31x faster when the leaving group is in the axial position for two reasons

1. The 1,3-diaxial interactions from the axial groups block the incoming nucleophile from the LUMO
2. The nucleophile must substitute into the axial position which is disfavourable

Question 14

Where will a nucleophile substitute onto when reacting with an epoxide on a cyclohexane ring?

Answer 14

Both epoxide bonds must be in the same direction without axial or equtiorial symmetry. Either way the nucleophile will substitute onto the epoxide in the opposite direction that the epoxide pointed in.

Question 15

When reforming an epoxide group on a cyclohexane with both carbons in the ring after the epoxide has be Nu attacked, what are the stereochemical requirements of the LG and the O^-?

Answer 15

After the substitution the groups will point different directions meaning the favoured conformation will be both groups equitorial. However for the ring closing to occur, there must be good overlap of orbitals which is only the case in the both axial conformation.

If the LG and O^- are cis (one axial, one equitorial), the elimination cannot occur as the LG LUMO is too far away from the O^-.

Question 16

What is the structure of a sp² hybridised group on a cyclohexane ring? Where does the sp² group move to when attacked by a nucleophile?

Answer 16

It points in the plane the sp² carbon makes with the 2 carbons it is attached to. This is between the axial and equitorial position but the group points up if it is attached to an up carbon.

The nucleophile can attack from either the equtiorial or axial position, wherever the attack is, the O^- will then go to the opposite position. Large nuceophiles will react at the equitorial position (no 1,3-diaxial blocking), small nucleophiles will react at the axial position.

Question 17

When reducing a ^tBu locked cyclohexyl carbonyl, which position will the hydride take for LiAlH₄ and Li(s-Bu)₃BH?

Answer 17

LiAlH₄ is a small so can react in either position, hence reacts preferentially at the axial postion. This places the OH into the equitorial position.

Li(s-Bu)₃BH is a large source of hydride so preferentially reacts at the equitorial position. This places the OH into the axial position.

Question 18

What is the conformation of cyclohexene? Give a sketch and name it. How will it react with Br₂ and bromonating agents?

Answer 18

It forms a half-chair.

The bromonium ion forms with both bonds pointing in the same direction, leaving the nucleophile to attack from the other side, forming the trans product.

Question 19

Sketch bicyclo[2.2.1]heptane and explain how it has been named.

Answer 19

1. Overall 7 carbons - hept
2. 2 rings - bicyclo
3. Identiy bridgehead carbons
4. List the number of carbons in each bridge from the bridgehead, largest to smallest, bicyclo rings must have 3 bridge labels, including a 0 if there isn’t a third

Question 20

How do reactions with more complicated ring systems need to be considered in terms of conformations?

Answer 20

The direction of attack of nucleophiles need to be considered since the confromation may block the reactants.

Question 21

What do rate equations and kinetic orders tell you about a reaction?

Answer 21

They tell you the molecularity (elementary stoichemometry) of the rate determining step.

Question 22

Give the definition and meaning of the activation entropy and enthalpy. How can these be calculated from experiment?

Answer 22

Activation entropy is a measure of the degrees of freedom in the system and is +ve for dissociative reactions and -ve for associative reactions.

Activation enthalpy measures the bond formation and dissociation, if equal numbers of bonds are made and broken, the value is < 100 kJ mol^-1. If more bonds broken than formed, the value is > 100 kJ mol^-1 and often much larger.

These are found from the Eyring equation by plotting ln(k/T) against 1/T.

Question 23

What are the limitations of activation parameters?

Answer 23

• Only work well in the gas phase, solvation often changes the enthalpy-entropy balance. It can still be used for solutions but it is harder to interpret the data.
• Provide information for the rate determining step only.
• Information provided relates to the transition state only, but this can be used as a high energy version of the intermediates/products.

Question 24

What is the basis of ‘extrathermodynamic’ relationship?

Answer 24

Different, seperate reactions (such as a hydrolysis and a reduction) are related provided they meet the following criteria:

1. Changes in the object (the changing part of the reaction) are relatively small
2. Changes in the object influence the process through a single effect
3. The overall function (multiple reactions) is additive of the the seperate components such as the changing part of the molecule and the process

Question 25

Why are ‘extrathermodynamic’ relationships important?

Answer 25

• They allow predictions to be made of the results of experiments
• The influence of objects on a process can be characterised
• Derivations give information about the objects

Question 26

How can the linear free energy relationship be used when comparing an amide substitution reaction and a hydrolysis reaction? What causes derivations from this?

Answer 26

The free energy of activation for each reaction is recorded for several substituents and a linear relationship between the points is seen. What is occuring is the line is a measure of the electronic effect that the substituent is having. This means with one data point, the free energy of activation for the other can be recorded.

Derivations are caused by changes other than the substituents electronic effects such as the mechnaism, or steric effects.

Question 27

How is Gibbs free energy of activation used practically in Hammett experiments?

What is the Hammet equation?

Answer 27

For irreversible reactions, log(k) is used. The Hammett equation is:

log(k_X/k_H) = ρ • σ

For reversible reactions, log(K) is used. The Hammett equation is:

log(K_X/K_H) = ρ • σ

ρ defines the properties of the reaction (e.g. forms +ve charge)

σ defines the properties of the substituent (e.g. has +I effect)

Question 28

What is the reference for ρ and σ?

Answer 28

They are both in reference to hydrogen as k_H/k_H = log(1) = 0. ρ is set to 1 for the ionisation of a substituted benzoic acid. Therefore σ must equal 0 for hydrogen.

Question 29

What is the physical meaning of σ for benzoic acids? Give an equation and what can be interpreted by the value of σ.

Answer 29

Using log rules, and the definitions of ρ = 1 for benzoic acid, following can be derived:

pK_H - pK_X = σ

For benzoic acid, pK_H = 4.2 thus:

σ = 4.2 - pK_X

Electron withdrawing substituents will stabilise the formed negative charge meaning that pK_X < pK_H (more acidic = more negative pK_a) and σ > 0.

Hence electron donating groups will destablise the negative charge, making pK_X more positive and σ < 0

Question 30

How do you measure ρ values?

Answer 30

Using the Hammett equation in the form log(k_X/k_H) = ρ • σ, an experiment is done to find the log(k_X/k_H) values for many substituents and the σ values for each is plotted. The gradient will give the ρ value and the intercep should be 0.

Question 31

What is the physical meaning of ρ? Give a numeric significance and how this can be interpreted.

Answer 31

By manipulating the Hammett equation, the following can be found:

k_X = k_H x 10^{ρ • σ}

For simplicity NMe₃⁺ is the substituent as σ = 1

Using different magnitudes of ρ, the effect of ρ on k_X can be seen.

ρ = 0.2 k_X = 1.6k_H

ρ = -2 k_X = 0.1k_H

Hence ρ is the sensitivity of a given reactions to the electronic effect of the substituent

ρ > 0 means the reaction is facilitated by electron withdrawing groups (-ve charge formed in the RDS)

ρ < 0 means the reaction is facilitated by electron donating groups (+ve charge formed in the RDS)

Question 32

What can be interepreted from the magnitude of ρ for a reaction?

Answer 32

Small values of ρ (< |0.5|) should be considered to be close to zero and not interepreted in terms of charge development. The magnitude of ρ depends on the distance between the charge and the substituent. This value increases if the charge can delocalise onto the ring.

Rough guide:

Charge on the ring ρ > 2
Charge next to ring, ρ ≈ 2

Charge one atom away from the ring ρ ≈ 1

Charge two atom away from the ring ρ ≈ 0

Question 33

How are σ values adapted for each position on the ring?

Answer 33

There is a seperate σ value for the para and meta positions, σ_p and σ_m respectively. There are no ortho σ parameters since the sterics become too dominant.

Question 34

How do the mesomeric and inductive effects of substituents affect the Hammett equation and when does the Hammett equation not work well?

Answer 34

The inductive effect is an inherent effect and the mesomeric effect is transmitted through the ring inductively. This is because the resonance will almost never make it off the ring so the inductive effect is from the aromatic carbon closest to it.

The normal Hammett equation doesn’t work well however when the reaction centre can delocalise onto the substituent.

Question 35

How can σ parameters be adapted for the direct conjugation of a charge from a reaction centre onto the ring?

Answer 35

σ⁺ and σ^- parameters have been developed which deal with direct conjugation onto a ring of a postive and negative charge respectively.

Question 36

How should σ parameters be practically used?

Answer 36

There are 3 scenarios:

1. The charge is not conjugated onto the ring - use σ_m and σ_p for meta and para substituents respectively on the same graph on the axis of σ.
2. A positive charge is conjugated on the ring - use σ_m for the meta substituents, σ⁺ for the +M para groups and σ_p for the other para substituents. The axis should be labelled σ⁺.
3. A negative charge is conjugated on the ring - use σ_m for the meta substituents, σ^- for the -M para groups and σ_p for the other para substituents. The axis should be labelled σ^-.

Question 37

Overall, how are the different σ parameters used? Describe the process.

Answer 37

The correlation against σ, σ⁺ and σ^- parameters can be used as a diagnostic tool to probe the location of charge.

1. Measure reaction rates/equlibrium constants with a range of substituents including +M and -M groups in the para position.
2. If log(k_X) shows the best correlation with σ, the charge is not conjugated with the substituent.
3. If log(k_X) shows the best correlation with σ⁺, a positive charge is conjugated with the substituent.
4. If log(k_X) shows the best correlation with σ^-, a negative charge is conjugated with the substituent.

Question 38

How can the Hammett equation be used for aliphatic compounds?

Why can the Hammett equation include both the inductive and mesomeric effects despite the fact they are different effects.

Answer 38

Only inductive effects will occur so a seperate set of σ parameters are required. Typically they are σ* and σ⁰.

They are additive effects so while they can be split into σ_I (for inductive) and σ_R (for resonance) it has no much effect.

Question 39

How do you account for multiple substituents in the Hammett equation?

Answer 39

You just sum the effect of the substituents.

Question 40

What would be the origin of a non-linear Hammett plot?

Answer 40

If there is competing mechanims which will be favoured differently by the substituents meaning that with certain substituents, different mechanisms will be favoured.

The general rule is that if the Hammett plot is concave downwards, the reaction mechanism changes as you change the substituents.

Question 41

What has occurred if a consecutive reactions Hammett plot has an upwards, concave shape?

Answer 41

The rate determining step changes from one step to another, hence the effects of the substituents changes.

Question 42

How can σ be used in conjuction with spectroscopic data?

Answer 42

It can show the effect the substituents have on measurements such as the chemical shift and bond strength of a group/nuclei.

Question 43

How do the strength of C-H and C-D bonds compare and by how much? Why is this the case?

Answer 43

The energyH for the vibration of the bonds are inversely proportional to the square root of the reduced mass of the atoms. Therefore, as D is double the mass of H, the C-D bond is a factor of √2 stronger than the C-H bond.

Question 44

What is the kinetic isotope effect and what is the maximum effect that is expected to be seen?

Answer 44

The normal way of reporting KIE is k_H/k_D and the maximum value this takes is approx. 8-9 when the bond is half-broken in the transition state.

Question 45

What is the primary KIE? Give an example of its effect.

Answer 45

The primary effect is where the X-H bond is broken in the rate-determining effect. k_H/k_D is always > 2.

Question 46

What does the magnitude of the KIE mean when deducing a mechanism?

Answer 46

It describes how much the C-H bond is broken during the reaction, different groups can affect this.

Question 47

What is the secondary KIE? Descibe it’s different effects and its magnitudes.

Answer 47

The secondary KIE is the k_H/k_D when the X-H bond is not broken in the RDS, it is always < 2. The alpha and beta affect are for when the H is attached to the alpha and beta carbons. The alpha effect is usual 0.8-1.4.

The normal effect is > 1 which comes from the reduction in frequency of the vibrations in an sp² carbon which is favourable.

The inverse effect is < 1 and is for an sp² to sp³ change which is unfavourable.