CHAPTER 3 RING STRAIN: REACTIONS AND UTILITY · rel k/s-1 a ∆h‡ ∆s‡ (phso 2) 2ch(ch 2)ncl (phso 2) 2c (ch 2)n-1 eto-na+ ch 2 etoh 3 4 5 1 6.7x10-6 1.2x10-2 20.5 21.8 16.3

CHAPTER 3 RING STRAIN: REACTIONS AND UTILITY

References: 1. Carey & Sundberg Part A, 3rd. ed. p.157 (4th. ed. p.162) (5th. ed. p.161) 2. Isaacs, p. 282 3. Carroll, p. 161, (130) 4. Miller, p. 160 5. C.J.M. Stirling Evaluation of the Effect of Strain on Reactivity Tetrahedron 41,

1613 (1985). 6. B. Halton, ed. Advances in Strain in Organic Chemistry QD 461.a33 1991

(particularly the introduction) 7. A. de Meijere and S. Blechert Strain and Its Implications in Organic Chemistry QD

461.S875 8. C. Galli and L. Mandolini Eur. J. Org. Chem. 2000, 3117-3125

I CYCLIZATIONS AND ACTIVATION PARAMETERS A BASIC TRENDS The concept of strain is something that we have encountered often in our

chemistry careers. It can be defined in many ways, but will be defined here as the energy

by which a molecule differs from its extended linear counterpart. Strain energy is usually

a combination of many contributors and the list includes angle strain, non-bonded

interactions (steric hindrance), torsional strain and bond extension strain. Our discussion

will focus primarily on angle strain as we observe its effect on reactivity and its

usefulness in mechanistic investigations.

ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014 3 - 1

http://www3.interscience.wiley.com/cgi-bin/fulltext/73001238/PDFSTART

Angle strain can be expected when atoms cannot adopt their usual bonding geometry as

required by their hybridization. Inevitably then, angle strain arises in small cycloalkanes

and heterocyclic analogs. The following table lists the strain energy of a number of

strained cyclic compounds. The information has been drawn from a number of sources

including Goumans and coworkers (Eur. J. Org. Chem. 2003, 2941), Borst and

coworkers (J. Org. Chem. 2005, 70, 8110), Beckhaus and coworkers (J. Am. Chem. Soc.

1995, 117, 11854) for some of the spiro fused compounds and J. Chem. Soc., Perkin 2,

2000, 793 for the cyclophanes. .

Cycle Strain Energy

(kcal/mol) Cycle Strain Energy

(kcal/mol)

28.4

56

NH

28.2

63

O 27.2 68.1

N 43.9 PH 21.4

S

19.8

56.1

32.7

32.2

26

34

28

6.7

O

25.3 S

19.6

chair: 0.6

boat: 6.8 7.0

9.6

6.3


98.5

84

138

149

167

84

98 I I

68

24.6

31.6

(CH2)n

n=5 n=6 n=7

27 17 14

(CH2)n(CH2)m

n=1, m=2 n=2, m=1 n=0, m=3 n=0, m=2

24 30 25 37

41.6

28.9

O

39.9

47.9

48.2

44.6


Many reactions can be accelerated by the release of ring strain. As an example, the

release of ring strain is vital to the mechanism of action of the penicillin family of

antibiotics. Attachment of the molecule indirectly to peptidoglycan molecule weakens the

cell wall of the bacterium, leading to its destruction.

N

S

OCOOH

HN

OPh

HN

SO

COOH

NHO

Ph

O linked to peptidoglycan molecules in bacteria

Angle strain can be used to some extent to explain rates of ring formation. Reactions

involving strained rings are based on free energy differences, those of ∆G. ∆G is

comprised of enthalpic and entropic terms as shown below.

∆G = ∆H - T∆S for a reaction

∆G‡ = ∆H‡ - T∆S‡ for a transition state Entropy can sometimes be a difficult parameter to explain. Simply bear in mind that

entropy always wishes to increase. Note that we have two different entropy terms here.

∆S: -the entropy of the reaction

-the overall entropy difference between products and reactants.

-if the reaction creates more molecules from fewer or if product(s) have more

flexibility than the reactants, look for the ∆S to increase and vice versa

∆S‡ -the activation entropy

-the difference in entropy between transition state and reactant (products are not

a consideration)

-look for changes in flexibility leading to the transition state


Although it is not the ideal case for demonstration, let us consider the following

lactonization reaction (p. 1628 of Stirling review) since it is the most commonly presented

data.

Br(CH2)nCO2-

(CH2)n

O CO

Kinetic Data for Lactonization Reactions

ring size n relative k ∆G‡ ∆H‡ ∆S‡

3 1 21.7 22.8 22.0 -2.5

4 2 2.4 X 104 19.3 17.7 -4.9

5 3 2.8 X 106 17.7 15.9 -5.6

6 4 2.6 X 104 18.6 17.2 -4.2

7 5 97.3 21.8 17.4 -13.6

8 6 1.00 24.8 21.8 -9.2

9 7 1.12 24.8 20.3 -13.9

10 8 3.35 24.0 17.4 -20.5

11 9 8.51 23.5 16.4 -21.9

12 10 10.6 23.3 17.6 -17.6

13 11 32.2 22.6 15.3 -22.5

14 12 41.9 22.4 14.8 -23.6

15 13 45.1 22.4 16.1 -19.5

16 14 52.0 22.3 16.8 -17.0

18 16 51.2 22.2 15.4 -21.2

23 21 60.4 22.2 14.5 -23.8 The general data indicate several trends. First, as indicated by the following graph, the 5

and 6 membered rings always form the fastest, whereas the smaller and larger rings are


the slowest to form. It should be noted here that the 4-membered ring closes relatively

fast in this reaction; often it is much slower.

Note that the there seems to be a general trend with the entropy. In none of the

examples does entropy favour the reaction. However, with the smaller rings, its resistive

influence is the least. The graph on page 3-6 depicts this. The solid line demonstrates

the influence of the entropy of activation for any general cyclization reaction.

The enthalpy of activation is most hindering of the 3- and 7-membered cyclizations. After

8-membered lactonization, the cyclizations of larger rings possess little enthalpic

variation.


This lactonization information is somewhat atypical and one reason may be that the

attacking nucleophilic group possesses an sp2 hybridized atom. Hence the more obtuse

O-C-C bond angle must constrict more during cyclization than is required for a series of

atoms that all retain sp3 hybridization. In some cyclization reactions, 3-membered ring

formation may be the fastest.

More typical data are in included in the following table.

Reaction ring size

rel k/s-1 a ∆H‡ ∆S‡

(PhSO2)2CH(CH2)nCl (PhSO2)2C(CH2)n-1

CH2EtO-Na+

EtOH

3 4 5

1

6.7X10-6

1.2X10-2

20.5 21.8 16.3

10 -9 -12

PhNH(CH2)nBr PhN(CH2)n-1

CH2K2CO3

EtOH-H2O

3 4 6

1 0.019 9.55

19.4 21.7 16.2

-11 -11 -17


a The rate constants for the different reactions are not related.

The reaction entropies of ring closure of simple alkanes have been calculated, with the

assistance of experimental data.

ring ∆S (kcal/mol) cyclopropane -7.7 cyclobutane -10.9 cyclopentane -13.3

ring ∆S (kcal/mol) cyclohexane -21.2 cycloheptane -19.8 cyclooctane -19.0

Note that these are reaction entropies, so the comparison is products to starting

materials. The ∆S for the larger rings are smaller than that for cyclohexane since

cyclohexane is a relatively stiff ring while the larger analogs have some flexibility.

B THE THORPE-INGOLD EFFECT Also known as the gem-dimethyl effect, the term has arisen as the name for an observed

acceleration of cyclization when one or two alkyl groups are added to one of the atoms

involved in the cyclization. The accompanying table for oxirane formation provides a

number of comparative examples.

Reactant relative rate Reactant relative rate

HOCH2CH2Cl 1 HOCH2CHClCH3

5.5

HOCHCH2ClCH3

21 HOCHCCl

CH3

CH3CH3

1360

HOCH2CClCH3

CH3

248 HOCCHCl

CH3

CH3 CH3

2040

HOCCH2ClCH3

CH3

252 C C

CH3

CH3

CH3

CH3

HO Cl

11,600


A number of explanations have been forwarded to account for the Thorpe-Ingold effect.

One favoured rationale invokes the steric strain of the methyls on the same carbon and

hence the bond angle change that those groups bring about. The methyl groups repel

each other and have a wider bond angle between them. The consequence is a tighter

bond angle where the ring is about to form. Hence, although the effect seems small, the

smaller bond angle means that there is a reduced angle difference between the starting

material and the transition state.

CH3

CH3

steric widening here leads tocontraction of the angle on the opposite side of the central carbon

contraction

Another possible explanation is based on reduced entropy of activation, since the

additional alkyl groups will inevitably lead to reduced bond rotation in the open chain

starting material.

A third explanation is that the

uncyclized starting material is

already a high energy

material. All of the gauche

interactions make it so and

since it possesses such a

high ground state position,

there is less energy required

to achieve the transitions

state for cyclization.

O-

HH

H H

Cl

O-

CH3H3C

H3C CH3

Cl

C CCH3

CH3

CH3

CH3

HO ClHOCH2CH2Cl

basebase

more gaucheinterations

s.m. 2s.m. 1


product2

product1

s.m.1

s.m.2

∆G‡(2) < ∆G‡(1)E

reaction coordinate

The concept of gem-disubstituted rings possessing less strain than unsubstituted

versions has been proposed (Ringer, A. L.; Magers, D. H. J. Org. Chem. 2007, 72, 2533-

2537) and subsequently challenged (S.M. Bachrach J. Org. Chem. 2008, 73, 2466-

2468).

II ENDOCYCLIC RESTRICTION TEST

P. Beak Acc. Chem. Res. 25, 215 (1992)

A unique and useful series of experiments has been popularized and exploited by

Professor Peter Beak of U. of Illinois. In general, these experiments utilize ring size and

strain in an intramolecular substitution or addition reaction, to probe the geometric

requirements of the regular bimolecular reactions.


http://pubs.acs.org/doi/pdf/10.1021/jo0624647

http://pubs.acs.org/doi/pdf/10.1021/jo0624647

http://pubs.acs.org/doi/pdf/10.1021/jo702665r

http://pubs.acs.org/doi/pdf/10.1021/jo702665r

Consider three general reactions:

X Y Z X ZY+ + [1]

X Y Z

[2]exocyclic

Y Z

X +

X Y Z X ZY

[3]endocyclic

Reaction [1] represents the reaction being analyzed. It is often assumed that the X-Y-Z

reaction is SN2-like and the atoms require a 180° angle in the transition state, but this is

not necessarily so.

Reaction [2] is an exocyclic reaction possessing a wide range of reaction geometries.

Hence its transition state, 180° or otherwise, can be obtained through a number of

conformations.

Reaction [3], which is an endocyclic model of [1] has rather specialized constraints. As

long as the reaction is truly an SN2 or proceeds through initial association of Z to Y, then

there will be a certain defined geometry for the reaction and only selected ring sizes will

undergo the reaction. If a certain substrate cannot undergo the reaction at hand, then

other competitive processes are expected to be observed, such as an intermolecular

version of the reaction.


It has been found that the following reaction proceeds very well.

SO

CH2

SO2Ar

CH3

OOS

O-

CH

O O

ArSO2 CH3

SO

CHSO2Ar

CH3

O O

-base

One can be readily misled into thinking this is an intramolecular reaction if one does not

perform some experiment to determine if the reaction was intermolecular. Indeed the

usual strategy is to prepare some of the substrate containing two different (usually

isotopic) labels. If a reaction is done with 50% unlabeled and 50% doubly labeled

material, two extreme outcomes may be expected. In the fully intramolecular case, as

exemplified by reaction [3], one will obtain product with no label and also there will be an

equal amount of product still containing both labels.

X Y Z X ZY

[3]endocyclic

X Y* Z* X Z*Y*

[3**]endocyclic

However, if there is predominantly an intermolecular component to the reaction, then one

will expect a significant amount of product (maximum: 50%) to contain a single label:

intermolecular +

X ZY* X Z*Y

+

X Y Z Z*X Y*


Hence in the above reaction with the methyl transfer from the sulfonate, it has been

proved through double labeling experiments that the reaction was wholly intermolecular.

The overall experiment lends support to the assumption that the methyl transfer requires

an angle between anionic carbon, methyl carbon and leaving oxygen that is significantly

larger than can be attained in the 6-membered endocyclic case.

To prove that the related chemistry can indeed proceed in an intramolecular sense, and

to ensure that no other unexpected force was preventing the chemistry, the following

exocyclic reaction was demonstrated

SO

CH2

SO2Ar

CH2I

OOS

O

C

O O

ArSO2 H

CH2

SO

CHSO2Ar

CH2I

O O

-base

More usefulness of the endocyclic restriction test was recently demonstrated (J. Am.

Chem. Soc. 118, 3426 (1996)). The goal was to learn more about the mechanism of

oxygen atom transfer from N to P in the ensuing reaction.

PPh2

N+O-

H tBu

PPh2

N

HtBu

O

100 °Ctoluene

The endocyclic restriction test on this substrate shows the reaction works and it was

repeated with a mixture of unlabeled and doubly labeled material. The doubly labeled

substrate bore a 13C at the imino C and an 18O at the oxygen.1 The reaction proved to

be fully intramolecular and the key piece of evidence is that the P does not do a simple

linear SN2 at O since this reaction is geometrically disallowed by the constraint of the 6-


membered ring transition state. A number of reasonable alternative mechanisms were

suggested and are shown here. In each case it was assumed that the first step of the

mechanism was rate determining.

PPh2

N+O-

H tBu

PPh2

N

HtBu

O

P

HNO

tBu

Ph Ph

PN

H

O

tBu

Ph Ph

H

PPh Ph

O-

NtBu

+

PN

H

O-

tBu

Ph Ph

+

P

N+

O

H tBu

Ph Ph-

A

B

C

At this stage the authors noted that pathways A and B involve the P atom acting as a

nucleophile while in route C, the P is electrophilic. It was decided that substituent effects

could differentiate the three possibilities above. So two additional substrates were

prepared, one with an EWG on the phosphinyl aryl groups and another with an EDG on

those groups. Rough (use of NMR) unimolecular rates of reaction were obtained and are

listed in the table.

1 The ratio of labeled to unlabeled substrate was 59:36. Using less of the doubly labeled material is not a problem and also incurs fewer obstacles since the rare isotopes are usually more expensive and can be synthetically difficult to incorporate into the molecule.


Since the system with the EWG is fastest and the molecule bearing the EDG reacts the

slowest, the experimental information

is consistent with an electrophilic

mechanism as shown by pathway C.

The overall mechanism involves

intramolecular attack of the P by the

PAr2

N+O-

H tBu

Ar of

Rate /s-1

CF3

259 ± 47

H

71.6 ± 5.4

OMe 57.2 ± 2.5

O with a bond angle appropriate for a

6-membered ring (ca. 110°). Of

particular note is that theoretical

calculations predicted a much larger

P-O-N angle.

III FREE-RADICAL CLOCKS

D. Griller and K.U. Ingold Acc. Chem. Res. 13, 317 (1980)

The determination of an absolute rate constant for a particular transformation can be

quite difficult. There exists a method however that is based on knowing only a few rate

constants and designing a system of employing the known rate constants to clock an

unknown rate constant. This method involves placing a known reaction in direct

competition with the uncalibrated reaction and following the outcome of the reaction. This

topic is suitable in this chapter since some key known clock reactions are based on

rearrangement of strained compounds. Yet to introduce the procedure involved, the

following explanation will utilize one of the most popular clock reactions, which is well

calibrated.

The 1-hexenyl rearrangement:


rate = 1.0 X 105 at 25 °C

As a means to understanding the autoxidation

of fats and oils it became important to learn

the rate constant for H atom abstraction from

a peroxide by a primary carbon radical. Under

the experimental conditions, the 1-hexenyl

radical was generated in the presence of a

known concentration of tbutyl hydroperoxide.

The product mixture was then isolated and

the various products identified. They were

attributed to either H transfer before

rearrangement or after rearrangement.

path 2

kr

ku

rearrangepath 1

tBuOOHtBuOOH

Rate1 = kr[1-hexenyl radical] =

if rearrangement occurs:

if rxn w/ peroxide occurs:

Rate2 = ku[1-hexenyl radical][peroxide] =

dprods1dt

dprods2dt

if we ratio the two rate expression significant cancellation occurs.

Rate1 Rate2 ku[1-hexenyl radical][peroxide]

dprods1dt

dprods2dt

kr[1-hexenyl radical] = =

becomes

==kr

prods2prods1

Rate2 ku[peroxide] Rate1


note that -kr is known since it is the clock reaction,

-[peroxide] is known since it is an experimental parameter and if the reaction is not

taken very far (toward completion), then it will stay essentially constant (a

version of pseudo-first order kinetics)

-the ratio of prods1/prods2 is simply determined through some sort of product

determination: usually either isolation or more accurately some quantitative

analytical chromatography technique.

ku is then simply calculated: ku prods2kr=

[peroxide] prods1

In this case the rate constant was found to be 2 x 104 L/mol.s at 50 ºC.

There are several items to note at this point. First the rate constant for the reaction of

cyclopentylmethyl radical with hydroperoxide is not part of the equation (although it is

likely to be close to 2 x 104 L/mol.s at 50 ºC, since it too is a primary radical). In fact, the

chemical fate of the cyclopentylmethyl radical does not matter at all except for product

analysis.

Second, the 1-hexenyl rearrangement is only useful here if product from both directions

can be detected. If only products from the unrearranged material had been detected,

then a different clock, with a more rapid and therefore competitive rearrangement would

have to be employed. Similarly, if only products from the rearrangement had been

recovered, then a slower clock would be required. Clearly, for the radical clock

methodology to be useful, there is a need for a number of clocks that can span a wide

range of rate constants.

Third, although the rate constant for the 1-hexenyl clock is listed only at 25 ºC, it is

nevertheless a well-calibrated rearrangement meaning kinetic parameters are known.


Hence the rate constant for the reaction at a temperature other than 25 ºC can readily be

calculated.

Finally, radical reactions are quite independent of solvent and solvent polarity, so a given

rate constant is often useful in a variety of solvents.

As an aside, tributyltin hydride and related compounds are very useful for simple radical

chemistry. The mechanism is as follows:

initiation

propagation

propagation

-in competition with direct reduction, the alkyl radical may also rearrange

nBuSn-H + init. nBuSn. + init-H

nBuSn. + R-X nBuSn-X + R.nBuSn-H + R. nBuSn. + R-H

R. R'.nBuSn-H + R'. nBuSn. + R'-H

The calibration of a radical clock is vitally important if that clock is to be useful in future

physical organic studies. A common procedure is to utilize a radical trap, which in itself is

a stable radical. One class of compounds is very obvious for this, the nitroxyl radicals.

The procedure for a fast benzylic radical clock

is outlined below (J. Org. Chem. 57, 4284

(1992)). This calibration is not a chain

reaction. It requires rapid and selective

generation of a radical at a particular atom,

followed by rearrangement of some of those

NO N

O

TEMPO ABNO


radicals and capture of both the radical and the rearranged form by the nitroxide. That

capturing reaction consumes the radical; there are no propagation steps.

Consider the following scheme.

NNO

OCMe3

Me3C 2 Me3CO + N2

Ph

Ph

Me3CO + Me3COHPh

Ph

kr

2.1.Ph

PhPh

Ph

T.

T.

Ph

Ph

T

PhPh

T

kTT.

= nitroxyl trap

1T 2T

Given these situations rate expressions can be set up:

kr[1.] = d[2T]

dtkT[1.][T.] =

d[1T]dt

=[2T][1T]dt

d[1T]kT[1.][T.] =

dtd[2T]

kr[1.] =

kr

kT[T.]=

ratioing these expression provides: which simplifies to:

kr = kT[T.] [1T][2T]


The kr constant can thus be obtained. Again the products are quantitated against one

another. The [T•] is a controllable and measurable variable. A very important point is that

some 2T and 1T must be detected and quantified in order to set up the ratio. The [T•]

can be varied to allow for this but if the rearrangement is very fast other

corrections/procedures must be invoked. The kT value may be known, or is so fast that it

is diffusion dependent, and the rate constant then depends on the solvent and

temperature.

Diffusion rate constant:

kdiff = 8RT

2000ηM-1sec-1

where η is the viscosity of the solution.

Selected Radical Clocks

Rearrangement k (at 25 °C)

1.0 x 105 s-1

NCN

3.9 x 103 s-1

1.3 x 108 s-1

OO

9.8 x 103 s-1

PhPh

Ph

Ph

3.6 x 108 s-1

(at 40 °C)


Atoms abstractions under radical conditions can be predicted using Bond Dissociation

Enthalpies (BDE’s). Some familiarity with these values is always useful. They should be

used with extreme caution in ionic reactions. See:

http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm

Baldwin’s rules for cyclizations

General tendencies for ring forming reactions were established some time ago. In 1976,

Baldwin published a series of observations that many people already recognized.

Cyclization rates depended on ring size, on hybridization of the atom and on the nature

of the reaction occurring.

As shown earlier in the Chapter, the competitive endocyclic and exocyclic modes of

reaction are as follows:

X Y Z X ZY

[3]endocyclic

X Y* Z* X Z*Y*

[3**]endocyclic

The preferences for one vs. the other were outlined by Baldwin (J.C.S. Chem Commun.

1976, 734) and are as follows:

exo endo

3-Tet y --

4-Tet y --

5-Tet y n

6-Tet y n

7-Tet y --

3-Trig y n

4-Trig y n

5-Trig y n

6-Trig y y

7-Trig y y


http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm

3-Dig n y

4-Dig n y

5-Dig y y

6-Dig y y

7-Dig y y

The reason offered in the paper relates to how close the accessible transition state bond

angle are to those in the final product.

For tet, α = 180 º

XNu-

αNu

α

For trig, α = 109.5 º

α

X

Nu-

NuX-

α

For dig, α = 120 º

+E

α

α

Nu-

Nu

Eα α

These are general rules and there are exceptions and orbital overlap actually determines

cyclization propensity. The rules have been refined somewhat for radical cyclization by

Beckwith (J.C.S. Chem Commun. 1980, 482), and since they are often applied to radical

chemistry, they are known as either the Baldwin rules or the Beckwith-Baldwin rules.

(See: http://www.chemtube3d.com/AdvancedTopics-Home.html for animations)


http://www.chemtube3d.com/AdvancedTopics-Home.html

IV MEASURING THE β-EFFECT OF SILICON

J.B. Lambert Tetrahedron 46, 2677 (1990)

Recall the general solvolysis reaction that we have seen already from Chapter 1. The

SN1 reaction is used extensively as a model for carbocation stability. In the past you

have seen that alkyl groups, phenyl rings and aromatic groups containing +R groups all

have the capability to accelerate a solvolysis reaction and this is interpreted to mean that

the attached groups also stabilize the carbocation. In this section a less familiar mode for

stabilizing cations is introduced and the use of ring strain to measure a conformational

dependence is also outlined.

It has been known for several years that a tetravalent silicon atom β to a carbon cation

has the ability to stabilize that positive charge. Model studies indicate that the presence

of an active silicon atom in the β position is worth a notable 21 kcal/mol of stabilization to

the cation.

21 kcal more stableC

SiMe

MeMe

+C

CMe

MeMe

+

Two modes of stabilization have been theorized. One is that the silicon can bridge and

demonstrated hypervalent character. Since the silicon is not a first row element, the

strain of the 3 membered ring is somewhat reduced. The other means of stabilization is

when the silicon atom can utilize its electron rich character and provide stabilization

through hyperconjugation. Note that hyperconjugation is simply conjugation utilizing a σ

bond rather than a π bond. Hyperconjugative stabilization is sometimes called vertical

stabilization. Several of the results presented herein are consistent with either the

hyperconjugative or the bridging mechanism. Subsequent research points toward

hyperconjugation (J. Am. Chem. Soc. 115, 1317(1993)).


bridging:

+

SiMe

MeMeSiMe

MeMe

+

hyperconjugation:

SiMe

MeMe +

+SiMe

MeMe

Si

electron density of the Si-C bond is shared with the empty p orbital of thetrivalent cationic carbon

Professor Joe Lambert of Northwestern University has prepared a number of substrates

capable of undergoing controlled solvolysis reactions. Each of these materials possesses

the requisite leaving group but they also have a silicon atom β to the imminent cationic

site. Further, the substrates’ silicon atom is constrained to a certain geometric

relationship, relative to the impending cationic centre, by the skeleton of the molecule.

Several compounds were prepared and analyzed in solvolysis reactions.2 They

represent systems with different dihedral angles between the silicon atom and the

leaving group. Some important examples are listed in the table nearby. Note that the

leaving group in each case was some sort of trifluoromethanesulfonate, trifluoroacetate

or benzoate derived from the parent alcohol. The authors compared the rates of

solvolysis of the substrates to the rate if the trimethylsilyl group was simply an H.

2 As with many physical organic rate studies, a number of corrections and/or internal checks had to be performed to allow one to have confidence in the experimental results. Although important to the Lambert work, they will not be outlined here.


Substrate dihedral

angle

rate enhancement over H-substituted

tBu

Lv

SiMe3

180º

1012

tBu

Lv

SiMe3

60º

104

Lv

SiMe3

120º

≥105

(estimate)

SiMe3

Lv

0º

105

The data were further broken down to provide a

component of stabilization by induction. It was

estimated that stabilization by a factor of 102 is

by induction alone and the remainder is by way

of the hyperconjugation. There is clearly a

dependence on dihedral angle and the authors

suggest that the effect mirrors the Karplus curve

used estimate vicinal 1H-1H NMR coupling

constants (CHEM*3760). 0° 90° 180°

increasing rateof solvolysis

Si-C-C-Lv dihedral angle


PRACTICE PROBLEMS 4 1. The cyclization rates for the six accompanying compounds have been determined (pyridine, benzene). Give the rate rankings from fastest to slowest for the following groups of these compounds. a) A, B b) B, D, E c) A, C, F

H2N(CH2)3Br H2N(CH2)4Br H2N(CH2)5Br

A B C H2N(CH2)6Br H2NC(CH2)3Br

H3C CH3

H2NCH2C(CH2)2BrH3C CH3

D E F

2. Radical 5 cyclizes to both radicals 6 and 7 with rate constants as follows:

k5-exo = 6 X 104 s-1and k6-endo = 4 X 103 s-1, at 25 ºC.

765

Si+

SiSi

If these radicals are generated and reduced by donation of H using the tin hydride reagent system, a) What is the ratio of the silicon heterocycles obtained from H donation to 6/7. b) Does that ratio depend on the concentration of tin hydride reagent present? 3. Regarding early studies pertaining to the endocyclic restriction test, a problem was occasionally observed when an intermolecular methyl transfer took place. Among the reactions are those exemplified below, where the nitrogen may not always be the nucleophile. Instead, the methyl may get transferred to the sulfonate oxygen by way of a competitive substitution.


one or other gains methyl group

has lost methyl group

acts as Nuloses methyl group

solvent

21

+

N+(CH3)3

SO2O-

N(CH3)2

SO2OCH3

N(CH3)2

SO2O-

+N(CH3)2

SO2O-

N(CH3)2

SO2OCH3

and/or

If the solvent is water, the reaction proceeds exclusively to afford 2. a) What change in product distribution would you expect if the solvent was changed

to benzene? b) If 1 is ever the exclusive product, then the reaction mixture looks exactly like the

starting materials. What can one do to ensure that any reaction has even taken place?

4. Radical 30 is known to ring open to an acyclic radical as shown. The rate constant was desired for this ring opening as it could serve as a useful radical clock. The researchers used nBu3SnH as a reducing agent to assist in their determination of the rate constant. The reaction was done with 1.5 molar nBu3SnH in cyclohexane. At this concentration the yield of 32 was 91% and of 31 was 9%. The reaction temperature was 30 °C. a) Given the rate constant for H donation from the tin reagent to a vinyl radical is 7.8 X 108 M-1 s-1, calculate the rate constant for the ring opening reaction. b) The concentration of nBu3SnH actually changes during this reaction. Is this a problem? c) The rate constant for nBu3SnH transferring an H to a tertiary alkyl radical (1.85 X 106

M-1 s-1 as per lecture material) is lower than to a vinyl radical at the same temperature. Offer a reason.


Ph

Br

Ph 30

Ph C

PhPh C

H H

H

31 32 5. The solvolysis of compounds 8, 9 and 10 was performed with the expectation that 8 and 10 would have additional stabilization. Draw the important resonance structures for the cation resulting from the solvolysis of each of 8 and 10 based on the type of stabilization suggested. a) for 8, a combination of conjugation and hyperconjugation b) for 10, double hyperconjugation

OMs

8

OMs

9

OMs

SnMe3

10


6. (From a previous midterm) Consider the radical conversion of 1 to 8. Using isotope labelling methods and other mechanistic tools, the accompanying mechanism is offered for this reaction

CHO

Ph

-H

Ph

O

O2

Ph

O

OO

Ph

O

OO

Ph

O

O

O

PhO

O2

PhO

OO

PhO

OOH

1 2 3

456

7

1

8

-CO2

The above mentioned isotope labelling experiments were done in the presence of a 50:50 mixture of (16O)2 (regular O2) and (18O)2. The addition of P(Ph)3 is a common method of converting a hydroperoxide to an alcohol (9) and that reaction was done to facilitate isolation and analysis. The next scheme outlines the products that were formed in the labelling experiment

PhO

OH

1 8P(Ph)3

1. O2/ *O2

t-BuOOH

PhO

OH

PhO

OH

PhO

OH

*

*

*

*

9-16O16O 9-16O16O

9-16O18O

+ (Ph)3PO

+ (Ph)3PO*

+

+

+

Answer this set of questions. a) Which step in the mechanism allows the whole reaction to be repeated, meaning that

the steps of 1 to 8 will be propagated over and over. b) What character of O2 allows its facile reaction with radicals 2 and 6? c) Identify a driving force for the conversion of 5 to 6. Provide a very brief explanation. d) Does the alkene of 1 become oxidized or reduced as the reaction proceeds? e) Given the mechanism of the reaction and the method of the labelling experiment,

state the theoretical ratio expected for 9-16O16O, 9-16O18O and 9-18O18O.


7. N-nitrenes can react with alkene to form aziridines. The transition state is usually quite symmetric as shown below, as long as the pendant chain allows.

(CH2)nNN

(CH2)nNN

(CH2)nNN

(CH2)nNN

δ-

δ+

(CH2)nNN

δ+

δ- or

Sometimes however, the transition state may involve a little skewing, with slight charge build-up, with two likely examples shown here. So knowing the information above, the following reaction was carried out in three different solvents.

4

3

solventPb(OAc)4

N

N

ON

NN

N

O

NN

N

ONH2

N

N

O

Solvent 3:4 ratiobenzene 2.8:1 CH2Cl2 3.4:1 CH3CN 4.7:1

Provide an explanation for the observations that is consistent with the chemistry occurring. DO NOT focus on any 1re vs 2re carbocation stability arguments.


8 (from a previous midterm). Consider the following reaction of substrates 4-7, which may be reversible depending on the magnitude of n.

O(CH2)n

O-

(CH2)n

O

-O

4, n=15, n=26, n=37, n=4

startingmaterial:

a) If the reaction is not reversible, which of substrates 5 and 6 will react the fastest? Which of 5 and 7 will react the fastest? No explanation is required in either case.

b) If there is an equilibrium set up between 5 and its product, which side of the equilibrium will be most populated and provide the reason for your answer. c) For which of the substrate(s) would it be impossible to tell if any reaction has occurred? Why? What could you do to overcome this problem? 9. Consider the three compounds shown (25-27) and their relative rates of solvolysis (rate determining loss of Cl). The solvent is HOAc in each case and the product is one of overall displacement of Cl by OAc. Explain their relative reactivity.

Compound: O Cl

25

Cl

26 O

Cl

27 Relative rate of

solvolysis: 0.14 1 48,500

10. (from a previous midterm). The first reaction demonstrates the Michael addition of a thiolate nucleophile on methyl acrylate, with conditions. The next three molecules show relative rate values for this same reaction on the substituted electrophiles drawn. a) Provide a reason why 18 reacts faster than 17. b) Provide a reason why 19 reacts slower than 18. c) What is the role of having PhSH present in addition to PhS-Na+?

H

CO2Me

H

CO2Me

H

CO2Me

H

CO2Me

PhSH, PhS-Na+

THF, r.t.

H

CO2MeH

PhS

krel 1 215 13

17 18 19


11. Using compound 20 as a standard, offer an explanation for the trend in solvolysis rates for the collection of cyclic enol triflates shown. Indicate for instance why 21 and 22 are so slow. Also explain why 24 solvolyzes faster than 20. Assume each substrate loses -OTf to give a carbocation (and think about the structure of that carbocation!!!) OTf = OSO2CF3, a very good leaving group.

Compound # Structure Relative rate of Solvolysis in EtOH/H2O = 50/50 (75 ºC)

20

CH3 CH3

OTf

1.0

21

OTf

5.9 X 10-6

22

OTf

7.1 X 10-5

23

OTf

0.25

24

OTf

2.6


SOLUTIONS TO PRACTICE PROBLEMS 4

1. a) B > A 5-membered ring formation is faster than 4 b) E > B > D 5-membered ring formation with Thorpe-Ingold effect is faster than

simple 5-membered ring formation which is faster than 6 c) A, C, F 5-membered ring formation with Thorpe-Ingold effect is faster than simple

6-membered ring formation which is faster than 4

H2N(CH2)3Br H2N(CH2)4Br H2N(CH2)5Br

A B C H2N(CH2)6Br H2NC(CH2)3Br

H3C CH3

H2NCH2C(CH2)2BrH3C CH3

D E F 2.

a) Since the formation of both 6 and 7 are unimolecular from 5, the ratio of products is

simply a ratio of the rate constants. Thus (five ring)/(six ring) = 6 X 104 s-1/4 X 103 s-1

= 15:1. b) Assuming that the cyclizations are not reversible, the ratio does not depend on the

concentration of tin hydride reagent present. Since the rate expressions for cyclization are k5-exo[5] and k6-endo[5] which do not contain a [tinhydride] term, the cyclizations are independent of the tin hydride.

3.

one or other gains methyl group


acts as Nuloses methyl group

solvent

21

+

N+(CH3)3

SO2O-

N(CH3)2

SO2OCH3

N(CH3)2

SO2O-

+N(CH3)2

SO2O-

N(CH3)2

SO2OCH3

and/or

a) Water to benzene is a polar to non-polar change. The formation of 2 in the reaction above requires substantial charge buildup in the TS, as the O assumes δ- and the N assumes δ+. Hence the reaction goes so well in water. Changing to benzene will destabilize that particular TS. However, 1 will form if a methyl group is simply


transferred from one oxygen to another. In this TS there is charge dispersion since the full negative on one O is shared between two O’s. Charge dispersion is favored by non-polar solvents. So on a change to benzene, one would expect compound 1 to be a detectable product. b) To measure the extent of the reaction that provides 1, one must do a double labeling experiment. A reaction has taken place if one of the labels is distributed from one molecule to another. For example:

N(CH3)2

SO2OCD3

D D

N(CH3)2

SO2O-

+

N(CH3)2

SO2O-

D D

N(CH3)2

SO2OCD3+

1

solvent

loses methyl group acts as Nu


4. a) Setting up two rate expression, each for the competitive fates of 30

d[31]

dt= kH [Bu3SnH][30]

d[32]

dt= kr [30]

ratio them:

d[31]

dt= kH [Bu3SnH][30]

d[32]

dtkr [30]

simplify, isolate and substitute

3132

=kH [Bu3SnH]

kr

kr = kH [Bu3SnH]3231

= 7.8 X 108 M-1s-1 [1.5 M]91

9


kr = 1.18 (or 1.2) X 1010 s-1

b) nBu3SnH will deplete as the reaction proceeds. This can be a problem, so one must so use it in excess and do not take the reaction to a very far extent. This way one can assume pseudo-first order kinetics and assume the [Bu3SnH] does not change.

c) H’s bonded to vinyl groups are stronger; just check the bond dissociation

enthalpies (BDE).

HH

stronger bond weaker bond

This is related to why tertiary radicals are stabilized and vinylic radicals are so reactive. 5.

+SnMe3

+

SnMe3+

+OMs

8

OMs

SnMe3

10

- -OMs - -OMs

6.

f) The conversion of 7 to 8. g) It is a radical and there is only a small barrier when two radical join to form a bond. h) i) One reason could be entropy. It will increase when CO2 is lost since two molecules

result from one. ii) Another reason is that a more stable radical is being made. Oxy radicals are highly reactive, but tertiary radicals are more stable.

i) oxidized e) 1 part 9-16O16O, 2 parts 9-16O18O and 1 part 9-18O18O


7. Let the benzene solvent ratio be our point of reference. In benzene, 2.8 molecules of 3 form for every 1 of 4. One would think that a larger tether would allow direct formation of the fused ring product (permissible by ring strain constraints), with minimal charge build-up in the TS, and that the formation of 3 requires a skewed TS because of the smaller tether. The additional rate/solvent data would appear to bear this out, since as one moves to a more polar solvent, the preference for 3, with the smaller tether increases. The more polar solvent would help stabilize a polar TS and would neither affect nor destabilize the fully neutral TS. So the formation of 3 requires the skewed transition state because of the strain constraints on the reaction and this TS can be stabilized in more polar solvent. The formation of 4 is more likely the symmetric TS involving no charge buildup, so as once moves to more polar solvents, its rate is not affected while the rate of competitive reactions involving a TS with polar character are increased. 8. Consider the following reaction of substrates 4-7, which may be reversible depending on the magnitude of n.

O(CH2)n

O-

(CH2)n

O

-O

4, n=15, n=26, n=37, n=4

startingmaterial:

a) 6 is faster than 5

7 is faster than 5 b) The equilibrium would lie to the right since the four-membered ring has slightly less strain than the three-membered ring. c) As the reaction is presented, it is impossible to tell if 4 has done the reaction, since the product and the starting material have the same structure. To overcome the problem, you could introduce an isotopic label in one of the oxygen or one CH2 group in other to differentiate the forward and reverse products.


9.

Compound: O Cl

25

Cl

26 O

Cl

27 Relative rate of

solvolysis: 0.14 1 48,500

Compound 26 defines the background rate of solvolysis of a cyclooctyl chloride. Compound 27 does the reaction faster because of transannular, intramolecular participation from the oxygen across the ring. The ability of the oxygen to assume some or most of the positive charge accelerates this reaction. The formation of two fused five membered rings is reasonable.

O

ClO+ + Cl-

AcOH-H+

O

OAc

Oδ+ δ+OR

With compound 25, the same assistance could occur, but the result would be a more strained 4-membered ring and hence there is less ability to do this. Why is the rate actually retarded? Since the intramolecular assistance cannot occur in this case, the oxygen a few atoms away acts as a –I EWG destabilizing cation formation as Cl- leaves.

O ClO+

much more strained

10. a) 18 reacts faster than 17 since upon S addition to the double bond, the hydridization of the receiving carbon changes from sp2 to sp3, releasing some of the ring strain. That is, sp2 carbons in a three-membered ring cause more strain than sp3 since 60º is farther from 120º (sp2) than from 109.5º (sp3).

b) 19 reacts slower than 18 because the methyls add a steric barrier to the approach of

the molecule. Remember requirement of a 109 º angle of approach? c) The PhSH provides the H to the anion that forms after addition of the PhS-.


11. The key to this question is knowing the hybridization of the cation that results when triflate leaves. The hybridization of the carbon as it assumes the cation becomes sp and hence the carbon becomes linear. You may know this a) by analogy whereby an sp3 carbon in benzyl chloride becomes sp2 when the chloride leaves or b) by using CHEM*1040 principals for determination of hybridization of atoms, or c) by recognizing what must be happening based on the data. So if the carbon losing the triflate becomes linear this creates geometry requirements that are different from the starting material. Three consecutive carbons must now be linear, as compared to the starting material which has only two carbons.

CH3 CH3

OTf

-OTf CH3

CH3+

CH2 CH2

OTf

δ+

δ-

So as the transition state for cation formation is approaching, the triflate is leaving and as the dark arrow shows, it will push itself up to the linear geometry. So, this effect will really create more strain in the strain free rings, since a 180º bond angle in 5 and 6 membered rings will really be tough to achieve. Hence the 5 and 6 membered rings will do the chemistry much slower. The 7 membered ring has some inherent strain and there is only a small change upon the solvolysis reaction. The 8-membered ring is interesting. 8, 9 and 10 membered rings have inherent strain mainly due to transannular interactions, so making the linear cation in the 8 membered ring actually stretches out the methylene groups and prevents them from interacting with each other. So, there is actually a reduction in ring strain effects by making the carbon cation and experimentally one observes a rate enhancement.


Documents

CHAPTER 3 RING STRAIN: REACTIONS AND UTILITY · rel k/s-1 a ∆h‡ ∆s‡ (phso 2) 2ch(ch 2)ncl (phso 2) 2c (ch 2)n-1 eto-na+ ch 2 etoh 3 4 5 1 6.7x10-6 1.2x10-2 20.5 21.8 16.3