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
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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
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.
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.
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.
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.
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
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.
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+?
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)
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
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
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
has lost methyl group
4. a) Setting up two rate expression, each for the competitive fates of 30
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.
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.