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CHAPTER 1 SUBSTITUENT EFFECTS ONORGANIC RATES AND EQUILIBRIA

I QUALITATIVE DESCRIPTION OF SUBSTITUENT EFFECTS

References:

1. Carey & Sundberg Part A, 3rd. ed. p.196 (4th Ed, p. 204) (5th Ed. pp. 297-318)

2. Isaacs, p. 135

3. Carroll, p. 366

4. J. Hine, Substituent Effects on Equilibria in Organic Chemistry, Wiley-Interscience, 1975, Chapt. 2. QD 503.H56

5. R.D. Topson.Acc. Chem. Res. 16, 292(1983)

[Lowry and Richardson and Miller texts have no non-quantitative analysis ofsubstituents.]

A INTRODUCTION

Substituent effects on the rates and equilibria of organic reactions are of two types:

a)Steric Effects. Steric substituent effects arise from the size (i.e., space requirements)

of the substituent and the fact that strong forces of repulsion result when two atoms are

forced to be closer than the sum of their van der Waals radii.

b) Electronic Effects. Electronic substituent effects are a result of changes in the

electronic distribution within a substrate, caused by the substituents on that substrate.

Electronic effects have been ascribed to several different processes, or mechanisms of

electron displacement, the principle ones being the following:

i) field effect: an electrostatic effect of a charge or dipole transmitted through space

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ii) inductive effect: the effect from differences in electronegativity between a substituent

X and the carbon to which it is attached, transmitted through bonds. For a

substituent such as Cl (i.e., more electronegative than C), the effect is often pictured

in the following way:

+

+

+

CH3CH2CH2CH2Cl

iii) resonance (or mesomeric) effects: delocalization of the electrons in a conjugated

system, affecting alternate positions.

Field effects and inductive effects are usually difficult to separate because they operate

in the same direction. The combined field and inductive effects are often termed polar

effects. Quantum-mechanical calculations indicate that inductive effects through -bonds should be negligible beyond the first two carbons. In a few specifically

constructed rigid systems (e.g., page 1-5), inductive and field mechanisms operate in

opposite directions so that their relative importance of such systems can be assessed.

These studies suggest that inductive effects through -bonds should be negligible

beyond the first two carbons.

B CLASSIFICATION OF SUBSTITUENTS (C.K. Ingold)

-substituents can be classified as + or - Iand + or - R.

- means electron withdrawing

+ means electron donating (or supplying or releasing)

I means polar effects (through bonds and space)

R means resonance effects (delocalization of electrons affecting alternate

positions)


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To summarize:

+R -donating; -R -withdrawing

+I-donating -I-withdrawing

Often the types of substituents are grouped together:

+R, +I +R, -I -R, -I

alkyl,

CH3-,

CH3CH2-,

trialkylsilyl-,

-NH2, -OH, -X, -SH

-NR2, -OR, -SR

-NH

CCH3

O

-OC

CH3

O

C

O

RC

O

ORC

O

OH

C N SO3H

SO2RNO2

examples: +R

OCH3 +

-

CH3O

etc.

-R

N-O O O

-

N-O

+

+

etc.

+

C EXAMPLES OF SUBSTITUENT EFFECTS ON BRNSTED ACIDITY

a) Polar Effects i) pKas of aliphatic carboxylic acids (H2O, 25 C)

Effect of Multiple Substitution on pKa.

Compound pKa single pKa total pKa

CH3CO2H 4.74 -- --

ClCH2CO2H 2.87 1.87 1.87

Cl2CHCO2H 1.29 1.58 3.45

CCl3CO2H 0.65 0.64 4.09


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pKas of Monosubstituted Carboxylic Acids, X-CH2-C(O)OH

X pKa X pKa

-NO2 1.68 -I 3.16

N(CH3)3+

1.83 C CH

1.84

-OCH3 3.53 -C(O)CH3 3.58

-SO2CH3 2.36 -SCH3 3.72

-CN 2.47 -C6H5 4.31

-F 2.66 -CH=CH2 4.26

-C(O)OH 2.83 -H 4.74

-Cl 2.87 -CO2- 5.69

-Br 2.90

Effect of Distance of Substituent on pKa

Compound pKa pKa

CH3CH2CH2CO2H 4.82 --

CH3CH2CHCO2HCl

2.84 1.98

CH3CHCH2CO2H

Cl

4.06 0.76

CH2CH2CH2CO2H

Cl

4.52 0.30


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ii) angular dependence of substituent effects

Grubbs and Firzgerald, Tetrahedron Letters, 4901(1968)(50% aq. ethanol, 25 C)

CO2H

ClCl

CO2H

ClCl

CO2H

pKa: 6.26 6.08 5.68

b) Resonance Effects

i) pKas of benzoic acids and phenols (H2O, 25 C)

-I, +R group:

4.474.094.18pKa:

-Ieffect only

+R effect of CH3O- groupputs negative charge on thcarboxyl group

CO2H

OCH3

CO2H

OCH3

CO2H

-I, -R group:

-Ieffect only

-R effect of nitro groupstabilizes negative chargeof the phenolate

7.158.409.92pKa:

OH

NO2

OH

NO2

OH


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c) Hydrogen Bonding

2.302.984.584.084.18pKa:

CO2H

OHHO

CO2H

OH

CO2H

OH

CO2H

OH

CO2H

CO2H

OH

pka: 2.98

CO2H

OMe

4.09

C

O

O OH

-

d) Steric Inhibition of Resonance

i) pKas of benzoic acids and phenols (H2O, 25 C)

3.253.914.344.244.18pKa:

CO2H

CH3H3C

CO2H

CH3

CO2H

CH3

CO2H

CH3

CO2H

3.46

CO2H

tBu

3.64

CO2H

iPr

3.79

CO2H

Et

3.91

CO2H

CH3

pKa:


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pKa: 10.60 7.16 10.17 8.25

OH

H3C CH3

OH

H3C CH3

NO2

OH

NO2

CH3H3C

OH

CH3H3C

pKa= 3.44 pKa= 1.79

ii) pKas of ammonium ions (H2O, 25 C)

= 2.79= 2.73

N

H+

N

H+

N(CH3)2

H+

10.587.795.06pKa:

e)Hyperconjugation

Another effect that will be called upon through the course should be introduced here.Whereas conjugationis the resonance shifting or distribution of electrons by way of bonds and lone pairs, hyperconjugationis the movement of electrons through bonds.As with normal resonance, it too requires alignment of the orbitals involved. The general

concept is introduced here and will be explained in more detail in later sections. Whennegative charge is involved, the effect is sometimes called negative hyperconjugation,whereas positive charge is stabilized by positive hyperconjugation. See King, Li, Cheng,& Dave, Heteroatom Chemistry, 2002, 13, 397 and Alabugin & Zeidan, J. Am. Chem.Soc. 2002, 124, 3175.

The MO energy level diagram shows why electronegative atoms as more willingparticipants in negative hyperconjugation.


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Orbital interaction diagrams comparing negative hyperconjugation in -C-C-X with (a) X =carbon and (b) X = oxygen. The left side of each diagram shows the interaction of thefilled p-orbital of the carbanion with each * orbital. The resulting energy lowering is

given by CCfor X = C and COfor X = O; CO> CC, the difference being due tothe lower energy of *CO vs *CC, the origin of which in turn is shown on the right side of

each diagram.


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II QUANTITATIVE CORRELATION OF RATES AND EQUILIBRIA

References:

1. Carey and Sundberg, Part A, pp. 196-209 (3rd ed.) (4th

Ed. pp. 204-215) (5th Ed.pp. 335-344)

2. Carroll, p. 371

3. Miller, p.122

4. L&R pp. 143-158

5. Isaacs Chapter 4

6. Anslyn/Dougherty, pp 445-451

6. J. Hine, Structural Effects on Equilibria in Organic Chemistry (see p.1)

7. C.D. Johnson, The Hammett Equation, Cambridge, Univ. Press, (QD 502 J63).

8. Hansch, Leo & Taft Chem. Rev. 91, 165(1991).

9. For a short biography on Louis Hammett, see J. Shorter, Prog. Phys. Org. Chem.17, 1(1990)

A THE HAMMETT EQUATION

The Hammett equation is one of several important linear free energy relationships. It was

developed as a correlation of reactivities (rates) and equilibria in reactions of meta- and

para-substituted benzene derivatives.

As early as 1912 it was noted that in reactions of m- and p-substituted benzene derivatives

there were regularities in substituent effects. For example, whatever effect a p-NO2grouphad on a rate or equilibrium constant, a m-Cl had an effect in the same direction but

smaller. Hammett (1935) found that for a number of reactions involving a series of m- and

p-substituted benzene derivatives, a plot of the log of the rate constant (k) or equilibrium

constant (K) for one reaction vs log kor log K for another reaction gave a fairly straight line.


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Examples

+-OH

k2+ EtOH

85% a . ethanol, 25 C

O Et

O

X

O-

O

X

1

+ H2OK

+ H3O+

(H2O, 25 C)

OH

O

X

O-

O

X

2

k2

+N

CH3

CH3CH3

X

3

N

CH3

CH3

X

(95% aq. acetone, 35 C

+ I-+ CH3I

A plot of log k2for reaction 1 vs.log K for reaction 2 gives a good straight line (slope ca.

2.6) if the points for the ortho substituents are omitted. Similarly, a plot of log k2for reaction

1 vs. log k2 for reaction 3 gives a reasonable straight line of slope ca -1.0 (see graphs,

starting on page 1-11).

In all such correlations the rate and equilibrium constants within a series must all be

measured under a single set of conditions (same temperature, solvent). Satisfactory linear

correlations are generally obtained as long as substituents are restricted to the meta and

para positions. Ortho substituents give poor correlations. For example, in the plot of log k2

(reaction 1) vs. log Ka, the ortho-substituted esters are consistently less reactive than

predicted by the correlation line for the meta and para substituents. Similarly, the same


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procedure applied to aliphatic systems generally does not give a good linear correlation. (A

plot of log k2for basic hydrolysis of XCH2CO2Et vs. log Kafor XCH2CO2H is shown.)

To put all such correlations on the same basis, log Kafor benzoic acids in H2O at

25C was taken as the standard reaction. For all other reactions we have a relationship ofthe type

log k= log Ka(benzoic acids, H2O, 25) + C

or log K = log Ka + C (1)

For no substituent on the benzene ring,

log ko or log Ko= log Ka+ C (2)

Subtracting (2) for (1),

log (k/ko) or log (K/Ko) = log(Ka/Ka) (3)

log Ka, benzoic acids, H2O, 25 C

logk

2 for

85% aq. ethanol, 25 C

+ -OHX

O

E

t

O


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log k2 for


+ -OHX

OEt

O

90% aqueous acetone, 35 C (k2in M-1min-1)

+ Me-IX

N

Me

Me

log k2 for


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+ -OHXOEt

O

log(k/ko)


for

log (K/Ko), ionization of X-CH2-CO2H, H2O, 25 C


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log Ka, benzoic acids, H2O, 25 C

+ Me-X

N

Me

Me

log k2


Equation (3) (p. 1-11)

is a linear free energy (LFE) relationship.

Multiplying both sides of eqn. (3) by -2.303 RT, we obtain

-RTln(k/ko) = -RTln(Ka/Ka)

G- Go= (G- Go)

or G= (G)

Stated in words, the change in the standard free energy of activation produced by a

substituent X is proportional to the change in the standard free energy produced by the

same substituent in ionization of benzoic acids in H2O at 25C.


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Though the effect of a substituent (e.g., a m-Cl) on the Kaof benzoic acid is exceedingly

complex and something that cannot be calculated from first principles, it is a good model

for the effect of the same substitution on the rate of basic hydrolysis of ethyl benzoates

(equally complex). LFE relationships, then, amount to using one process that is too

complex to understand completely as an experimental model for some other process that

is too complex to understand completely.

define: (log Ka- log Ka) or log (Ka/Ka) (4)

(Kaand Kameasured in H2O at 25 C)

Substituting into (3): log(k/ko) = (rate) (5)

or log (K/Ko) = (equilibrium)

Equation (5) is known as the Hammett equation.

is fixed by the substituent (substituent constant) and measures the effect of a substituent

on the Kaof benzoic acid (H2O, 25 C). Since electron withdrawing substituents increase

the acidity of benzoic acids, substituents having a net electron-withdrawing effect

(compared to H) have positive values, and those with a net electron donating effect

(compared to H) have negative values.

Some values (from Hine) are tabulated on page 1-16. Some of these values were

obtained directly from the Kavalues of the corresponding substituted benzoic acids in H2O

at 25 C using equation (4). These are called primary values. Sometimes the values

cannot be obtained in this way, either because the substituent reacts with water, or (more

commonly) because the substituted benzoic acid is not soluble enough in water. In such

cases a secondary reaction of known values can be used, with the expression = 1/log

(K/Ko) or = 1/ log (k/ko). For example, the Ka's of benzoic acids in 50% aqueous

ethanol at 25 C ( = 1.52) is a secondary reaction. Values obtained in this way are

secondary values. Still other values are statistical values that are averages of values

obtained from two or more secondary reactions.


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Substituent Constants, substituent substituent

meta para meta para

Me -0.07 -0.17 NHCO2Et 0.09c 0.13c

Et -0.07 -0.15 NHAc 0.21 0.00iPr -0.07b -0.15 NHCHO 0.22c 0.05c

tBu -0.10 -0.20 NHC(O)CF3 0.35c 0.14c

CH2Ph -0.18c -0.11c NHSO2Me 0.21c 0.05c

CCH 0.20d 0.23d OH 0.12 -0.37Ph 0.06 -0.01 OMe 0.12 -0.27

Picryl 0.27e 0.31e OEt 0.10 -0.24

CH2SiMe3 -0.16 -0.21 OPh 0.25 -0.32n

CH2OMe

0.02c

0.03c

OCF

3

0.40h

0.35h

CH2OPh 0.03c 0.07c OAc 0.39 0.31

CHO 0.36f 0.44f OSO2Me 0.39c 0.37c

C(O)Ph 0.36c 0.44c F 0.34 0.06

CO2- -0.10 -0.00 SiMe3 -0.04 -0.07

CO2Et 0.37 0.45 PO3H- 0.20 0.26

CN 0.56 0.66 SH 0.25 0.15

CH2CN 0.15c 0.17c SMe 015 0.00

CH2I

0.07c

0.09c

SCF

3

0.0h

0.50h

CH2Br 0.11c 0.12c SAc 0.39 0.44

CH2Cl 0.09c 0.12c S(O)Me 0.52 0.49

CF3 0.43 0.54 SO2Me 0.60 0.72

CF(CF3)2 0.37h 0.53h SO2CF3 0.79h 0.93h

NH2 -0.16 -0.66 SO2NH2 0.46 0.57

NMe2 -0.15I -0.83 SO3- 0.05 0.09

N(CF3)2 0.40j 0.53j SMe2+ 1.00 0.90

NMe3+ 0.88 0.82 SF5 0.61 0.68N3 0.37k 0.08k Cl 0.37 0.23

N=NPh 0.30l 0.35l Br 0.39 0.23

N2+ 1.76m 1.91m I 0.35 0.28

NO2 0.71 0.78 Ac 0.38 0.50


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a These substituent constants are based on the ionization constants of benzoic acids taken from the compilation of McDanieland Brown and rounded off, unless otherwise stated. D.H. McDaniel and H.C. Brown, J. Org. Chem., 23, 420 (1958).

b H. van Bekkum, P.E. Verkade, and B.M. Wepster, Rec. Trav. Chim. Pays-Bas, 78, 815 (1959).c Based on a pK value in 50% aqueous ethanol and the corresponding value (1.52).d J.A. Landgrebe and R.H. Rynbrandt, J. Org. Chem., 31, 2585 (1966).e D.J. Glover, J. Org. Chem., 31, 1660 (1966).f A.A. Humffray, J.J. Ryan, J.P. Warren, and Y.H. Yung, Chem. Commun., 610 (1965).

g Values of for electrically charged groups are relatively unreliable.h W.A. Sheppard, J. Am. Chem. Soc., 85, 1314 (1963); 87, 2410 (1965).i J.C. Howard and J.P. Lewis, J. Org. Chem., 31, 2005 (1966).

j F.S. Fawcett and W.A. Sheppard, J. Am. Chem. Soc., 87, 4341 (1965).k P.A.S. Smith, J.H. Hall, and R.O. Kan, J. Am. Chem. Soc., 84, 485 (1962).l M. Syz and H. Zollinger, Helv. Chim. Acta, 48, 383 (1965).m E.S. Lewis and M.D. Johnson, J. Am. Chem. Soc., 81, 2070 (1959).n This value is almost implausibly small, much smaller than the value 0.14 reported in 50% ethanol. See footnote c.

Table taken from J. Hine, Structural Effects on Equilibria in Organic Chemistry, Wiley-Interscience, 1974, p.66.

In examining the fit of a set of experimental rate or equilibrium constants to a

Hammett relationship, log kor log K is plotted vs. , and the best fit slope and intercept

obtained by statistical analysis. The value is the statistical slope, while ko(or Ko) is the

statistical intercept approximately k(or K) for the unsubstituted compound. A recent paper

indicates how easily the relative data can be obtained (J. Am. Chem. Soc. 2000, 122,

6357) although most researchers are much more rigorous, and garner more precise data.

is fixed by the reaction (reaction constant) and the reaction conditions (temperature,

solvent). measures the sensitivity of the rate or equilibrium constant to substitution

(relative to the sensitivity of the rate or equilibrium constant to substitution (relative to the

sensitivity of the Kaof benzoic acids in water at 25 C). The definition log (Ka/Ka) for

benzoic acids (H2O, 25 C) fixes = 1.00 for this reaction. Reactions with a positive

value respond qualitatively to substitution in the same way as the Ka's of benzoic acids do,

i.e., electron withdrawing groups increase k or K. If > 1.00, the rate or equilibrium

constant is more sensitive to substitution than Ka.


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a) Qualitative interpretation of the sign of

positive

equilibrium: The product has a higher electron density at the reaction site than the

reactant.rate: The transition state has a higher electron density at the reaction site than the

reactant.

(By reactant we mean the substituted benzene derivative.)

negative

equilibrium: The product has a lower electron density at the reaction site than the

reactant.

rate: The transition state has a lower electron density at the reaction site than the

reactant.

(The reactant again means the substituted benzene derivative.)

b) Interpretation of Values

Let us examine the values in the large table (p. 1-16) in the light of our qualitative ideas

about the electrical effects of substituents, based on polar and resonance effects (earlier).

R.W. Taft and others have attempted to separate the total electronic effect of a substituent,

as measured by its value, into polar (inductive and field) and resonance contributions.

We will not discuss these attempts in detail at this point, but we will use his ideas and

conclusions to examine some values.


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Assumptions

(i) measures the sum of polar and resonance contributions, and can be expressed

as a weighted sum of two position-independent parameters, I and R, which

measure the polar and resonance properties of the substituent, respectively.

(ii) The polar effect of a substituent is about the same in the meta and para positions.

As partial justification for this assumption we note that for the +N(CH3)3 group,

which should not exert an appreciable resonance effect, m/p = 0.88/0.82 1.

More recent experimental evidence puts this ratio near 1. (JOC2007, 72, 5327)

(iii) The resonance effect of a substituent in the meta position is some constant fraction,

, of its effect in the para position.p = I + R

m = I + R

That is: where I & R depend onthe substituent only and < 1.

It should be noted that it has been determined (JCS, Perkin Trans. II, 133 (1988)) that this

assumption is not a good one).

The reason that the resonance effect is not zero in the meta position is because resonance

can supply or withdraw electrons to or from the positions ortho to the reacting group, from

which the effect can be relayed by the inductive and field mechanisms.

e.g.

G

NH2

G

NH2+

-

G

NH2+

-

Ivalues can be estimated from substituent effects in some rigid saturated systems. Taft

found that by using a value of 0.33 for , a Rvalue could be obtained for each substituent

that reproduced each mand pvalue fairly well.


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Thus, R3/2(p- m) and I1/2 (3m- p)

We can now examine the values for some selected substituents using this separation of

polar and resonance effects to see if the results are in reasonable agreement with our

qualitative ideas.

(i) NH2group (-I, +R) (very strong +R)

m= -0.16 The negative value of mshows that the +R effect

p= -0.66 is the dominant effect, even in the meta position.

R3/2 (-0.66 + 0.16) = -0.75 m= 0.09 - 0.75/3 = -0.16

I1/2(-0.48 + 0.66) = 0.09 p= 0.09 - 0.75 = -0.66

(ii) Other -I, +R groups

group p m p - m( 2/3R) approx. I approx. R

-OH -0.37 0.12 -0.49 0.37 -0.74

-OMe -0.27 0.12 -0.39 0.32 -0.59

With the OH and OR groups the -I effect dominates in the meta position and the +R effect

dominates in the para position.

Halogens: -Ieffect: F > Cl > Br > I +R effect: F > Cl > Br > I

halogen p m p - m( 2/3R) approx. I approx. R

F 0.06 0.34 -0.28 0.48 -0.42

Cl 0.23 0.37 -0.14 0.44 -0.21

Br 0.23 0.39 -0.16 0.47 -0.24

I 0.28 0.35 -0.07 0.39 -0.11


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(iii) -I, -R groups

group p m p- m( 2/3R) approx. I approx. R

-CO2Et 0.45 0.37 0.08 0.33 0.12

-C(O)Me 0.38 0.50 0.12 0.32 0.18

-CN 0.56 0.66 0.10 0.51 0.15

-NO2 0.71 0.78 0.07 0.67 0.11

(iv) Alkyl Groups (+I, +R)

group p m p- m( 2/3R) approx. I approx. R

-CH3 -0.07 -0.17 -0.10 -0.02 -0.15

PRACTICE PROBLEMS 1

1.a) Using the table provided on p. 1-16 of your notes, determine the and R

values for i) the diazonium group and ii) the trifluoromethyl groupb) Having performed the exercise in part a) classify those substituents as +R, +I, -Rand/or -I.

2. Provide two reasons why compound 2 is a much stronger base than N,N-dimethylaniline (1).

NCH3H3C

OCH3CH3O

NEt2Et2N

pKa of conjugate acid = 5.1 pKa of conjugate acid = 16.3

1 2

3. Rank the following functional groups according to the strength of their +R effect.Rank from strongest to weakest. Provide explanations for your rankings.


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N CH3

O

CH3N CH3

N

CH3

CH3N

CH2CH3

CH3N CH3

N

CH3

CH3CH3 +a) b) c) d)

4.Please explain the difference in acidity of the underlined H of the two isomersdrawn.

N

H H

N

H H

pKa= 30.1 pKa= 26.7

OH

OMeO

OH

OMeO

1 2

5. (From a previous midterm) Please indicate whichof compounds 1or 2has the lower pKaand indicatethe reason for your choice.

6.A number of N-protonated amides have pKas in the range of 0-1 as indicated by

these examples:

N

OH

+ A

1.00.41.00.1

N

O

H CH3

+N

O

H H

+CH3 N

OH+

CH3 N CH3

O

CH3

H+

Explain why compoundAhas a pKaof 5.33.

7. The pKavalues for methine hydrogen of triphenylmethane (4) and the substitutedanalogs are as indicated: 4: 30.8; 5, 16.8; 6, 14.4; 7, 12.7. Note that the introductionof one para-nitro group enhances the acidity by about 14 pKaunits, but theintroduction of more para-nitro groups offers substantially less effect (about 2 units).Please provide an explanation; use structures liberally.


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C

H

C

H

NO2 C

H

NO2

O2N

C

H

NO2

O2N

O2N

4 5 6 7

8.Compounds 9and 10are similar acetic acids in that they both bear thedeprotonated form of an acid at the -position. Nevertheless one has a pKagreaterthan 4.74 and the other has a pKathat is less than 4.74. Please provide anexplanation for the relatively large difference in pKavalues, despite the similarities in

ucture.str

CO2-Na+

O

HOSO3

-Na+O

HO

9, pKa= 5.59 10, pKa= 4.05

9. Compound 1 is quite acidic, with an experimental pKa of about 1.

a) nGiven the extensive functionality within the molecule and given that more thaone of the various atoms or groups may contribute independent I and Rmodes of stabilization of anion 2, identify eight stabilization factors that lower the

pKa to 1, a value significant lower than 16. Draw useful resonance structures forou invoke.

b) Identify the significance of 16.any resonance efforts (i.e., -R) y

O O

OH

Cl

SCH3

O O

O-

Cl

SCH3

pKa= -1

1 2


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10. Amidine 3 is a strong organic base with two basic nitrogens. One of the nitrogenaccepts and hold protons preferentially.

a) Identify which nitrogen is more basic and justify your

answer with the aid of a drawing(s).b) Does the pKa of the conjugate acid of 3increase ordecrease if a nitro group is added to the paraposition of one of the aryl groups?

c) Given your answer in b) part, will the effect ofadding a p-nitro group be more significant on the N-aryl group or the C-aryl group? Justify your answer. 3

N NHH3C

C aryl

N aryl

SOLUTIONSTO PRACTICE PROBLEMS 1

1.= (3m-p) R= 3/2(p-m)

for diazonium: (-I, -R)= (3m-p) R= 3/2(p-m)

= (3(1.76) - 1.91) = 3/2(1.91 - 1.76)

= 1.69 = 0.23

for the trifluoromethyl group: (-I, -R)= (3m-p) R= 3/2(p-m)

= (3(0.43) - 0.54) = 3/2(0.54 - 0.43)

= 0.38 = 0.16

2. Compound 1is a simple aromaticamine. It is a weak base and itsconjugate acid is comparativelystrong for an ammonium ion. The

reason is that the lone pair will shareits electrons, by resonance, with thearomatic ring when those electronsare free. To do this the C-N-C atomsmust be coplanar with the ring.

NCH3H3C

NCH3H3C +

-

1


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In 2the two methoxy groups are offering minimal electronic forces but are actingas a steric barrier. These prevent the C-N-C atoms of the Et2N group from gettingcoplanar with the rings and thus when the lone pair is free on the nitrogens of 2,there is essentially no +R donation of electron density into the ring. Thus the lonepairs are more available to accept a hydrogen (and thus be a stronger base and

weaker acid). This is one of the reasons and is an example of steric inhibition ofresonance.

Since the lone pair are not in line with thearomatic p orbitals, they can pick up an Hand in one certain arrangement, that H isdirected right at the other nitrogen lonepair, creating a perfect situation for H-bonding. So the second reason for thehigh basicity of 2is the ability for H-bonding, whereby the H is simply captured

perfectly between the nitrogens and thenitrogens share the positive charge.

CH3O O 3

N N

EtEt EtEtH

CH++

protonated 2

3. A measure of the +R strength of these compounds is tied to the availability of the

lone pair of electrons on the N which is common to all the groups. Below is a tablewhich ranks the compounds and describes the role played by each Ns lone pair.

Rank Compound Explanationstrongest d)

N

CH2CH3

CH3

N lone pair is fully free to donate and the alkyl groups

enhance this effect a little with their +I character

2nd b)

N CH3

N

CH3

CH3

N CH3

NCH3

CH3

N CH3

NCH3

CH3+

-

The Ns lonepair participates in resonance similar to normal amideresonance except a nitrogen, slightly less

electronegative than an O, is not quite as good ataccepting electron density.3rd a)

N CH3

O

CH3

N CH3

O

CH3

N CH3

O

CH3+

-

The Ns lonepair participates in normal amide resonance and


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therefore is less available than for a normal amine.Since the oxygen is more electronegative than N, thisplaces this molecule 3rdinstead of 2nd.

weakest c)

N CH3

N

CH3

CH3CH3 +

N CH3

N

CH3

CH3CH3 +

N CH3

N

CH3

CH3CH3

+

The Ns lonepair is rather tied up trying to assist the other N with theburden of sharing a positive charge. In doing so, thatlone pair is not available to donate elsewhere. Thepositive charge on the other N makes it a particularlygood -R group.

4.Deprotonation makes the anions shown. We have assumed that inductive effectsare about the same whether the influencing group is in the meta or the para position.Hence the pKa difference rests on resonance. Looking at all of the resonancestructures below, only one offers particular stabilization over the others, and this onecaused the enhanced acidity of the particular isomer.

N

H H

pKa= 30.1

-H+ N

H

N

H

N

H

N

H

N

H H

pKa= 26.7

-H+N

H

N

H

N

H

N

Hparticular stabilizationof the -ve charge on theelectronegative atom

5.a) Compound 1


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b)

O-

OMeO

O-

OMeO

1 2

O

O-MeO

O

O-MeO

steric hindrance preventsfull planarity and hencefull conjugation

6.

N

OH

+ A

1.00.41.00.1

N

O

H CH3

+NO

H H

+CH3 N

OH

+CH3 N CH3

O

CH3

H

+

Loss of H+ from any of the four typical examples results in typical amides that havethe usual amide resonance with the carbonyl. Since when the H is lost, the resultinglone pair can establish resonance with the carbonyl, as shown for one example.

CH3 N

O H+CH3 N

O

-H+CH3 N

O-

+

However, with compoundA, when the H is removed, the lone pair cannot do typicalamide resonance, because the lone pair of the N cannot align properly with theorbitals of the carbonyl group. So, if the free lone pair does not participate inresonance with the carbonyl, the electrons have little else to do, so they are less

inclined to lose the proton. Less acidic means a higher pKa value.

+N

OH -H

+ N

O

X N

O -

+


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If nomenclature is based on reactive character, compoundA should really be namedan amino ketone rather than an amide.

7. If the anions of any of these deprotonated compounds are to be stabilized by wayof resonance with the three rings, they all have to lay (lie?) in one plane. Because ofthe ortho hydrogens of adjacent rings bumping into one another (i.e., stericcrowding), this full three-ring stabilization does not occur completely.

C

H

4

C

HHH

H

H

H

crowding H's

-H+

Only one ring is allowed full resonance, while the other two are twisted out of theplane to make room for the ortho Hs.

C

HH

crowding H'scauses the otherrings to twist outof the plane

So when you add one nitro, you get a strong and complete -R and -I stabilizing effectwhich lowers the pKa substantially. The addition of more NO2 groups to the otherrings does not add any more -R stabilization, since the twisting of the other two ringsprevents transmission of the resonance effects. The addition of those other two nitrogroups adds only a -I effect and the increment is of the size that we are used toseeing.

This is an example of steric inhibition of resonance.


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8.

CO2-Na+

O

HOSO3

-Na+O

HO

9, pKa= 5.59 10, pKa= 4.05

As discussed in class, the CO2-M+ group is an electron donating one and you cancorroborate this by looking at itsvalues. Hence this EDG hinders ionization of theother carboxyl group.

The pKa of 4.05, being less than 4.74, means that the SO3-M+ must be an EWG,despite carrying a charge. It was important to mention this in your solution.

The reasons could be a combination of many:

1. As many said, the extra oxygen offers a third resonance structure to helpstabilize the charge. The sulfur is also electronegative and adds someinductive withdrawal.

2. This one was not in anybodys solution: having three oxygens assuming someof the charge means there are three sites for water to H-bond to, in order toassist in the stabilization or distribution of charge.

3. As some people did, you could inspect the values for SO3-M+ (and maybecalculate its I value) and learn that it is a EWG inductively.

9 a)

b) 16 is the pKa of methanol, which could be viewed as the starting point onto which allthe groups and atoms of 1are introduced. Methanol is the simplest alcohol, so 16 is thebase pKa of the simplest alcohol.


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10 a)

b) Adding an EWG to a +-charged

species will destabilize it. Thus ithas less inclination to maintain its+-charge and therefore will be moreacidic. The pKa of the conjugateacid will decrease.

c)


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c) Reaction Constants,

Some examples of reaction constants are shown in the table. Further examples are given

in O.. Exner, In Advances in Linear Free Energy Relationships, Edited by N.B. Chapman

and J. Shorter, Plenum Press, 1972, Chapt. 1, and in the references cited at the end of thesubstituents constants table.

Reaction

ArCO2H ArCO2- + H+, w ater 1.00

ArCO2

H

ArCO2

-

+ H+, EtOH 1.57

ArCH2

CO2

H

ArCH2CO2- + H

+, water 0.56

ArCH2CH2CO2H ArCH 2CH 2CO 2

- + H

+, water 0.24

ArOH

ArO

-+ H

+, w ater 2.26

ArNH3

+

ArNH2

+ H+, water 3.19

ArCH2

NH3

+ArCH

2

NH2 + H+, w ater 1.05

ArCO2Et + -OH ArCO2- + EtO 2.61

ArCH2CO2Et +

-

OH

ArCH2CO2- + EtOH 1.00

ArCH 2Cl + H 2O ArCH 2OH + HCl -1.31

ArC(Me) 2Cl + H 2O ArC(Me) 2OH + HCl -4.48ArNH2 + PhC(O)Cl ArNHC(O)Ph + H -3.21

ArC(O)Ph + Me-Li

+ArC(O

-)(Me)Ph 0.94

Data taken from P.R. Wells, Linear Free Energy Relationships, Academic Press, New York, 1968, pp 12-13,except for last entry which comes from J. Org. Chem. 2002, 67, 4370.

The sign and size of for a reaction rate is one kind of experimental probe of transitionstate structure. In particular, it can provide some indication of the difference in electron

density at the reaction site between the reactant and the transition state. In order to be

able to interpret values in this way, it is necessary to understand some of the factors that

affect .

i) Transmission of Electrical Effects to the Reaction Site

Since is a measure of the electrical effect of the substituent, must depend in part on

how much of this effect is transmitted to the reaction site. This factor is illustrated by

values for several acid-base equilibria.


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Acid , Ka(H2O, 25 C) Acid , Ka(H2O, 25 C)

CO2H

X

1.00 C CCOOHH

H

X

0.42

CH2CO2H

X

0.56SCH2CO2H

X

0.30

CH2CH2CO2H

X

0.24NH3

X

+

3.2

OH

X

2.3

ii) Solvent Effect on

The magnitude of increases with decreasing solvent polarity.

Some examples:

solvent H2O 1.00

50:50 EtOH/H2O 1.57

EtOH 1.96

CO2H

X

Ka, 25 C

N2

Ph Ph+

C

O

OCHPh2

X

CO2H

X

solvent MeOH, 30 C 0.844

toluene, 25 C 2.22


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B FAILURE OF THE SIMPLE RELATIONSHIP

The failure of correlations involving reactions of ortho-substituted benzene derivatives and

aliphatic compounds has been mentioned. Such failures are reasonably attributed to steric

effects. Three other kinds of failures will now be discussed.

a) Ka's of Phenols and Anilinium Ions

+ H2O + H3O+

OH

X

O-

X= 2.26

Ka

+ H2O + H3O+

NH3+

X

NH2

X= 3.19

Ka

If we examine the plots for these equilibria we see that the meta substituents give a

reasonably good linear correlation but that certain kinds of substituents deviate badly and

in a systematic way from the best fit line for the para substituents. For example, in the plotof log (Ka/Ka)(phenols) vs. log (Ka/Ka)(benzoic acids) (i.e., ) we see that the para -R

group points all lie above the line. For such groups the substituted phenol is more acidic

than predicted. For example, p-NO2and p-Ac-phenol are ca. 1 log unit (10X) more acidic

than predicted.

Reason: -R groups have a special effectiveness when conjugated with a +R group (i.e., -O-

or -NH2).Called through resonance or: direct resonance


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NH2

C

N

+

-

NH2

C

N

O

N-O O

-+

O-

N-O O

+

No similar direct resonance interaction is possible in the p-NO2 or p-CN substituted

benzoate ions, consequently the values for these groups measure mostly a polar (-I)

effect (refer to the IRdissection of values, page 1-19). Stated another way, ionization

of benzoic acids is a poor model for ionization of phenols when we have strong -R

substituents.

We need two sets of constants for strong -R groups: one where through resonance (in

reactant, product, or TS) is possible and one where no through resonance is possible. The

special p values for use with reactions of the -OH group in phenols, of the NH2group in

anilines (and other reactions where direct resonance interaction with an unshared electron

pair is possible) are called - constants.


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b) Ionization of Phenols and Anilinium Ions

log Ka,

phenols

H2O, 25 C


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log Ka,

anilinium ions,

H2O, 25 C

Table of corrections

log Kaphenols

scale

(p-CN) -CN

group p p- (p

- - p)

p-C(O)CH3 0.50 0.82 0.32

p-C(O)OEt 0.45 0.74 0.29p-CN 0.66 0.99 0.33

p-NO2 0.78 1.23 0.45

log (K/Ko) for anilinium ions

1

2.77log (K/Ko) for phenols or

1

2.11-=


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-values for para substituents

substituent - substituent

-CCPh 0.36 N2

+ 3.20CHO 0.98 NO2 1.23Ac 0.82 SiMe3 0.17CO2H 0.73 S(O)Me 0.73CO2ME 0.74 SO2Me 1.05CO2Et 0.74 SO2CF3 1.36C(O)NH2 0.62 SO2NH2 0.94CN 0.99 SO2F 1.32CF3 0.56 SF6 0.70NNPh 0.69 SMe2

+ 1.16

c) Solvolysis of 2-Aryl-2-chloropropanes (t-cumyl chlorides)

H.C. Brown and Y. Okamoto, J. Am. Chem. Soc. 80, 4979 (1958).

C CH3CH3

Cl

X

+ HClC CH3

CH3

OH

X

90% aq. acetone

25 C

A plot of log kvs gives a reasonable straight line for all meta substituents and for -R para

substituents, with = -4.54. The points for p-F, p-Cl, p-Br, p-I, p-CH3and p-OCH3all lie

above this line (more reactive than predicted by factors of ca 10 and 100 (1 and 2 log units)

respectively (see plot nearby). The transition state for this reaction resembles the 2-aryl-2-

propyl cation. It is clear that we should not expect Ka's of benzoic acids to provide a good

measure of how well a group can supply electrons by resonance on extreme electron

demand. Although a through-resonance +R effect of p-CH3O- and other +R groups is

measured to some extent in their values, the electron demand by the carboxyl group in

benzoic acids is much less than that of the carbocation-like group in the solvolysis

transition state.


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+values forpara substituents

substituent +

Me -0.32

Et -0.31

i-Pr -0.29

t-Bu -0.27 log k/kHPh -0.21 solvolysisNH2 -1.47 of t-cumylNMe2 -1.67 chloridesNHAc -0.58

OH -0.91 90% aq.OMe -0.79 acetoneOPh -0.50 25 CO- -2.30

F -0.08SMe -0.62

Cl 0.11

Br 0.14

I 0.13

Using the best fit straight line defined by the meta substituents H.C. Brown obtained a set

of + values for use in reactions where through-resonance between a +R group and a

carbocation-like reaction site is possible.

log k

scale

(p-OMe)+(p-OMe)

log (k/ko) 1

4.54

+= -

for t-cumyl chloride solvolysis,90% aq. acetone, 25 C


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PRACTICE PROBLEMS 2

1.The for benzoic acid dissociation at 25 C in DMSO (dimethyl sulfoxide) is 2.60.Why is this value different from the same equilibrium in water?

2. How many different values have you been introduced to? Explain exactly whateach of them is.

3. The reaction shown below gives a very strange and non-linear Hammett plot, alsoshown below. This occurrence usually means that the reaction is actually proceedingby two different mechanisms, depending on the identity and more specifically, the

electronic character of X. Please provide two different mechanisms and identifywhere you would invoke these mechanisms.

100 oC

benzene

X

N

CH2

X

N

+

CH2Cl +

log (kX/kH)

+


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4. The rates of chlorination of acetylenes are sensitive to substituents:

Xp-OMe 19, 500

p-Me 190p-F 14.9H 10.6

p-Cl 4.15p-Br 2.81

m-NO2 0.0165

p-NO2 0.00325

C C H

X

+ Cl2 C C

Cl

Cl

HX Do two Hammett analyses on the data in order to learn if the data better correlatedwith or with +?Based on your answer, sketch a structure for the transition state ofthe rate determining step? Hint, Cl2reacts similarly to Br2.

5. In the accompanying scheme, when X = Cl, compounds 13are known as O-isopropvlbenzohydroximoyl chlorides. These compound undergo methoxy substitution whentreated with methoxide ion (48 C).

Y

XN

OiPr NaOMe10% MeOH/90% DMSO

Y

XN

OiPrMeO

-

Y

MeON

OiPr

13

-Cl-

STEP 1 STEP 2

a) Given a value of 2.20, using regular values for this reaction, please indicate,with an explanation, which of the two steps is rate-determining. The explanationshould address both steps of the mechanism.

b) Which reaction presented in the list on page 1-22 of the notes closely resemblesthe reaction at hand in both general mechanism and approximate value.

c) The overall transformation also occurs when compounds 13(X = Cl) are exposedto MeOH, but the mechanism is different. That mechanism involves initial rate-determining loss of chloride and the value based on +values was found to be-2.40. Provide a mechanism for this particular reaction and identify importantresonance structures involved, including those consistent with the use of +

values.


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d) Reverting back to the scheme shown, if X = OEt, the reaction does not occur,and the starting material is recovered, partially isomerized. Provide anexplanation for the isomerization of the C=N bond and for the lack of reaction.

6. Which set of 's would be needed to handle the substituent effects on theionization of para-X-2,6-dimethyl-N,N-dimethylanilinium ions?

7. In the solvolysis of trifluoroacetates and chlorides, the methoxy oximino group hasbeen demonstrated to be a rate accelerating functionality when placed at the properposition.Examples:

Me2C C

Cl

N OMe

H

Me2C CH3Cl

14 15

Rate ratio 14/15= 10 in CF3CH2OH

H

N OMe

OC(O)CF3 OC(O)CF3

CH3

16 17

Rate ratio 16/17= 7 in CF3CH2OH

a) Explain with structures how the oximino group accelerates the rate of thesolvolysis reaction in these cases.

b) Explain why the oximino group in theexamples below severely retards the rate ofthe solvolysis reaction.

Rate ratio 18/19 = 6.1 X 10-5 in CF3CH2OH. OMs OMsH

N OMe

18 19

8. Predict the sign of the Hammett equation in each of the following reactions.a)

+ CH3CH2 I + IOCH2CH3X

O

-

X

98% EtOH in H2O

70 C

b)

N(CH3)3X

+

CH3OCH3+N(CH3)2X

-OCH3

+

CH3CN

25 C

Is this a circumstance where one could invoke through-resonance?


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9.In a J. Org. Chem.(JOC1993, 58, 5434) article, there is a mechanistic palladiumcatalyzed cross-coupling reaction.

OSO2CF3tBu + X SnBu3

tBu XPd2(dba)3

AsPh3, 333 KNMP (solvent)

The following rate data on the following page were obtained.

Parasubstituent log (kx/ko)

CF3 -0.22

Cl -0.18

H 0.0

OMe 0.42

NMe2 0.90

Now it is recognized that a 4thyr.undergraduate is not particularly familiar withthis reaction, but you now have beenintroduced to a tool to provide some light onthe mechanism. The paper presents a plot ofthese rate data vs. , which gave asomewhat linear plot with = -.89 and r2=.954.

a) Prepare a graph using these data and confirm the slope and regression datapreviously obtained. What role are the electron donating substituents playing in thetransition state of the rate determining step of the reaction.

b) Have these authors done a good job in this particular aspect of their study? Is thereany experiment that should be done to gain more detail regarding the role of thesubstituents? Is so, perform that experiment.


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SOLUTIONSTO PRACTICE PROBLEMS 2

1. The for benzoic acid dissociation at 25 oC in H2O is defined as 1.00. Thestabilization of the dissociated form of the molecule arises from substituents and from

interaction with water, much of which is H-bonding. The in DMSO (dimethylsulfoxide) is 2.60, which implies that substituents are involved to a much largerextent. The reason is that the solvent is much less involved. The dipolar character ofDMSO has no effective means of stabilizing the negative charge of the carboxylateand so, there is greater need for participation by the substituents.

2. -general usage when there in no opportunity for through resonance, but thereis still some simple +R and -R interactions of the substituents with the aromatic ring

I -used when there is absolutely no chance for resonance of any kind. Typically the

situation is when the atoms separating reaction site and substituents are saturated

R -not really used at all, since use would imply no induction going on and it isdifficult to envision a situation where induction can be eliminated

+ -should be employed when +R substituents have a chance to stabilize startingmaterial, transition state, intermediate or product

- -should be employed when -R substituents have a chance to stabilize startingmaterial, transition state, intermediate or product

3. The two mechanisms are SN1 and SN2.

+

+

CH2-Cl

CH2-Cl CH2+

XN

X

N

CH2

X

N

r.d.s.

r.d.s.

The data can be interpreted as follows.SN1 region. Since the rate plot is flat, substituents have no influence on the reaction,and it follows that since the pyridine possesses the substituents, that the pyridine hasno bearing on the reaction. That is, the pyridine is not part of the rate equation. An


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SN1 reaction is consistent with this data and is invoked herein. The flat section is onthe more right region of the plot and includes mildly electron donating and all types ofelectron withdrawing substituents. These pyridines are not a strong enoughnucleophile to attack the benzylic position before the Cl-leaves by itself.

SN2 region. With the powerful electron donating substituents, the pyridine is now astronger nucleophile and can perform its chemistry BEFORE the Cl-leaves by itself.The left side of the graph has the strong donating substituents. Now, since thepyridine is part of the rate equation, we would expect to see a non-zero slope, sincesubstituents have an opportunity to exert their influence. The slope is negative,consistent with the fact that the strongest electron donators can activate the pyridinethe most and prompt it to react as a nucleophile. + values were used since throughresonance can be invoked on the pyridines bearing strong +R substituents.

log (KX/KH)

+

SN1 region, loss of Cl is r.d.s.

SN2 region,pyridine is most nucleophili

4. The data table is expanded to the following:

X + log (k/ko)

p-OMe -.27 -.79 19, 500 3.26p-Me -.17 -.32 190 1.25p-F .06 -.08 14.9 0.148H 0 0 10.6 0.00

p-Cl .23 .11 4.15 -0.407

p-Br .23 .14 2.81 -0.576m-NO2 .71 .71 0.0165 -2.81p-NO2 ,78 .78 0.00325 -3.51

C C HX

For an analysis of the data vs.


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Y = M0 + M1*XM0 0.73677M1 -5.4389 = r 0.96817

For an analysis of the data vs. +Y = M0 + M1*XM0 -0.044166M1 -4.1667 = r 0.99814

The two Hammett analyses indicate that the data is better correlated with + Thefollowing mechanism is consistent with the result.

or

Cl-

T.S. to this intermediatecan be stabilized bythrough resonance

+

X C C

Cl

HCl ClXC C H

C C

Cl

Cl

HX

r.d.s.

5.a)The transition state (T.S.) for STEP 1 involves the addition of a methoxide, with itsadditional electron density to the carbon to the aryl ring. In the reaction intermediatethat follows the T.S., that charge ends up on the atom twice removed from the aryl ring.In the T.S., the few atoms around the aryl ring have more electron density than thestarting neutral material and should be positive. Its magnitude is about right for anucleophilic attack on a carbon so close to the ring. (see part b) of this question). I

cannot see any room for through resonance, so simple values are suitable.

In STEP 2, the substrate goes from an electron rich negatively charged compound to aneutral. I would expect a negative value here.

Based on the given value of 2.20, STEP 1 is rate determining.

b)


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The following entry is very similar to the reaction under analysis. It is the 8thentry onpage 1-22. Be sure that you appreciate the mechanistic analogies between thesystems.

ArCO2Et +-OH ArCO2

- + EtO 2.61

c)In this case, the p value means we are making a cation from a neutral and the

correlation with + values means we should invoke through resonance in the transitionstate for chloride loss. The reaction is SN1 and Cl- loss is rate determining. EventualMeOH addition provides the product.

Drawing the intermediate after complete Cl- loss will serve as an indication of howthrough resonance stabilizes the T.S.

C

Cl

NOiPr

Y

MeO

NOiPr

-Cl-C N

OiPr+

EDG EDG

CN

OiPr

EDG

+CN

OiPr

EDG+

MeOH

-H

d)An easy, but wrong explanation is the following. The first step of the reaction occurs,but the last step does not. This is because alkoxide is a poor leaving group compared tohalide. We know the first step occurs since it is the only method of isomerizing the C=N

bond. Another way of saying all this is to say STEP 2 is now rate determining and itdetermines that any further reaction will not even occur.


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Y

EtO

NOiPr

NaOMe10% MeOH/90% DMSO

Y

OEt

NOiPr

MeO

-

Y

MeO

NOiPr

13

X

Y

OEt

N

OiPr

MeO

-

bond rotation

loss of MeO-

Y

EtO

N

OiPr

A analogy of this has already been introduced to you. Recall how the Meisenheimercomplex as isolated in Chem 3750 lecture during the aromatic addition eliminationmechanism. THIS IS WRONG because there is no reason why MeO- should leave witha preference over EtO-. Hence if MeO- can leave and revert back to starting material,why cant EtO- leave to provide product? (Alkyl groups are boring)

So the better explanation must require that MeO- does not even add to the substrate.So why is 13(X = Et) so much less reactive that 13(X = Cl)? I think the answer lies inthe comparison of the +R/-I capacities of EtO and of Cl. Cl is primarily a I substituent,permitting and promoting attack of an Nu on the C=N bond. EtO is primarily a +Rsubstituent which hinders attack by the resonance structure shown:

Y

EtO

NOiPr

13 Y

EtO

NOiPr-

+

Such a structure clearly reduces the electrophilicity of the C=N feeding the nitrogensome of the electron density it is already striving for. Think about the reactivitydifferences between a carboxylic acid chloride and a carboxylic ester.

More importantly, the increased C-N single bond character permits bond rotation tothe other isomer.


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Y

EtO

NOiPr

13 Y

EtO

NOiPr-

+

Y

EtO

N

OiPr

-

+

Y

EtO

N

OiPr

bond rotation

This is offered as the mechanism by which the isomerization can occur.

6. An initial analysis would suggest that through resonance would be appropriate,

since in the non-protonated form, there is through resonance of the Ns lone pair ontoa para-X group when the X is a -R group. This would mean that - values would givethe best correlation. However, it should be noted that the 2 and 6 methyl groupscreate a steric barrier to through resonance as they force the dimethylamino group totwist out of resonance and relieve steric strain. Hence on both the left side and rightside of the equilibrium, there is no opportunity for through resonance.

NCH3

CH3CH3

CH3

X

N

CH3

CH3CH3

CH3

X

-H+

NCH3

CH3CH3

CH3H

X

+

probable rotationalisomer

steric problems

7. a)The intermediates arising from compounds 14and 16benefit from resonancestabilization as shown in the diagrams. The MeO is a +R group that is capable of

transferring its electrons to the carbocation via resonance.

Please note that the diagram shows stabilization of the cationic intermediate. This istaken as an indication that there is also significant resonance stabilization in thetransition state for loss of the leaving group.


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Me2C C

Cl

N OMe

H

Me2C CH3Cl

14

15

H

N OMe

OC(O)CF3

OC(O)CF3

CH3

16

17

Me2C CN OMe

H

+Me2C C

N OMe

H

+SN1

Me2C CH3SN1 +

SN1

SN1

CH3

H

NOMe

+

+

H

NOMe+

b)In the case of 18, the methoximino group is one carbon removed from the reaction siteand is separated by a saturated, insulating carbon. Hence the resoance as shownabove is not possible. Furthermore, the bicyclic nature of the homoadamantyl carbonbackbone of 18does not permit intramolecular backside attack by the N or O of themethoximino group.

So, the methoximino group as a conglomeration of electronegative atoms can only actas a -I group and retard the solvolysis as compared to compound 19.

8. Predict the sign of the Hammett equation in each of the following reactions.a)

+ CH3CH2 I + IOCH2CH3X

O-

X

98% EtOH in H2O

70 C

should be negative, the oxygen is going from high electron density to lower electrondensity.

b)


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+ -OCH3 N(CH3)2X

+ CH3OCH3N(CH3)3X

+heat

should be positive since the nitrogen gains electron density as the reaction proceeds.Through resonance is likely since the lone pair which is getting restored onto thenitrogen can be delocalized onto the X groups when they are -R groups in the paraposition.

9.The J. Org. Chem.(JOC1993, 58, 5434) article gives a Hammett plot using andthe data simply does not correlate very well.

OSO2CF3tBu + X SnBu3

tBu XPd2(dba)3

AsPh3, 333 KNMP (solvent)

So, I have plotted on the same graph the log (kx/ko) vs + and I get a better fit and areduced value. The plot of both curves is on the next page.

Parasubstituent + log (kx/ko)

CF3 .54 .54 0.90

Cl .23 .11 0.42

H 0 0 0.0

OMe -.27 -.79 -0.18

NMe2 -.83 -1.67 -0.22

Following the paper: = -.89 and r2= .946.

Using+: = -.54 and r2= .979.


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The new data suggest that thereis reduced substituentinvolvement, but that when thesubstituents do participate, the+R groups among them have an

opportunity to utilize their throughresonance capabilities.

The authors were very neglectfulin this analysis. The weakcorrelation coefficient shouldhave prompted them to try +values in an effort to get a betterfit.

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Chapter 1 4720