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Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

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Page 1: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Chemistry 125: Lecture 63March 31, 2010

Aromatic Substitutionand Triphenylmethyl

This

For copyright notice see final page of this file

Page 2: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

an important intermediate for aromatic synthesisHow to make it?

N2 is a spectacular leaving group.

Diazotization I: N-Nitrosoamine (sec. 14.7)

(anhydride)

H-O-N=O(nitrous acid)

Why not cut out the middle man?

Slow

Slow

Fast

at modest acidity

Rate [HONO]2 [PhNH2]0[PhNH2]02 0

rate-determining step

Page 3: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Diazotization II: Higher Acidity

“allylic” proton

Fast

at higher acidity

Rate [HONO]2 [PhNH2]1

H

+

Slower

1

Slow

Slow

depends on small

concentration of free aniline

rate-determining step

Page 4: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Diazotization II: Higher Acidity

increasing acidity

Complex!Shifts of rate-determining step,

and even of mechanism.

Many roads lead to benzenediazonium.

Allylic Rearrangement

Page 5: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Diazonium Uses (sec. 14.7)

Page 6: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Diazonium Uses (sec. 14.7)

What use is this

reaction?

Good way to make toluene?

Page 7: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Preparation ProblemsNO2

Cl

NO2

Cl2

FeCl3

NO2

Cl

Cl2

FeCl3

(Cl+ FeCl4-)

NH2

Cl

HNO3

H2SO4

OH

Cl

“reduction”

(Zn/HCl?)

H2O

100°C

HONO

(NaNO2/HCl)

N2

Cl

+

diazonium salt

Cl-

or :HalCNNO2

H

Cl

both o,p-directing!

Page 8: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Yale ‘98S

Page 9: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

CH3

CH3H3C

NH2

Indirect Synthesis

HNO3

H2SO4 AlCl3

CH3Cl

“reduction”

(Zn/HCl)

HONO

(NaNO2/HCl)

CH3 CH3

CH3H3C

NO2

CH3

NH2

CH3

CH3C

O

AlCl3

CH3Cl

CH3C

O

N2

+Cl-

CH3

CH3H3C

H3PO2 !

wrong group?wrong place?

if necessary

stronger o,p-director than CH3

Page 10: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Nucleophilic Aromatic SubstitutionNO2

F

NO2

Sanger’s Regent

H2N-R

NO2

NO2

HN-RUsed to visualize fragments with exposed amine

groups in 2D chromatography. Amino acid sequence

of insulin (1955)

NHRFNO2

NO2

+HWhy F?

(generally a poorleaving group)

SlowStep

Identify the amino acid at the end of a chain

or fragment of a chain1958 Nobel Prize to

Frederick Sanger

activated by two NO2

groups

activated by F

also won 1980 Nobel for DNA sequencing!

Page 11: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

NAD+ NADH (p. 679)

(closely enough balanced to run both ways)

key molecules in biological

Oxidation & Reduction(H- acceptor) (H- donor)

Page 12: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Benzylic Reactivity - Anion - pKa (sec. 13.12)

Ph2CH-H pKa = 33.5

~109 Ka Alkyl-H ~12 kcal/molefrom benzylic

~108 Ka PhCH2-H ~11 kcal/molefrom 2nd benzylic

102 Ka Ph2CH-H ~3 kcal/molefrom 3rd benzylic

Ph3C-H pKa = 31.5

Steric hindrance in triphenylmethyl causes

twists that reduce overlap with 2pC by 25%

from diphenylmethyl.

PhCH2-H pKa = 41

H2C=CH-CH2-H pKa = 43 ~107 Ka Alkyl-H ~9 kcal/molefrom allylic stabilization

Page 13: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

(C6 H5 )3 C

TriphenylmethylFree Radical

110 years old

Page 14: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Transcript

Chemistry

Metallurgy Geology &

Mineralogy Scientific Free-Hand Drawing

ChemicalLiteraturein German& French

French RhetoricLogicPsychology

Physics Math

b. Elizabetgrad, Ukraine (1866) University of Michigan (1886) Chicago (1884)

Speaker at Dedication of SCL (1923)

Page 15: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

MG as Student

Gomberg in the Analytical Laboratory

After freshman year

Gomberg averaged9.5 Chemistry

Labs per Week.2/3 of them wereAnalytical

Labs(not hours!)

(no spectroscopy)

Page 16: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Adolf v. Baeyerage 61

R. Willstätterage 24

Munich 1896

Moses Gombergage 30

Johannes Thieleage 31

"Moses Gomberg was Thiele's coworker in the student laboratory. He was very reserved and modest, kept entirely to himself, and never chatted in or out of the laboratory. Some years later the work he carried out in the United States on the triphenylmethyl radical, a case of trivalent carbon, became famous."

"This brilliant Experiment, one of the most beautiful in organic chemistry and one which few people credited at first, gave great impetus to chemistry and would have been worthy of any distinction." R. Willstätter

Page 17: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Thiele & Heuser (January 1896)

Hydrazine Derivatives of Isobutyric Acid

AIBN 50%

Gomberg (Ann. 300, 59) Under the sponsorship of Prof. Thiele I have followed up these reactions...

CH3

C

CH3

C

CN CN

CH3

CH3

N N

CH3

C

CH3

C

CN CN

CH3

CH3

Page 18: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Victor Meyer 9/8/48 - 8/8/97“Geliebte Frau! Geliebte Kinder! Lebt wohl! Meine Nerven sind zerstört; ich kann nicht mehr.” Victor Meyer

(1848 - 1897)

Third TermHeidelberg

with

who hadintroduced the idea ofSteric

Hindrance

Page 19: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Friedel-Craftsor

Ph2Mg

Tetraphenylmethane (1897)

“I have tried to solve this problem in a completely different way.”

Ph C

Ph

Ph

Ph?

Ph C

Ph

Ph

PhN N

8 g

110°

Cu

0.3 gSolubilityAnalysis : C 93.32 (93.75) H 6.36 (6.25)

100 mg for Mol. Wt. : 0.289° 306 (320 calc.)(by solvent b.p. elevation)

Page 20: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

O2N- - 3C- + EtOHO2N- - 3C-H EtO-

Back in Ann Arbor (1898-9)

Confirmed Mol. Wt. 0.285° 318 (320 calc.)

“Unlike the trinitrotriphenylmethane… it does not dissolve in sodium ethylate, nor does it give any coloration…”

How about O2N- - 3 C-C - -NO2 3 ?

Prepared O2N- - 4C (99.5% yield)

Page 21: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Prepared "Hexaphenylethane"

C+H = 93.97 , 94.20 , 94.00 , 94.57%from first 4 methods.

Reported more than 17 methods.

Prepared authentic peroxide from Na2O2 .

Prepared hydrocarbon in CO2 atmosphereusing special apparatus with ground glass

joints.

Ph3C-ClZn

Ph3C-CPh3Ph3CO-OCPh3O2 ( C+H = 93.82 )

Page 22: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

Free Radical! (1900)

Highly “Unsaturated” (O2, Cl2, Br2, I2!)

Launched an American Century of Chemistry

Page 23: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

October, 1900

Page 24: Chemistry 125: Lecture 63 March 31, 2010 Aromatic Substitution and Triphenylmethyl This For copyright notice see final page of this file

End of Lecture 63March 31, 2010

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The following attribution may be used when reusing material that is not identified as third-party content: J. M. McBride, Chem 125. License: Creative Commons BY-NC-SA 3.0