Chemistry 125: Lecture 64April 1, 2011

TriphenylmethylCarbonyl Compounds:

Overview This

For copyright notice see final page of this file

Ph3CPropeller

Conformation

Ph

H-H contactpreventsbonding

Two Ph3CPropellers

HopelesslyRemote

Steric hindrance in triphenylmethyl causes twists that reduce overlap with 2pC by 25% from diphenylmethyl.

It also makes tetraphenylmethane very difficult to prepare,

not to mention hexaphenylethane !

Friedel-Craftsor

Ph2Mg

Tetraphenylmethane (1897)

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

?

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)

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)

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 (SN2).

Prepared hydrocarbon in CO2 atmosphereusing special apparatus with ground glass joints.

Ph3C-Cl Zn Ph3C-CPh3Ph3CO-OCPh3O2 ( C+H = 93.82% )failed for preparing

Ph-CPh3

Free Radical! (1900)

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

Launched an American Century of Chemistry

October, 1900

Par

is (W

urtz

- M

édic

ine)

Cor

nell

MIT

MIT

Par

is (F

ried

el -

Min

es)

Pre

side

nt

AlCl3/C-Electrophile: The Friedel-Crafts Reaction

1877Gibbs

Equilibrium 1876-8

age

Moses Gomberg Publications (1888-1942)

Mun

ich-

Hei

delb

erg

Two Gomberg papers from this period contained more graphs than all 4290 pages of Berichte in 1900.

Launching the American Chemical Century

CPh4(1896)

•CPh3(1900)

Second Thoughtson

Friedel-Crafts

This

What if CH3 gets stuck halfway?

Rearrangement in Friedel-Crafts Alkylation

AlCl3+ Cl +

Cl

AlCl3

+

Hgives

n-PrPh product

gives i-PrPh product

Cl

AlCl3

+

CH3

givesn-PrPh product

Hydride Shiftwhy not

Methide Shift?

Protonated Cyclopropane(stability between1° and 2° cations)

Still

H+

3/7 2/72/7Deno (1968)

D+

Which of these givesthe n-PrPh product in Friedel- Crafts?

CH3+

+

Nu

with one D

Where?

PROBLEM:How to decide?

givesn-PrPh product!

Still

Carbonyl Compounds

This

e.g. J&F Chapters 16-19(268 pages!)

Good News:Much of this is review!

Chapter 16: Aldehydes & Ketones

C=O Stable, but Reactive!

Average Bond Energies(kcal/mole)

C-C 83

C=C 146

C-O 86

C=O 176 (aldehyde) 179 (ketone)

“second bond” 63 9093 (more substituted sp2

C)

100 MHz 13CMR SpectrumProton Decoupled

(from Chem 220)

HCO

CHn

199.6 31.2

CHn

201.8 37.6 5.2

CH3

201.6 45.7 15.7 13.3

206.6 45.2 17.5 13.529.3

CH3

206.3 36.4 7.628.8

CHn

202.0

13.422.1

23.943.3

205.1 30.230.2

CO

CHnCHn

209.3 35.3 7.37.3 35.3

CH3

13CMR

Use table to identify this compound

0.93

1.351.592.429.76

400 MHz PMR Spectrum(from Chem 220, corrected with d from SDBS#10637)

20 HzJ =

1.8 Hz

J =6.8 Hz

same compound

Carbonyl Reactivity

O

Nu

1) Nucleophilic Addition (Bürgi-Dunitz Angle)

Chapter 16

O

Nu

O

Nu

(C=C prefers “Electrophilic” addition)

2) Nucleophilic Substitution of “Acid Derivatives”

(A/D, like Aromatic)

Chapter 18

LL

*H+

especially interesting for alcohol synthesis

Nu = “R-” (e.g. CH3Li)

Nu = “H-” (e.g. LiAlH4)

Secs. 16.13,16.16

HO Hydrate (gem-diol) RO Hemiacetal ( Acetal)RNH Carbinolamine ( Imine)HOSO2 Bisulfite addition product NC Cyanohydrin etc. Secs. 16.6-16.11

Carbonyl Reactivity

O

H

AB

3) Electrophilic Addition

n (acid catalysis)

OH

(Easier than for C=C)

OH

4) Allylic Rearrangement Ketone to Enol

OH

AAO

H

AO

5) Substitution (Electrophilic)

-protonNu

Chapter 19

H

then Nucleophile

HO Nu

the enol is a carbon

nucleophile

Enols, Enolates and Enolization

O

>1013

OH

-3

OH

O O9 3

O

O

pKa

19

Kenol formation

510-9O

H

(ketone ~11 kcal below enol)

(ketone enolhelp from conjugation,

H-bonding)

(ketone >17 kcal above enol help from aromaticity)

pKa = 10

(anion only ~13 kcal/mol above phenol)

HB

(base catalysis) Chapter 19

pKa~11!

RCOOH Reactions (Chapter 17)

O HR C

O

Hsubstitutionat -C

substitutionat C

Rsubstitution

at O

R

addition A

Nu

Mechanism for Acid-Catalyzed Hydrolysis of Acetal

RO

ROCH2

+H

HOH

:

:

RO

ROCH2

+ H ROH

RO-CH2

+

HO

ROCH2+

H

First remove RO, and replace it by HO.

HO

ROCH2

Now remove second RO, then H (from HO) +H

:HO

ROCH2

+ H

RO=CH2

+

cation unusually stable;thus easily formed

ROH

H-O-CH2

+

O=CH2O=CH2

ROH

ROH

RO

RO

CH2 OH

H

:Overall Transformation:

H2O + Acetal Carbonyl + 2 ROHH+

(pp. 785-787)

(hemiacetal)

Good News: Much of this is review! e.g. secs. 16.9-16.10

Ph3CPropeller

Conformation

Ph Ph3CMirror

Conformation

Ph

Ph3C-PhTwo Mirrors

rotated a bit

End of Lecture 64April 1, 2011

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