Chemistry 125: Lecture 64 April 1, 2011 Triphenylmethyl Carbonyl Compounds: Overview This For copyright notice see final page of this file
Chemistry 125: Lecture 64April 1, 2011
TriphenylmethylCarbonyl Compounds:
Overview This
For copyright notice see final page of this file
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
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)
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
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
End of Lecture 64April 1, 2011
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