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

Ph 3 C Propeller Conformation Ph H-H contact prevents bonding

Two Ph 3 C Propellers Hopelessly Remote

Steric hindrance in triphenylmethyl causes twists that reduce overlap with 2p C by 25% from diphenylmethyl. It also makes tetraphenylmethane very difficult to prepare, not to mention hexaphenylethane !

Friedel-Crafts or Ph 2 Mg Tetraphenylmethane (1897) “I have tried to solve this problem in a completely different way.” ? 8 g 110° Cu 0.3 g Solubility Analysis : C (93.75) H 6.36 (6.25) 100 mg for Mol. Wt. : 0.289° 306 (320 calc.) (by solvent b.p. elevation)

O 2 N- - 3 C - + EtOH O 2 N- - 3 C-H EtO - Back in Ann Arbor (1898-9) Confirmed Mol. Wt ° 318 (320 calc.) “Unlike the trinitrotriphenylmethane… it does not dissolve in sodium ethylate, nor does it give any coloration…” How about O 2 N- - 3 C-C - -NO 2 3 ? Prepared O 2 N- - 4 C (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 Na 2 O 2 (S N 2). Prepared hydrocarbon in CO 2 atmosphere using special apparatus with ground glass joints. Ph 3 C - Cl Zn Ph 3 C - CPh 3 Ph 3 CO - OCPh 3 O2O2 ( C+H = 93.82% )

Free Radical! (1900) Highly “Unsaturated” (O 2, Cl 2, Br 2, even I 2 !) Launched an American Century of Chemistry

October, 1900

Paris (Wurtz - Médicine) Cornell MIT Paris ( Friedel - Mines) President AlCl 3 /C-Electrophile: The Friedel-Crafts Reaction 1877 Gibbs Equilibrium

age Moses Gomberg Publications ( ) Munich-Heidelberg Two Gomberg papers from this period contained more graphs than all 4290 pages of Berichte in Launching the American Chemical Century CPh 4 (1896) CPh 3 (1900)

Second Thoughts on Friedel-Crafts This

What if CH 3 gets stuck halfway? Rearrangement in Friedel-Crafts Alkylation AlCl 3 + Cl + AlCl 3 + H gives n-PrPh product gives i-PrPh product Cl AlCl 3 + CH 3 gives n-PrPh product Hydride Shift why not Methide Shift? Protonated Cyclopropane (stability between 1° and 2° cations) Still H + 3/72/7 Deno (1968) D+D+ Which of these gives the n-PrPh product in Friedel- Crafts? CH Nu with one D Where? PROBLEM: How to decide? gives n-PrPh product! Still

Carbonyl Compounds This e.g. J&F Chapters (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” (more substituted sp 2 C )

100 MHz 13 CMR Spectrum Proton Decoupled (from Chem 220) HC O CH n CH n CH CH CH n C O CH n CH 3 13 CMR Use table to identify this compound

MHz PMR Spectrum (from Chem 220, corrected with  from SDBS#10637)SDBS# Hz J = 1.8 Hz J = 6.8 Hz same compound

Carbonyl Reactivity O Nu 1)Nucleophilic Addition (Bürgi-Dunitz Angle) Chapter 16 O Nu O (C=C prefers “Electrophilic” addition) 2) Nucleophilic Substitution of “Acid Derivatives” (A/D, like Aromatic) Chapter 18 L L ** H+H+ especially interesting for alcohol synthesis Nu = “R - ” (e.g. CH 3 Li) Nu = “H - ” (e.g. LiAlH 4 ) Secs ,16.16

HO Hydrate (gem-diol) RO Hemiacetal (  Acetal) RNH Carbinolamine (  Imine) HOSO 2 Bisulfite addition product NC Cyanohydrin etc. Secs Carbonyl Reactivity O H A B 3) Electrophilic Addition n (acid catalysis) O H (Easier than for C=C) O H 4) Allylic Rearrangement Ketone to Enol O H A A O H A O 5)  Substitution (Electrophilic)  -proton Nu Chapter 19 H then Nucleophile HO Nu the enol is a carbon nucleophile

Enols, Enolates and Enolization O >10 13 OH -3 O H O O 93 O O pK a 19 K enol formation 5  O H (ketone ~11 kcal below enol) (ketone  enol help from conjugation, H-bonding) (ketone >17 kcal above enol help from aromaticity) pK a = 10 (anion only ~13 kcal/mol above phenol) H B (base catalysis) Chapter 19 pK a ~11!

RCOOH Reactions (Chapter 17) O H R C O H substitution at  -C substitution at C R substitution at O R addition A Nu

Mechanism for Acid-Catalyzed Hydrolysis of Acetal RO CH 2 + H HOH : : RO CH 2 + HROH RO-CH 2 + HO RO CH 2 + H First remove RO, and replace it by HO. HO RO CH 2 Now remove second RO, then H (from HO) + H : HO RO CH 2 + H RO=CH 2 + cation unusually stable; thus easily formed ROH H-O-CH 2 + O=CH 2 ROH RO CH 2 O H H  : Overall Transformation: H 2 O + Acetal Carbonyl + 2 ROH H+H+ (pp ) (hemiacetal) Good News: Much of this is review! e.g. secs

Ph 3 C Propeller Conformation Ph Ph 3 C Mirror Conformation Ph Ph 3 C - Ph Two Mirrors rotated a bit

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