1. Chemistry 125: Lecture 74 April 27, 2011 The Structure of Glucose and Synthesis of Two Un-Natural Products This For copyright notice see final page of this file

2. Couper 1858 CHOH HOH “I may be allowed to denote compounds [with] carbon plus hydrogen and oxygen in the same ratio that obtains in water…as carbohydrates” Schmidt (1844) Carbohydrate (CHOH) n γλε ῦ κος gleukos “sweetness” Glucose Dumas (1838) “ose” soonbecame the generic suffix for sugars Grape Sugar (1792)

3. Carbohydrate (CHOH) n CHOH C=O CHOH H H aldose (e.g. glucose) CHOH C=O CHOH H H C=O CHOH H H ketose (e.g. fructose)

4. no aldehyde or ketone! http://riodb01.ibase.aist.go.jp/sdbs/ We know the structure of glucose by spectroscopy

5. 13 C-NMR Crystalline Glucose upon dissolution in D 2 O 1 H-NMR no aldehyde or ketone no aldehyde Cf. J. E. Gurst, J. Chem. Ed. 1003 (1991) H H J 7.9Hz J 3.7Hz gauche anti After 1 day in D 2 O [  ] D = 112° “dextrose” [  ] D = 53° [  ] D = 19° hemiketal with H + or OH -

6. We know the structure of glucose by spectroscopy and X-ray Diffraction. But how could they know?

7. Fructose Sucrose Arabinose reduced oxidized Gum Arabic Mannitol Manna Chirality - van’t Hoff (1877) Galactose H + (1876) (1873) Mannose (1888) HNO 3

8. Heinrich Kiliani 1855-1945

9. Berlin 1885 While a large number of compounds are very easily formed upon oxidation of dextrose by dilute nitric acid or by halogens, these molecules retain 6 carbon bound together in a chain. Under the same conditions laevulose yields substances containing chains with a smaller number of carbons (glycolic acid and inactive tartaric acid). Here oxidation causes immediate splitting of the molecule, a fact which means that laevulose is a ketone. 1 ) Bearing in mind further the fact that laevulose in transformed into mannitol by nascent hydrogen 2 ), one comes to the conclusion that laevulose must be adjudged to have one of the following two constitutional formulae:

10. One could hope to distinguish definitively between one and the other formula, by succeeding in adding hydrocyanic acid to the ketone radical of laevulose and transforming the cyanhydrin into the corresponding carboxylic acid. Since the carboxylic acid from compound I above must upon exhaustive reduction by concentrated hydriodic acid yield methylbutylacetic acid, and on the contrary the carboxylic acid from compound II under the same conditions leads to ethylpropylacetic acid. 1) HCN 2)HCl (fuming) 1) HCN 2)HCl (fuming) HI / P “which means that my heptanoic acid is identical with [unknown] ethylpropylacetic acid.” “does not agree with the description that Hecht gives of the Ca salt of methylbutyacetic acid,” ? ? KOH (STRONG) KOH (STRONG) NOT from Laevulose! Kiliani 1886 Ca, Ba, Sr, Pb salts all agree NOT ketone! RCO 2 - + CH 3 CO 2 -

11. Heinrich Kiliani: On the Composition and Constitution of Arabinosecarboxylic Acid and of Arabinose In a report dated 27 November 1886 I showed on one hand that arabaric acid formed by oxidation of arabinose has the formula C 5 H 10 O 6, but on the other hand described several derivatives of arabinose carboxylic acid with the formula C 7 H 14 O 8, since at that time I had no basis, in truth, to dispute the generally accepted formula for arabinose – C 6 H 12 O 6 – and my analytical results did not contradict either assumption. arabinose mannitol 1887 HCN arabinose carboxylic acid H+H2OH+H2O Na/Hg mixture

12. Na/Hg Emil Fischer 1852-1919 Kiliani-Fischer Synthesis elongates an aldose Na/Hg  K-F

13. Emil Fischer 1852-1919 Phenylhydrazine - Fischer’s First Chemical Love Osazone crystalline!

14. The D-Aldoses Hexoses Pentoses Tetroses D-Glyceraldehyde Glucose Mannose Galactose Gulose Arabinose Xylose Erythrose Kiliani HO H H H OH H HC=O HH Abbreviate as Must differ in configuration Which is which? How many are there?

15. Fischer’s Evidence 1a) Glucaric Diacid is chiral (both enantiomers known) 2a) Mannitol & Mannonic Acid are chiral 3a) Arabinose gives active Arabitol and Arabaric Diacid Xylose gives inactive Xylitol and Xylaric Diacid 1) Glucose Glucaric Diacid “Gulose” HNO 3 2) Na (Hg)  2) Glucose & Mannose give same Osazone; Arabinose Gluconic & Mannonic acids Fructose Glucitol & Mannitol Kiliani Na (Hg) 3) Arabinose Glucose (& Mannose) Xylose Gulose (& Idose?) K-F syn

16. The D-Aldoses Hexoses Pentoses Glucose Mannose Galactose Gulose Arabinose Xylose 1a) Glucaric Diacid is chiral (both enantiomers known) 1) Glucose Glucaric Diacid “Gulose” HNO 3 2) Na (Hg)  1)XX XX1a)

17. The D-Aldoses Hexoses Pentoses Glucose Mannose Galactose Gulose Arabinose Xylose 2) MMMM 1)XX XX1a) 2) Glucose & Mannose give same Osazone Arabinose Gluconic & Mannonic acids Fructose Glucitol & Mannitol Kiliani Na (Hg) 2a) Mannitol & Mannonic Acid are chiral 2a) XX

18. The D-Aldoses Hexoses Pentoses Glucose Mannose Galactose Gulose 2) MMMM 1)XX XX1a) 2a) XX 3) Arabinose Glucose (& Mannose) Xylose Gulose (& Idose?) K-F syn 3a) Arabinose gives active Arabitol and Arabaric Diacid Xylose gives inactive Xylitol and Xylaric Diacid 3) 3a) X Arabinose Xylose

19. The D-Aldoses Hexoses Pentoses Tetroses D-Glyceraldehyde Glucose Mannose Galactose Gulose Erythrose Arabinose XyloseLyxoseRibose Threose Glucose Mannose Galactose Gulose Allose Altrose IdoseTalose All Altruists Gladly Make Gum In Gallon Tanks Arabinose Xylose Erythrose

20. REVIEW 1: The Synthesis of Two Unnatural Products (in order to settle a question in the theory of organic chemistry)

21. Is cyclobutadiene antiaromatic (4n)? Diels- Alder O O O O + O=C=O h h (must be disrotatory) Make it and see. (2  +2  forbidden thermally) very strained (2  +2  forbidden thermally, but it happens anyway) Presumptive Evidence of its Existence. Spectroscopy?

22. Make one molecule per cage Making & Studying “antiaromatic” Cyclobutadiene (for solubility) O CH CH 2 Ph mouth Cram, Tanner, and Thomas (1991)

23. Preparing Dihydrocinnamaldehyde O CH CH 2 Ph O CH Ph O CH CH 3 O CH Ph

24. (as mixture with tetra- substituted and two disubstituted analogues) Start with Hemisphere O H + H + etc. The electrophilic aromatic substitution is reversible, and ultimately the desired “tetramer” stereoisomer precipitates from the equilibrating mixture in 69% yield based on hydrocinnamaldehyde. O-CH 2 -O bonds by S N 2 Ar-O - with CH 2 BrCl How to form the C 16 ring? 1) “Br + ” / -H + (3 moles) Br 2) - - ClCH 2 Br (from benzaldehyde see above) Resorcinol Lucky! H+H+ (OH is activating, o,p-directing) (by chromatography; 5% from tetramer) Hydrocinnam- aldehyde

25. Joining Hemispheres 1) Br + /-H + (3 moles) Br 3) BuLi 4) B(OR) 3 5) HOO - Li B(OR) 2 OH O-CH 2 -O bonds by S N 2 Ar-O - with CH 2 BrCl 2) - - ~40% (1% overall) HO O - O-B(OR) 2 HO - - O-O- 6) O-CH 2 -O bonds by S N 2 Ar-O - with CH 2 BrCl (add “Ar - ” to B ; lose RO - ) (insert O between C and B. Cf. hydroboration/oxidation; (halogen-metal exchange  more stable “Ar - anion”) Note: the purpose of 1,3,4,5 is to “hide” an OH group between the OH groups of resorcinol, and then reveal it. HO - lose most stable ArO - anion)

26. CHCl 3 CH3CNCH3CN CH3CNCH3CN CH3CNCH3CNCH3CNCH3CN HC-N(CH 3 ) 2 held within molecule. O Stereo Pair X-Ray View JACS, 113, 7717 (1991) (easier to see without a viewer if you make it small) CHCl 3 & CH 3 CN are held between adjacent molecules in crystal but lost with t 1/2 = 34 hrs at 140°C.

27. ......... as guestBenzene Antiaromatic upfield shift? Most shift comes from other rings, still ~1.5 ppm above benzene Normal Replace DMF by  -Pyranone above center of 8 benzene rings.... O O O O.... O O.... Proton NMR

28. End of Lecture 74 April 27, 20101 Copyright © J. M. McBride 2011. Some rights reserved. Except for cited third-party materials, and those used by visiting speakers, all content is licensed under a Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0).Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0) Use of this content constitutes your acceptance of the noted license and the terms and conditions of use. Materials from Wikimedia Commons are denoted by the symbol. Third party materials may be subject to additional intellectual property notices, information, or restrictions. 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

29. The Benzoin Condensation (prob. 19.90) Ph C OH C N CH 3 OH H C N nucleophile like C=O an  -activator leaving group H C Ph O also an  -activator (benzylic) Ph C O C N H H C Ph OH N C base - HCN CN “reverses the polarity” of O=C  + to C - (“umpolung”) need Ph-C O to attack O=CH-Ph H+H+ C N where we’re going: what we have: not basic enough to pull off H +.  pK a > 30