Organic Chemistry

1. Organic Chemistry Mario Lintz Mario.Lintz@ucdenver.edu 303-946-5838 1

2. Organic Chemistry I • Functional Groups • Molecular Structure • Hydrocarbons • Substitution and Elimination • Oxygen Containing Compounds • Amines 2

3. Functional GroupsList #1- Critical for the MCAT • Alkane • Alkene • Alkyne • Alcohol • Ether • Amine • Aldehyde • Ketone • Carboxylic Acid • Ester • Amide 3

4. Alkyl Halogen Gem-dihalide Vic dihalide Hydroxyl Alkoxy Hemiacetal Hemiaketal Mesyl group Tosyl group Carbonyl Acetal Acyl Anhydride Aryl Benzyl Hydrazine Hydrazone Vinyl Vinylic Allyl Nitrile Epoxide Enamine Imine Nitro Nitroso Functional GroupsList #2- Memorize as well 4

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6. Bonds • Types: • Ionic: complete transfer of electrons • Covalent: shared electrons • Coordinate covalent bonds- One atom provides both electrons in a shared pair. • Polar covalent: unequal sharing of electrons • Hydrogen Bonds: bonds between polar molecules containing H and O, N, or F 6

7. Bonds • In the pi bond of an alkene, the electron pair have: • 33% p character and are at a lower energy level than the electron pair in the o bond. • 33% p character and are at a higher energy level than the electron pair in the o bond. • 100% p character and are at a lower energy level than the electron pair in the o bond. • 100% p character and are at a higher energy level than the electron pair in the o bond. 7

8. Covalent Bonds • Sigma s • Pi P 8

9. Covalent Bonds • Sigma s • Between s orbitals • Small, strong, lots of rotation • Pi P 9

10. Covalent Bonds • Sigma s • Between s orbitals • Small, strong, lots of rotation • Pi P • Between p orbitals • Discreet structure, weaker than sigma, no rotation 10

11. Covalent Bonds • Sigma s • Between s orbitals • Small, strong, lots of rotation • Pi P • Between p orbitals • Discreet structure, weaker than sigma, no rotation • Always add to sigma bonds creating a stronger bond 11

12. When albuterol I dissolved in water, which of the following hydrogen-bonded structures does NOT contribute to its water solubility? 12

13. Dipole Moments(Solely responsible for Intermolecular Attractions) • Charge distribution of bond is unequal • Molecule with dipole moment = polar • Molecule without dipole moment = nonpolar • Possible to have nonpolar molecules with polar bonds • Induced Dipoles • Spontaneous formation of dipole moment in nonpolar molecule • Occurs via: polar molecule, ion, or electric field • Instantaneous Dipole • Due to random e- movement • Hydrogen Bonds • Strongest dipole-dipole interaction • Responsible for high BP of water • London Dispersion Forces • Between 2 instantaneous dipoles • Responsible for phase change of nonpolar molecules 13

14. Lewis Dot Structures • Rules for writing • Find total # valence e- • 1 e- pair = 1 bond • Arrange remaining e- to satisfy duet and octet rules • Exceptions • Atoms containing more than an octet must come from the 3rd period, (vacant d orbital required for hybridization) • Not very popular on the MCAT • Formal Charge • # valence e- (isolated atom) - # valence e- (lewis structure) • Sum of formal charge for each atom is the total charge on the molecule • (actual charge distribution depends on electronegativity) 14

15. Structural Formulas • Dash Formula • Condensed Formula • Bond-line Formula • Fischer projection • Newman projection • Dash-line-wedge • Ball and stick • All Images courtesy of Exam Krackers 15

16. Hybridization 16

17. Hybrid Bonds 17

18. Hybrid Bonds 18

19. Hybrid Bonding in Oxygen and Nitrogen • Nitrogen- • Lone pair occupies more space than N-H • Causes compression of the bond angle. Bond angles are 107.3 as opposed to 109.5 • Oxygen- • 2 sets of lone pair electrons • Causes greater compression than in Nitrogen. H2O bond angles are 104.5 vs 109.5. 19

20. For the molecule 1,4 pentadiene, what type of hybridization is present in carbons # 1 and # 3 respectively? • A) sp2, sp2 • B) sp2, sp3 • C) sp3, sp3 • D) sp3, sp2 20

21. VSEPR • valance shell electron pair repulsion • Prediction of shape • Minimize electron repulsion 1. Draw the Lewis dot structure for the molecule or ion 2. Place electron pairs as far apart as possible, then large atoms, then small atoms 3. Name the molecular structure based on the position of the atoms (ignore electron pairs) 21

22. VSEPR 1. Draw the Lewis dot structure for the molecule or ion 2. Place electron pairs as far apart as possible, then large atoms, then small atoms 3. Name the molecular structure based on the position of the atoms (ignore electron pairs) 22

23. Delocalized e- and Resonancepassage 25 • Resonance forms differ only in the placement of pi bond and nonbonding e- • Does not suggest that the bonds alternate between positions • Neither represent the actual molecule, rather the real e assignment is the intermediate of the resonant structures. The real structure is called a resonance hybrid (cannot be seen on paper) 23

24. Organic Acids and Bases • Organic Acids- Presence of positively charged H+ • Two kinds • present on a OH such as methyl alcohol • present on a C next to a C=O such as acetone • Organic Bases- Presence of lone pair e to bond to H • Nitrogen containing molecules are most common • Oxygen containing molecules act as bases in presence of strong acids 24

25. Stereochemistry • Isomers: same elements, same proportions. Different spatial arrangements => different properties. • Structural (constitutional): Different connectivity. • Isobutane vs n-butane • Both C4H10 Conformational (rotational): Different spatial arrangement of same molecule • Chair vs. boat • Gauche vs Eclispsed vs Antistaggered vs Fully Eclipsed 25

26. Stereochemistry-isomers • Stereoisomers: different 3D arrangement • Enantiomers: mirror images, non-superimposable. • Same physical properties (MP, BP, density, solubility, etc.) except rotation of light and reactions with other chiral compounds • May function differently; e.g. thalidomide, sugars, AA • Have chiral centers 26

27. Stereochemistry-isomers • Stereoisomers: different 3D arrangement • Diastereomers: not mirror images (cis/trans) • Different physical properties (usually), • Can be separated • Chiral diastereomers have opposite configurations at one or more chiral centers, but have the same configuration at others. 27

28. Stereochemistry-isomers • What kind of isomers are the two compounds below? • A. Configurational diastereomers • B. Enantiomers • C. Constitutional isomers • D. Cis -trans diastereomers 28

29. Stereochemistry-polarization of light • Excess of one enantiomer causes rotation of plane-polarized light. • Right, clockwise, dextrarotary (d), or + • Left, counterclockwise, levarotary (l), or – • Racemic: 50:50 mixture of 2 enantiomers, no net rotation of light • RELATIVE Configuration: configuration of one molecule relative to another. Two molecules have the same relative configuration about a carbon if they differ by only one substituent and the other substituents are oriented identically about the carbon. • Specific rotation [a]: normalization for path length (l) and sample density (d). ocm3/g [a] = a / (l*d) 29

30. Stereochemistry-Chiral moleculespassage 27 • Achiral=plane or center of symmetry • ABSOLUTE Configuration: physical orientation of atoms around a chiral center • R and S: 1. Assign priority, 1 highest, 4 lowest • H < C < O < F higher atomic #, higher priority • If attachments are the same, look at the b atoms (ethyl beats methyl) 2. Orient 4 away from the observer 3. Draw a circular arrow from 1 to 2 to 3 • R = clockwise • S = counterclockwise • This has nothing to do with the rotation of light! • E and Z: Different than cis and trans • Z= same side of high priority groups • E=opposite side of high priority groups 30

31. IUPAC Naming Conventions • IUPAC Rules for Alkane Nomenclature 1.   Find and name the longest continuous carbon chain.  2.   Identify and name groups attached to this chain. 3.   Number the chain consecutively, starting at the end nearest a substituent group. 4.   Designate the location of each substituent group by an appropriate number and name. 5.   Assemble the name, listing groups in alphabetical order. The prefixes di, tri, tetra etc., used to designate several groups of the same kind, are not considered when alphabetizing. 31

32. Hydrocarbons 32

33. Hydrocarbons • Saturated: CnH(2n+2)\ • Unsaturated: CnH[2(n-u+1)] ; u is the # of sites of unsaturation • Primary, secondary, tertiary, and quaternary carbons • Know and be able to recognize the following structures n-propyl Iso-propyl n-butylsec-butyl iso-butyltert-butyl 33

34. Alkanes • Physical Properties: • Straight chains: MP and BP increase with length (increased van Der Waals interactions) • C1-4: gas • C5-17: liquid • C18+: solid • Branched chains: • BP decreases (less surface area, fewer vDW) • When compared to the straight chain analog, the straight chain will have a higher MP than the branched molecule. BUT, amongst branched molecules, the greater the branching, the higher the MP. 34

35. Alkanes-Important ReactionsVery Unreactive • Combustion: • Alkane + Oxygen + High energy input (fire) • Products: H2O, CO2, Heat • Halogenation • Initiation with UV light • Homolytic cleavage of diatomic halogen • Yields a free radical • Propagation (chain reaction mechanisms) • Halogen radical removes H from alkyl • Yields an alkyl radical • Termination • Radical bonds to wall of container or another radical • Reactivity of halogens: F > Cl > Br >>> I • Selectivity of halogens (How selective is the halogen in choosing a position on an alkane): • I > Br > Cl > F • more electronegative (Cl) means less selective (Br) • Stability of free radicals: more highly substituted = more stable • aryl>>>alkene> 3o > 2o > 1o >methyl 35

36. Halogenation • In the halogenation of an alkane, which of the following halogens will give the greatest percent yield of a tertiary alkyl halide when reacted with 2-methylpentane in the presence of UV light. • F2 • Cl2 • Br2 • 2-methylpentane will not yield a tertiary product 36

37. Cycloalkanes • General formula: (CH2)n or CnH2n • As MW increases BP increases though MP fluctuates irregularly because different shapes of cycloalkanes effects the efficiency in which molecules pack together in crystals. • Ring strain in cyclic compounds: • Bicyclic Molecules: 37 http://www.chem.uh.edu/Courses/Thummel/Chem3331/Notes/Chap3/

38. CycloalkanesNaming • Find parent • Count C’s in ring vs longest chain. If # in ring is equal to or greater than chain, then name as a cycloalkane. • Number the substituents and write the name • Start at point of attachment and number so that subsequent substituents have the lowest # assignment • If two or more different alkyl groups are present, number them by alphabetic priority • If halogens are present, treat them like alkyl groups • Cis vs Trans • Think of a ring as having a top and bottom • If two substituents both on top: cis • It two substituents and 1 top, 1 bottom: trans 38

39. Cycloalkanes • Ring Strain • Zero for cyclohexane (All C-C-C bond angles: 111.5°) • Increases as rings become smaller or larger (up to cyclononane) • Cyclohexane • Exist as chair and boat conformations • Chair conformation preferred because it is at the lowest energy. • Hydrogens occupy axial and equatorial positions. • Axia (6)l- perpendicular to the ring • Equatorial (6)- roughly in the plane of the ring • Neither energetically favored • When the ring reverses its conformation, substituents reverse their conformation • Substituents favor equatorial positions because crowding occurs most often in the axial position. 39

40. Cyclohexanes • In a sample of cis-1,2-dimethylcyclohexane at room temperature, the methyl groups will: • Both be equatorial whenever the molecule is in the chair conformation. • Both be axial whenever the molecule is in the chair conformation. • Alternate between both equatorial and both axial whenever the molecule is in the chair conformation • Both alternate between equatorial and axial but will never exist both axial or both equatorial at the same time 40

41. Substitutions • Substitution: one functional group replaces another • Electrophile: wants electrons, has partial + charge • Nucleophile: donates electrons, has partial – charge 41

42. Substitution • SN1: substitution, nucleophilic, unimolecular • Rate depends only on the substrate (i.e. leaving group) • R=k[reactant] • Occurs when Nu has bulky side groups, stable carbocation (3o), weak Nu (good leaving group) • Carbocation rearrangement • Two step reaction • 1.spontaneous formation of carbocation (SLOW) • 2. Nucleophile attacks carbocation (chiral reactants yield racemic product mixtures) 42

43. Substitution • SN2: substitution, nucleophilic, bimolecular • Rate depends on the substrate and the nucleophile • R=k[Nu][E] • Inversion of configuration • Occurs with poor leaving groups (1o or 2o) • One step reaction • 1. Nu attacks the C with a partial + charge 43 http://www.mhhe.com/physsci/chemistry/carey5e/Ch08/ch8-4.html

44. Which of the following carbocations is the most stable? 44

45. Benzene • Undergoes substitution not addition • Flat molecule • Stabilized by resonance • Electron donating groups activate the ring and are ortho-para directors • Electron withdrawing groups deactivate the ring and are meta directors • Halogens are electron withdrawing, however, are ortho-para directors 45

46. BenzeneSubstituent Effects 46

47. Oxygen Containing Compounds • Alcohols • Aldehydes and Ketones • Carboxylic Acids • Acid Derivatives • Acid Chlorides • Anhydrides • Amides • Keto Acids and Esters 47

48. Alcohols One of the most common reactions of alcohols is nucleophilic substitution. Which of the following are TRUE in regards to SN2 reactions: • Inversion of configuration occurs • Racemic mixture of products results • Reaction rate = k [S][nucleophile] • I only • II only • I and III only • I, II, and III 48

49. Alcohols • Physical Properties: • Polar • High MP and BP (H bonding) • More substituted = more basic • (CH3)3COH: pKa = 18.00 • CH3CH2OH: pKa = 16.00 • CH3OH: pKa = 15.54 • Electron withdrawing substituents stabilize alkoxide ion and lower pKa. • Tert-butyl alcohol: pKa = 18.00 • Nonafluoro-tert-butyl alcohol: pKa = 5.4 • IR absorption of OH at ~3400 cm- • General principles • H bonding • Acidity: weak relative to other O containing compounds (CH groups are e- donating = destabilize deprotonated species) • Branching: lowers BP and MP 49

50. AlcoholsNaming • Select longest C chain containing the hydroxyl group and derive the parent name by replacing –e ending of the corresponding alkane with –ol. • Number the chain beginning at the end nearest the –OH group. • Number the substituents according to their position on the chain, and write the name listing the substituents in alphabetical order. 50