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Fischer-Rosanoff Convention Before 1951, only relative configurations could be known. Sugars and amino acids with same relative configuration as (+)-glyceraldehyde.

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Presentation on theme: "Fischer-Rosanoff Convention Before 1951, only relative configurations could be known. Sugars and amino acids with same relative configuration as (+)-glyceraldehyde."— Presentation transcript:

1 Fischer-Rosanoff Convention Before 1951, only relative configurations could be known. Sugars and amino acids with same relative configuration as (+)-glyceraldehyde were assigned D and same as (-)- glyceraldehyde were assigned L. With X-ray crystallography, now know absolute configurations: D is (R) and L is (S). No relationship to dextro- or levorotatory. =>

2 D and L Assignments * * * =>

3 Properties of Diastereomers Diastereomers have different physical properties: m.p., b.p. They can be separated easily. Enantiomers differ only in reaction with other chiral molecules and the direction in which polarized light is rotated. Enantiomers are difficult to separate. =>

4 Resolution of Enantiomers React a racemic mixture with a chiral compound to form diastereomers, which can be separated. =>

5 Chromatographic Resolution of Enantiomers =>

6

7 Chapter 6 Alkyl Halides: Nucleophilic Substitution and Elimination Organic Chemistry, 5 th Edition L. G. Wade, Jr.

8 Classes of Halides Alkyl: Halogen, X, is directly bonded to sp 3 carbon. => Vinyl: X is bonded to sp 2 carbon of alkene. Aryl: X is bonded to sp 2 carbon on benzene ring. Examples:

9 Polarity and Reactivity Halogens are more electronegative than C. Carbon-halogen bond is polar, so carbon has partial positive charge. Carbon can be attacked by a nucleophile. Halogen can leave with the electron pair. =>

10 Classes of Alkyl Halides Methyl halides: only one C, CH 3 X Primary: C to which X is bonded has only one C-C bond. Secondary: C to which X is bonded has two C-C bonds. Tertiary: C to which X is bonded has three C-C bonds. =>

11 Classify These: =>

12 Dihalides Geminal dihalide: two halogen atoms are bonded to the same carbon => Vicinal dihalide: two halogen atoms are bonded to adjacent carbons.

13 IUPAC Nomenclature Name as haloalkane. Choose the longest carbon chain, even if the halogen is not bonded to any of those C’s. Use lowest possible numbers for position. =>

14 Systematic Common Names Name as alkyl halide. Useful only for small alkyl groups. Name these: =>

15 “Trivial” Names CH 2 X 2 called methylene halide.. CHX 3 is a haloform. CX 4 is carbon tetrahalide. Examples: –CH 2 Cl 2 is methylene chloride –CHCl 3 is chloroform -CHI 3 is iodoform –CCl 4 is carbon tetrachloride

16 Preparation of RX Free radical halogenation (Chapter 4) - REVIEW Free radical allylic halogenation –produces alkyl halide with double bond on the neighboring carbon. LATER =>

17 Substitution Reactions The halogen atom on the alkyl halide is replaced with another group. Since the halogen is more electronegative than carbon, the C-X bond breaks heterolytically and X - leaves. The group replacing X - is a nucleophile. =>

18 Elimination Reactions The alkyl halide loses halogen as a halide ion, and also loses H + on the adjacent carbon to a base. The alkyl halide loses halogen as a halide ion, and also loses H + on the adjacent carbon to a base. A pi bond is formed. Product is alkene.  Also called dehydrohalogenation (-HX).

19 Ingold

20 Sir Christopher Father of Physical Organic Chemistry Coined such names and symbols as: SN1, SN2, E1, E2, nucleophile, electrophile resonance effect, inductive effect/ In print, he often attacked enemies vigorously and sometimes in vitrolic manner. Ingold

21 S N 2 Mechanism Rate is first order in each reactant Both reactants are involved in RDS Note: one-step reaction with no intermediate Bimolecular nuleophilic substitution. Concerted reaction: new bond forming and old bond breaking at same time INVERSION OF CONFIGURATION

22 S N 2 Energy Diagram One-step reaction. Transition state is highest in energy. =>

23 Uses for S N 2 Reactions Synthesis of other classes of compounds. Halogen exchange reaction. =>

24 S N 2: Nucleophilic Strength Stronger nucleophiles react faster. Strong bases are strong nucleophiles, but not all strong nucleophiles are basic. =>

25 Trends in Nuc. Strength Of a conjugate acid-base pair, the base is stronger: OH - > H 2 O, NH 2 - > NH 3 Decreases left to right on Periodic Table. More electronegative atoms less likely to form new bond: OH - > F -, NH 3 > H 2 O  Increases down Periodic Table, as size and polarizability increase: I - > Br - > Cl -

26 Polarizability Effect =>

27 Bulky Nucleophiles Sterically hindered for attack on carbon, so weaker nucleophiles. =>

28 Solvent Effects (1) Polar protic solvents (O-H or N-H) reduce the strength of the nucleophile. Hydrogen bonds must be broken before nucleophile can attack the carbon. =>

29 Solvent Effects (2) Polar aprotic solvents (no O-H or N-H) do not form hydrogen bonds with nucleophile Examples:

30 Crown Ethers Solvate the cation, so nucleophilic strength of the anion increases. Fluoride anion becomes a good nucleophile

31 S N 2: Reactivity of Substrate Carbon must be partially positive. => Must have a good leaving group Carbon must not be sterically hindered.

32 Leaving Group Ability Electron-withdrawing => Stable once it has left (not a strong base) Polarizable to stabilize the transition state.

33 Structure of Substrate Relative rates for S N 2: CH 3 X > 1° > 2° >> 3° Tertiary halides do not react via the S N 2 mechanism, due to steric hindrance. =>

34 Effect of Beta Branching Propyl Isobutyl bromide Neopentyl bromide NOTE: ALL ABOVE ARE PRIMARY

35 Miscellaneous Substrate All have sterically hindered backsides - No SN2 reactivity

36 Stereochemistry of S N 2 Walden inversion =>

37 EXAMPLES S R

38 EXAMPLES 2 MESO

39 EXAMPLE 3

40 EXAMPLE 4 How would you prepare the following from an alkyl halide? retrosynthetic analysis

41 S N 1 Reaction Unimolecular nucleophilic substitution. Two step reaction with carbocation intermediate. Rate is first order in the alkyl halide, zero order in the nucleophile. Racemization occurs.

42 S N 1 Mechanism (1) Formation of carbocation (slow) =>

43 S N 1 Mechanism (2) Nucleophilic attack => Loss of H + (if needed)

44 S N 1 Energy Diagram Forming the carbocation is endothermic Carbocation intermediate is in an energy well. =>

45 Rates of S N 1 Reactions 3° > 2° > 1° >> CH 3 X –Order follows stability of carbocations (opposite to S N 2) –More stable ion requires less energy to form Better leaving group, faster reaction (like S N 2) Polar protic solvent best: It solvates ions strongly with hydrogen bonding

46 Stereochemistry of S N 1 Racemization: inversion and retention =>

47 Rearrangements Carbocations can rearrange to form a more stable carbocation. Methyl shift: CH 3 - moves from adjacent carbon if no H’s are available. Hydride shift: H - on adjacent carbon bonds with C +.

48 Hydride Shift =>

49 Methyl Shift =>

50 Interesting Rearrangement = PUSH PULL push pull -AgBr

51 S N 2 or S N 1? Primary or methyl Tertiary Strong nucleophile Polar aprotic solvent Rate = k[halide][Nuc] Inversion at chiral carbon No rearrangements Weak nucleophile (may also be solvent ) Polar protic solvent, Ag+ Rate = k[halide ] Racemization of optically active compound  Rearranged products

52 E1 Reaction Unimolecular elimination Two groups lost (usually X - and H + ) Nucleophile acts as base)  Also have S N 1 products (mixture S N 1 and E1 have common first step.

53 E1 EXAMPLES +

54 E1 Mechanism Halide ion leaves, forming carbocation. Base removes H + from adjacent carbon. Pi bond forms.

55 A Closer Look =>

56 E1 Energy Diagram Note: first step is same as S N 1 =>

57 E2 Reaction Bimolecular elimination Requires a strong base Halide leaving and proton abstraction happens simultaneously - no intermediate

58 E2 Examples

59 E2 Mechanism =>

60 Saytzeff’s Rule If more than one elimination product is possible, the most- substituted alkene is the major product (most stable). R 2 C=CR 2 >R 2 C=CHR>RHC=CHR>H 2 C=CHR tetra > tri > di > mono => minor major

61 E2 Stereochemistry =>

62 E1 or E2? Tertiary > Secondary Weak base Good ionizing solvent Rate = k[halide] Saytzeff product No required geometry Rearranged products Tertiary > Secondary Strong base required Solvent polarity not important Rate = k[halide][base] Saytzeff product Coplanar leaving groups (usually anti) No rearrangements =>

63 End of Chapter 6


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