1. Alkenes C n H 2n “unsaturated” hydrocarbons C 2 H 4 ethylene Functional group = carbon-carbon double bond sp 2 hybridization => flat, 120 o bond angles σ bond & π bond => H 2 C=CH 2 No rotation about double bond!

2. C 3 H 6 propylene CH 3 CH=CH 2 C 4 H 8 butylenesCH 3 CH 2 CH=CH 2 α-butylene 1-butene CH 3 CH 3 CH=CHCH 3 CH 3 C=CH 2 β-butylene isobutylene 2-butene2-methylpropene

3. there are two 2-butenes: cis-2-butene trans-2-butene “geometric isomers” (diastereomers)

4. C=C are called “vinyl” carbons If either vinyl carbon is bonded to two equivalent groups, then no geometric isomerism exists. CH 3 CH=CHCH 3 CH 3 CH 2 CH=CH 2 yesno CH 3 (CH 3 ) 2 C=CHCH 3 CH 3 CH=CCH 2 CH 3 no yes

5. Confusion about the use of cis- and trans-. According to IUPAC rules it refers to the parent chain. “cis-” ????????

6. E/Z system is now recommended by IUPAC for the designation of geometric isomerism. 1.Use the sequence rules to assign the higher priority * to the two groups attached to each vinyl carbon. 2. * * * * (Z)- “zusammen” (E)- “entgegen” together opposite

7. * ** * (Z)- (E)-

8. Nomenclature, alkenes: 1.Parent chain = longest continuous carbon chain that contains the C=C. alkane => change –ane to –ene prefix a locant for the carbon-carbon double bond using the principle of lower number. 2.Etc. 3.If a geometric isomer, use E/Z (or cis/trans) to indicate which isomer it is.

9. * ** * (Z)-3-methyl-2-pentene ( 3-methyl-cis-2-pentene ) (E)-1-bromo-1-chloropropene

10. CH 3 CH 3 CH 2 CHCH 2 CH 3 \ / C = C3-ethyl-5-methyl-3-heptene / \ CH 3 CH 2 H (not a geometric isomer)

11. -ol takes precedence over –ene CH 2 =CHCH 2 -OH2-propen-1-ol CH 3 CHCH=CH 2 3-buten-2-ol OH

12. Physical properties: non-polar or weakly polar no hydrogen bonding relatively low mp/bp ~ alkanes water insoluble Importance: common group in biological molecules starting material for synthesis of many plastics

13. Syntheses, alkenes: 1.dehydrohalogenation of alkyl halides 2. dehydration of alcohols 3.dehalogenation of vicinal dihalide 4. (later)

14. 3.dehalogenation of vicinal dihalides | | | | — C — C — + Zn  — C = C — + ZnX 2 | | X X eg. CH 3 CH 2 CHCH 2 + Zn  CH 3 CH 2 CH=CH 2 + ZnBr 2 Br Br Not generally useful as vicinal dihalides are usually made from alkenes. May be used to “protect” a carbon-carbon double bond.

15. 1.dehydrohalogenation of alkyl halides | | | | — C — C — + KOH(alc.)  — C = C — + KX + H 2 O | | H X a)RX: 3 o > 2 o > 1 o b)no rearragement c)may yield mixtures  d)Saytzeff orientation e)element effect f)isotope effect g)rate = k [RX] [KOH] h)Mechanism = E2

16. rate = k [RX] [KOH] => both RX & KOH in RDS R-I > R-Br > R-Cl “element effect” => C—X broken in RDS R-H > R-D “isotope effect” => C—H broken in RDS  Concerted reaction: both the C—X and C—H bonds are broken in the rate determining step.

17. Mechanism = elimination, bimolecular E2 One step! “Concerted” reaction.

18. CH 3 CHCH 3 + KOH(alc)  CH 3 CH=CH 2 Br isopropyl bromidepropylene CH 3 CH 2 CH 2 CH 2 -Br + KOH(alc)  CH 3 CH 2 CH=CH 2 n-butyl bromide 1-butene CH 3 CH 2 CHCH 3 + KOH(alc)  CH 3 CH 2 CH=CH 2 Br 1-butene 19% sec-butyl bromide + CH 3 CH=CHCH 3 2-butene 81%

19. Problem 8.6. What akyl halide (if any) would yield each of the following pure alkenes upon dehydrohalogenation by strong base? CH 3 CH 3 isobutylene  KOH(alc) + CH 3 CCH 3 or CH 3 CHCH 2 -X X 1-pentene  KOH(alc) + CH 3 CH 2 CH 2 CH 2 CH 2 -X note: CH 3 CH 2 CH 2 CHCH 3 would yield a mixture!  X 2-pentene  KOH(alc) + CH 3 CH 2 CHCH 2 CH 3 X 2-methyl-2-butene  KOH(alc) + NONE!

21. Saytzeff orientation: Ease of formation of alkenes: R 2 C=CR 2 > R 2 C=CHR > R 2 C=CH 2, RCH=CHR > RCH=CH 2 > CH 2 =CH 2 Stability of alkenes: R 2 C=CR 2 > R 2 C=CHR > R 2 C=CH 2, RCH=CHR > RCH=CH 2 > CH 2 =CH 2 CH 3 CH 2 CHCH 3 + KOH(alc)  CH 3 CH 2 CH=CH 2 RCH=CH 2 Br 1-butene 19% sec-butyl bromide+ CH 3 CH=CHCH 3 RCH=CHR 2-butene 81%

22. KOH (alc) CH 3 CH 2 CH 2 CHBrCH 3  CH 3 CH 2 CH=CHCH 3 + CH 3 CH 2 CH 2 CH=CH 2 71% 29% CH 3 CH 3 CH 3 CH 3 CH 2 CCH 3 + KOH(alc)  CH 3 CH=CCH 3 + CH 3 CH 2 C=CH 2 Br 71% 29% CH 3 CH 3 CH 3 CH 3 CHCHCH 3 + KOH(alc)  CH 2 =CHCHCH 3 + CH 3 CH=CCH 3 Br major product

23. Order of reactivity in E2: 3 o > 2 o > 1 o CH 3 CH 2 -X  CH 2 =CH 2 3 adj. H’s CH 3 CHCH 3  CH 3 CH=CH 2 6 adj. H’s & more stable Xalkene CH 3 CH 3 CH 3 CCH 3  CH=CCH 3 9 adj. H’s & most stable Xalkene

24. Elimination unimolecular E1

25. Elimination, unimolecular E1 a) RX: 3 o > 2 o > 1 o b) rearragement possible  c) may yield mixtures  d) Saytzeff orientation e) element effect f) no isotope effect g) rate = k [RW]

26. E1 : Rate = k [RW] => only RW involved in RDS R-I > R-Br > R-Cl “element effect” => C—X is broken in RDS R-H  R-D no “isotope effect” => C—H is not broken in the RDS

27. Elimination, unimolecular E1 a)RX: 3 o > 2 o > 1 o carbocation b)rearragement possible “ c) may yield mixtures d) Saytzeff orientation e) element effectC—W broken in RDS f) no isotope effect C—H not broken in RDS g) rate = k [RW] only R-W in RDS

28. alkyl halide + base  substitution or elimination?

29. R-X + base  ???????? 1) If strong, conc. base: CH 3 > 1 o => S N 2  R-Z 3 o > 2 o => E2  alkene(s) 2)If weak, dilute base: 3 o > 2 o > 1 o => S N 1 and E1  R-Z + alkene(s)  3)If KOH(alc.) 3 o > 2 o > 1 o => E2  alkene(s)

30. S N 2 CH 3 CH 2 CH 2 -Br + NaOCH 3  CH 3 CH 2 CH 2 -O-CH 3 1 o CH 3 E2CH 3 CH 3 CCH 3 + NaOCH 3  CH 3 C=CH 2 + HOCH 3 Br 3 o E2 CH 3 CH 2 CH 2 -Br + KOH(alc)  CH 3 CH=CH 2

31. CH 3 CH 3 CH 3 CHCHCH 3 + dilute OH -  CH 3 CCH 2 CH 3 S N 1 Br OH  CH 3 + CH 3 C=CHCH 2 E1 CH 3 CH 3 CHCHCH 3 CH 3  + CH 2 =CCH 2 CH 3 E1  [1,2-H]  CH 3 CH 3 CCH 2 CH 3 

32. 2.dehydration of alcohols: | | | | — C — C — acid, heat  — C = C — + H 2 O | | H OH a)ROH: 3 o > 2 o > 1 o b)acid is a catalyst c)rearrangements are possible  d)mixtures are possible  e)Saytzeff f)mechanism is E1 note: reaction #3 for alcohols!

34. CH 3 CH 2 -OH + 95% H 2 SO 4, 170 o C  CH 2 =CH 2 CH 3 CH 3 CH 3 CCH 3 + 20% H 2 SO 4, 85-90 o C  CH 3 C=CH 2 OH CH 3 CH 2 CHCH 3 + 60% H 2 SO 4, 100 o C  CH 3 CH=CHCH 3 OH + CH 3 CH 2 CH=CH 2 CH 3 CH 2 CH 2 CH 2 -OH + H +, 140 o C  CH 3 CH 2 CH=CH 2 rearrangement!  + CH 3 CH=CHCH 3

35. Synthesis of 1-butene from 1-butanol: CH 3 CH 2 CH 2 CH 2 -OH + HBr  CH 3 CH 2 CH 2 CH 2 -Br S N 2 E2  KOH(alc) CH 3 CH 2 CH=CH 2 only! To avoid the rearrangement in the dehydration of the alcohol the alcohol is first converted into an alkyl halide.

36. Syntheses, alkenes: 1.dehydrohalogenation of alkyl halides E2 2. dehydration of alcohols E1 3.dehalogenation of vicinal dihalide 4. (later)

37. R-OH R-X Alkene vicinal dihalide H + Zn KOH (alc.)

38. Alkyl halides: nomenclature syntheses: 1. from alcohols a) HX b) PX 3 2. halogenation of certain alkanes 3. 4. 5. halide exchange for iodide reactions: 1. nucleophilic substitution 2. dehydrohalgenation 3. formation of Grignard reagent 4. reduction

39. Alcohols: nomenclature syntheses later reactions 1. HX 2. PX 3 3. dehydration 4. as acids 5. ester formation 6. oxidation