H H C C. Alkenes C n H 2n unsaturated hydrocarbons. C 2 H 4 ethylene. Functional group = carbon-carbon double bond

Transcription

1 Alkenes C n H 2n unsaturated hydrocarbons C 2 H 4 ethylene H H C C H H 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 butylenes CH 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-butene 2-methylpropene

3 there are two 2-butenes: H H C C H 3 C CH 3 H CH 3 C C H 3 C H 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 yes no 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. H 3 C CH 2 CH 3 C C H CH 3 cis- H 3 C Cl C C H Br????????

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 together (E)- entgegen opposite *

7 * * H 3 C CH 2 CH 3 C C H CH 3 (Z)- * H 3 C Cl C C H Br * (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 2. Etc. the principle of lower number. 3. If a geometric isomer, use E/Z (or cis/trans) to indicate which isomer it is.

9 * * H 3 C CH 2 CH 3 C C H CH 3 * H 3 C Cl C C H Br * (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 = C 3-ethyl-5-methyl-3-heptene / \ CH 3 CH 2 H (not a geometric isomer)

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

12 Physical properties: non-polar or weakly polar no hydrogen bonding relatively low mp/bp water insoluble ~ alkanes 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 eg. C C + Zn C = C + ZnX 2 X X 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 W C C RDS C C + H:base + :W base: H One step! Concerted reaction.

18 CH 3 CHCH 3 + KOH(alc) CH 3 CH=CH 2 Br isopropyl bromide propylene 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!

20 KOH(alc) CH 3????????? CH 3 CH CCH 3 PURE! 2-methyl-2-butene CH 3 CH 3 CHCHCH 3 X KOH(alc) CH 3 CH 3 H 2 C CHCHCH 3 + CH 3 CH CCH 3 CH 3 CH 3 CH 2 CCH 3 X KOH(alc) CH 3 CH 2 C CH 2 CH 3 CH 3 + CH 3 CH CCH 3 No alkyl halide will yield 2-methyl-2-butene as the only product of dehydrohalogenation

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 X alkene CH 3 CH 3 CH 3 CCH 3 CH=CCH 3 9 adj. H s & most stable X alkene

24 Elimination unimolecular E1 1) C C RDS C C + :W H W H 2) C C H - H C C

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 effect C 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? C C H X S N 2 E2 :Z

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 E2 CH 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!

33 Mechanism for dehydration of an alcohol = E1 1) C C H OH + H C C H OH 2 2) C C H OH 2 RDS C C H + H 2 O 3) C C C C + H H

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 % H 2 SO 4, o C CH 3 C=CH 2 OH CH 3 CH 2 CHCH % 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 H + R-X KOH (alc.) Alkene Zn vicinal dihalide