CHAPTER 5 STRUCTURE AND PREPARATION OF ALKENES ELM|NAT| ON REACT○NS SOLUTIONS TO TEXT PROBLEMS 5.1 (b) Writing the structure in more detail, we see that the longest continuous chain contains four carbon atoms H3 The double bond is located at the end of the chain and so the alkene is named as a derivative f 1-butene. Two methyl groups are substituents at C-3. The correct IUPAC name is 3, 3- (c) Expanding the structural formula reveals the molecule to be a methyl-substituted derivative of CH3-C=CHCH,CH,CH 2-Methyl-2-hexene numbering the longest carbon chain. CH,=CHCH, CHCH Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
CHAPTER 5 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS SOLUTIONS TO TEXT PROBLEMS 5.1 (b) Writing the structure in more detail, we see that the longest continuous chain contains four carbon atoms. The double bond is located at the end of the chain, and so the alkene is named as a derivative of 1-butene. Two methyl groups are substituents at C-3. The correct IUPAC name is 3,3- dimethyl-1-butene. (c) Expanding the structural formula reveals the molecule to be a methyl-substituted derivative of hexene. (d) In compounds containing a double bond and a halogen, the double bond takes precedence in numbering the longest carbon chain. Cl CH2 CHCH2CHCH3 1 2 3 4 5 4-Chloro-1-pentene CH3 CH3 C CHCH2CH2CH3 1 2 3 4 5 6 2-Methyl-2-hexene CH3 CH CH3 CH3 C CH2 4 32 1 90 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS (e) When a hydroxyl group is present in a compound containing a double bond, the hydroxyl takes precedence over the double bond in numbering the longest carbon chain H2=CHCH, CHCH3 4-Penten-2-ol 5.2 There are three sets of nonequivalent positions on a cyclopentene ring, identified as a, b, and c on the cyclopentene structure shown Thus, there are three different monochloro-substituted derivatives of cyclopentene. The carbons that bear the double bond are numbered C-I and C-2 in each isomer, and the other positions are num bered in sequence in the direction that gives the chlorine-bearing carbon its lower locant. I-Chlorocyclopentene 3-Chlorocyclopentene 4-Chlorocyclopentene 5.3(b) The alkene is a derivative of 3-hexene regardless of whether the chain is numbered from left to right or from right to left. Number it in the direction that gives the lower number to the substituent 3-Ethyl-3-hexene (c) There are only two sp-hybridized carbons, the two connected by the double bond. All other carbons(six) are sp-hybridized (d) There are three sp - o bonds and three sp-sp o bonds. 5.4 Consider first the c he alkenes that have an unbranched carbon chain I-Pent 2-Pentene trans-2-Pentene There are three additional isomers. These have a four-carbon chain with a methyl substituent 2-Methyl-1-butene 5.5 First, identify the constitution of 9-tricosene. Referring back to Table 2. 4 in Section 2.8 of the text, we see that tricosane is the unbranched alkane containing 23 carbon atoms. 9-Tricosene therefore contains an unbranched chain of 23 carbons with a double bond between C-9 and c-10. Since the Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(e) When a hydroxyl group is present in a compound containing a double bond, the hydroxyl takes precedence over the double bond in numbering the longest carbon chain. 5.2 There are three sets of nonequivalent positions on a cyclopentene ring, identified as a, b, and c on the cyclopentene structure shown: Thus, there are three different monochloro-substituted derivatives of cyclopentene. The carbons that bear the double bond are numbered C-1 and C-2 in each isomer, and the other positions are numbered in sequence in the direction that gives the chlorine-bearing carbon its lower locant. 5.3 (b) The alkene is a derivative of 3-hexene regardless of whether the chain is numbered from left to right or from right to left. Number it in the direction that gives the lower number to the substituent. (c) There are only two sp2 -hybridized carbons, the two connected by the double bond. All other carbons (six) are sp3 -hybridized. (d) There are three sp2 –sp3 bonds and three sp3 –sp3 bonds. 5.4 Consider first the C5H10 alkenes that have an unbranched carbon chain: There are three additional isomers. These have a four-carbon chain with a methyl substituent. 5.5 First, identify the constitution of 9-tricosene. Referring back to Table 2.4 in Section 2.8 of the text, we see that tricosane is the unbranched alkane containing 23 carbon atoms. 9-Tricosene, therefore, contains an unbranched chain of 23 carbons with a double bond between C-9 and C-10. Since the 2-Methyl-1-butene 2-Methyl-2-butene 3-Methyl-1-butene 1-Pentene cis-2-Pentene trans-2-Pentene 5 4 3 2 1 6 3-Ethyl-3-hexene Cl 1 2 3 4 5 Cl 2 1 5 4 3 Cl 1 2 3 4 5 1-Chlorocyclopentene 3-Chlorocyclopentene 4-Chlorocyclopentene a a b c b OH CH2 CHCH2CHCH3 5 4 3 2 1 4-Penten-2-ol STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 91 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
92 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS problem specifies that the pheromone has the cis configuration, the first 8 carbons and the last 13 must be on the same side of the c-9-C-10 double bond (CH,)12CH H H cis-9.Tricosene 5.6 (b) One of the carbons of the double bond bears a methyl group and a hydrogen; methyl is of higher rank than hydrogen. The other doubly bonded carbon bears the groups -CH,CH F and-CH,CH,CH CH3. At the first point of difference between these two, fluorine is of igher atomic number than carbon, and so-CH, CH, F is of higher precedence Higher CH CH,CH,F Higher CH,CH,CH,CH Higher ranked substituents are on the same side of the double bond the alkene has the Z con- guration (c) One of the carbons of the double bond bears a methyl group and a hydrogen; as we have seen methyl is of higher rank. The other doubly bonded carbon bears-CH,CH,OH and-C(CH Lets analyze these two groups to determine their order of precedence C(C, H, H C(CCC Lower priority Higher priority We examine the atoms one by one at the point of attachment before proceeding down the chain. Therefore, -C(CH;) outranks-CH,CH,OH Higher CH; CHCHOH Lower C(CH3)3 Higher Higher ranked groups are on opposite sides; the configuration of the alkene is E. (d) The cyclopropyl ring is attached to the double bond by a carbon that bears the atoms( C, C, h) and is therefore of higher precedence than an ethyl group-C(C, H, H) Lower Higher ranked groups are on opposite sides; the configuration of the alkene is E 5.7 A trisubstituted alkene has three carbons directly attached to the doubly bonded carbons. There are three trisubstituted CHi isomers, two of which are stereoisomers H3 CH,CH CH2CH3 2-Methyl-2-penter (E)3-Methyl-2-pentene (Z)-3-Methyl-2-pentene Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
92 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS problem specifies that the pheromone has the cis configuration, the first 8 carbons and the last 13 must be on the same side of the C-9–C-10 double bond. 5.6 (b) One of the carbons of the double bond bears a methyl group and a hydrogen; methyl is of higher rank than hydrogen. The other doubly bonded carbon bears the groups @CH2CH2F and @CH2CH2CH2CH3. At the first point of difference between these two, fluorine is of higher atomic number than carbon, and so @CH2CH2F is of higher precedence. Higher ranked substituents are on the same side of the double bond; the alkene has the Z con- figuration. (c) One of the carbons of the double bond bears a methyl group and a hydrogen; as we have seen, methyl is of higher rank. The other doubly bonded carbon bears @CH2CH2OH and @C(CH3)3. Let’s analyze these two groups to determine their order of precedence. We examine the atoms one by one at the point of attachment before proceeding down the chain. Therefore, @C(CH3)3 outranks @CH2CH2OH. Higher ranked groups are on opposite sides; the configuration of the alkene is E. (d) The cyclopropyl ring is attached to the double bond by a carbon that bears the atoms (C, C, H) and is therefore of higher precedence than an ethyl group @C(C, H, H). Higher ranked groups are on opposite sides; the configuration of the alkene is E. 5.7 A trisubstituted alkene has three carbons directly attached to the doubly bonded carbons. There are three trisubstituted C6H12 isomers, two of which are stereoisomers. C C CH3 CH3 CH2CH3 H C C CH3 H CH3 CH2CH3 C C CH3 H CH2CH3 CH3 2-Methyl-2-pentene (E)-3-Methyl-2-pentene (Z)-3-Methyl-2-pentene C C CH3CH2 H CH3 Higher Lower Lower Higher C C H CH3 CH2CH2OH C(CH3)3 Higher Lower Higher Lower CH2CH2OH C(C,H,H) Lower priority C(CH3)3 C(C,C,C) Higher priority C C H CH3 CH2CH2F CH2CH2CH2CH3 Higher Lower Higher Lower C C H CH3(CH2)7 H (CH2)12CH3 cis-9-Tricosene Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
STRUCTURE AND PRE TION OF ALKENES: ELIMINATION REACTIONS 5.8 The most stable CH, alkene has a tetrasubstituted double bond CH CH 5.9 Apply the two general rules for alkene stability to rank these compounds. First, more highly substi tuted double bonds are more stable than less substituted ones second when two de bonds are imilarly constituted, the trans stereoisomer is more stable than the cis. The predicted order of de creasing stability is therefore: H CHCH CH,CH,CH CHCH 2-Methyl-2-butene (E)-2-Pentene (disubstituted most stable 5.10 Begin by writing the structural formula corresponding to the IUPAC name given in the problem. A bond-line depiction is useful here 3,4-Di-1ert-butyl-2, 2, 5. 5-tetramethyl-3-hexene The alkene is extremely crowded and destabilized by van der Waals strain Bulky tert-butyl groups are cis to one another on each side of the double bond. Highly strained compounds are often quite difficult to synthesize, and this alkene is a good 5.11 Use the zigzag arrangement of bonds in the parent skeleton figure to place E and Z bonds as appro- priate for each part of the problem. From the sample solution to parts(a)and(b), the ring carbons have the higher priorities. Thus, an E double bond will have ring carbons arranged and a Z double bond (Z)-3-Methylcyclodecene (Z)-5-Methylcyclodecene H H CH (E)-3-Methylcyclodecene E)-5-Methylcyclodecene 5.12 Write out the structure of the alcohol, recognizing that the alkene is formed by loss of a hydrogen nd a hydroxyl group from adjacent carbons Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
5.8 The most stable C6H12 alkene has a tetrasubstituted double bond: 5.9 Apply the two general rules for alkene stability to rank these compounds. First, more highly substituted double bonds are more stable than less substituted ones. Second, when two double bonds are similarly constituted, the trans stereoisomer is more stable than the cis. The predicted order of decreasing stability is therefore: 5.10 Begin by writing the structural formula corresponding to the IUPAC name given in the problem. A bond-line depiction is useful here. The alkene is extremely crowded and destabilized by van der Waals strain. Bulky tert-butyl groups are cis to one another on each side of the double bond. Highly strained compounds are often quite difficult to synthesize, and this alkene is a good example. 5.11 Use the zigzag arrangement of bonds in the parent skeleton figure to place E and Z bonds as appropriate for each part of the problem. From the sample solution to parts (a) and (b), the ring carbons have the higher priorities. Thus, an E double bond will have ring carbons arranged and a Z double bond . 5.12 Write out the structure of the alcohol, recognizing that the alkene is formed by loss of a hydrogen and a hydroxyl group from adjacent carbons. H H CH3 CH3 1 2 H H 2 3 5 4 1 3 (Z)-3-Methylcyclodecene (E)-3-Methylcyclodecene (E)-5-Methylcyclodecene (Z)-5-Methylcyclodecene H 1 2 3 4 5 H CH3 H H CH3 2 3 1 (c) (d) (e) (f) 3,4-Di-tert-butyl-2,2,5,5-tetramethyl-3-hexene C C CH3 CH3 CH3 H 2-Methyl-2-butene (trisubstituted): most stable C C CH3 CH2CH3 H H (E)-2-Pentene (disubstituted) C C CH3 CH2CH3 H H (Z)-2-Pentene (disubstituted) C C CH2CH2CH3 H H H 1-Pentene (monosubstituted): least stable C C CH3 CH3 CH3 CH3 2,3-Dimethyl-2-butene STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 93 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
94 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS (b, c) Both 1-propanol and 2-propanol give propene on acid-catalyzed dehydration OH 1-Propanol Propene (d) Carbon-3 has no hydrogens in 2, 3, 3-trimethyl-2-butanol. Elimination can involve only the hydroxyl group at C-2 and a hydrogen at C-1 CHCH CH CH CH -CH3 2, 3, 3-Tnimethyl-2-butanol 2.3.3-Trimethyl-1-butene 5.13(b) Elimination can involve loss of a hydrogen from the methyl group or from C-2 of the ring in 1-methylcyclohexanol CH I-Methylcyclohexanol disubstituted clohexane I- Methyl ed alkene kene According to the Zaitsev rule, the major alkene is the one corresponding to loss of a hydrogen from the alkyl group that has the smaller number of hydrogens. Thus hydrogen is removed from the methylene group in the ring rather than from the methyl group, and l-methylcyclo- hexene is formed in greater amounts than methylenecyclohexane (c) The two alkenes formed are as shown in the equation double bond double bond The more highly substituted alkene is formed in greater amounts, as predicted by Zaitsev's rule 5.14 2-Pentanol can undergo dehydration in two different directions, giving either 1-pentene or 2-pentene. 2-Pentene is formed as a mixture of the cis and trans stereoisomers CH H CHaCHCH, CH, CH3 CH =CHCHCH CH,+ CHCM x A 2-Pentanol 1-Pentene cis-2-Pentene trans.2-Pentene Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(b, c) Both 1-propanol and 2-propanol give propene on acid-catalyzed dehydration. (d) Carbon-3 has no hydrogens in 2,3,3-trimethyl-2-butanol. Elimination can involve only the hydroxyl group at C-2 and a hydrogen at C-1. 5.13 (b) Elimination can involve loss of a hydrogen from the methyl group or from C-2 of the ring in 1-methylcyclohexanol. According to the Zaitsev rule, the major alkene is the one corresponding to loss of a hydrogen from the alkyl group that has the smaller number of hydrogens. Thus hydrogen is removed from the methylene group in the ring rather than from the methyl group, and 1-methylcyclohexene is formed in greater amounts than methylenecyclohexane. (c) The two alkenes formed are as shown in the equation. The more highly substituted alkene is formed in greater amounts, as predicted by Zaitsev’s rule. 5.14 2-Pentanol can undergo dehydration in two different directions, giving either 1-pentene or 2-pentene. 2-Pentene is formed as a mixture of the cis and trans stereoisomers. 1-Pentene cis-2-Pentene CH2 CHCH2CH2CH3 C C CH3 H CH2CH3 H trans-2-Pentene C C CH3 H H CH2CH3 H heat 2-Pentanol OH CH3CHCH2CH2CH3 OH H H Compound has a trisubstituted double bond Compound has a tetrasubstituted double bond; more stable H2O H2O H3C OH 1-Methylcyclohexanol CH2 Methylenecyclohexane (a disubstituted alkene; minor product) CH3 1-Methylcyclohexene (a trisubstituted alkene; major product) H heat 2,3,3-Trimethyl-2-butanol No hydrogens on this carbon 2,3,3-Trimethyl-1-butene C HO CH3 CH3 H3C C CH3 C C CH3 CH3 CH3 CH3 CH3 H2C H heat H heat 1-Propanol Propene CH3CH2CH2OH CH3CH CH2 2-Propanol CH3CHCH3 OH 94 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 5.15 (b) The site of positive charge in the carbocation is the carbon atom that bears the hydroxyl group the starting alcohol HC、 t h,o I-Methylcyclohexanol Water may remove a proton from the methyl group, as shown in the following equation HO: H-CH Loss of a proton from the ring gives the major product l-methylcyclohexene C H,O: H H, O++ 1-Methylcyclohexene c) Loss of the hydroxyl group under conditions of acid catalysis yields a tertiary carbocation HSO Water may remove a proton from an adjacent methylene group to give a trisubstituted alkene HO HH +H3O+ Removal of the methine proton gives a tetrasubstituted alkene 一 hO H OH 5.16 In writing mechanisms for acid-catalyzed dehydration of alcohols, begin with formation of the carbocation intermediate CH H Dimethy cyclohexanol 2, 2-Dimethylcyclohexyl cation Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
5.15 (b) The site of positive charge in the carbocation is the carbon atom that bears the hydroxyl group in the starting alcohol. Water may remove a proton from the methyl group, as shown in the following equation: Loss of a proton from the ring gives the major product 1-methylcyclohexene. (c) Loss of the hydroxyl group under conditions of acid catalysis yields a tertiary carbocation. Water may remove a proton from an adjacent methylene group to give a trisubstituted alkene. Removal of the methine proton gives a tetrasubstituted alkene. 5.16 In writing mechanisms for acid-catalyzed dehydration of alcohols, begin with formation of the carbocation intermediate: CH3 CH3 CH3 H CH3 OH H 2,2-Dimethylcyclohexanol 2,2-Dimethylcyclohexyl cation H H2O H H3O OH2 H H H H H H3O H2O OH H H H2SO4 H2O CH3 H H 1-Methylcyclohexene H3O CH3 H H2O Methylenecyclohexane H3O H2O H CH2 CH2 1-Methylcyclohexanol H CH3 H2O H3C OH STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 95 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
96 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTION: This secondary carbocation can rearrange to a more stable tertiary carbocation by a methyl group 2, 2-Dimethylcyclohexyl cation 2. Dimet yertia hexyl cation Loss of a proton from the 1, 2-dimethylcyclohexyl cation intermediate yields 1, 2-dimethylcyclo- hexene HO 1, 2-Dimethylcyclohexyl cation 1. 2-Dimethylcyclohexene 5.17 (b) All the hydrogens of tert-butyl chloride are equivalent. Loss of any of these hydrogens alon with the chlorine substituent yields 2-methylpropene as the only alkene CH C、 CH,CCI HC 2-Methylpropene (c) All the B hydrogens of 3-bromo-3-ethylpentane are equivalent, so that B-elimination can give only 3-ethyl-2-pentene CHCH CH2CH,C—B CHCH=C CHCH (d) There are two possible modes of B-elimination from 2-bromo-3-methylbutane Elimination in one direction provides 3-methyl-1-butene; elimination in the other gives 2-methyl-2-butene CH3 CHCH(CH3) CH,=CHCH(CH3)2+ CH_CHC(CH3) 2-Bromo-3-methylbutane Methy l-2 The major product is the more highly substituted alkene, 2-methyl-2-butene. It is the more stable alkene and corresponds to removal of a hydrogen from the carbon that has the fewer (e) Regioselectivity is not an issue here, because 3-methyl-1-butene is the only alkene that can be formed by B-elimination from l-bromo-3-methylbutane BrCH, CH, CH(CH,)2- CH2CHCH(CH,) 1-Bromo-3-methylbutene 3-Methyl-l-butene Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
This secondary carbocation can rearrange to a more stable tertiary carbocation by a methyl group shift. Loss of a proton from the 1,2-dimethylcyclohexyl cation intermediate yields 1,2-dimethylcyclohexene. 5.17 (b) All the hydrogens of tert-butyl chloride are equivalent. Loss of any of these hydrogens along with the chlorine substituent yields 2-methylpropene as the only alkene. (c) All the hydrogens of 3-bromo-3-ethylpentane are equivalent, so that -elimination can give only 3-ethyl-2-pentene. (d) There are two possible modes of -elimination from 2-bromo-3-methylbutane. Elimination in one direction provides 3-methyl-1-butene; elimination in the other gives 2-methyl-2-butene. The major product is the more highly substituted alkene, 2-methyl-2-butene. It is the more stable alkene and corresponds to removal of a hydrogen from the carbon that has the fewer hydrogens. (e) Regioselectivity is not an issue here, because 3-methyl-1-butene is the only alkene that can be formed by -elimination from 1-bromo-3-methylbutane. 1-Bromo-3-methylbutene 3-Methyl-1-butene BrCH2CH2CH(CH3)2 CH2 CHCH(CH3)2 2-Bromo-3-methylbutane 3-Methyl-1-butene (monosubstituted) 2-Methyl-2-butene (trisubstituted) CH3CHCH(CH3)2 Br CH2 CHCH(CH3)2 CH3CH C(CH3) 2 3-Bromo-3-ethylpentane 3-Ethyl-2-pentene CH3CH2 CH2CH3 CH2CH3 C Br CH3CH CH2CH3 CH2CH3 C C H3C H3C CH3CCl CH2 CH3 CH3 tert-Butyl chloride 2-Methylpropene H3O OH2 H CH3 CH3 CH3 CH3 H 1,2-Dimethylcyclohexyl cation 1,2-Dimethylcyclohexene CH3 CH3 CH3 H CH3 H 2,2-Dimethylcyclohexyl cation (secondary) 1,2-Dimethylcyclohexyl cation (tertiary) 96 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS (f Two alkenes may be formed here. The more highly substituted one is 1-methylcyclohexene and this is predicted to be the major product in accordance with Zaitsev's rule H I-lodo-1-methylcyclohexane methylenecyclohexane 1-Methylcyclohexene (disubstituted) 5.18 Elimination in 2-bromobutane can take place between C-1 and C-2 or between C-2 and C-3. Ther are three alkenes capable of being formed: 1-butene and the stereoisomers cis-2-butene and trar 2-butene HC CH3 H3C CH3 CHCH,CH3 CH2=CHCH,CH3+ 2-Bromobutane cis-2-Butene As predicted by Zaitsev's rule, the most stable alkene predominates. The major product is trans 2-butene 5.19 An unshared electron pair of the base methoxide(CHao )abstracts a proton from carbon. The pair of electrons in this C-H bond becomes the T component of the double bond of the alkene. The pair of electrons in the C-Cl bond becomes an unshared electron pair of chloride ion CH,O: H CH CH一H+H2C=CCH3)2+Q! 5.20 The most stable conformation of cis-4-tert-butylcyclohexyl bromide has the bromine substituent in an axial orientation. The hydrogen that is removed by the base is an axial proton at C-2. This hydrogen and the bromine are anti periplanar to each other in the most stable conformation H (CH3)C 5.21(a) 1-Heptene is CH2=CH(CH2)4CH, (b)3-Ethyl-2-pentene is CH_ CH=C(CH_CH3) (c) cis-3-Octene is CH, CH CH, CHCHCH (d) trans-1, 4-Dichloro-2-butene is CICH, CHCI Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
( f ) Two alkenes may be formed here. The more highly substituted one is 1-methylcyclohexene, and this is predicted to be the major product in accordance with Zaitsev’s rule. 5.18 Elimination in 2-bromobutane can take place between C-1 and C-2 or between C-2 and C-3. There are three alkenes capable of being formed: 1-butene and the stereoisomers cis-2-butene and trans- 2-butene. As predicted by Zaitsev’s rule, the most stable alkene predominates. The major product is trans- 2-butene. 5.19 An unshared electron pair of the base methoxide (CH3O) abstracts a proton from carbon. The pair of electrons in this C@H bond becomes the component of the double bond of the alkene. The pair of electrons in the C@Cl bond becomes an unshared electron pair of chloride ion. 5.20 The most stable conformation of cis-4-tert-butylcyclohexyl bromide has the bromine substituent in an axial orientation. The hydrogen that is removed by the base is an axial proton at C-2. This hydrogen and the bromine are anti periplanar to each other in the most stable conformation. 5.21 (a) 1-Heptene is . (b) 3-Ethyl-2-pentene is . (c) cis-3-Octene is (d) trans-1,4-Dichloro-2-butene is C ClCH2 H C H CH2Cl C CH3CH2 H C CH2CH2CH2CH3 H CH3CH C(CH2CH3)2 CH2 CH(CH2)4CH3 Br H (CH3)3C (CH3)3C O H H C C CH3 H CH3 Cl CH Cl 3O H H2C C(CH3)2 CH3O 1-Butene cis-2-Butene CH2 CHCH2CH3 C C H3C H CH3 H trans-2-Butene C C H3C H H CH3 2-Bromobutane Br CH3CHCH2CH3 H3C I 1-Iodo-1-methylcyclohexane Methylenecyclohexane (disubstituted) CH2 CH3 1-Methylcyclohexene (trisubstituted; major product) STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 97 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
98 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS (e)(z)-3-Methyl-2-hexene is H,C CH, CH, CH3 (f)(E)-3-Chloro-2-hexene is H,C, CH,CH,CI (g) 1-Bromo-3-methylcyclohexene is H3C (h) 1-Bromo-6-methylcyclohexene is (i) 4-Methyl-4-penten-2-ol is H3 CH CHCHC=CH ( Vinylcycloheptane is k) An allyl group is -, CH=CH2. 1, 1-Diallylcyclopropane is CH () An isopropenyl substituent is -C=CH,. trans-1-Isopropenyl-3-methylcyclohexane is CH CH 5.22 Alkenes with tetrasubstituted double bonds have four alkyl groups attached to the doubly bonded carbons. There is only one alkene of molecular formula ChH that has a tetrasubstituted double bond, 2, 3-dimethyl-2-pentene H H, C CHCH 5.23 (a) The longest chain that includes the double bond in(CH_CH,),C=CHCH, contains five carbon atoms, and so the parent alkene is a pentene. The numbering scheme that gives the double bond the lowest number is The compound is 3-ethy l-2-pentene Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(e) (Z)-3-Methyl-2-hexene is ( f ) (E)-3-Chloro-2-hexene is (g) 1-Bromo-3-methylcyclohexene is (h) 1-Bromo-6-methylcyclohexene is (i) 4-Methyl-4-penten-2-ol is ( j) Vinylcycloheptane is (k) An allyl group is @CH2CH?CH2. 1,1-Diallylcyclopropane is (l) An isopropenyl substituent is . trans-1-Isopropenyl-3-methylcyclohexane is therefore 5.22 Alkenes with tetrasubstituted double bonds have four alkyl groups attached to the doubly bonded carbons. There is only one alkene of molecular formula C7H14 that has a tetrasubstituted double bond, 2,3-dimethyl-2-pentene. 5.23 (a) The longest chain that includes the double bond in (CH3CH2)2C?CHCH3 contains five carbon atoms, and so the parent alkene is a pentene. The numbering scheme that gives the double bond the lowest number is The compound is 3-ethyl-2-pentene. H 5 4 1 3 2 C H3C H3C C CH3 CH2CH3 2,3-Dimethyl-2-pentene CH3 CH2 CH3 C CH2 CH3 H CH3CHCH2C CH2 CH3 OH Br CH3 Br H3C C H3C H C CH2CH2CH3 Cl C H3C H C CH2CH2CH3 CH3 98 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS (b) Write out the structure in detail, and identify the longest continuous chain that includes the double bond CH CH2 CH, CH3 CHCH H,CH The longest chain contains six carbon atoms and the double bond is between C-3 and c-4 The compound is named as a derivative of 3-hexene. There are ethyl substituents at C-3 and C-4. The complete name is 3, 4-diethyl-3-hexene (c) Write out the structure completel H The longest carbon chain contains four carbons. Number the chain so as to give the lowest numbers to the doubly bonded carbons, and list the substituents in alphabetical order. This compound is 1, I-dichloro-3, 3-dimethyl-l-butene. (d) The longest chain has five carbon atoms, the double bond is at C-1, and there are two methyl substituents. The compound is 4, 4-dimethyl-l-pentene (e) We number this trimethylcyclobutene derivative so as to provide the lowest number for the substituent at the first point of difference. We therefore number H C H. rather than H,C The correct IUPAC name is 1, 4, 4-trimethylcyclobutene, not 2, 3, 3-trimethylcyclobutene (f) The cyclohexane ring has a 1, 2-cis arrangement of vinyl substituents. The compound cis-1, 2-divinylcyclohexane (g) Name this compound as a derivative of cyclohexene. It is 1, 2-divinylcyclohexene 5.24 (a) Go to the end of the name, because this tells you how many carbon atoms are present in the longest chain. In the hydrocarbon name 2, 6, 10, 14-tetramethyl-2-pentadecene, the suffix pentadecane"reveals that the longest continuous chain has 15 carbon atoms and that there Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(b) Write out the structure in detail, and identify the longest continuous chain that includes the double bond. The longest chain contains six carbon atoms, and the double bond is between C-3 and C-4. The compound is named as a derivative of 3-hexene. There are ethyl substituents at C-3 and C-4. The complete name is 3,4-diethyl-3-hexene. (c) Write out the structure completely. The longest carbon chain contains four carbons. Number the chain so as to give the lowest numbers to the doubly bonded carbons, and list the substituents in alphabetical order. This compound is 1,1-dichloro-3,3-dimethyl-1-butene. (d ) The longest chain has five carbon atoms, the double bond is at C-1, and there are two methyl substituents. The compound is 4,4-dimethyl-1-pentene. (e) We number this trimethylcyclobutene derivative so as to provide the lowest number for the substituent at the first point of difference. We therefore number The correct IUPAC name is 1,4,4-trimethylcyclobutene, not 2,3,3-trimethylcyclobutene. ( f ) The cyclohexane ring has a 1,2-cis arrangement of vinyl substituents. The compound is cis-1,2-divinylcyclohexane. (g) Name this compound as a derivative of cyclohexene. It is 1,2-divinylcyclohexene. 5.24 (a) Go to the end of the name, because this tells you how many carbon atoms are present in the longest chain. In the hydrocarbon name 2,6,10,14-tetramethyl-2-pentadecene, the suffix “2-pentadecene” reveals that the longest continuous chain has 15 carbon atoms and that there H H H3C H3C H3C 1 4 2 3 H3C H3C H3C 2 3 1 4 rather than 1 2 3 4 5 C H3C H3C CH3 C H C Cl Cl 3 4 2 1 C CH3CH2 CH2CH3 CH3CH2 CH2CH3 C 1 2 5 6 3 4 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 99 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
