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