CHAPTER 15. Practice exercises 15.1

Transcription

1 CAPTER 15 Practice exercises (a) C 9 18 : isobutylcyclopentane C : sec-butylcycloheptane C 6 2 : 1-ethyl-1-methylcyclopropane 15.5 (a) 3,3-dimethyl-1-pentene 2,3-dimethyl-2-butene 3,3-dimethyl-1-butyne 15.7 (a) (E)-1-chloro-2,3-dimethyl-2-pentene (Z)-1-bromo-1-chloropropene (E)-2,3,4-trimethyl-3-heptene 15.9 cis,trans-2,4-heptadiene cis,cis-2,4-heptadiene (a) 2-iodopropane

2 1-iodo-1-methylcyclohexane C 3 I C 3 I Step 1: Protonation of the alkene to give the most stable 3 carbocation intermediate: C 3 I slow step rate-determining C 3 I step Step 2: Nucleophilic attack of the iodide anion on the 3 carbocation intermediate to give the product: C 3 I C 3 I O C 3 C 3 3 O O Step 1: Protonation of the alkene to give the most stable 3 carbocation intermediate: slow step C 3 O C 3 rate-determining step O Step 2: Nucleophilic attack of the water on the 3 carbocation intermediate to give the protonated alcohol:

3 C 3 O O C 3 Step 3: The protonated alcohol loses a proton to form the product: O C 3 O C 3 O 3 O (a) 2-phenyl-2-propanol (E)-3,4-diphenyl-3-hexene 3-methylbenzoic acid or m-methylbenzoic acid Review questions 15.1 A hydrocarbon is a compound composed only of hydrogen and carbon atoms In saturated hydrocarbons, each carbon is bonded to four other atoms, either hydrogen or carbon atoms. Unsaturated hydrocarbons have carbon atoms that have a double or triple bond to another carbon atom (a) (d) (e) (f)

4 15.7 (C 2 ) 2 C 3 (a) C 3 (C 2 ) 4 C(C 3 ) 2 C 3 (C 2 ) 2 C(C 2 ) 2 C 3 (C 2 ) 2 C 3 C 3 (C 2 ) 2 C(C 2 ) 4 C 3 C (a) different compounds constitutional isomers constitutional isomers (d) different compounds (e) constitutional isomers (f) constitutional isomers (a) 2-methylpentane 2,5-dimethylhexane 3-ethyloctane (d) 2,2,3-trimethylbutane (e) isobutylcyclopentane (f) 1-tert-butyl-2,4-dimethyl-cyclohexane (a) 1,3-dimethylbutane The longest chain is pentane. The IUPAC name is 2-methylpentane. 4-methylpentane The pentane chain is numbered incorrectly. The IUPAC name is 2-methylpentane. 2,2-diethylbutane The longest chain in pentane. The IUPAC name is 3-ethyl-3-methylpentane. (d) 2-ethyl-3-methylpentane The longest chain is hexane. The IUPAC name is 3,4-dimethylhexane.

5 (e) 2-propylpentane The longest chain is heptane. The IUPAC name is 4-methylheptane. (f) 2,2-diethylheptane The longest chain is octane. The IUPAC name is 3-ethyl-3-methyloctane. (g) 2,2-dimethylcyclopropane The ring is numbered incorrectly. The IUPAC name is 1,1-dimethylcyclopropane. (h) 1-ethyl-5-methylcyclohexane The ring is numbered incorrectly. The IUPAC name is 1-ethyl-3-methylcyclohexane No, because alkanes do not have rings or C C double bonds and so all conformations are usually interconvertable by rotation about a C C single bond. owever, there are some extremely crowded molecules which are locked into specific conformers. These are called conformational isomers or locked conformational isomers and are not correctly defined by cis-trans nomenclature There are two enantiomers of the trans-1,2-dimethylcyclopropane.. C 3 C 2 C 3 cyclopentane methylcyclobutane ethylcyclopropane

6 C 3 C 3 C 3 C 3 C 3 trans-1,2-dimethylcyclopropane 1,1-dimethylcyclopropane C 3 cis-1,2-dimethylcyclopropane (a) trans-2-methylhex-3-ene 2-methyl-3-hex-3-yne 2-methylbut-1-ene (d) 3-ethyl-3-methylpent-1-yne (e) 2,3-dimethylbut-2-ene (f) cis-pent-2-ene (a) 2-isobutylhept-1-ene 1,4,4-trimethylcyclopentene 1,3-cyclopentadiene (d) 3,3-dimethylbut-1-yne (e) 2,4-dimethylpent-2-ene (f) oct-1-yne (g) 2,2,5-dimethylhex-3-yne (h) 3-methylpent-1-yne (a) 2,2-dimethylhex-3-yne oct-2,5-diyne 3,6-dimethylhept-2-ene-4-yne hept-1,4-diyne (a) Correct name: but-2-ene.you must select the longest carbon chain containing the ene functional group but-2-ene 4 Not: 2 3 C methylpropene Correct name: pent-2-ene. You must number the chain to give the first carbon of the double bond the lowest possible number.

7 Not: pent-2-ene pent-3-ene Correct name: 1-methylcyclohexene. By default, the first carbon of the double bond in a ring is given the number 1. In this, case this can also be the ring carbon with the methyl group attached methylcyclohexene Not: 2-methylcyclohexene 2 1 (d) (e) (f) This name is not incorrect even though there is no indication of where the double bond is positioned. The indication of the two methyls on the third carbon means that the double bond is attached to the first carbon. Technically, the correct name is 3,3-dimethylpent-1-ene. Correct name: 4-hexyne. As with the chain carbons are numbered to give the first carbon of the double bond the lowest possible number C 3 C 2 C 2 C CC 2 C Correct name: 2-isopropyl-2-butene. The longest chain containing the function group is a pentene (a) No Yes Yes (d) No (e) Yes (f) No

8 (e) For alkenes to exist as a pair of cis-trans isomers, both carbons of the double bond must have two different substituents. Thus, only and (d) can exist as a cis or trans isomer. 3 C cis-bromoprop-1-ene 3 C trans-bromoprop-1-ene (d) Molecules (a) and do not show cis-trans isomerism. (d) (a) Alkenes that do not show cis-trans isomerism are: pent-1-ene 2-methylbut-2-ene 3-methylbut-1-ene 2-methylbut-1-ene Alkenes that do show cis-trans isomerism are: trans-pent-2-ene cis-pent-2-ene

9 Cycloalkanes that do not show cis-trans isomerism are: C 2 C 3 C3 3 C C 3 cyclopentane methylcyclobutane ethylcyclopropane 1,1-dimethyl cyclopropane (d) Cycloalkanes that do show cis-trans isomerism are: 3 C C 3 3 C C 3 cis-1,2-dimethylcyclopropane trans-1,2-dimethylcyclopropane (a) C CC is more stable. C 3 3 C 3 C C C is more stable C 3 C 3 C=C 2 6O 2 4CO O Ni C C O 2 SO 4 C 3 C 3 C 3 C 2 O 2 C=C 2 C 3 C 2 2 (d) 2 C C 2 C 2 C 2 (e) 2 C C 2 C 3 C 2

10 / Pd/C 2 Lindlar catalyst 2 Na/N (a) 1-bromo-2-chloro-4-ethylbenzene 4-iodo-1,2-dimethylbenzene I 2,4,6-trinitrotoluene (d) 4-phenylpentan-2-ol O 2 N NO 2 O NO 2

11 (e) p-cresol (f) 2,4-dichlorophenol O O (g) 1-phenylcyclopropanol (h) styrene (phenylethene) O (i) m-bromophenol (j) 2,4-dibromoaniline O N 2 (k) isobutylbenzene (l) m-xylene isopropylbenzene

12 (1) /Al 3 (2) / 2 SO 4 (3) O / 2 SO 4 Review problems (1) Alkanes are less dense than water. (2) As alkane molar mass increases, density increases. (3) Constitutional isomers have similar densities Boiling points of unbranched alkanes are related to their surface area: the larger the surface area, the greater the strength of the dispersion forces and the higher the boiling point. The relative increase in molecular size per C 2 group is greatest between C 4 and C 3 C 3 and becomes progressively smaller as molecular mass increases. Therefore, the increase in boiling point per C 2 group is greatest between C 4 and C 3 C 3, and becomes progressively smaller for the higher alkanes (a) No Yes Yes (d) It is a liquid. (e) It is less dense than water The three structures with molecular formula C22 are 1,1-dibromoethane, cis-1,2- dibromoethane and trans-1,2-dibromoethane. The dipole moment of the C bond is shown in the structures below by a solid arrow. trans-1,2-dibromoethane has no dipole moment because the two C dipoles point in opposite directions and therefore cancel each other out. owever, 1,1-dibromoethane and cis-1,2-dibromoethane both have a net dipole moment (in the direction shown by the dotted arrow) since the two C dipoles do not point in opposite directions. C C C C C C No dipole moment 1,1-dibromoethane cis-1,2-dibromoethane trans-1,2-dibromoethane (a)

13 C 3 C 3 C C 3 C C cis-4-methylpent-2-ene C 3 C 3 C C C C 3 trans-4-methylpent-2-ene C 3 C 3 CC 2 C C 2 4-methylpent-1-ene (a) 2,2-dichlorobutane

14 butan-2-ol 2,2,3,3-tetrabromobutane A hydrocarbon of formula C 5 8 must have a triple bond, or two double bonds, or a combination of a double bond and a ring. Since it only reacts with one mole of 2 indicates that the compound is an alkyne. Lindlar s catalyst gives a cis alkene, which in turn is brominated to give a dibromoalkane which has two chiral carbons. Only the final possibility shown below meets this last criterion. OR OR (a) The formation of the more stable tertiary carbocation intermediate proceeds at a faster rate, therefore 2-methylbut-2-ene reacts faster than trans-but-2-ene.

15 I trans-but-2-ene secondary carbocation 2-iodobutane I 2-methylbut-2-ene tertiary carbocation 2-iodo-2-methyl butane (major product) The formation of the more stable tertiary carbocation intermediate proceeds at a faster rate, therefore 1-methylcyclohexene reacts faster than cyclohexene. I cyclohexene secondary carbocation iodocyclohexane C C 3 C 3 3 I 1-methylcyclohexene tertiary carbocation 1-iodo-1-methyl cyclohexane (major product) The first step in the reaction is the protonation of the alkene to generate a carbocation. The reaction path that produces the more stable carbocation occurs at a faster rate, thus producing the observed regioselectivity. (a) C 3 C 3 C CC 2 C 3 I C 3 C 3 CC 2 C 2 C 3 I C 3 2 O C 3 C 3 C CC 2 C 3 C 3 CC 2 C 2 C 3 2 SO 4 O

16 15.69 (a) Chlorine undergoes an anti-addition to alkenes. (a) 2 2 C 3 3 C 2 2 (d) C Both cis- and trans-hex-3-ene yield hexan-3-ol upon acid-catalysed hydration because they both form the same carbocation intermediate that leads to hexan-3-ol according to the partial mechanism: O O O (a) 2 O 3 O O

17 C 3 C 3 2 O 3 O C 3 O C 3 2 O O C 3 3 O C 3 (d) 3 C 2 O 3 O 3 C O Step 1: Protonation of the alkene gives a stable 3 carbocation intermediate: C 3 C 3 C C 2 O C3 3 C C 3 C C 3 O C 3 Step 2: Nucleophilic attack of methanol on the carbocation intermediate yields a protonated ether intermediate. 3 C C 3 C 3 C O C 3 3 C C 3 C O C 3 C 3 Step 3: Proton loss by the protonated ether intermediate yields the ether and regenerates the acid catalyst. 3 C C 3 C O C 3 C 2 C 3 O 3 C C 3 C O C 3 C 3 O C The skeleton and location of the double bonds in hydrocarbon A are identified from the structure of the brominated product:

18 2 mol 2 ydrocarbon A (C 5 8 ) 2-methylbuta-1,3-diene 1,2,3,4-tetrabromo-2-methylbutane (a) 2 2 O 2 SO 4 O ( d ) 2 P t (a) or 2 O 2 SO 4 O 2

19 15.85 Resonance accounts for the three equivalent structures of naphthalene: (a) - 8 π electrons - Non-aromatic - 14 π electrons - Not planar - Non-aromatic O N - 10 π electrons - Planar - Aromatic (d) B (e) O (f) O - 6 π electrons - Planar - Aromatic - 6 π electrons - Not a 2p orbital on every atom - Non-aromatic - 8 π electrons - Non-aromatic Although compound contains 14 π electrons (a ückel number), a closer inspection of the molecule reveals that the protons inside the ring are held too close together and force the ring to be non-planar. Molecular modelling of structure confirms its non-planarity and it therefore non-aromatic. In compound, only one of the electron pairs on oxygen is in a 2p orbital, the other electron pair is in an sp 2 hybridised orbital, perpendicular to the π bonds. The nitrogen lone pair in compound is also in a 2p orbital for a total of 10 π electrons in a planar ring making it aromatic. In compound (d), the boron atom contributes an empty 2p orbital, giving an aromatic compound with six π electrons in a seven-membered ring with seven 2p orbitals. Compound (f), is similar to compound where the oxygen has one electron pair in a 2p orbital and the other in an sp 2 hybridised orbital with a total of eight π electrons, giving a non-aromatic compound Two monochloronaphthalenes are possible when naphthalene is treated with 2 /Al 3 : 2 Al 3

20 15.91 Al 2 3 C 2 2 C 2 Step 1: Formation of Lewis acid-lewis base complex: C 2 Al C 2 Al Step 2: Nucleophilic attack of benzene on the electrophilic Lewis acid-lewis base complex and the formation of a resonance-stabilised carbocation (two more resonance structures exist): C 2 Al C 2 Al 4 - Step 3: Deprotonation of the carbocation to give benzyl chloride,, and Al 3 : C 2 Al C 2 Al 3 Step 4: Formation of Lewis acid-lewis base complex between benzyl chloride and Al 3 : C 2 Al C 2 Al Step 5: Dissociation of the complex to give a resonance-stabilised benzyl cation (only one out of five contributing structures is shown) and Al 4 : C 2 Al C 2 Al

21 Step 6: Nucleophilic attack of the second molecule of benzene on the benzylic cation to form another resonance-stabilised carbocation (only one structure is shown): C C 2 2 Step 7: Deprotonation of the carbocation intermediate to regenerate the aromatic ring (diphenylmethane),, and Al 3 : Al C 2 C 2 Al 3 Additional exercises A molecular model of cyclohexane shows hindered rotation about the C C single bonds, allowing the possibility of cis-trans isomers in disubstituted six-membered rings. The much larger cyclododecane ring is flexible enough to allow unrestricted rotation about the C C single bonds, thus no cis-trans isomerisation exists. C 3 C 3 C 3 C 3 C 3 cis-1,2-dimethylcyclohexane trans-1,2-dimethylcyclohexane C 3 1,2-dimethylcyclododecane (a) o 120 o C 2 O 120 o 180 o C C C C 120 o (d) 120 o

22 o O o o (a) 180 o (d) 120 o 120 o o (a) Carbons 1 and 3 are sp 2 hybridised. Carbon 2 is sp hybridised. π 2pz -2p z y-axis z-axis π 2py -2p y Each of the terminal carbons is sp 2 hybridised and thus has three sp 2 orbitals oriented 120 apart. Two of these sp 2 hybrid orbitals form sigma bonds with hydrogen atoms and the third forms a sigma bond with the central carbon. Each terminal carbon has one 2p orbital (C1 has a 2p z orbital and C3 has a 2p y orbital) that each form a π bond with the 2p z and 2p y orbitals on the central carbon atom. Because these two 2p orbitals on the centre carbon atom are at right angles to each other, the 2p orbitals on the terminal carbons must be at right angles to each other to permit full orbital overlap Molecules of the trans isomer can pack together more tightly than those of the cis isomers, which means that the dispersion forces controlling intermolecular bonding are greater for the trans than for the less symmetrical cis isomer. COO COO COO COO COO COO The trans isomer of octadec-9-enoic acid can pack together closely and so it has a higher melting point (44 45 C) than the cis isomer.

23 COO COO COO COO (a) O 2 O 2 enzyme catalyst O 2 2 O heat O O Yes