CHAPTER 4 ALCOHOLS AND ALKYL HALIDES SOLUTIONS TO TEXT PROBLEMS 4.1 There are four C4H, alkyl groups, and so there are four C4HOCI alkyl chlorides. Each may be named by both the functional class and substitutive methods. The functional class name uses the name of the alkyl group followed by the halide as a second word. The substitutive name modifies the name of the corresponding alkane to show the location of the halogen atom. nctional class name ve name CH, CH, CH, CH,CI n-Butyl chloride 1-Chlorobutane (Butyl chloride) CHaCHCH, CH3 sec-Butyl chloride 2-Chlorobutane (1-Methylpropyl chloride) CH, CHCH,CI Isobutyl chloride 1-Chloro-2-methylpropane (2-Methylpropyl chloride) CH3CCH tert-Butyl chloride 2-Chloro-2-methylpropane (1, 1-Dimethylethyl chloride) 4.2 Alcohols may also be named using both the functional class and substitutive methods, as in the 67 Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
67 CHAPTER 4 ALCOHOLS AND ALKYL HALIDES SOLUTIONS TO TEXT PROBLEMS 4.1 There are four C4H9 alkyl groups, and so there are four C4H9Cl alkyl chlorides. Each may be named by both the functional class and substitutive methods. The functional class name uses the name of the alkyl group followed by the halide as a second word. The substitutive name modifies the name of the corresponding alkane to show the location of the halogen atom. Functional class name Substitutive name CH3CH2CH2CH2Cl n-Butyl chloride 1-Chlorobutane (Butyl chloride) sec-Butyl chloride 2-Chlorobutane (1-Methylpropyl chloride) Isobutyl chloride 1-Chloro-2-methylpropane (2-Methylpropyl chloride) tert-Butyl chloride 2-Chloro-2-methylpropane (1,1-Dimethylethyl chloride) 4.2 Alcohols may also be named using both the functional class and substitutive methods, as in the previous problem. CH3CHCH2CH3 Cl CH3CHCH2Cl CH3 CH3 CH3CCH3 Cl Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
68 ALCOHOLS AND ALKYL HALIDES Functional class name bstitutive name CHCH, CH,CH,OH n-Butyl alcohol 1-Butanol CH3 CHCH, CH3 sec-Butyl alcohol 2-Butanol (l-Methylpropyl alcohol) CH3 CHCH,OH Isobutyl alcohol 2-Methyl-1-propanol (2-Methylpropyl alcohol) CH CCH tert-Butyl alcohol 2-Methyl-2-propanol (1, I-Dimethylethyl alcohol) 4.3 Alcohols are classified as primary, secondary, or tertiary according to the number of carbon substituents attached to the carbon that bears the hydroxyl group. H..CH CH3C—CH2CH3 Primary alcohol econdary alcohol (one alkyl group bonded to--CH, OH) (two alkyl groups bonded to/CHOH) CH (CH3)2CH—C-OH CH -C-OH Primary kyl group bonded to-CH, OH)(three alkyl groups bonded to - COH) 4.4 Dipole moment is the product of charge and distance. Although the electron distribution in the carbon-chlorine bond is more polarized than that in the carbon-bromine bond, this effect is counterbalanced by the longer carbon-bromine bond distance Distance Charge CH -CI CH--Br Methyl chloride (greater value of e) 8D 4.5 All the hydrogens in dimethyl ether(CH3OCH3) are bonded to carbon; therefore, intermolecular hydrogen bonding between dimethyl ether molecules does not take place, and its boiling point is lower than that of ethanol (CHa CH,OH), where hydrogen bonding involving the -OH group is 4.6 Ammonia is a base and abstracts(accepts)a proton from the acid (proton donor) hydrogen chloride H3N NH, cl: B Acid Conjugate Conjugate Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
68 ALCOHOLS AND ALKYL HALIDES Functional class name Substitutive name CH3CH2CH2CH2OH n-Butyl alcohol 1-Butanol (Butyl alcohol) sec-Butyl alcohol 2-Butanol (1-Methylpropyl alcohol) Isobutyl alcohol 2-Methyl-1-propanol (2-Methylpropyl alcohol) tert-Butyl alcohol 2-Methyl-2-propanol (1,1-Dimethylethyl alcohol) 4.3 Alcohols are classified as primary, secondary, or tertiary according to the number of carbon substituents attached to the carbon that bears the hydroxyl group. 4.4 Dipole moment is the product of charge and distance. Although the electron distribution in the carbon–chlorine bond is more polarized than that in the carbon–bromine bond, this effect is counterbalanced by the longer carbon–bromine bond distance. 4.5 All the hydrogens in dimethyl ether (CH3OCH3) are bonded to carbon; therefore, intermolecular hydrogen bonding between dimethyl ether molecules does not take place, and its boiling point is lower than that of ethanol (CH3CH2OH), where hydrogen bonding involving the @OH group is important. 4.6 Ammonia is a base and abstracts (accepts) a proton from the acid (proton donor) hydrogen chloride. H Cl 3N H NH4 Cl Base Acid Conjugate acid Conjugate base e d Charge Dipole moment Distance CH3 Cl Methyl chloride (greater value of e) 1.9 D CH3 Br Methyl bromide (greater value of d) 1.8 D (CH3)2CH C H H OH Primary alcohol (one alkyl group bonded to CH2OH) CH3 CH3 CH3 C OH Tertiary alcohol (three alkyl groups bonded to COH) CH3CH2CH2 C H H OH Primary alcohol (one alkyl group bonded to CH2OH) CH3 C OH H CH2CH3 Secondary alcohol (two alkyl groups bonded to CHOH) CH3CHCH2CH3 OH CH3CHCH2OH CH3 CH3 CH3CCH3 OH Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
ALCOHOLS AND ALKYL HALIDES 69 4.7 Since the pk, of HCN is given as 9.1, its K=10.. In more conventional notation, K 8X10. Hydrogen cyanide is a weak acid 4.8 Hydrogen cyanide is a weak acid, but it is a stronger acid than water(pK= 15.7). Since HCN is a stronger acid than water, its conjugate base(Cn )is a weaker base than hydroxide(Ho), which the conjugate base of water. 4.9 An unshared electron pair on oxygen abstracts the proton from hydrogen chloride CHD),C. co +HC Co-H + :C: B Acid Conjugate 4.10 In any proton-transfer process, the position of equilibrium favors formation of the weaker acid and the weaker base from the stronger acid and base. Alkyloxonium ions(ROH, t) have approximately the same acidity as hydronium ion(H,o, pk=-1.7). Thus hydrogen chloride(pK,"-7)is the stronger acid. tert-Butyl alcohol is the stronger base because it is the conjugate of the weaker acid (tert-butyloxonium ion) (CH3)3COH HCI (CH3)3COH2 CI The equilibrium constant for proton transfer from hydrogen chloride to tert-butyl alcohol is much greater than 1 4.11 The proton being transferred is partially bonded to the oxygen of tert-butyl alcohol and to chloride at the transition state 8-+-c 4.12 (b) Hydrogen chloride converts tertiary alcohols to tertiary alkyl chlorides (CH,CH,)COH (CH,CH2)3CCI H,O 3-Ethyl-3-pentanol 3-Chloro-3-ethylpentane (c) 1-Tetradecanol is a primary alcohol having an unbranched 14-carbon chain. Hydrogen bromide reacts with primary alcohols to give the corresponding primary alkyl bromide CHa(CH2)12CH,OH HBr CHa(CH)) CH,Br H,O Hydrogen 4.13 The order of carbocation stability is tertiary secondary primary. There is only one CsH,car- bocation that is tertiary, and so that is the most stable one. CH3,C+ 1, 1-Dimethylpropyl cation Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
4.7 Since the pKa of HCN is given as 9.1, its Ka 109.1. In more conventional notation, Ka 8 1010. Hydrogen cyanide is a weak acid. 4.8 Hydrogen cyanide is a weak acid, but it is a stronger acid than water (pKa 15.7). Since HCN is a stronger acid than water, its conjugate base (CN) is a weaker base than hydroxide (HO), which is the conjugate base of water. 4.9 An unshared electron pair on oxygen abstracts the proton from hydrogen chloride. 4.10 In any proton-transfer process, the position of equilibrium favors formation of the weaker acid and the weaker base from the stronger acid and base. Alkyloxonium ions (ROH2 ) have approximately the same acidity as hydronium ion (H3O, pKa 1.7). Thus hydrogen chloride (pKa 7) is the stronger acid. tert-Butyl alcohol is the stronger base because it is the conjugate of the weaker acid (tert-butyloxonium ion). The equilibrium constant for proton transfer from hydrogen chloride to tert-butyl alcohol is much greater than 1. 4.11 The proton being transferred is partially bonded to the oxygen of tert-butyl alcohol and to chloride at the transition state. 4.12 (b) Hydrogen chloride converts tertiary alcohols to tertiary alkyl chlorides. (c) 1-Tetradecanol is a primary alcohol having an unbranched 14-carbon chain. Hydrogen bromide reacts with primary alcohols to give the corresponding primary alkyl bromide. 4.13 The order of carbocation stability is tertiary secondary primary. There is only one C5H11 carbocation that is tertiary, and so that is the most stable one. 1,1-Dimethylpropyl cation CH3CH2C CH3 CH3 CH3(CH2)12CH2OH HBr 1-Tetradecanol CH3(CH2)12CH2Br Hydrogen 1-Bromotetradecane bromide H2O Water (CH3CH2)3COH HCl 3-Ethyl-3-pentanol (CH3CH2)3CCl Hydrogen 3-Chloro-3-ethylpentane chloride H2O Water O H Cl H (CH3)3C (CH3)3COH HCl Cl Stronger base (pKa 7) Stronger acid (CH3)3COH2 (pKa 1.7) Weaker acid Weaker base Acid H Cl H Conjugate base Cl Conjugate acid O (CH3)3C H Base O (CH3)3C H ALCOHOLS AND ALKYL HALIDES 69 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
ALCOHOLS AND ALKYL HALIDES 4.14 1-Butanol is a primary alcohol; 2-butanol is a secondary alcohol. A carbocation intermediate is possible in the reaction of 2-butanol with hydrogen bromide but not in the corresponding reaction of 1-butanol The mechanism of the reaction of 1-butanol with hydrogen bromide proceeds by displacement of water by bromide ion from the protonated form of the alcohol(the alkyloxonium ion) Protonation of the alcohol CH CH, CH,CH,O:+ H →CH2CH CHO++ Hydroge Bromide Displacement of water by bromide: CHCHCH CH,O+ CH CH Br o The slow step, displacement of water by bromide from the oxonium ion, is bimolecular. The reaction of 1-butanol with hydrogen bromide follows the Sn2 mechanism. The reaction of 2-butanol with hydrogen bromide involves a carbocation intermediate Protonation of the alcohol CH3 CH CHCH3+ CH3,CHCH3 Br: 2-Butanol sec-Butyloxonium ion Bromide io Dissociation of the oxonium ion CH3, CHCH3 CHaCH,CHCH Butyloxonium io sec-Butyl cation Water Capture of sec-butyl cation by bromide: CHCH Br CHCH2—CH3CH2CHCH3 Bromide ion sec-Butyl cation 2-Bromobutane The slow step, dissociation of the oxonium ion, is unimolecular. The reaction of 2-butanol with 4.15 The most stable alkyl free radicals are tertiary. The tertiary free radical having the formula C has the same skeleton as the carbocation in Problem 4.13 CH CH CH2-C Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
4.14 1-Butanol is a primary alcohol; 2-butanol is a secondary alcohol. A carbocation intermediate is possible in the reaction of 2-butanol with hydrogen bromide but not in the corresponding reaction of 1-butanol. The mechanism of the reaction of 1-butanol with hydrogen bromide proceeds by displacement of water by bromide ion from the protonated form of the alcohol (the alkyloxonium ion). Protonation of the alcohol: Displacement of water by bromide: The slow step, displacement of water by bromide from the oxonium ion, is bimolecular. The reaction of 1-butanol with hydrogen bromide follows the SN2 mechanism. The reaction of 2-butanol with hydrogen bromide involves a carbocation intermediate. Protonation of the alcohol: Dissociation of the oxonium ion: Capture of sec-butyl cation by bromide: The slow step, dissociation of the oxonium ion, is unimolecular. The reaction of 2-butanol with hydrogen bromide follows the SN1 mechanism. 4.15 The most stable alkyl free radicals are tertiary. The tertiary free radical having the formula C5H11 has the same skeleton as the carbocation in Problem 4.13. CH3CH2 C CH3 CH3 Bromide ion sec-Butyl cation 2-Bromobutane CH3CH2CHCH3 Br CHCH3 CH3CH2 Br CH3CH2CHCH3 O H H CH3CH2CHCH3 O H H sec-Butyloxonium ion sec-Butyl cation Water slow CH3CH2CHCH3 O H Br 2-Butanol Hydrogen bromide sec-Butyloxonium ion Bromide ion H Br CH3CH2CHCH3 O H H CH3CH2CH2CH2Br 1-Bromobutane O Water slow Br Bromide ion CH3CH2CH2 CH2 O Butyloxonium ion H H H H CH3CH2CH2CH2O H 1-Butanol H Br Hydrogen bromide Br Butyloxonium ion Bromide CH3CH2CH2CH2O H H 70 ALCOHOLS AND ALKYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
ALCOHOLS AND ALKYL HALIDES 4.16(b riting the equations for carbon-carbon bond cleavage in propane and in 2-methylpropane a primary ethyl radical is produced by a cleavage of propane wh of 2-methylprop CH,CHO -CH3 CHa CH2 CHCHCH CH3 CHCH3 . CH 2-Methylpropane Isopropyl radical Methyl radical A secondary radical is more stable than a primary one, and so carbon-carbon bond cleavage of 2-methylpropane requires less energy than carbon-carbon bond cleavage of propa (c) Carbon-carbon bond cleavage of 2, 2-dimethylpropane gives a tertiary radical / CH CH3 CH CCH3 CH- H CH 2, 2-Dimethylpropane tert-Butyl radical Methyl radical As noted in part(b), a secondary radical is produced on carbon-carbon bond cleavage of 2-methylpropane. We therefore expect a lower carbon-carbon bond dissociation energy for .2-dimethylpropane than for 2-methylpropane, since a tertiary radical is more stable than a 4.17 First write the equation for the overall reaction CHC Chloromethane Chlorine The initiation step is dissociation of chlorine to two chlorine atoms Chlorine 2 Chlorine atoms A chlorine atom abstracts a hydrogen atom from chloromethane in the first propagation step Cl—C.+ H H Chloromethane Chlorine atom Chloromethyl radical Hydrogen ch Chloromethyl radical reacts with Cl, in the next propagation step H g!1g:一Cl-C-g: nloromethyl radical Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
4.16 (b) Writing the equations for carbon–carbon bond cleavage in propane and in 2-methylpropane, we see that a primary ethyl radical is produced by a cleavage of propane whereas a secondary isopropyl radical is produced by cleavage of 2-methylpropane. A secondary radical is more stable than a primary one, and so carbon–carbon bond cleavage of 2-methylpropane requires less energy than carbon–carbon bond cleavage of propane. (c) Carbon–carbon bond cleavage of 2,2-dimethylpropane gives a tertiary radical. As noted in part (b), a secondary radical is produced on carbon–carbon bond cleavage of 2-methylpropane. We therefore expect a lower carbon–carbon bond dissociation energy for 2,2-dimethylpropane than for 2-methylpropane, since a tertiary radical is more stable than a secondary one. 4.17 First write the equation for the overall reaction. The initiation step is dissociation of chlorine to two chlorine atoms. A chlorine atom abstracts a hydrogen atom from chloromethane in the first propagation step. Chloromethyl radical reacts with Cl2 in the next propagation step. Chlorine Cl Dichloromethane Chlorine atom C Cl H H C Cl Cl Cl H H Cl Chloromethyl radical Cl Chlorine atom H Cl Chloromethyl radical Hydrogen chloride C H H Cl H C H H Cl Chloromethane Cl Chlorine 2 Chlorine atoms Cl Cl Cl CH3Cl Cl 2 Chloromethane CH2Cl2 Chlorine Dichloromethane HCl Hydrogen chloride CH3CCH3 CH3 CH3 CH3 CH3 C CH3 2,2-Dimethylpropane tert-Butyl radical Methyl radical CH3 CH3CH2 CH3 CH3CH2 CH3 Propane Ethyl radical Methyl radical CH3CHCH3 CH3CHCH3 CH3 2-Methylpropane Isopropyl radical Methyl radical CH3 ALCOHOLS AND ALKYL HALIDES 71 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website