67 CHAPTER 4 ALCOHOLS AND ALKYL HALIDES SOLUTIONS TO TEXT PROBLEMS 4.1 There are four C 4 H 9 alkyl groups, and so there are four C 4 H 9 Cl 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 CH 3 CH 2 CH 2 CH 2 Cl 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. CH 3 CHCH 2 CH 3 Cl CH 3 CHCH 2 Cl CH 3 CH 3 CH 3 CCH 3 Cl Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 68 ALCOHOLS AND ALKYL HALIDES Functional class name Substitutive name CH 3 CH 2 CH 2 CH 2 OH 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 (CH 3 OCH 3 ) 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 (CH 3 CH 2 OH), 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. ClH 3 NHH11001H11001NH 4 H11001 Cl H11002 Base Acid Conjugate acid Conjugate base H9262 H11005 e H11080 d Charge Distance Dipole moment CH 3 Cl Methyl chloride (greater value of e) H9262 1.9 D CH 3 Br Methyl bromide (greater value of d) H9262 1.8 D (CH 3 ) 2 CH C H H OH Primary alcohol (one alkyl group bonded to CH 2 OH) CH 3 CH 3 CH 3 C OH Tertiary alcohol (three alkyl groups bonded to COH) CH 3 CH 2 CH 2 C H H OH Primary alcohol (one alkyl group bonded to CH 2 OH) CH 3 C OH H CH 2 CH 3 Secondary alcohol (two alkyl groups bonded to CHOH) CH 3 CHCH 2 CH 3 OH CH 3 CHCH 2 OH CH 3 CH 3 CH 3 CCH 3 OH Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 4.7 Since the pK a of HCN is given as 9.1, its K a H11005 10 H110029.1 . In more conventional notation, K a H11005 8H11003 10 H1100210 . Hydrogen cyanide is a weak acid. 4.8 Hydrogen cyanide is a weak acid, but it is a stronger acid than water (pK a H11005 15.7). Since HCN is a stronger acid than water, its conjugate base (CN H11002 ) is a weaker base than hydroxide (HO H11002 ), 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 (ROH 2 H11001 ) have approximately the same acidity as hydronium ion (H 3 O H11001 ,pK a H11005H110021.7). Thus hydrogen chloride (pK a H33360H110027) 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 H11022 secondary H11022 primary. There is only one C 5 H 11 H11001 car- bocation that is tertiary, and so that is the most stable one. 1,1-Dimethylpropyl cation CH 3 CH 2 C H11001 CH 3 CH 3 CH 3 (CH 2 ) 12 CH 2 OH HBrH11001H11001 1-Tetradecanol CH 3 (CH 2 ) 12 CH 2 Br 1-BromotetradecaneHydrogen bromide H 2 O Water (CH 3 CH 2 ) 3 COH HClH11001H11001 3-Ethyl-3-pentanol (CH 3 CH 2 ) 3 CCl 3-Chloro-3-ethylpentaneHydrogen chloride H 2 O Water OHCl H (CH 3 ) 3 C H9254H11002H9254H11001 (CH 3 ) 3 COH HClH11001H11001Cl H11002 Stronger base (pK a H33360 H110027) Stronger acid (CH 3 ) 3 COH 2 H11001 (pK a H33360 H110021.7) Weaker acid Weaker base H11001H11001 Acid HHCl Conjugate base Cl H11002 Conjugate acid O H11001 (CH 3 ) 3 C H Base O (CH 3 ) 3 C H ALCOHOLS AND ALKYL HALIDES 69 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 S N 2 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 S N 1 mechanism. 4.15 The most stable alkyl free radicals are tertiary. The tertiary free radical having the formula C 5 H 11 has the same skeleton as the carbocation in Problem 4.13. CH 3 CH 2 C CH 3 CH 3 sec-Butyl cationBromide ion 2-Bromobutane CH 3 CH 2 CHCH 3 Br CHCH 3 CH 3 CH 2 H11001 H11001 H11002 Br CH 3 CH 2 CHCH 3 O HH H11001 H11001CH 3 CH 2 CHCH 3 O HH H11001 sec-Butyloxonium ion sec-Butyl cation Water slow CH 3 CH 2 CHCH 3 O H H11002 BrH11001H11001 2-Butanol Hydrogen bromide sec-Butyloxonium ion Bromide ion HBr CH 3 CH 2 CHCH 3 H11001 O H H H11001CH 3 CH 2 CH 2 CH 2 Br 1-Bromobutane O Water slow Br H11002 Bromide ion CH 3 CH 2 CH 2 CH 2 O H11001 Butyloxonium ion H H H H CH 3 CH 2 CH 2 CH 2 O H H11001H11001 1-Butanol HBr Hydrogen bromide Br H11002 BromideButyloxonium ion CH 3 CH 2 CH 2 CH 2 O H11001 H H 70 ALCOHOLS AND ALKYL HALIDES 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 Cl 2 in the next propagation step. H11001 H11001 Chlorine Cl Chlorine atomDichloromethane ClC H H ClClClC H H Cl Chloromethyl radical H11001 H11001Cl Chlorine atom ClH Hydrogen chlorideChloromethyl radical C H H Cl H C H H Cl Chloromethane Cl H11001 Chlorine 2 Chlorine atoms ClCl Cl CH 3 Cl Cl 2 H11001H11001 Chloromethane CH 2 Cl 2 DichloromethaneChlorine HCl Hydrogen chloride CH 3 CCH 3 CH 3 CH 3 CH 3 CH 3 C CH 3 2,2-Dimethylpropane tert-Butyl radical Methyl radical CH 3 H11001 CH 3 CH 2 CH 3 CH 2 CH 3 CH 3 H11001 Propane Ethyl radical Methyl radical CH 3 CHCH 3 CH 3 CHCH 3 CH 3 H11001 2-Methylpropane Isopropyl radical Methyl radical CH 3 ALCOHOLS AND ALKYL HALIDES 71 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 4.18 Writing the structural formula for ethyl chloride reveals that there are two nonequivalent sets of hydrogen atoms, in either of which a hydrogen is capable of being replaced by chlorine. The two dichlorides are 1,1-dichloroethane and 1,2-dichloroethane. 4.19 Propane has six primary hydrogens and two secondary. In the chlorination of propane, the relative proportions of hydrogen atom removal are given by the product of the statistical distribution and the relative rate per hydrogen. Given that a secondary hydrogen is abstracted 3.9 times faster than a pri- mary one, we write the expression for the amount of chlorination at the primary relative to that at the secondary position as: H11005H11005 Thus, the percentage of propyl chloride formed is 0.77H208621.77, or 43%, and that of isopropyl chloride is 57%. (The amounts actually observed are propyl 45%, isopropyl 55%.) 4.20 (b) In contrast with free-radical chlorination, alkane bromination is a highly selective process. The major organic product will be the alkyl bromide formed by substitution of a tertiary hydrogen with a bromine. (c) As in part (b), bromination results in substitution of a tertiary hydrogen. 4.21 (a) Cyclobutanol has a hydroxyl group attached to a four-membered ring. (b) sec-Butyl alcohol is the functional class name for 2-butanol. (c) The hydroxyl group is at C-3 of an unbranched seven-carbon chain in 3-heptanol. OH CH 3 CH 2 CHCH 2 CH 2 CH 2 CH 3 3-Heptanol OH CH 3 CHCH 2 CH 3 sec-Butyl alcohol OH Cyclobutanol CH 3 CCH 2 CHCH 3 2,2,4-Trimethylpentane 2-Bromo-2,4,4-trimethylpentane light Br 2 CH 3 CH 3 CH 3 CH 3 CCH 2 CCH 3 CH 3 CH 3 CH 3 Br CH(CH 3 ) 2 CH 3 C(CH 3 ) 2 CH 3 Br Tertiary hydrogen Br 2 light 1-(1-Bromo-1-methylethyl)- 1-methylcyclopentane 1-Isopropyl-1- methylcyclopentane 0.77 H5007 1.00 6 H11003 1 H5007 2 H11003 3.9 Number of primary hydrogens H11003 rate of abstraction of primary hydrogen H5007H5007H5007H5007H5007H5007H5007H5007 Number of secondary hydrogens H11003 rate of abstraction of a secondary hydrogen CH 3 CH 2 Cl H11001 Ethyl chloride CH 3 CHCl 2 1,1-Dichloroethane 1,2-Dichloroethane ClCH 2 CH 2 Cl light or heat Cl 2 72 ALCOHOLS AND ALKYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) A chlorine at C-2 is on the opposite side of the ring from the C-1 hydroxyl group in trans- 2-chlorocyclopentanol. Note that it is not necessary to assign a number to the carbon that bears the hydroxyl group; naming the compound as a derivative of cyclopentanol automatically requires the hydroxyl group to be located at C-1. (e) This compound is an alcohol in which the longest continuous chain that incorporates the hydroxyl function has eight carbons. It bears chlorine substituents at C-2 and C-6 and methyl and hydroxyl groups at C-4. ( f ) The hydroxyl group is at C-1 in trans-4-tert-butylcyclohexanol; the tert-butyl group is at C-4. The structures of the compound can be represented as shown at the left; the structure at the right depicts it in its most stable conformation. (g) The cyclopropyl group is on the same carbon as the hydroxyl group in 1-cyclopropylethanol. (h) The cyclopropyl group and the hydroxyl group are on adjacent carbons in 2-cyclopro- pylethanol. 4.22 (a) This compound has a five-carbon chain that bears a methyl substituent and a bromine. The numbering scheme that gives the lower number to the substituent closest to the end of the chain is chosen. Bromine is therefore at C-1, and methyl is a substituent at C-4. CH 3 CH 3 CHCH 2 CH 2 CH 2 Br 1-Bromo-4-methylpentane CH 2 CH 2 OH 2-Cyclopropylethanol CHOH CH 3 1-Cyclopropylethanol trans-4-tert-Butylcyclohexanol (CH 3 ) 3 C OH (CH 3 ) 3 C H H OH 2,6-Dichloro-4-methyl-4-octanol OH CH 3 CHCH 2 CCH 2 CHCH 2 CH 3 ClCl CH 3 OH Cl trans-2-Chlorocyclopentanol ALCOHOLS AND ALKYL HALIDES 73 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) This compound has the same carbon skeleton as the compound in part (a) but bears a hydroxyl group in place of the bromine and so is named as a derivative of 1-pentanol. (c) This molecule is a derivative of ethane and bears three chlorines and one bromine. The name 2-bromo-1,1,1-trichloroethane gives a lower number at the first point of difference than 1-bromo-2,2,2-trichloroethane. (d) This compound is a constitutional isomer of the preceding one. Regardless of which carbon the numbering begins at, the substitution pattern is 1,1,2,2. Alphabetical ranking of the halo- gens therefore dictates the direction of numbering. Begin with the carbon that bears bromine. (e) This is a trifluoro derivative of ethanol. The direction of numbering is dictated by the hydroxyl group, which is at C-1 in ethanol. ( f ) Here the compound is named as a derivative of cyclohexanol, and so numbering begins at the carbon that bears the hydroxyl group. (g) This alcohol has its hydroxyl group attached to C-2 of a three-carbon continuous chain; it is named as a derivative of 2-propanol. (h) The six carbons that form the longest continuous chain have substituents at C-2, C-3, and C-5 when numbering proceeds in the direction that gives the lowest locants to substituents at the first point of difference. The substituents are cited in alphabetical order. Had numbering begun in the opposite direction, the locants would be 2,4,5 rather than 2,3,5. 1 2 3 4 5 6 Br 5-Bromo-2,3-dimethylhexane CH 3 CH 3 OH 2-Cyclopentyl-2-propanol cis-3-tert-Butylcyclohexanol OH 2,2,2-Trifluoroethanol CF 3 CH 2 OH 1-Bromo-1,2,2-trichloroethane Cl 2 CHCHBr Cl Cl 3 CCH 2 Br 2-Bromo-1,1,1-trichloroethane CH 3 CH 3 CHCH 2 CH 2 CH 2 OH 4-Methyl-1-pentanol 74 ALCOHOLS AND ALKYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (i) Hydroxyl controls the numbering because the compound is named as an alcohol. 4.23 Primary alcohols are alcohols in which the hydroxyl group is attached to a carbon atom which has one alkyl substituent and two hydrogens. Four primary alcohols have the molecular formula C 5 H 12 O. The functional class name for each compound is given in parentheses. Secondary alcohols are alcohols in which the hydroxyl group is attached to a carbon atom which has two alkyl substituents and one hydrogen. There are three secondary alcohols of molecular formula C 5 H 12 O: Only 2-methyl-2-butanol is a tertiary alcohol (three alkyl substituents on the hydroxyl-bearing carbon): 4.24 The first methylcyclohexanol to be considered is 1-methylcyclohexanol. The preferred chair confor- mation will have the larger methyl group in an equatorial orientation, whereas the smaller hydroxyl group will be axial. OH CH 3 Most stable conformation of 1-methylcyclohexanol CH 3 OH CH 3 CCH 2 CH 3 2-Methyl-2-butanol (1,1-Dimethylpropyl alcohol) OH CH 3 CHCH 2 CH 2 CH 3 2-Pentanol (1-Methylbutyl alcohol) OH CH 3 CH 2 CHCH 2 CH 3 3-Pentanol (1-Ethylpropyl alcohol) CH 3 OH CH 3 CHCHCH 3 3-Methyl-2-butanol (1,2-Dimethylpropyl alcohol) CH 3 CH 3 CH 2 CHCH 2 OH 2-Methyl-1-butanol (2-Methylbutyl alcohol) CH 3 CH 3 CHCH 2 CH 2 OH 3-Methyl-1-butanol (3-Methylbutyl alcohol) CH 3 CH 3 CH 3 CCH 2 OH 2,2-Dimethyl-1-propanol (2,2-Dimethylpropyl alcohol) CH 3 CH 2 CH 2 CH 2 CH 2 OH 1-Pentanol (Pentyl alcohol) 4,5-Dimethyl-2-hexanol 6 5 4 3 2 1 OH ALCOHOLS AND ALKYL HALIDES 75 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website In the other isomers methyl and hydroxyl will be in a 1,2, 1,3, or 1,4 relationship and can be cis or trans in each. We can write the preferred conformation by recognizing that the methyl group will always be equatorial and the hydroxyl either equatorial or axial. 4.25 The assumption is incorrect for the 3-methylcyclohexanols. cis-3-Methylcyclohexanol is more stable than trans-3-methylcyclohexanol because the methyl group and the hydroxyl group are both equa- torial in the cis isomer, whereas one substituent must be axial in the trans. 4.26 (a) The most stable conformation will be the one with all the substituents equatorial. The hydroxyl group is trans to the isopropyl group and cis to the methyl group. (b) All three substituents need not always be equatorial; instead, one or two of them may be axial. Since neomenthol is the second most stable stereoisomer, we choose the structure with one axial substituent. Furthermore, we choose the structure with the smallest substituent (the hydroxyl group) as the axial one. Neomenthol is shown as follows: 4.27 In all these reactions the negatively charged atom abstracts a proton from an acid. (a) (b) CH 3 COHH11001CH 3 CH 2 O H11002 Ethoxide ion: base Acetic acid: acid (stronger acid, K a H11015 10 H110025 ) CH 3 CH 2 OH Ethanol: conjugate acid (weaker acid, K a H11015 10 H1100216 ) O H11001 CH 3 CO H11002 Acetate ion: conjugate base O HI H11001H11001HO H11002 H 2 O Hydrogen iodide: acid (stronger acid, K a H11015 10 10 ) Hydroxide ion: base I H11002 Iodide ion: conjugate base Water: conjugate acid (weaker acid, K a H11015 10 H1100216 ) OH CH(CH 3 ) 2 H 3 C OH CH(CH 3 ) 2 H 3 C OH CH 3 cis-3-Methylcyclohexanol more stable; smaller heat of combustion trans-3-Methylcyclohexanol less stable; larger heat of combustion OH CH 3 OH CH 3 CH 3 H 3 C cis-2-Methylcyclohexanol trans-3-Methylcyclohexanol cis-4-Methylcyclohexanol OH OH OH CH 3 OH CH 3 OH H 3 C trans-2-Methylcyclohexanol cis-3-Methylcyclohexanol trans-4-Methylcyclohexanol 76 ALCOHOLS AND ALKYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) (d) (e) ( f ) (g) 4.28 (a) The proton-transfer transition state can represent the following reaction, or its reverse: When the reaction proceeds as drawn, the stronger acid (hydrogen bromide) is on the left, the weaker acid (isopropyl alcohol) is on the right, and the equilibrium lies to the right. (b) Hydroxide is a strong base; methyloxonium ion is a strong acid. 4.29 (a) This problem reviews the relationship between logarithms and exponential numbers. We need to determine K a , given pK a . The equation that relates the two is pK a H11005 H11002log 10 K a Therefore K a H11005 10 H11002pK a H11005 10 H110023.48 H11005 3.3 H11003 10 H110024 H O H CH 3 H11001 H11001 O H CH 3 H11001 H11002 HO HOH Hydroxide ion (base) Methyloxonium ion (acid) Water (conjugate acid) Methanol (conjugate base) K H11022 1 (CH 3 ) 2 CH (CH 3 ) 2 CHO OH OHBrH H11002 H11001 Base (CH 3 ) 2 CH Conjugate acid (weaker acid) K a H11005 10 H1100217 Conjugate base Acid (stronger acid) K a H11005 10 9 H9254H11002 Br H9254H11002 Br H11002 H11001 H 2 SO 4 H11001H11001HSO 4 H11002 Fluoride ion: base Sulfuric acid: acid (stronger acid, K a H11015 10 5 ) Hydrogen fluoride: conjugate acid (weaker acid, K a H11015 10 H110024 ) Hydrogen sulfate ion: conjugate base HFF H11002 (CH 3 ) 2 CHOH H 2 NH11001H11001 H11002 Isopropyl alcohol: acid (stronger acid, K a H11015 10 H1100217 ) Amide ion: base (CH 3 ) 2 CHO H11002 Isopropoxide ion: conjugate base H 3 N Ammonia: conjugate acid (weaker acid, K a H11015 10 H1100236 ) H 2 OH11001H11001HO H11002 tert-Butoxide ion: base Water: acid (stronger acid, K a H11015 10 H1100216 ) tert-Butyl alcohol: conjugate acid (weaker acid, K a H11015 10 H1100218 ) Hydroxide ion: conjugate base (CH 3 ) 3 CO (CH 3 ) 3 COH H11002 HClH11001H11001Cl H11002 Acetate ion: base Hydrogen chloride: acid (stronger acid, K a H11015 10 7 ) Acetic acid: conjugate acid (weaker acid, K a H11015 10 H110025 ) Chloride ion: conjugate base CH 3 COH O CH 3 CO H11002 O H 2 NH11001H11001 H11002 HF Hydrogen fluoride: acid (stronger acid, K a H11015 10 H110024 ) Amide ion: base F H11002 Fluoride ion: conjugate base H 3 N Ammonia: conjugate acid (weaker acid, K a H11015 10 H1100236 ) ALCOHOLS AND ALKYL HALIDES 77 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) As described in part (a), K a H11005 10 H11002pK a , therefore K a for vitamin C is given by the expression: K a H11005 10 H110024.17 H11005 6.7 H11003 10 H110025 (c) Similarly, K a H11005 1.8 H11003 10 H110024 for formic acid (pK a 3.75). (d) K a H11005 6.5 H11003 10 H110022 for oxalic acid (pK a 1.19). In ranking the acids in order of decreasing acidity, remember that the larger the equilibrium constant K a , the stronger the acid; and the lower the pK a value, the stronger the acid. Acid K a pK a Oxalic (strongest) 6.5 H11003 10 H110022 1.19 Aspirin 3.3 H11003 10 H110024 3.48 Formic acid 1.8 H11003 10 H110024 3.75 Vitamin C (weakest) 6.7 H11003 10 H110025 4.17 4.30 Because the pK a of CH 3 SH (11) is smaller than that of CH 3 OH (16), CH 3 SH is the stronger acid of the two. Its conjugate base (as in KSCH 3 ) is therefore weaker than the conjugate base of CH 3 OH (as in KOCH 3 ). 4.31 This problem illustrates the reactions of a primary alcohol with the reagents described in the chapter. (a) (b) (c) (d) (e) 4.32 (a) This reaction was used to convert the primary alcohol to the corresponding bromide in 60% yield. (b) Thionyl chloride treatment of this secondary alcohol gave the chloro derivative in 59% yield. COCH 2 CH 3 CH 3 O OH COCH 2 CH 3 CH 3 O Cl SOCl 2 pyridine CH 2 CH 2 Br PBr 3 pyridine CH 2 CH 2 OH CH 3 CH 2 CH 2 CH 2 OH CH 3 CH 2 CH 2 CH 2 Cl 1-Chlorobutane SOCl 2 CH 3 CH 2 CH 2 CH 2 OH CH 3 CH 2 CH 2 CH 2 Br 1-Bromobutane PBr 3 CH 3 CH 2 CH 2 CH 2 OH CH 3 CH 2 CH 2 CH 2 Br 1-Bromobutane NaBr, H 2 SO 4 heat CH 3 CH 2 CH 2 CH 2 OH CH 3 CH 2 CH 2 CH 2 Br 1-Bromobutane HBr heat CH 3 CH 2 CH 2 CH 2 OH H11001 NaNH 2 CH 3 CH 2 CH 2 CH 2 O H11002 Na H11001 H11001 NH 3 Sodium butoxide 78 ALCOHOLS AND ALKYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) The starting material is a tertiary alcohol and reacted readily with hydrogen chloride to form the corresponding chloride in 67% yield. (d) Both primary alcohol functional groups were converted to primary bromides; the yield was 88%. (e) This molecule is called adamantane. It has six equivalent CH 2 groups and four equivalent CH groups. Bromination is selective for tertiary hydrogens, so a hydrogen of one of the CH groups is replaced. The product shown was isolated in 76% yield. 4.33 The order of reactivity of alcohols with hydrogen halides is tertiary H11022 secondary H11022 primary H11022 methyl. Reactivity of Alcohols with Hydrogen Bromide: Part More reactive Less reactive (a) (b) (c) 2-Butanol: secondary CH 3 CHCH 2 CH 3 OH 2-Methyl-2-butanol: tertiary (CH 3 ) 2 CCH 2 CH 3 OH 2-Methyl-1-butanol: primary CH 3 CH 3 CH 2 CHCH 2 OH 2-Butanol: secondary CH 3 CHCH 2 CH 3 OH 1-Butanol: primary CH 3 CH 2 CH 2 CH 2 OH 2-Butanol: secondary CH 3 CHCH 2 CH 3 OH HBr RBr H 2 OROH H11001H11001 Br C 10 H 15 Br Br 2 , light 100H11034C HBr heat HOCH 2 CH 2 CH 2 CH 2 OH BrCH 2 CH 2 CH 2 CH 2 Br COH CH 3 CH 3 Br CCl CH 3 CH 3 Br HCl ALCOHOLS AND ALKYL HALIDES 79 (continued) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Part More reactive Less reactive (d) (e) ( f ) (g) 4.34 The unimolecular step in the reaction of cyclohexanol with hydrogen bromide to give cyclohexyl bromide is the dissociation of the oxonium ion to a carbocation. 4.35 The nucleophile that attacks the oxonium ion in the reaction of 1-hexanol with hydrogen bromide is bromide ion. 4.36 (a) Both the methyl group and the hydroxyl group are equatorial in the most stable conformation of trans-4-methylcyclohexanol. H 3 C OH trans-4-Methylcyclohexanol CH 3 (CH 2 ) 4 CH 2 O H HHO H H11001 H11001 H11001 H11002 Br Bromide ion Hexyloxonium ion CH 3 (CH 2 ) 4 CH 2 Br 1-Bromohexane Water O H H H HHHO H11001 H11001 H11001 Cyclohexyloxonium ion Cyclohexyl cation Water CHCH 3 OH 1-Cyclopentylethanol: secondary CH 3 CH 2 OH 1-Ethylcyclopentanol: tertiary trans-2-Methylcyclopentanol: secondary OH CH 3 H H CH 3 OH 1-Methylcyclopentanol: tertiary Cyclohexanol: secondary OHHCH 3 OH 1-Methylcyclopentanol: tertiary 2-Methylbutane: not an alcohol; does not react with HBr CH 3 CHCH 2 CH 3 CH 3 2-Butanol CH 3 CHCH 2 CH 3 OH 80 ALCOHOLS AND ALKYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) The positively charged carbon in the carbocation intermediate is sp 2 -hybridized, and is planar. (c) Bromide ion attacks the carbocation from both above and below, giving rise to two stereo- isomers, cis- and trans-1-bromo-4-methylcyclohexane. 4.37 Examine the equations to ascertain which bonds are made and which are broken. Then use the bond dissociation energies in Table 4.3 to calculate H9004H° for each reaction. (a) H9004H° H11005 energy cost of breaking bonds H11002 energy given off in making bonds H11005 953 kJ/mol H11002 936 kJ/mol (228 kcal/mol H11002 224 kcal/mol) H11005 H1100117 kJ/mol (H110014 kcal/mol) The reaction of isopropyl alcohol with hydrogen fluoride is endothermic. (b) H9004H° H11005 energy cost of breaking bonds H11002 energy given off in making bonds H11005 816 kJ/mol H11002 836 kJ/mol (195 kcal/mol H11002 200 kcal/mol) H11005 H1100220 kJ/mol (H110025 kcal/mol) The reaction of isopropyl alcohol with hydrogen chloride is exothermic. (c) H9004H° H11005 energy cost of breaking bonds H11002 energy given off in making bonds H11005 828 kJ/mol H11002 774 kJ/mol (198 kcal/mol H11002 185 kcal/mol) H11005 H1100154 kJ/mol (H1100113 kcal/mol) The reaction of propane with hydrogen chloride is endothermic. 4.38 In the statement of the problem you are told that the starting material is 2,2-dimethylpropane, that the reaction is one of fluorination, meaning that F 2 is a reactant, and that the product is (CF 3 ) 4 C. You CH 3 CHCH 3 CH 3 CHCH 3 HCl 397 kJ/mol (95 kcal/mol) H11001 HCl 431 kJ/mol (103 kcal/mol) H11001 HH 435 kJ/mol (104 kcal/mol) 339 kJ/mol (81 kcal/mol) Bond breaking: 828 kJ/mol (198 kcal/mol) Bond making: 774 kJ/mol (185 kcal/mol) HOH(CH 3 ) 2 CH Cl 431 kJ/mol (103 kcal/mol) 385 kJ/mol (92 kcal/mol) H11001 HCl(CH 3 ) 2 CH OH 497 kJ/mol (119 kcal/mol) 339 kJ/mol (81 kcal/mol) H11001 Bond breaking: 816 kJ/mol (195 kcal/mol) Bond making: 836 kJ/mol (200 kcal/mol) HOH(CH 3 ) 2 CH F 568 kJ/mol (136 kcal/mol) 385 kJ/mol (92 kcal/mol) H11001 HF(CH 3 ) 2 CH OH 497 kJ/mol (119 kcal/mol) 439 kJ/mol (105 kcal/mol) H11001 Bond breaking: 953 kJ/mol (228 kcal/mol) Bond making: 936 kJ/mol (224 kcal/mol) H 3 C Br cis-1-Bromo-4-methylcyclohexane trans-1-Bromo-4-methylcyclohexane H 3 C Br Carbocation intermediate H 3 C H H11001 ALCOHOLS AND ALKYL HALIDES 81 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website need to complete the equation by realizing that HF is also formed in the fluorination of alkanes. The balanced equation is therefore: 4.39 The reaction is free-radical chlorination, and substitution occurs at all possible positions that bear a replaceable hydrogen. Write the structure of the starting material, and identify the nonequivalent hydrogens. The problem states that one of the products is 1,2,3-trichloro-1,1-difluoropropane. This compound arises by substitution of one of the methyl hydrogens by chlorine. We are told that the other product is an isomer of 1,2,3-trichloro-1,1-difluoropropane; therefore, it must be formed by replacement of the hydrogen at C-2. 4.40 Free-radical chlorination leads to substitution at each carbon that bears a hydrogen. This problem es- sentially requires you to recognize structures that possess various numbers of nonequivalent hydro- gens. The easiest way to determine the number of constitutional isomers that can be formed by chlo- rination of a particular compound is to replace one hydrogen with chlorine and assign an IUPAC name to the product. Continue by replacing one hydrogen on each carbon in the compound, and compare names to identify duplicates. (a) 2,2-Dimethylpropane is the C 5 H 12 isomer that gives a single monochloride, since all the hydro- gens are equivalent. (b) The C 5 H 12 isomer that has three nonequivalent sets of hydrogens is pentane. It yields three isomeric monochlorides on free-radical chlorination. CH 3 CH 2 CH 2 CH 2 CH 3 Cl 2 ClCH 2 CH 2 CH 2 CH 2 CH 3 1-Chloropentane Pentane CH 3 CHCH 2 CH 2 CH 3 Cl 2-Chloropentane CH 3 CH 2 CHCH 2 CH 3 Cl 3-Chloropentane CH 3 CH 3 CCH 3 CH 3 2,2-Dimethylpropane CH 3 CH 3 CCH 2 Cl CH 3 1-Chloro-2,2-dimethylpropane Cl 2 light Cl F C F Cl C H CH 2 Cl 1,2,3-Trichloro-1,1-difluoropropane Cl F C F Cl C Cl CH 3 1,2,2-Trichloro-1,1-difluoropropane Cl F C F Cl C H CH 3 1,2-Dichloro-1,1-difluoropropane (CH 3 ) 4 C (CF 3 ) 4 CH11001 12F 2 H11001 12HF 82 ALCOHOLS AND ALKYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) 2-Methylbutane forms four different monochlorides. (d) For only two dichlorides to be formed, the starting alkane must have a structure that is rather symmetrical; that is, one in which most (or all) of the hydrogens are equivalent. 2,2-Di- methylpropane satisfies this requirement. 4.41 (a) Heptane has five methylene groups, which on chlorination together contribute 85% of the total monochlorinated product. Since the problem specifies that attack at each methylene group is equally probable, the five methylene groups each give rise to 85H208625, or 17%, of the monochloride product. Since C-2 and C-6 of heptane are equivalent, we calculate that 2-chloroheptane will con- stitute 34% of the monochloride fraction. Similarly, C-3 and C-5 are equivalent, and so there should be 34% 3-chloroheptane. The remainder, 17%, is 4-chloroheptane. These predictions are very close to the observed proportions. Calculated, % Observed, % 2-Chloro 34 35 3-Chloro 34 34 4-Chloro 17 16 (b) There are a total of 20 methylene hydrogens in dodecane, CH 3 (CH 2 ) 10 CH 3 . The 19% 2-chlorododecane that is formed arises by substitution of any of the four equivalent methyl- ene hydrogens at C-2 and C-11. The total amount of substitution of methylene hydrogens must therefore be: H11003 19% H11005 95% 20 H5007 4 CH 3 (CH 2 ) 5 CH 3 CH 3 (CH 2 ) 5 CH 2 Cl (2-chloro H11001 3-chloro H11001 4-chloro)H11001 15% 85% CH 3 CH 3 CCH 3 CH 3 2,2-Dimethylpropane CH 3 CH 3 CCHCl 2 CH 3 1,1-Dichloro-2,2- dimethylpropane CH 3 ClCH 2 CCH 2 Cl CH 3 1,3-Dichloro-2,2- dimethylpropane H11001 2Cl 2 light Cl 2 1-Chloro-2-methylbutane ClCH 2 CHCH 2 CH 3 CH 3 CH 3 CHCH 2 CH 3 CH 3 (CH 3 ) 2 CCH 2 CH 3 Cl 2-Chloro-2-methylbutane (CH 3 ) 2 CHCHCH 3 Cl 2-Chloro-3-methylbutane (CH 3 ) 2 CHCH 2 CH 2 Cl 1-Chloro-3-methylbutane 2-Methylbutane ALCOHOLS AND ALKYL HALIDES 83 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The remaining 5% corresponds to substitution of methyl hydrogens at C-1 and C-12. The proportion of 1-chlorododecane in the monochloride fraction is 5%. 4.42 (a) Two of the monochlorides derived from chlorination of 2,2,4-trimethylpentane are primary chlorides: The two remaining isomers are a secondary chloride and a tertiary chloride: (b) Substitution of any one of the nine hydrogens designated as x in the structural diagram yields 1-chloro-2,2,4-trimethylpentane. Substitution of any one of the six hydrogens designated as y gives 1-chloro-2,4,4-trimethylpentane. Assuming equal reactivity of a single x hydrogen and a single y hydrogen, the ratio of the two isomers is then expected to be 9:6. Since together the two primary chlorides total 65% of the monochloride fraction, there will be 39% 1-chloro-2,2,4-trimethylpentane (substitution of x) and 26% 1-chloro-2,4,4-trimethylpentane (substitution of y). 4.43 The three monochlorides are shown in the equation Pentane has six primary hydrogens (two CH 3 groups) and six secondary hydrogens (three CH 2 groups). Since a single secondary hydrogen is abstracted three times faster than a single primary hydrogen and there are equal numbers of secondary and primary hydrogens, the product mixture should contain three times as much of the secondary chloride isomers as the primary chloride. The primary chloride 1-chloropentane, therefore, is expected to constitute 25% of the product mixture. The secondary chlorides 2-chloropentane and 3-chloropentane are not formed in equal amounts. Rather, 2-chloropentane may be formed by replacement of a hydrogen at C-2 or at C-4, whereas 3-chloropentane is formed only when a C-3 hydrogen is replaced. The amount of 2-chloropentane is therefore 50%, and that of 3-chloropentane is 25%. We predict the major product to be 2-chloropen- tane, and the predicted proportion of 50% corresponds closely to the observed 46%. CH 3 CH 2 CH 2 CH 2 CH 3 Pentane CH 3 CH 2 CH 2 CH 2 CH 2 Cl H11001H11001CH 3 CHCH 2 CH 2 CH 3 Cl CH 3 CH 2 CHCH 2 CH 3 Cl 2-Chloropentane 3-Chloropentane1-Chloropentane Cl 2 light CH 3 CCH 2 CHCH 3 CH 3 CH 3 CH 3 x x x y y 3-Chloro-2,2,4-trimethylpentane CH 3 CCH 2 CCH 3 CH 3 CH 3 Cl CH 3 2-Chloro-2,4,4-trimethylpentane CH 3 C CHCHCH 3 CH 3 CH 3 Cl CH 3 ClCH 2 CCH 2 CHCH 3 CH 3 CH 3 CH 3 1-Chloro-2,2,4-trimethylpentane CH 3 CCH 2 CHCH 2 Cl CH 3 CH 3 CH 3 1-Chloro-2,4,4-trimethylpentane 84 ALCOHOLS AND ALKYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 4.44 The equation for the reaction is The reaction begins with the initiation step in which a chlorine molecule dissociates to two chlorine atoms. A chlorine atom abstracts a hydrogen atom from cyclopropane in the first propagation step. Cyclopropyl radical reacts with Cl 2 in the next propagation step. 4.45 (a) Acid-catalyzed hydrogen–deuterium exchange takes place by a pair of Br?nsted acid–base reactions. (b) Base-catalyzed hydrogen–deuterium exchange occurs by a different pair of Br?nsted acid–base equilibria. H11001H11001 AcidBase DODR O H11002 Conjugate base DO H11002 Conjugate acid RDO H11001 DOH Conjugate acid RHO H11001 Acid Base OD H11002 Conjugate base R O H11002 RHO H11001 RDDO DOH11001 H11001H11001 HD Acid Conjugate base Conjugate acid OD 2 Base OD 2 RHDO H11001 RHD 2 O D O H11001 H11001 H11001 Base Acid Conjugate acid Conjugate base H11001H Cyclopropyl radical Chlorine atom Cl Chlorine ClClH11001 Cyclopropyl chloride H Cl H11001 H11001 H H Cyclopropane Cl Chlorine atom H Cyclopropyl radical ClH Hydrogen chloride Chlorine ClCl 2 Chlorine atoms H11001Cl Cl H H Cyclopropane H Cl Cyclopropyl chloride H11001 Chlorine Cl 2 H11001 Hydrogen chloride HCl ALCOHOLS AND ALKYL HALIDES 85 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website SELF-TEST PART A A-1. Give the correct substitutive IUPAC name for each of the following compounds: A-2. Draw the structures of the following substances: (a) 2-Chloro-1-iodo-2-methylheptane (b) cis-3-Isopropylcyclohexanol A-3. Give both a functional class and a substitutive IUPAC name for each of the following compounds: A-4. What are the structures of the conjugate acid and the conjugate base of CH 3 OH? A-5. Supply the missing component for each of the following reactions: A-6. (a) Write the products of the acid–base reaction that follows, and identify the stronger acid and base and the conjugate of each. Will the equilibrium lie to the left (K H11021 1) or to the right (K H11022 1)? The approximate pK a of NH 3 is 36; that of CH 3 CH 2 OH is 16. (b) Draw a representation of the transition state of the elementary step of the reaction in part (a). A-7. (a) How many different free radicals can possibly be produced in the reaction between chlorine atoms and 2,4-dimethylpentane? (b) Write their structures. (c) Which is the most stable? Which is the least stable? A-8. Write a balanced chemical equation for the reaction of chlorine with the pentane isomer that gives only one product on monochlorination. A-9. Write the propagation steps for the light-initiated reaction of bromine with methylcyclo- hexane. A-10. Using the data in Table B-1 of this Study Guide, calculate the heat of reaction (H9004H°) for the light-initiated reaction of bromine (Br 2 ) with 2-methylpropane to give 2-bromo-2- methylpropane and hydrogen bromide. A-11. (a) Write out each of the elementary steps in the reaction of tert-butyl alcohol with hydrogen bromide. Use curved arrows to show electron movement in each step. (b) Draw the structure of the transition state representing the unimolecular dissociation of the alkyloxonium ion in the preceding reaction. CH 3 CH 2 O H11002 NH 3 H11001 (a)CH 3 CH 2 CH 2 OH ? SOCl 2 (b)? CH 3 CH 2 C(CH 3 ) 2 HBr Br OH Cl (a) (b) CH 3 Br (a) (b) CH 3 CH 2 CHCH 2 CHCH 3 CH 2 OH CH 2 CH 3 86 ALCOHOLS AND ALKYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) How does the mechanism of the reaction between 1-butanol and hydrogen bromide differ from the reaction in part (a)? A-12. (Choose the correct response for each part.) Which species or compound: (a) Reacts faster with sodium bromide and sulfuric acid? 2-methyl-3-pentanol or 3-methyl-3-pentanol (b) Is a stronger base? KOC(CH 3 ) 3 or HOC(CH 3 ) 3 (c) Reacts more vigorously with cyclohexane? Fluorine or iodine (d) Has an odd number of electrons? Ethoxide ion or ethyl radical (e) Undergoes bond cleavage in the initiation step in the reaction by which methane is converted to chloromethane? CH 4 or Cl 2 PART B B-1. A certain alcohol has the functional class IUPAC name 1-ethyl-3-methylbutyl alcohol. What is its substitutive name? (a) 1-Ethyl-3-methyl-1-butanol (d) 2-Methyl-4-hexanol (b) 2-Methyl-1-hexanol (e) 5-Methyl-3-hexanol (c) 3-Methyl-1-hexanol B-2. Rank the following substances in order of increasing boiling point (lowest → highest): CH 3 CH 2 CH 2 CH 2 OH (CH 3 ) 2 CHOCH 3 (CH 3 ) 3 COH (CH 3 ) 4 C 12 (a)1 H11021 3 H11021 2 H11021 4(c)4 H11021 2 H11021 3 H11021 1(e)4 H11021 3 H11021 2 H11021 1 (b)2 H11021 4 H11021 3 H11021 1(d)2 H11021 3 H11021 1 H11021 4 B-3. Which one of the following reacts with HBr at the fastest rate? B-4. What is the decreasing stability order (most stable → least stable) of the following carbo- cations? (a)3 H11022 2 H11022 1 H11022 4 H11022 5(c)3 H11022 2 H11015 5 H11022 1 H11015 4 (b)1H11015 4 H11022 2 H11015 5 H11022 3(d)3 H11022 1 H11015 4 H11022 2 H11015 5 H11001 H11001 H11001 H11001 H11001 123 4 5 OH (a) OH CH 3 (b) CH 3 OH (c) H 3 C OH (d) H 3 C OH (e) ALCOHOLS AND ALKYL HALIDES 87 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-5. Rank the bond dissociation energies (BDEs) of the bonds indicated with the arrows from smallest to largest. (a)1 H11021 2 H11021 3(d)1 H11021 3 H11021 2 (b)3 H11021 2 H11021 1(e)3 H11021 1 H11021 2 (c)2 H11021 3 H11021 1 B-6. What are the chain-propagating steps in the free-radical chlorination of methane? (a)2,4 (b)1,2 (c) 3, 5 (d) 1,3,5 (e) A combination different from those listed B-7. Which of the following is least able to serve as a nucleophile in a chemical reaction? (a)Br H11002 (b)OH H11002 (c)NH 3 (d)CH 3 H11001 B-8. Thiols are alcohol analogs in which the oxygen has been replaced by sulfur (e.g., CH 3 SH). Given the fact that the S@H bond is less polar than the O@H bond, which of the following statements comparing thiols and alcohols is correct? (a) Hydrogen bonding forces are weaker in thiols. (b) Hydrogen bonding forces are stronger in thiols. (c) Hydrogen bonding forces would be the same. (d) No comparison can be made without additional information. B-9. Rank the transition states that occur during the following reaction steps in order of increas- ing stability (least → most stable): (a)1 H11021 2 H11021 3(b)2 H11021 3 H11021 1(c)1 H11021 3 H11021 2(d)2 H11021 1 H11021 3 B-10. Using the data from Appendix B (Table B-1), calculate the heat of reaction H9004Ho for the following: (a) H1100169 kJ/mol (H1100116.5 kcal/mol) (b) H1100269 kJ/mol (H1100216.5 kcal/mol) (c) H1100144 kJ/mol (H1100110.5 kcal/mol) (d) H1100244 kJ/mol (H1100210.5 kcal/mol) B-11. An alkane with a molecular formula C 6 H 14 reacts with chlorine in the presence of light and heat to give four constitutionally isomeric monochlorides of molecular formula C 6 H 13 Cl. What is the most reasonable structure for the starting alkane? (a)CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 (d) (CH 3 ) 3 CCH 2 CH 3 (b) (CH 3 ) 2 CHCH 2 CH 2 CH 3 (e) (CH 3 ) 2 CHCH(CH 3 ) 2 (c)CH 3 CH(CH 2 CH 3 ) 2 B-12. The species shown in the box represents of the reaction between isopropyl alcohol and hydrogen bromide. (a) the alkyloxonium ion intermediate (b) the transition state of the bimolecular proton transfer step (CH 3 ) 2 CH H9254H11001 OH 2 H9254H11001 CH 3 CH 3 BrH11001H11001CH 3 CH 2 HBr CH 3 H11001 H 2 OH11001 H 2 OH110012. (CH 3 ) 3 C (CH 3 ) 3 C H11001 H110013. (CH 3 ) 2 CH (CH 3 ) 2 CH H11001 H 2 O 1. CH 3 OH 2 H11001 OH 2 H11001 OH 2 H11001 1. Cl 2 2Cl H11001H110012. Cl CH 4 CH 3 Cl H H11001H110013. Cl CH 4 CH 3 HCl 4. H H11001 Cl 2 HCl H11001 Cl 5. H11001CH 3 Cl 2 H11001CH 3 Cl Cl H11001CH 3 CH 4 6. H11001CH 4 CH 3 H CH 2 H 1 2 3 H 88 ALCOHOLS AND ALKYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) the transition state of the capture of the carbocation by a nucleophile (d) the carbocation intermediate (e) the transition state of the unimolecular dissociation step For the remaining four questions, consider the following free-radical reaction: B-13. Light is involved in which of the following reaction steps? (a) Initiation only (b) Propagation only (c) Termination only (d) Initiation and propagation B-14. Which of the following statements about the reaction is not true? (a) Halogen atoms are consumed in the first propagation step. (b) Halogen atoms are regenerated in the second propagation step. (c) Hydrogen atoms are produced in the first propagation step. (d) Chain termination occurs when two radicals react with each other. B-15. How many monohalogenation products are possible. (Do not consider stereoisomers.) (a)2 (b)3 (c)4 (d)5 B-16. Which halogen (X 2 ) will give the best yield of a single monohalogenation product? (a)F 2 (b)Cl 2 (c)Br 2 (d)I 2 H11001 X 2 monohalogenation product light ALCOHOLS AND ALKYL HALIDES 89 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website