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Ch8 problem solutions

Ch8 problem solutions - Solutions to Problems I 137 I...

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Unformatted text preview: Solutions to Problems I 137 I ' lrmfi that follow. The practice will not only help you develop and improve your ability to analyze synthesis ' problems, it will also help you become more and more familiar with all the reactions and reagents. You will have to know them in the end. SUMMARY CHART Synthesis of alcohols from carbonyl compounds _______.___________——-— Alcohol product from reaction with Carbonyl compounds NaBHd or LiAlI-I4 R”Li 01' R"MgX _______,____.___._____.._———— Formaldehyde methanol 1° ‘alcohol (HCI-IO) (CH3OH) t (K'CHzOH) Aldehyde 1° alcohol 2° alcohol (RCHO) (RCH20H) R\ /CHOH E 3° alcohol Ketone 2° alcohol I? R R \ lg ( CO) ( \CHOH) R (EOH l' / f / I! n & ____________—__.._._..——-—-—-———-——- Solutions to Problems 21. (a) 2~Butanol, 2° (1)) S-Bromo-3—hexanol, 2° (c) 2-Propyl—l-pentanol, 1° ((1) (S)-l—Chloro-2—'propanol, 2° (e) l—Ethylcyclobutanol, 3° (1‘) (1R,2R)—2—Bromocyclodecanol, 2° (g) 2,2—Bis(hydroxymethyl)—1,3-propanediol, 1° [“Bis” is used as the prefix instead of “di” when the name that follows is complicated enough to be in parentheses] (h) meso-l,2,3,4—Butanetetraol, 1° on C1 and C4, 2° on C2 and C3 (i) (1R,2R)-2-(2—Hydroxyethyl)cyclopcntanol, 2° on ring, 1° on side chain (j) (R)—2—Chloro-2-methyl—l-butanol, 1° CH3 22. (a) (enggocnzcnzon (b) D< OH (Izmcngz CHZCH2CH3 (c) CHgCHOHCHCH2CH2CH3 (d) H+OH CH3 (e) OH Br Br 23. (a) Cyclohexanol > chlorocyclohexane > cyclohexane (polarity) (b) 2-Heptanol > 2-methy1-2-hexanol > 2,3-dimethyl-2—pentanol (branching) I38 I Chapter 8 HYDROXY FUNCTIONAL GROUP: ALCOHOLS: PROPERTIES, PREPARATION, AND STRATEGY OF SYNTHESIS 24. (a) Ethanol hydrogen bonds to water. Chloroethane is attracted to water by dipole forces. Ethane is nonpolar and attracted least to water. (b) Solubility decreases as the relative size of the nonpolar portion of a molecule increases. 25. lntramolecular hydrogen bonding stabilizes the gauche conformation of 1,2-ethauediol, _..H shown at right, but not the anti form. In 2—chloroethanol, similar hydrogen bonding can H H0 6 occur (although it is weaker because of poorer overlap between the large Cl 3p lone pair orbital with the small hydrogen): HQH H .H C1" | H 0 So the conformational ratio of 2-chloroethanol should be more like that of 1,2-ethanediol than 1,2-dichloroethane, in which hydrogen bonding is absent. H H 26. (a) The diequatorial conformation of the diol is shown below, left. It is stabilized in two ways, relative to the dismal form. First, the —OH groups are larger than —H, and therefore sterics favors the diequatorial. Second, the two hydroxy groups are close enough in the trans—1,2—diequatorial con- formation to form an internal hydrogen bond (below, right). The energy of this conformation is further lowered by the strength of this hydrogen bond, about 2 kcal moi—1. H H /,H OH 0 .1 OH QT‘H H ' H trans-1,2-Cyclohexanediol (both -OH groups equatorial) (b) As the models of the diol in part (a) show, adjacent diequatorial substituents are in rather close proximity. A Newman projection View further illustrates this point, revealing that the groups are in a gauche relationship. Replacement of the two hydroxy hydrogens with very bulky silyl groups makes the diequatorial form less stable than the diaxial. because the 1,2—silyl—silyl steric interfer- ence is greater than the alternative. two pairs of 1,3—silyl-hydrogen diaxial interactions. In addition, with the hydrogens gone from the oxygen atoms no hydrogen bonding is possible to help lower the energy of the diequatorial form. Compare the structures: H . H Wosrcmcmms osacmcng) 1 _ _ OSi[CH(CI—2133)2]3 — H OSIICH(CH3)2]3 H Diequatorial H 081[CH(CH3)3]3 H H OSi[CH(CH3)2]3 (I)Si[CH(CH3)2]3 H H OSi[CH(CH3)2]3 H H Diaxial I I l. Solutions to Problems 0 139 27. Three factors are involved: the electronegativity of the electron—withdrawing atoms, how many there are, and their distance from the hydroxy group. (a) CH3CHC1CH20H > CH3CHBrCH20H > BrCHZCHzCI-IZOH (b) CCISCHZOH > CH3CC12CH20H > (CH3)2CC1CH20H (c) (CF3)2CHOH > (CC13)2CHOH > (CH3)2CHOH dd H+ A d, (”‘10— .(a) (CH3J2CHOH2 (AWL (CH3)ZCHOH MM (CH3)2CHO' 2—Propanol 13 both a weaker acid and a weaker base than methanol (Tables 8-2 and 8-3). + H+ H0 A (b) CH3CHFCH20H2 <—~+~m CH3CHFCH20H & CH3CHFCH20_ + H+ H0’,(_H+) _ (c) CC13CH20H2 <— CC13CH20H —> CC13CH20 The last two alcohols are both stronger acids and weaker bases than methanol. In each, the alkoxide is stabilized and the alkyloxonium ion destabilized by the electronegative halogens. 29. (a) Halfway between the two pKa values: pH 6.7. (Compare H20: At pH 7, equal amounts of H30l" and H0' are present.) (b) pH -2.2 (c) pH +155 30. No. Recall that stabilization of carbocations by hyperconjugation involves 1T overlap between a bonding hybrid orbital and an empty p orbital on carbon (Section 7-5). There aren’t any empty p orbitals on oxygen in alkyloxonium ions; therefore such overlap is not possible. 3'! . (a) Worthless. H20 is a very poor nucleophile in 8N2 reactions. (b) Good. Excellent 3N2 reaction—CH3 attached to sulfonate leaving group. (c) Not so good. Bases give much elimination with 2° haloalkanes. ((1) Good, but slow, via an 8N1 mechanism. (e) Worthless. ‘CN is a bad leaving group. (1') Worthless. ‘OCH3 is a bad leaving group. (g) Good. SNl first step. (h) Not so good. Branching reduces 3N2 reactivity, and E2 occurs. 32. (a) CH3CH2CHOHCH3 (b) CH3CHOHCH2CH2CHOHCH3 H (c) OLCHon (d) O OH CH3 (e) (CH3)2CH From addition of hydride to less sterically hindered (bottom) face of ring 140 ' Chapter 8 HYDROXY FUNCTIONAL GROUP: ALCOHOLS: PROPERTIES, PREPARATION, AND STRATEGY OF SYNTHESIS (f) Make a model. Here the main steric interference with hydride addition is by the axial hydrogens on the “top” side of the ring in the chair conformation: O \I: H H H Less hindered Predominant alcohol diastereomer 33. To the right. H2 is a weaker acid than H20, and H07 is a weaker base than H‘. 34. (a) CH3CHDOH (b) CH3CH20D, from reaction of CH3CH20_ with D+ (c) CH3CH2D, from 8N2 reaction. (LiAlD4 serves as a source of “deuteride” nucleophile, D7, just as LiAlH4 is a source of hydride nucleophile, HT.) 1\i/IgCl 35. (a) CH3(CH2)5CHCH3 (b) CH3(CH2)5CHDCH3 Li H0 on (e) CH3CH2CH2MgCl (r) chcmcmcm CH3 OH OH (a) U (h) CH3CCH2CH2CCH3 do 36. The desired reaction is CH3MgI + C6H5CHO —’ CH3CH(OH)C6H5, a synthesis of a secondary alcohol by addition of a Grignard reagent to an aldehyde. The unexpected and undesired side reaction is the addition of methylmagnesium iodide to acetone: CH3MgI + CH3C0CH3 —> (CH3)3COH, forming 2-methyl-2—pr0panol (tert~butyl alcohol). 37. Only the compounds pictured in parts (a) and (c) will form Grignard reagents in a problem-free manner. In (13) the —OH group contains an acidic hydrogen that will destroy the carbon—metal bond of any Grignard reagent that forms; similarly, the tenninal hydrogen on the alkyne function in (e) is acidic enough to do the same (the pKEL is about 25, although all you need to know—from the information in Problem 42 of Chapter l—is that such hydrogens are much more acidic than hydrogens on alkane carbons). Finally, in (d) the carbonyl function contains a strongly electrophilic carbon and will interfere. The ether function in (c) is not a problem, because it does not contain any acidic hydrogen atoms. Solufions #0 Problems I 141 38. Products after hydrolysis are given. (a) D‘CHZOH (b) (CH3)2CHCH2CHOHCH3 OH (c) C5H5CH2CHOHC6H5 (d) O CH(CH3)2 (e) O'CHOHCH(CH2CH3)2 39. (a) The C-wMg bond is source of the nucleophilic electron pair 0)) (CH3)2CHC}rz—MgCI 8 H\ KN CH3/ 5+ .. (C) Céflscqzflfl 8 H\ C6H5/ 8+ . . (d) (CH3)2CH '—MgCl 5.. :6) _, HE??? —* Val:— - H (3:03 —> (CH3)2CHCH2—CH-“Q :- +MgCl —> Q: +MgBr fl, D—CHzfioH + H0‘ +MgBr mp OH \ CH3 (CH3)2CHCH2—CHfi:C:)H + H0“ +Mgc1 l CH3 Ar“ H—OH 0:0: —> C6H5CH2— CPI—o:— +Li —> CHHS CEHSCHZWCH—éfl + H0‘ +Li I C6H5 (CH3)ZCH _o: “ +MgBr_> H 0“ i: (CH3)2CH 6H 6 + H0 +MgBr 142 0 Chapter 8 HYDROXY FUNCTIONAL GROUP: ALCOHOLS: PROPERTIES, PREPARATION, AND STRATEGY OF SYNTHESIS (a) H 6: 5-H/’—\\(% . ©<H/H /+MC\1HQ)H _, . . g _ _, MgCl W In (b) and (c) the hydroxy groups are located on newly formed stereocenters. The product in (b) forms as a racemic mixture of enantiomers. The starting material in (0) already possessed one stereocenter; addition of a second means that a total of four stereoisomers (two diastereomeric pairs of enantiomers) are formed. OH OH OH OH ((1) W (e) W (D Ab (g) W The products in (e) and (g) will both be racemic mixtures of enantiomers. HO - CH CH XCHgCHg 3 ,5H 41. (a) CH3CH2C02H. The carboxylic acid is the major product (from overoxidation) when primary alcohols are treated with NaZCrgoq in aqueous H+. H (b) (CH3)2CHCHO (e) O C02H H «00 Go CHO 42. No need for you to write anything but the final product, but for your information the compounds formed after each step in (a) and (b) are given here. (a) 1. (CH3)2C:O; 2. (CH3)2(|3—OMgBr; 3. (CH3)2(l}—OH (Final product) CHZCH3 CHZCHZi Solutions to Problems 0 143 Li+ to HCI) 3. CH3CH2CH2CHC 4. CHgCHgCHgCHC (Final product) t t1 (c) 1, CH3CH2CH2C 3. CH3CH2CH2(|I‘<j (Final product) D t 43. The Wurtz reaction constitutes the direct coupling of the nucleophilic carbon of an alkylsodium compound and the electrophilic carbon in a haloalkane. It is an unselective reaction. There is no way to control it to prevent coupling of two “like” alkyl groups while attempting to couple two different ones. In other words, the reaction between chloroethane and l—chloropropane gives a statistical mixture of butane {from two ethyls), pentane (from an ethyl and a propyl), and hexane (from two propyls). 44. A = BngCHZCHZCHZCHZMgBr (A “bis-Grignard” reagent) 45- (a) cm 3.2% CH3C1 B = CH3CHOHCHZCHZCH2CH2CHOHCH3 (From addition of each end to aldehyde) Mai, CH 30H (b) Same as (a), starting with ethane. (c) Same as (a), starting with propane. Not very good. (d) CH3CH2CI-I3 % CH3CHB1'CH3 i CH3CHOHCH3 (e) Same as (a), starting with butane. Not very good. (f) Same as (d), starting with butane. Much better. (g) (CH3>3CH & (CH3)3CBr 51% (CH3)3C0H Beginning with alkanes, the only possible first step is halogenation. Functionalization of 1° carbons is difficult, too: Recall that radical halogenation favors 3° and 2" positions. 46. Note: The halide (X) in all Grignard reagents may be Cl, Br, or I. (a) From aldehydes, use RCHO M RCHon for (a) R 2 H, (b) R = CH3, (c) R = CH3CH2, and (e) R = CH3CH2CH2. Alternatively, for all but the first one, use HCHO M RCHZOH with (b) R = CH3, (0) R : CH3CH2, and (e) R = CH3CH2CH2. Finally, CHgMgX or CHgLi ——_F.§ (d) CH3CHO CH3CI-IOHCH3 and CH3CH2MgX or CH3CH2Li _____e—__> (f) either CH3CHO CHgCHOHCHZCH3 0r CHgMgX or CH3Li —_._r—) CHgCHgCHO CH3CHZCHOHCH3 144 ' Chapter 8 HYDROXY FUNCTIONAL GROUP: ALCOHOLS: PROPERTIES, PREPARATION, AND STRATEGY OF SYNTHESIS (ii) From ketones, use 0 II R CR I NaBH4 or LiAlH4 RCHOHR’ for (d) both R and R’ = CH3 and (f) R : CH3, R’ = CH3CH2. Also 0 (g) CH3(II:CH3 M (CH3)3COH A suitable solvent for all these reactions is (CH3CH2)20, and protonation of all alkoxide products with aqueous acid is required to generate each final alcohol. 47- (a) I:>'OH W (b) CHgCHgCHZCI-IZCHZOH W (e) <:>—CH on ——~——>*PCC’CH2C12 2 (d) (CH3)2CHCHOHCH3 ——~——r-—->N“‘2C“07’ H250“ H20 C (e) CH3CH20H —————>*PC ’CHZCIZ *Use PCC to avoid overoxidation. 0H 48. Target molecule is CHg—(EfiCHZCH’ZCHZCHg. at O (a) CHgdtcn3 + CH3CH2CH2CH2Li, followed by H“, H20. 0 (b) CH3(II3CH2CH2CH2CH‘3 + CHgLi, followed by H+, H20. ((2) This is a bit more tricky. Analyzing retrosynthetically, we look at the bonds in the product that need to be formed if we start with the 5-carborl aldehyde: ‘1“ ii i)“ b CHgiNfiCHzCHZCHZCHg, => CHg-i-CCH2CH2CH2CH3 => CHg—E-CHCHZCHZCHZCHg a CH3 Starting with an aldehyde, we will need to make two bonds (“‘a” and “b”) by adding two methyl groups. Working backward, bond “at” is addition of a methyl nucleophile to a ketone. Where does the ketone come from? It must be via oxidation of a secondary alcohol. The secondary alcohol then comes from formation of bond “b” to the aldehyde. The answer: CH3CH2CH2CH2CHO + CHgLi gives CH3CH2CH2CH2CHOHCH3. Oxidation of this secondary alcohol with Cr(VI) in aqueous HT gives the fir Solutions to Problems . 145 ketone CH3CH2CH2CH2COCH3. Addition of a second CH3Li to this ketone ends in the desired tertiary alcohol. This sequence may be written in the following way: 1. CH3Li 2. H+, H20 1. CHaLl 3.NCO,HSO,HO 2.H+,H0 CH3CHZCH2CH2CHO M CH3CH2CH2CH2COCH3 ——2~—> product Grignards may be used in place of alkyllithiums in any of these. The solvent in each case is (CH3CH2)2O> 49. (a) CH3CH2COCH2CH2CHZCH2CH3 + LiAlH4 or NaBH4, CH3CH20H (b) CH3CH2CHO + CH3CH2CH2CH2CH2MgBr (Lithium reagent okay, tbo) (c) CH3CH2CH2CH2CH2CHO Jr CH3CH2MgBr (Lithium reagent okay, too) The solvent in each case is (CH3CH2)20, unless indicated otherwise. 50. Transformation A is replacement of a tertiary hydrogen by bromine. There are two tertiary hydrogens in the molecule, but they are at equivalent positions because of symmetry. All the other hydrogens in the compound are secondary. Therefore, this step can be achieved by a simple radical bromination, which we know is highly selective for tertiary over secondary hydrogens (Chapter 3). H 7 Br BIZ ,125°C H H Transformation B requires carbon—carbon bond formation, and generation of alcohol functionality, for which we have a new strategy in this chapter: addition of an organometallic reagent (Grignard or lithium) to a car— bonyl compound. 80 the question becomes, which organometallic reagent and which carbonyl compound? The answers are to be found in the structures already given in the problem. The required new carbon—carbon bond goes from the bromine—bearing carbon to a hydroxy-beaiing carbon of a new 2—carbon substituent. Based upon the models in this chapter, the solution is to convert the brominated product of transformation A into an organometallic reagent, and to react this species with a 2—carbon carbonyl compound: 2-carbon carbonyl compound (aldehyde) convert to organometallic reagent 0 / CH3 OH BI/ MgBI‘ H / new C—C bond 1. H3C—C—H Mg 2. H30+ H H (i 51 - CH3(CH2)14CO_ + I(CH2)15CH3 52. (a) CH3CH20H H? (H) (b) CH3CHCOH O 0 Only the ketone carbonyl is reduced in these. || II (c) Hoccnzcrrcorr l OH ———_—T 146 ' Chapfer 8 HYDROXY FUNCTIONAL GROUP: ALCOHOLS: PROPERTEES, PREPARATION, AND STRATEGY OF SYNTHESIS 53. OH OH I I O (3., C,‘ II H/ \ ‘CH3 and H/ \ ‘CHZCOH (IlOH (IlOH o o 54. The bottom (or) face of the steroid is less hindered. (b) 55. The hydrogen atoms in hydroxy groups are acidic enough to react virtually instantaneously with the highly basic metal—bearing carbon atom of a Grignard or a lithium reagent. This process, which is an acid— base reaction, destroys the organometallic reagent by protonating this carbon and forming an inert alkane. The alcohol group is converted to its conjugate base, an alkorride, with a positively charged magnesium or lithium counterion. Such acid-base reactions are much faster than addition reactions of organometallic reagents to carbonyl group C=O double bonds. Therefore, enough excess organometallic species has to be introduced to compensate for the amount that is destroyed by this acid-base process, with enough left over to then add to the carbonyl groups in the substrate molecule. 56. Design each strategy beginning with disconnection of one of the bonds to the alcohol carbon atom (the “strategic bonds"). 0 \ OH O fix/3% a C b CHgMgBI‘ + <———- 0 Z} + CH3CH2MgBr Us MgBr +)Ol\/ Next, consider how readily available (and if available, how expensive) are the starting materials arising from each disconnection. For Grignard reagents, consider the corresponding halides; for ketones, consider the corresponding alcohols: Disconnection “a” Disconnection “b” Disconnection “c” CHgBI' $600/kg CH3CH2BI $30/kg CH3CHOHCH2CH3 $32/kg 0H 0H not Br available $200125 g 0’ $72/kg Solutions to Problems 0 147 Although l—cyclohexyl—l—propanol could certainly be prepared by oxidation of cyclohexylmethanol to the aldehyde and addition of CH3CH2MgBr, the extra effort and expense makes route “a” a poor choice. Route “b” is not bad, but “c ” is certainly the most efficient overall. Thus: M gBr Mg, ether OH UMgBr ether NazCr207 H2804 jv 2 H+ H20 _____, 57. (b) ...
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