Ch7 problem solutions - mag =_ _ r I I 'l'l6 ' Chapter 7...

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Unformatted text preview: mag =_ _ r I I 'l'l6 ' Chapter 7 FURTHER REACT IONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION Substitution or elimination? Analysis: Factor Favors ____________—_———-——-—— SUBSTRATE is primary: unhindered Substitution NUCLEOPHILE is ‘NH2: strongly basic Elimination NUCLEOPHILE is ’NHZ: statically unhindered Substitution Result: Substitution, to form CH3CH2CH2CH2NH2. _ The text summarizes the preferences for El, E2, SNI, and 8N2 reactions for 1°, 2°, and 3° haloalkanes, as a function of reaction conditions, in quite a bit of detail. The chart that follows repeats the same material, again somewhat oversimplified for clarity (solvent effects are not included, for instance). ‘ SUMMARY CHART Maior reactions of haloalkunes with nucleophiles 3 E Type of nucleophile : _—_——F—_—_—___—_____—__—n—- 5 I Weakly basic Strongly basic Strongly basic ‘ I Poor good unhindered hindered 1 ' Type of nucleophile nucleophile nucleophile nucleophile : halide like H20 like 1— like CH30_ like (CI-193C0— I Methyl No reaction 3N2 8N2 5N2 1° No reaction 8N2 8N2 E2 3 ‘ 2° Slow SNl 3N2 E2 E2 _ g | 3° 3N1 and E1 3N1 and E1 E2 E2 3 o g l Solutlons to Problems 1 (|)CH2CF3 i ‘ 22. (a) (CH3)3COCH2CH3 (b) (CH3)2CCH2CH3 - I j , 1 CH3an OCH3 (EH3 (I? g (c) 6 (d) Oc—ocn ' CH . fl 3 :3! :i H (e) (CH3)3COD (f) (CH3)3C#O¢O 23. (a) For the answer to this part we show each step on a separate line. CH3 CH3 I I CHngIQI' ——> CI"I3—(l:Jr 'I‘ Br— CH3 CH3 Solutions to Problems O 117 9% ‘6“3 H -- + CH3—(|:+ CH3CH29H ——> CH3—(l:ao< CH3 CH3 CHZCH3 CH3 H CH3 | +/ | .. ens—meatyD —> CHBM-CwQ—CHECH3 + H+ l \CH2CH3 I CH3 CH3 In the last step of problems like this students frequently make the‘mistake of removing an alkyl group from the oxygen, and giving an alcohol as the final reaction product. That process does not happen in solvolyses in alcohol solvents: Loss of a proton is far more favorable than loss of an unstable alkyl cation (which would be a primary ethyl cation in the example above). (b) For the remaining parts, we still show each step separately, but we connect the steps in a continuous sequence. The full structure of each intermediate is still shown. Br .. q +/—CFBCEQ_H I —_> CHgfiC—CH2CH3 _——"—> | ‘3' l I CH3 CH3 | H CHZCF3 \Q / : $ + C|)CH2CF3 CH3—$—CH2CH3 T CH2CH3 CH3 CH3 (P Q CHZCH3 CH2CH3 Cflhgf CHZCH3 CH3‘e'ci: CHZCH3 (a) é @ —> "1—) ~—> —Cl_ —H+ (d) (ii—CH3 ? (lj—CH3 ——> CH3 CH3 H " H O O \CQE‘H \C¢ 547/ / :(i)+ :C!): OW T (ti-CH3 CH3 C 3 OK, you’re asking, why did that oxygen attach to the carbocation instead of the other one? Makes it look a lot more complicated, too. The reason is that the species you get is resonance stabilized gm— 118 0 Chapter 7 FURTHER REACTIONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION (see below). When we cover carboxylic acids 3 dozen chapters from now you’ll see more of this. Sorry—reality is reality. H H H + 0 o 0 \C/ \H \C+/..\H \C¢ \H //2 / / C|)+ Ki): :Cl): w OW “w CH3 CH3 CH3 CH3 CH3 ., CH3 CH3 I p I 6:22—13 | +23 I .. (e) CH3—(i3—C1 in? CH3_(lj+ __+ giro—9:3 T57 CH3—(iZ—Q—D CH3 CH3 CH3 D CH3 CH3 CH3 I f) I D—QH (f) CHg—C—Cl in? (:Hg—(l:+ ————> CH3 CH3 CH3 CH3 H 24. CH3 CH3 CH3 OCH3 H OCH3 H CH3 (3) Two steps: _H+ —* Product with trans CH3’S _H"' —> Product with CiS CH3’S Solutions to Problems 0 'I I9 The nucleophile can attach to either face of the planar cationic carbon, reactions yielding the two products shown. B CH3 By reattachment of BF to the opposite side of cationic carbon. (Reversal of the dissociation step.) (b) CH3CH20 H ‘ CH3 C5H5 CH3 C5H5 25. Two products: 12g and 8 CH3 C6H5 C6H5 CH3 H OCHZCH3 From attachment of nucleophile to either face of the planar cation. 26. (a) H20 will speed up all of the reactions except for 22((1), because it is more polar than any of the other solvolysis solvents. It will also compete for the carbocations, forming alcohols as products. (b) Ionic salts strongly increase polarity and accelerate SNI reactions (see Problem 45, however). The main products will be iodoalkanes. (c) Same as (b); azide ion is a strong nucleophile and products will be azidoalkanes (alkyl azides, R—N3). (d) This solvent should reduce polarity and slow down all the solvolyses. + CH3 H CH3 H CH2 + 27c > > + (Tertiary) (Secondary) (Primary) 28. (a) (CH3)2CC1CH2CH3 > (CH3)2CHCHC1CH3 > (CH3)2CHCH2CH2C1 (3° > 2° > 1°) 0 ll (b) RC1 > ROCCH3 > ROH (Order of leaving group ability) Cl «Q «yoga CH3 29. (a) A secondary system with an excellent leaving group and a poor nucleophile :> SN] reaction. D 7 + cngcnz—on (CH3)2CH 0502CF3 —) + (CH3)ZCH ——> H l? (CH3)2CH9CH2CH3 —+ H+ + (CH3)2CHOCH2CH3 120 ' Chapter 7 FURTHER REACTIONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION (b) A tertiary halide in a polar solvent :- SNI reaction ‘t S [flBr —Br’ CH3 SCH3 ———*-> ~H+ CH3 (c) A primary halide with a good nucleophile in an aprotic solvent :> 8N2 reaction. A u + CH3CH2CH2CH2-Br + (C6H5)3P —> CH3CH2CH2CH2P(C5H5)3 Br* ((1) Similar to (c), except a secondary halide :> still an 8N2 reaction. as + I- —)‘ CH3CH2CI‘HCH2CH3 V 30. First decide What the most likely mechanism is for each reaction. Then write the product. Finally, recall that 8N2 reactions are faster in polar aprotic solvents, whereas SNI reactions are faster in polar protic solvents because of the greater stabilization of the transition state for cation—anion dissociation. (a) Primary substrate :9 8N2 to give CH3CH2CHZCH2CN', best in aprotic solvent. (b) Branched, but still primary, and nucleophile is not a strong base 2 8N2 again. Product is (CH3)2CHCH2N3; aprotic solvent is again best. (c) Tertiary substrate :> 8N1 substitution to form (CH3)3CSCH2CH3; best in protic solvent. ((1) Secondary substrate with an excellent leaving group and a weak nucleophile .-_> SNl is the most likely mechanism from this combination, forming (CH3)2CHOCH(CH3)2; fastest in protic solvent. 31. Two successive 8N2 inversion steps are necessary to give the desired net result of stereochemical retention: ' KB , DMSO (R)—2—chlorobutane ~’—> NaN ,DMSO (S)—2-bromobutane —3—-+ (R)—2-azidobutane 0 || 32. (l) Racernic CH3CH2CH(OCH)CH3 will be formed via an SN] process (a solvolysis). The solvent (a carboxylic acid) is very polar and protic, but a weak nucleophile. 0 || (2) (R)—CH3CH2CH(OCH)CH3 will be formed via an 3N2 process (good nucleophilc, aprotic solvent). Note the very different conditions. Solutions to Problems 0 121 33. The first reaction is an uncomplicated 8N2 displacement. The second is 8N2 as well, but there is a complication. Let’s take a look at how its product can react further, and then perhaps we can see why it happens in the second case but not in the first. . The pathway to the byproduct, (CH3CH2CH2CH2)2S, must involve reaction between the product of the first displacement, CH3CH2CH2CHZSH, and another molecule of the initial starting material CH3CH2CH2CH2BI': This is the only reasonable way to get a second butyl group on the sulfur. Since butyl is primary, we are limited to the 8N2 mechanism. The simplest way to get there is to use the product molecule itself as a nucleophile: F) CH3CH2CH2CH2 \+ CH3CH2CH2CH2§H + CH3CH2CH2—CH2—Br $> ‘ /_S_%H CH3CH2CH2CH2 Then loss of H+ from sulfur completes the sequence. Why doesn’t the same thing happen in the first reaction, the formation of the alcohol? What do you know about the differences in nucleophile strength between S and 0? Sulfur is far better, especially in protic solvent (Section 6-8). Alcohols are too weak as nucleophiles to carry out 5N2 displacements, while thiols can achieve this transformation readily. Did you consider an alternative mechanism, one in which the SH group of the thiol is deprotonated before nucleophilic attack? That is, CH3CH2CH2CH2:SH :2 CH3CH2CH2CH2:S::‘ + H+ 1 followed by K) CH3CH2CH2CH2 \ CH3CH2CH2CH2§ : _ + CH3CH2CH2— CH2— BI‘ ? CH3CH2CH2CH2 This mechanism is qualitatively plausible, but because the pKa of the thiol SH bond is around 10, in the absence of base the equilibrium of the initial deprotonation is too unfavorable for this sequence to be competitive: The concentration of the conjugate base is too low. 34. (1a) (CH3)2C=CH2 (1c) CH3CH=<j CH3CH2@ (1d) (CH3)2C=<:> CH2=C(CH3)O (1e), (1f) Same as (1a) 35. (a) The base is a very strong one (the pKa of NH3 is way up there, around 35; it is a very weak acid), and that favors E2 rather than E1. The exclusive product is CH3CH2 \ / H /c$c\ CH3CH2 CH3 E—_— _ .__ I22 - Chapter 7 FURTHER REACTIONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELlMINATION This product forms no matter which of the six possible protons next to the reactive carbon is lost: CH3 H \ ll‘_ H2 /g\ /J CH3CH2\ /H H (C ———> NH3 + Czc + Br / \ CH3CH2 CHZCH3 CH3CH2 CH3 (b) E2 CH3CH2 ,H \c’ C1 / \ a H ‘4 C —> (CH3)3C—OH + CH3CH2CHZCH2 + C1— (CH3)3C—§=* (e) E2 ((1) Either E1 or E2 can occur, and two products can form from either: 0 CH3 and O: CH2. Example of E1 mechanism: C] H C 1 CH3 —‘—> (:1- + + CH3 WM") H+ + Q‘CH3 Example of E2 mechanism: CC1 Mom + CHggr —> CH30H + cr + <:>=CH2 H<———-/ 36. As in Problems 29 and 30, first things first: Categorize the substrate (primary, secondary, etc.) in order to identify the mechanistic pathways available to it. l—Bromobutane is an unbranched primary haloalkane; Solutions to Problems 0 123 the mechanistic possibilities are 5N2 and E2. 8N2 will occur with good nncleophiles, but poor nucleophiles don’t react. Strong, hindered bases give E2. (3) and (c) l—Chlorobutane (b) l—Iodobutane (d) CH3CH2CH2CH2NH3+ Br" (e) CH3CH2CHZCH20CH2CH3 All of the above form via 8N2 mechanisms. (f) No reaction. Alcohols are poor nucleophiles. (g) CH3CH2CH=CH2 (E2 reaction) (h) CH3CH2CH2CH2P(CH3)3+ Br' (5N2) (i) No reaction. Carboxylic acids are poor nucleophiles. (No, we never told you that, but you do know that nucleophilicity goes along with basicity, and carboxylic acids are not terribly likely to be strong bases. Make sense?) 37. Now the substrate, 2-bromobutane, is secondary. A11 mechanisms are possible. (a) 2-Chlorobutane (b) 2—Iodobutane DMF is a polar aprotic solvent. (a) and (b) both are 8N2 reactions. (c) 2-Chlorobutane, but now by the 8N1 pathway. Nitromethane is a nonnucleophilic solvent with good ionizing power. ((1) CHSCHZCH(NH3)CH3+ Br— 8N2: NH3 is a good but not strongly basic nucleophile. (e) and (g) CH3CH2CH=CH2 and CH3CH:CHCH2 (E2 products—the problem states that the reagents are in excess, thus favoring E2 over E1) (f) CH3CH2CH(OCH2CH3)CH3 Secondary substrate undergoes SNl reaction with the poorly nucleophilic alcohol (solvolysis). (h) CH3CH2CPI[P(CH3)3]CH3+ Br_ (8N2) (i) CH3CH2CH(02CCH3)CH3 Another SN] solvolysis, as in (f). 38. The substrate, 2-bromo-2—methy1propane, is now tertiary. The 8N2 pathway is now eliminated as a possibility. (a) and (b) N 0 reaction. The usual polar aprotic solvents have rather poor ionizing power. DMF is somewhat better than acetone (a really poor SNl solvent), but substitution will still be very slow. (c) 2-Chloro-2—methy1propane (SNl) (d) Ammonia is a borderline case in terms of base strength, giving more substitution with secondary substrates (where 8N2 is an option), but more elimination (E2) with tertiary haloalkanes: (CH3)2C=CH2 (e) and (g) (CH3)2C=CH2 (E2) (f) (CH3)3C—O—CH2CH3 (h) (CH3)3CP(CH3)3+ Br‘ (5N1), together with (CH3)2C:CH2 (E1) (i) (CH3)3C02CCH3 (8N1) 39. (a) (1) (CH3)3CSH + HCl (2) (CH3)3C02CCH3 + (CH3)2C=CH2 + KC] + CH3COOH (CH3COO'K+ is basic enough to give some elimination from tertiary halides.) (3) (CH3)2C=CH2 + KC1+ CH30H I24 0 Chapter 7 FURTHER REACTIONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION (b) Rates of (l) and (2) will be the same, or nearly so. (1) is SNl and (2) is SN} and E1, and they have the same rate—determining step; if the reaction mixtures are similar in polarity, the rates will be very close. Rate of (3) will depend on the concentration of the base because of the occurrence of bimolecular elimination. 40. (a) OZCHz E2 (b) No reaction. Poor leaving group. (c) Racernic CH3CH2CHOHCH3 SNl (:1) 0 E2 (6) (CH3)2CHCH2CH20CH2CH3 SNZ I | Racemic CH3CH2C(CH3)CH2CH2CH3 SNI (g) No reaction, except for reversible proton transfer. 01) OCHQCHZCHZCN and NCOCHZCHZCHZCN E2 and sN2 OCH3 (i) (S)—CH3CH2CHSHCH3 3N2 (j) 0 SN] CHZCHs (k) (CH3)CCH2CH2 E2 (I) No reaction. Poor nucleophile. 41 I Reagent Haloalkane H20 NaSeCH3 NaOCH3 K0C(CH3)3 CH3C1 no reaction CHgSeCH3 CH3OCH3 J} S 2 CH30C(CH3)3 8N2 CH3CH2CH2C] no reaction CH3CH2CHZSeCH3 3N2 CH3CH2CH20CH3 CH3CH=CH2 (CH3)2CHC1 (CH3)2CHOH S l (CH3)2CHSeCH3 CH3CH=CH2 } E2 CH3CH=CH2 E2 N and and (CH3)2C=CH2}E1 (CH3)2C:CH2}E1 42. See answer to Problem 41. Secondary halides give higher EZ/El ratios than do tertiary halides (secondary carbocations form less readily than tertiary). 43. (a) Poorly. CH3CH=CHCH3 and CH3CH2CH=CH3 are important products. (b) Not at all. No reaction; poor nucleophile. Solutions to Problems 0 125 (c) Not at all. No appreciable reaction besides SNl with the solvent. ((1) Well. An “intramolecular” (internal) SNl reaction. (e) Well, eventually, but very, very slowly. (f) Well. (g) Well. (h) Not at all. No reaction because of the very poor nucleophile. (i) Not at all. No reaction. (j) Not at all. No reaction because of the very poor leaving group. ~ (k) Poorly. (CH3)2CHCH2CH20CH2CH3 forms also. (I) Poorly. Good nucleophile gives mainly CH3CH2CH2CH20CH2CH3. Brz, A KI, DMSO 44. (a) CH3CH2CH2CH3 ——> CHgCHQCHBrCI—lg. —~——> (:chnzcrncn3 C12, hv N31, acetone CH3CH2CH2CH3 _'—) SOHIC CH3CH2CH2CH2C1 —"'——“‘> CH3CH2CH2CH21 C12, hv KOH, H20 Brz, A (6) CH4 H—-> CH3C1 —d—> CHgOH; then (CH3)3CH —"—> CH OH (CH3)3CBr —3—+ (CH3)3COCH3 Br (d) 0 Br;, hv O” KOCHgCH3,CH3CH20H 0 Br OH (e) From (d), 0/ m 0/ A better method will be presented in Chapter 8. (f) NaZS in alcohol: B B Cl r C1 r CH2 82— CH; / K.__./ \.._/' \ . CH2 CH2 Four 8N2 dlsplacements! \ f—‘N (sz 82— (13H; (Br qBr 45. (a) Rate = k[RBr], so 2 >< 10“4 = k(0.1), and therefore I: = 2 x 10’3 5‘1. Product is ©C(CH3)2—OH (ROH). (b) New “kLiCI” = 4 X 10—3 8—1. Addition of LiCl increases the polarity of the solution by adding ions, and this speeds up the rate—determining ionization step in the solvolysis process. I26 - Chapter 7 FURTHER REACTIONS OF HALOALKANES: UNI-MOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION (c) In this case the added salt contains Br‘, which is also the leaving group in the solvolysis reaction. This leads to a decrease in rate because the first step in solvolysis is reversible, and recombination of R+ and Br’ is occurring in competition with reaction of R+ with H20: Ionization H20 Br_ + R+ *-—> ROH RBr Recombination 46. (a) The solvent is an alcohol and, therefore, already not very nucleophilic. In addition, however, the three fluorine atoms in the structure, CF3CH20H, are very electronegative and electron—attracting. The combined effect of their electron-attracting power reduces the ability of the oxygen atom to serve as an electron-donating nucleophilic atom. This is an example of an inductive effect, which We will discuss more fully in the next chapter. For now all we need to recognize is that this alco- hol is a much poorer nucleophile than “normal” alcohols (ones lacking halogen substituents). (b) In the usual SNl reaction mechanism the first step is rate-limiting; that is, rate] < ratez. In the case of the reaction in this problem, the reduced nucleophilicity of the solvent diminishes rate; so that it becomes slow and rate-limiting. Thus the carbocation intermediate has the opportunity to build up in concentration, because it is formed faster than it is consumed by attachment to the nucleophile. (c) As the carbocation stability increases and the solvent nucleophilicity decreases, rate] should in- crease further relative to rate;, and the carbocation intermediate should exhibit a more extended lifetime. This principle was used by George Olah of the University of Southern California in his pioneering studies associated with direct observation of carbocations and the direct measurement of their properties in solvent environments of extremely low nucleophilicity. (d) as A + I l CH2CF3 | (I) I —fl> CH30 E£Ki>r0cn3 $32» CHgOQEQOCHg 47. (a) A tertiary halide 2 SNl reaction, which will be two simple steps as in reaction profile (iii). -. a" a E = (CH3)3C:--C F = (CH3)3CJr , 8+ 5 + G : (CH3)3C"'P(C6H5)3 H = (CH3)3CP(C6H5)3 (b) A secondary halide being displaced by another halide :> SNZ reaction. The product and starting material are comparable in stability: reaction profile (ii). H C g_ CI 8 D CH CHB r r —( 3)2 1‘ CH3 CH3 Solutions to Problems ' I27 (c) A tertiary halide with a poor nucleophiie :> SNl (solvolysis), but with one more step than (a) because of proton loss from an intermediate alkyloxonium ion: reaction profile (iv). 3+ 3" I = (CH3)3C---Br J = (CH3)3C+ 5+ 5+ + K = (CH3)3C-‘-(?CH2CH3 L = (CH3)3C(‘)CH2CH3 H H 5+ M = (CH3)3CQCH2CH3 N = (CH3)3COCH2CH3 8+ (d) A primary halide and a good nucleophile :> 8N2 reaction, but the product is much more stable than the starting material (C~QO bonds are stronger than C—Br bonds): reaction profile (i). H H at \ / a“ A = B = CH3CH20CH2CH3 CH3 48. Neutral polar conditions are ideal for an intramolecular SNl reaction: (ED +m. (CH3)2CCH2CH2CH20H —'"’ (CH3)2CCH2CH2CH20H CH3 CH3 on + M [X CH CH a? 3 O 3 H Basic conditions promote elimination leading to alkenes. Two isomeric alkenes are actually possible: CH2: and 49. (a) El rate = (1.4 x 10—4 s")(2 >< iii—QM) = 2.8 x 10"6 mol L“ 3’1 E2 rate = (1.9 x 10'4 L moi" s")(2 >< 10'2M)(5 x 10*] M) : 1.9 x 10*6 mol L“ sfI El rate is faster, so E1 reaction predominates. (b) E1 rate : 2.8 X 10—6 mol L“ 5—1 (no change) E2 rate : (1.9 x 10‘4 L mor1 3")(2 X 10—2 M)(2 M) = 7.6 x 10‘6 mol 1:1 s,—1 E2 rate is now faster, so E2 reaction predominates. (c) Solve for [NaOCHg] when E1 rate : E2 rate: 2.8 x 10'6 mol L'1 5—1 = (1.9 x 1044 L moi”1 5—1) (2 x 10‘2 M) [NaOCH3]. [NaOCH31 = 0.74 M. 50. The first task is to examine both starting materials and products in order to classify the reaction that has taken place. A methyl group has been removed from the substrate and converted to iodomethane. In m 128 0 Chapter 7 FURTHER REACTIONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION addition, an oxygen—carbon covalent bond has become an ionic bond between oxygen and lithium. Does this fit any pattern we’ve seen? Yes. If we go back to Chapter 6 and examine the components of substitution reactions, we can see two familiar elements: a nucleophile (iodide) and an organic substrate to the left of the arrow. Having defined these, it is then possible to recognize iodornethane as the product of substitution, and the lithium salt structure as containing the leaving group. Having defined the process as a substitution, what remains is to consider experiments to identify whether the operative mechanism is SNl or 8N2. A kinetic study to reveal whether or not the iodide ion appears in the rate equation is one possibility. Changing the solvent from pyridine (polar, aprotic) to a protic solvent (e.g., alcohol) and observing the effect on the rate is another. Of course, you are not operating entirely in the dark: Given that the displacement is occurring at a methyl carbon, it is reasonable to predict that the answer will be 8N2. 51. As in Problem 50, first classify the reaction. Not only does a carbon—oxygen bond break, but a hydrogen is lost from the terr-butyl group, apparently shifting to the oxygen to give an alcohol (cyclohexanol) as one product, and leaving an alkene as a second organic product. Loss of both hydrogen and oxygen from two adjacent carbon atoms to give a carbon—carbon double bond resembles elimination, but it differs from the eliminations from haioalkanes that we have encountered previously, in which one of the atoms lost was a halogen. Halogens, in the form of halide ions, are good leaving groups. Oxygen is another story, however, because negatively charged oxygen (as in hydroxide, for example) is usually strongly basic and not easily displaced. Where do we turn for a rationalization, then? Consider the reaction conditions. If this is indeed an elimination, it is an unusual one: The eliminations in this chapter occur in the presence of base, not acid. Taking that anomaly one step further, what role could the acid serve? Where on the substrate could a proton conceivably go? There is only one answer: the oxygen atom. The product is an oxonium ion, an alkyl-substituted analog of the hydronium ion. Looking at the relationship between the structure of this ion and the products, notice that departure of cyclohexanol to leave a tert—butyl carbocation is a reasonable next step. Cyclohexanol is a neutral, weakly basic molecule and should be a good leaving group. The ten—butyl cation is stable, as carbocations go, and it can lose a proton to give the observed alkene product. Overall, this is a variation on the E1 process, using the acid to turn a poor leaving group into a better one. .. (EH3 11: (fHB M \+ P/ H IQ—(llmCI-Ig + H+ —> Q—(FI—CHg —> OH + /C—IC1 + 2 CH3 CH3 H3C Oxonium ion Cyclohexanol / . H C + 3 \ H + /C=CH3 H3C 52. (a), (d), (e), (g), (h), (i) Mainly 8N2. These nucleophiles are all weak bases, so elimination is not favored. (c), (f) Mixture of 8N2 and E2. (b) Mainly E2. Strong bases favor elimination. 53. In both A and B the necessary anti alignment between H and Cl (both of which are in axial positions) is present, so E2 elimination will give the desired alkene. In C the Cl is equatorial and anti elimination Solutions to Problems ' 129 cannot occur (make a model). Instead, very slow eliminations will proceed via syn geometries to give a mixture containing the desired alkene together with the isomer shown below. CH3 \etc. CH3 H0” 54. Look at the conformations first [the deuterium atoms for (b) are also written in]. @ CH3 CH3 D ~ vs. D Br G) CH3 H CH3 (i-d) (ii-d) J (a) and (b) Compound i reacts much faster because it already possesses the necessary anti alignment of Br and H (on the neighboring carbon, circled). In the deuterated example shown, the E2 I reaction results in loss of I-[Br only; all the D is retained because the D atom is not capable of adopting an anti alignment relative to bromine. The conformation shown is the reactive one. Compound ii possesses no hydrogen anti to Br in the conformation shown. However, if the left—hand ring adopts a boatlike conformation resembling that shown below, E2 elimination via an anti transition state can readily occur. GD CH3 H ‘0" @ CH3 As indicated, all the D should be lost, aeoording to this mechanism, and that is the result observed. 55. As we have been doing, characterize each subsn'ate—nucleophile (base) combination before going any further. We have: A primary; B and C secondary; and D tertiary bromoalkanes, reacting with (a) a good nucleophile that is a moderately weak base; (b) a strong, bulky base; (c) a good nucleophile that is a moderately strong, nonbulky base; (d) like (a), a good but not particularly strongly basic nucleophile; and (e) a weak and essentially nonbasic nucleophile. The problem is best solved by taking each bromoalkane in turn and evaluating the most reasonable mechanism that will take plaCe upon its reaction under each of the conditions (a) through (6). We’ll do the easy ones first. E__——________— I30 0 Chapter 7 FURTHER REACTIONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION A l-Bromobutane (primary) will give 8N2 products, CH3CH2CH2CH2NU, under all conditions except (b) (E2 with strong bulky base to give CH3CH2CH= CH2) and (e) (no reaction with poor nucleophile). D 2-Brorno-2-methylpropane (tertiary) with methanol (e) will give the SNl product, (CH3)3COCH3, accompanied by minor amounts of (CH3)2C= CH2, the E1 side product. The SNl products (CH3)3CN3 and (CH3)3C02CCH3 will form under conditions (a) and ((1), respectively, together with increased amounts of (CH3)2C= CH2, which now arises through E2 reaction with the moderately basic azide and acetate nucleophiles. Under strongly basic conditions (0) and (b) only E2 takes place. B and C 2-Bromobutane (secondary) will give 8N2 products under conditions (a) and ((1) (good nucleophiles that are not very strong bases), 3 mixture of 8N2 and E2 with hydroxide (c) (at the borderline of strong base for substitution versus elimination with secondary substrates), and E2 with LDA (b). Methanol (e) gives mostly 8N1, some E1. The presence of two diastereomeric deuterium-substituted 2—bromobutanes raises stereochemical questions, which the mechanistic descriptions that follow explore: Substitution reactions 8N2: Inversion at reaction site Nu ‘ Nu:— 1.3 1.3 1.3 1? AV —* /S\S:/ /S\S:/ "*4"? AV BR Nu RD Nu B C SNl: Mixture of stereoisomers at reaction site obtained Either B or C —> mixture of (2S,3S) and (2R,3S) substitution products. Elimination reactions E2: Anti conformation of hydrogen and leaving group in transition state. Each substrate may undergo reaction to give alkenes with the double bond either between C1 and C2 or between C2 and C3; the latter may form from either of two conformations. Base:\ Base:"\ DH H H. awn/y .. Rx R) Br H B H The factors that result in one E2 pathway being preferred over another will be presented in Chapter 11. Solutions to Problems ' 131 I2 H I.) t I / M ——> 5 Br Baseffl‘ Base:l\ D H D \\ D 1'31“ H Br H I E1: Mixture of all alkene isomers obtained from either substrate ‘ 56. (d) 57. (b) 58. (b) 59. (b) 60. (a) ...
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This note was uploaded on 09/30/2009 for the course CHEM 3A taught by Professor Fretchet during the Spring '08 term at University of California, Berkeley.

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Ch7 problem solutions - mag =_ _ r I I 'l'l6 ' Chapter 7...

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