Ch6 problems - Problems 245 1 the submifllr I VIM. Easily...

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Unformatted text preview: Problems 245 1 the submifllr I VIM. Easily displaced, causing the rate of reaction for (iii) to be greater. Problem 49 re- " In.— reasoning. rlllll“ “'fi' a; mu: substrates (iii) and (v) according to the IUPAC system. I". ' ' ' I “a” I[;'1'{If_.l|_\l'. - “fl HI,“ bL-L-tinna 2-5 and 4-1 if necessary. Iliii I—Hrontr- 3-methylbutane (v) (2-Chloroethyl)cyclopentane gas? |fl‘|}fllTF!i" Concepts L A huh-mnmm, commonly termed an alkyl halide, con— 6. An orbital description of the SNZ transition state includes “'51,, m" an alkyl group and a halogen. an spZ-hybridized carbon center, partial bond-making be- tween the nuclEophile and the electrophilic carbon, and simultaneous partial bond-breaking between that carbon and the leaving group. Both the nucleophile and the leav- E. 'I'ht- physical properties of the haloalkanes are strongly :Il'i'eL-[ct'l by the polarization of the C—X bond and the po— |.II'iJ.:LiJ|']if1-' of X. woofer mil; ing group bear partial charges. r. (The acmfil 3, Reagents bearing lone electron pairs called nucleo- 7. Leaving group ability, a measure of the ease of dis_ phiile when they attack pos1uvely polarized centers (other . . . _ placement, IS roughly proportional to the strength of the tindflred mu" llh'lll protons). The latter are called electrophlhc. When . t .d E . n d] . ak ' ~ - ' leads to displacement of a substituent it configa 8 am ' Spec“! y goo eavmg groups are we [6 Prequel “l “Eh J readlofl. .. . . . ’ bases such as chloride, bromide, iodide, and the sulfonates. is :i unelenpmhr substitution. The group being displaced : is more hm; by the nucleophile is the leaving group. 8. Nucleophilicity increases (a) with negative charge, (b) for elements farther to the left and down the periodic table, and (c) in polar aprotic solvents. late is larger. 4. The kineilun of the reaction of nucleophiles with primary olvent (cone. rml most secondary) haloalkanes are second order, indicae e Strenth Iii. lie-.- of a bimolecular mechanism. This process is called 9. Polar aprotic solvents accelerate 8N2 reactions because atzms “I'f‘ll‘l‘ bimoleruiar nucleophilic substitution {8N2 reaction). It the nucleophiles are well separated from their counteri- re Zolfowfl is a concerted reaction, one in which bonds are simulta— ons but are not tightly solvated. an - ,.. annual}- broken and formed. Curved arrows are typically Gala-ates in." used to depict the flow of electrons as the reaction proceeds. 10. Branching at the reacting carbon or at the carbon next atom (trim. to it in the substrate leads to steric hindrance in the 8N2 5. Tee SNZ reaction is stereospecific and proceeds by back- transition state and decreases the rate of bimolecular side displacement, thereby producing inversion of con- substitution. ’2 manner ->1|. l'llfltifiililfili at the reacting center. 7 F' shirts F 17. tune the following molecules according to the IUPAC system. Q CH,CH3 33 fihtlw structures for each of the following molecules. (at) 3—Ethyl—2-iodopentane; ACN: nucleophile. ’tso We “H 3-br0mo-l,l—dichlorobutane; (c) cis-1~(bromomethyl)-2-(2-chloroethyl)cyclobutane; gig: _ it!) (trichloromethyl)cyclopropane; (e) 1,2,3-trichloro-2—methy1propane. iii) contains @V-‘Fflw and name all possible structural isomers having the formula C3H6BrC1. . ' the . [1:33:le DHIW and name all structurally isomeric compounds havmg the formula CsHuBr. splacemt'fll @Fnl' each structural isomer in Problems 29 and 30, identify all stereocenters and give the lulhl number of stereoisomers that can exist for the structure. azide will @th each reaction in Table 6-3, identify the nucleophile, its nucleophilic atom (draw its [JP-V55 structure first), the electrophilic atom in the substrate, and the leaving group. I Jasecond Lewis structure can be drawn for one of the nucleophiles in Problem 32. (a) Iden- lckside 0-1:" I”? it and draw its alternate structure (which is simply a second resonance form). (b) Is Brefore, W "1'3 '1' a second nucleophilic atom in the nucleophile? If so, rewrite the reaction of Problem an chlofiti'i 31-, using the new nucleophilic atom, and write a correct Lewis structure for the product. 246 thapteré PROPERTIES AND REACTIIJNS 0F HMMIJiANES 34. ‘ior each reaction shown here, identify the nucleophile, its nucleophilic atom, the elec~ trophilic atom in the substrate molecule, and the leaving group. Write the organic prod- /\/0\ /CF3 'i' NaI —> not of the reaction. ([3) 031- + NaSH —> (c) /S\ (a) - + NaN3 —> O 0 \Cl I (a) CH31+ NaNHg —> (e) CHgCl + /\N/\ —> I (f) + KSeCN —> CH3 @\ solution containing 0.1 M CH3Cl and 0.1 M KSCN in DMF reacts to give CHgSCN and KC] with an initial rate of 2 X 10—8 mol L’1 s7]. (a) What is the rate constant for this reaction? 0)) Calculate the initial reaction rate for each of the following sets of re- CH3C (i) CH3C 4]. Show hot {it} {R)—C C actant concentrations: (i) [CHgCl] = 0.2 M, [KSCN] = 0.1 M; (ii) [CHgCll = 0.2 M, ll .41.. Rank the [KSCN] = 0.3 M; (iii) [CH3C1] ? 0.4 M, [KSCNI = 0.4 M. IuIcEcoph 36. Vrite the product of each of the following bimolecular substitutions. The solvent is l indicated above the reaction arrow. ’ | . . CE 0115 “I If” (a) CH3CH2CH2Br + Na+I‘ —A——‘—> (b) (CH3)2CHCHZI + Na+ *CN 953% (C)CH31+Na+’OCH(CH3)2 (CH’)’CH°H (a) CH3CH2Br+Na+'SCH2CH3 “30H I if“ €ch 3 2 2 I (e) Oicflzm + CH3CH28eCH2CH3 % (f) (CH3)2CHOSOZCH3 + N(CH3)3 (CH air I 33;; 37} etermine the R/S designations for both staning materials and products in the following I {u [_ / 8N2 reactions. Which of the products are optically active? I-‘C H Cl H O (3) CH3 C1 + Br' (b) H3C” ‘CH3 + 21‘ I H Br 1-} . L CH4“ CH2CH3 - 0 Cl 0 {C1 0 I l H H is} C‘ch (c) + "OCCH3 (d) + "OCCH3 H0 H0 [El [1192‘ For each‘reaction presented in Problems 36 and 37, write out the mechanism using M- 'I'I'IL' silher curved-arrow notation. iiuuill a: n It _ . @List the product(s) of the reaction of l-bromopropane with each of the following I It: reagents. Write “no reaction” where appropriate. (Hint: Carefully evaluate the nucleo- Ilnp'lJELI'IJ'EJ philic potential of each reagent.) h”, m} i‘i' I '~ (a) H20 (b) H2804 (c) KOH (1!) CSI (e) NaCN “m. dig in: (f) HCI (g) (CH3)2S (h) NH; (i) 012 (j) KF ....,;,M W emulate the potential product of- each of the following reactions. As you did in mi How u. Problem 39, write “no reaction” where appropriate. (Hint: Identify the expected leavifllr' I "l'ilrnpfnli group in each of the substrates and evaluate its ability to undergo displacement.) I 15_ I 3.... 15. all (a) CH3CH2CH2CH2Br + K+‘OI-I W» (b) CH3CH21+ K+cr M) [with new 1“ 'Illltl |I1 an CH.CH.0H cHaffiL. “3' d'ml len- (c) CH2C1+ L1 OCHZCHg, ——> (d) (CH3)2CHCH2Br + CST ——»" mud-“Ir H frfi'III|:-.-:-.i.-i-r. - — 30H ll'.| - -. (e) CH3CH2CH2C1+ K+ 'SCN C—HIE’fl» (o CH3CH2F + LNCI 95—? sy L tom, the flier. = organic pin-“i + —+ ive CHgSCN : constant for lg sets of re- 51] = 0.2 M, solvent is DMSO ""—> CH OH 3 3 (CHaCHflJJ _ 3)3 the followingr 11 using rwing he nucleo- lid in cted leaving ent.) :MF ——> _ CH DE 3’1 —’-> CH H ,0 [III + K+I7 m 0 in caicazoCHchfi Natton fl ®Eum how each of the following transformations might be achieved. (€502CH3 1T3 CH3 CH3 .3, (R)-CH3CHCH2CH3 —~—> (S)—CH3CHCHZCH3 (b) H Br H CN CH3O H CH3O H CH3 CH3 H H O ‘ O —) , --Br -—> SCH3 d + n} qD- ( ) 171 /N\ H H CH3 CH3 CH3 @aunk the members of each of the following groups of species in the order of basicity, nunkophilfcity, and leaving-group ability. Briefly explain your answers. (3) H20, HOC, :11 [;C02'; (b) Br‘, Cl‘. F‘, I’; (c) ‘NHZ, NH3, 'PHZ; (d) ‘OCN, "SCN; (2)13", Irfl', ‘SCI-I3; (f) H20, H28, NH3. @9111}: the product(s) of each of the following reactions. Write “no reaction” as your - i 5 war, if appropriate. no CH3CH2CH2CH3 + Na+cr 51% Br Cl H3C H I A m (c) +NH+I_ flan—E) + Na+7SCH3 CB ne H3C H CH3CH2 CH3 H (3” OSOZCHg (e) + Na+ ‘CN ‘~—) + CH3CH20H OSOZCHg p CH I l orncnzon / 3 (g) CH3CHCH3 + Na+ CN ——~a- (h) H3C fiOCHZCHZCH O (i) CH3CH2NH2+Na+Bf w» (j) CH3I+Na+'Nl-I2 fl> 44. The substance 1-butyl~3-n1ethylimidazolium (BMIM) hexafluorophosphate (margin) is a liquid at room temperature, even though it is a salt composed of positive and negative ions. BMIM and other ionic liquids constitute a new class of solvents for organic reac- CH3CH2CH2CH tions, because they are capable of dissolving both organic and inorganic substances. More important, they are relatively benign environmentally, or “green,” because they can be easily separated from reaction products and reused virtually indefinitely. Therefore they do not constitute a waste-disposal problem, unlike conventional solvents. (a) How would you characterize BMIM as a solvent? Polar or nonpolar? Protic or aprotic? (b) How would changing the solvent from ethanol to BMIM affect the rate of the nucleophilic substitution reaction between sodium cyanide and 1-chloropentane‘? : (25,3S)-3-Hydroxyleucine is an amino acid (Chapter 26) that is a key component in the structures of many “depsipeptide” antibiotics, such as sanjoinine (margin, p. 248). (3) Find the part of the sanjoinine molecule that is derived structurally from (2S,3S)-3— hydroxyleucine. (b) Although many depsipeptide antibiotics occur in nature, the quantities available are too small to be useful pharrnaceutically; thus these molecules must be Synthesized. (2S,3S)—3—Hydroxy1eucine, which is also not available in quantity from nature, must be synthesized as well. Possible starting materials are the four diastereomers of (h) CH31+ Na+’SCH3 % H (.i) CHgCHZI-l- K+"OCCH3 PE» 0:) CH3CH2C1+ Na+’0CH3 C—H‘O-‘L CH3 Problems 247 + K+’SCN -—>CH’°H m 2 PF; ca3 l-Butyl-3-methylimidazolium (BMIM) hexaflunrophosphate NH; (25,3S)-3-Hydroxyleucine OH 243 Chaptem PROPERTIES AND REACTIONS Ill HALOALKANES HO 2-bromo-3-hydroxy—4-methylpentanoic acid (margin). Draw structural formulas for each of these diastereomers and identify which of the four should be the best starting mate- [m 31 rial for a preparation of (2S,3S)-3-hydroxyleucine. NH ' O >7 Iodoalkanes are readily prepared from the corresponding chloro compounds 0 _ O NH y SNZ reaction with sodium iodide in acetone. This particular procedure is especially ‘ useful because the inorganic by-product, sodium chloride, is insoluble in acetone; its pre- I“ B] J cipitation drives the equilibrium in the desired direction. Thus, it is not necessary to use , excess NaI, and the process goes to completion in a very short time. Because of its great i NH convenience, this method is named after its developer (the Finkelstein reaction). In an at- up O tempt to synthesize optically pure (R)-2-iodoheptane, a student prepared a solution of ‘ I‘mEfi— tion ft transit (S)—2~chloroheptane in acetone. In order to ensure success, he added excess sodium iodide (CH3)2N and allowed the mixture to stir over the weekend. His yield of 2-iodoheptane was high, sanjoinine but, to his dismay, he found that his product was racemic. Explain. 5-1. Nun'ch OH 0 47. Using the information in Chapters 3 and 6, propose the best possible syn- on an) thesis of each of the following compounds with propane as your organic starting mate- OH rial and any other reagents needed. [I-Iint: On the basis of the information in Section 3-7, “Wm you should not expect to find very good answers for (a), (c), and (e). One general ap- Team Pr Br preach is best, however] ,3 Hum 2-Bmmo-3-hydr0xy-fl-methyl- (a) l-Chloropropane (b) 2-Chloropropane (c) l-Bromopropane L1H ‘l Pentanmc and (d) 2-Bromopropane (e) l-Iodopropane (f) 2-Iodopropane lg“; ‘ “SCI-13 48. repose two syntheses of trans-l-methyl-2-(methylfliio)cyclohexane (shown in the "'3 d3 margin), beginning with the starting compound (a) cis—l-chloro-2~methy1cyclohexane; "Ii'fir‘: 1 CH3 (b) rmns—1~chloro-2-methylcyclohexane. 2‘0” 511 each of the following sets of molecules in order of increasing 5N2 reactivity. ' Tm" (a) CHgCHgBr, cnger, (CH3)2CHBr 0”” (b) (CH3)2CHCH2CH2C1, (CH3)2CHCH2C1, (CH3)2CHC1 * Corr (c) CH3CH2C1,CH3CH21,QCI - eon- figm a. 1] (d) (CH3CH2)2CHCH2Br, CH3CH2CH2(|3HBr, (CH3)2CHCHgBr CH3 “x . , , I‘I'fln‘nfl: 50. Predict the effect of the changes given below on the rate of the reaction {I5 .HE S . . } CH 0 (in w CH3C1+ ‘OCH3 ——w_»3 H CH30CH3 + or. m (a) Change substrate from CH3C1 to CHgI; (b) change nucleophile from CH300 to 51 T1]: In CH38_; (c) change substrate from CH3C1 to (CH3)2CHC1; ((1) change solvent from mm m CH30H to (CH3)ZSO. mm} 51. The following table presents rate data for the reactions of CH31 with three different nu- cleophiles in two different solvents. What is the significance of these results regarding relative reactivity of nucleophiles under different conditions? £ ¥ a] ;_fi Id) 1.”! Nucleophile km], CH30H km, DMF : I I ‘ 3- 1Winch CI’ 1 1.2 x 106 is: r-' Br 20 6 x 105 51* Gm I. NCSe’ 4000 6 X 105 which @Explain the outcome of the following transformations mechanistically. H” (a) HSCHZCHZBr-l-NaOH w v “'3 :l' W Problems 249 as for em: CH OH ting mas:— lh.’ firrnzcl-IZCHECHZCHQBr + NaOH —>3 O cempoul‘ulr. P6931” | H I;.FH2CH2CHZCH2CHZBr+NH3 ————>CH3CH=°H ne; its i try to US: )f its great _ H 'In a“ 3" ‘1 ill! 8N2 reactions of halocyclopropane and halocyclobutane substrates are very “on. 0f_ ' ' “ind; shin-er than those of analogous acyclic secondary haloalkanes. Suggest an explana- um lo'dldi' Linn fur this finding. (Hint: Consider the effect of bond-angle strain on the energy of the V33 hlgfl‘ ,l-mmitiun state; see Figure 6-4.) _ _ 5* Muck-whine attack on halocyclohexanes is also somewhat retarded compared With that 35MB SYI'I‘ ml “FL-Jit- secondary haloalkanes, even though in this case bond-angle strain is not an 1118 ma“;- important factor. Explain. (Hint: Make a model, and refer to Chapter 4 and Section 6-9.) lection 3-91 cm ap' .. -- m Pruhicm 55‘ imnpounds A through H are isomeric bromoalkanes with the molecular formula ' '5“ I ,3: With your team, draw all eight constitutional isomers. Indicate any stereocen— texts]. but do not label it (them) as R or S until you have completed your analysis. Using 1 the the data below, assign structures to A through H. Divide the problem into equal parts to hexane; shale the effort of finding a solution. Reconvene and discuss your analysis. At this point, you should indicate the stereochemistry with wedged and dashed lines as appropriate. MW. ' - Trentmem of compounds A through G with NaCN in DMF followed second- ordei kinetics and showed the following relative rates: | AEB>C>DEE>F>>G I Compound H does not undergo the 8N2 reaction under the preceding conditions. - Compounds C, D, and F were found to be optically active, each having S absolute con— figumlion at the stereocenter. Substitution reactions of D and F with NaCN in DMF pro- ceeded with inversion of configuration, while treatment of C in the same way proceeded wiili retention of configuration. I’r'cprot’fisionai Problems iii. The 8N2 reaction mechanism best applies to In: cyclopropane and Hz (b) l-chlorobutane and aqueous NaOH it'll KOH and NaOH (d) ethane and H20 Dfr to 5?. 'i'he reaction CH3C1+ OH' ——> CH30H + Cl‘ is first order in both chloromethane 0m IIIlll hydroxide. Given the rate constant k = 3.5 X 10'3 mol L'1s’1, what is the ob- scri'i'd rate at the following concentrations? ierent nu- :garding [CH3C1]= 0.50 mol L—1 [OH‘] = 0.015 mol 1:1 {at 2.6 ><10r5 mol I:1 s'1 (b) 2.6 ><10'6 mol L“ s*1 (c) 2.6 ><10w3 mol 1:1 s—1 “I: 1.75 ><10—3 mol Lr1 s’1 (e) 1.75 ><10'5 molL_ls_1 53' which ion is the strongest nucleophile in aqueous solution? la] Fi (b) Cl' (c) Br' (d) 1— (e) all of these are equally strong 59. Uniy one of the following processes will occur measurably at room temperature. Which one? o. .. in} :.F_——(':_1: (b) :NE(:?/“>*c3t-I3 in =Nr=-N:/“>\CH3£‘:1:: (a) ::<:i=o::P\t:H2=r“cr+12 ...
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Ch6 problems - Problems 245 1 the submifllr I VIM. Easily...

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