lab_solutions-1-6

lab_solutions-1-6 - Chem 322bL/325bL Period 1 JBE ANSWERS...

Info iconThis preview shows pages 1–18. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 18
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chem 322bL/325bL Period 1 JBE ANSWERS to QUESTIONS for GRIGNARD SYNTHESIS OF 2—METHYL—2—HEXANOL 1. One could make the title product from 2-hexanone and methyl bromide. But usually among aliphatics, species with four or fewer C's are more readily available than those withumore. Cf Solm p 525ff. a) N (b) (d) Iodine helps start a Grignard by (1) etching the surface of the metal, and (2) very likely by acting as a radical chain carrier (12 very easily dissociates to I- atoms). Grignard formation probably involves some kind of radical chain reaction at the metal surface. R-MgBr + -----> R-I + MgBrI I2 Add a cone solution of R-Br initially but only a small part of the total amount of halide. Once reaction starts, dilute the reaction mixture with ether and then add the rest of R-Br soln slowly so that at any instant the cone of unreacted R-Br in the rxn mixt is low. For the desired reaction: R-Br + Mg -—---> R-MgBr For the coupling reaction, the R-MgBr reacts with as yet unreacted R-Br: R-MgBr + R-Br ---——> R-R + MgBr2 Thus, the overall reaction for the coupling is the sum of these two reactions: 3. (a) CH3 (CH2) 3C (CH3) (OH) CH3 (b) (c) (d) (e) Page 2 - Chem 322bL/325bL Grignard (Pd 1) Answers 2 ReBr + Mg R—R + -—---> MgBr2 Thus, in the desired reaction, the molar stoichiometry is 1:1 RBr:Mg; for the coupling rxn, it is 2:1.' Since one is taking the SM's 1:1, if only the desired rxn occurs, all the metal is just consumed. The coupling reaction con- sumes only half as much metal. The product alcohol will dehydrate to alkenes: -————> CH3CH2CH2C=C(CH3)2 + 1-isomer With HCl, one will obtain the corresponding alkyl chloride. Magnesium(II) plus hydroxide produces a slurry of Mg(OH)2, i.e., milk of magnesia (MOM). Separation of the product from this gooey semi—solid will be difficult. Thus, one needs acid to keep the Mg+2 dissolved. The problem with having any residual acid in the ether soln of the product arises during the distillation. While cold, dilute acid is not a problem, during the distilla- tion it becomes hot, (more) conc acid. The distillation then becomes the steam distillation of alkenes; cf (a). K2C03 is not only a DA, it is also basin. Thus any acid which escaped the earlier washings will be neutralized and precipitated as its potassium salt. ‘ This is an extra safeguard; use of a basic DA assures removal of all acid. Chem 322b/325bL Period 2 ' JBE ANSWERS to QUESTIONS for FRIEDEL-CRAFTS PREP 1. Alkylations with an alkyl halide plus a Lewis acid are simi— lar to those using an alcohol and a protonic acid in that the key intermediate, a carbocation, is the same, i.e., the reactions are mechanistically similar, broadly speaking. 2. (a) See cited Solm problem. EtOH is much more nucleophilic than anisole. (b) The Ph3COH/EtOH product must form irreygxsihlx, but the Ph3COH/H0Ac product must form zgyersibly. a . (c) (1) Ph3C-O-g-Me; no other reasonable structures. £1 0 an 39 . (2) Ph3c—0=p-Me <-———> ph3c-o—c-Me <—---> Ph3C-O-S-Me HO HO H0 65 Second ion is more stable; only it can delocalize charge. Whole mechanism is analogous to that of (a). (d) Assuming the overall reaction Ag's are comparable, Ag of W, AGt, for both the formation and W of the ester, Ph3C—0Ac, will be less than for the ether, Ph3C-OEt, because of greater IM stability. Under the conditions of the reaction, the ester but not the ether can furnish the required Ph3C+. Page 2 — Chem 322/5b Answers to Pd 2 (F-C Alkylation) 3. (a) Answers to question S(a)(b) from the Appendix: (a) The rate ratio = exp[(—Ea'rxnl + Ea'rxn2)/RT], which taking rxnl to have the lower Ea, gives e2 (~7.4) at the lower T and e1 (~2.7) at the higher T, i.e., a given difference in Ea makes a bigger rate difference at the lower T. (b) (1) Since the desired reaction has the lower Ea, to favor it relative to path 2, one should go to lower T. (2) Using the equation at the bottom of p 4f of the Appendix gives for Ba = 20 kcal/mole (- 20,000 cal/mole) ln(rate) I tion is slowed by a factor of 400 since antiln(-6) = (-60-20000)/I(2-100000) = -6, i.e., the reac- [exp(-3)]2 = [1/e3]2 = [1/2012. The same calculation for an Ea = 30 kcal/mole gives a slowing of the reaction by a factor of 8000 (e'9). Thus, if path 1 and path 2 have the same rates at 316°, path 1 becomes Iglagiygly favored by a factor of 20 (8000/400) at 60° 1912:. (3) If at the original T, the reaction is complete in one minute, it is practical to run it at the lower T, the reaction will be 400 minutes, about 7 hours, But if the reaction originally required about a day for completion, at the lower T it will require over a year (400 days) for completion. Better find another way if this is the case. Page 3 - Chem 322/5b Answers to Pd 2 (F-C Alkylation) 3, (b) Given that the desired reaction has the lower 33, one has ‘OK{VL a larger fraction of the SM's forming product at the lower T (with the longer reaction time). But this result also means less side products. If, as here, the side products are removed by a process (rextln here) that also causes some loss of product, one also gets a higher practical yield of pure material if one gets more product Thus, and less undesired by—product(s). one wins two ways—-by getting more desired product formed and by greater ease of product purification. (c) Look back at the answer to cited Background question. 4. The high viscosity and high bp will make the solvent diffi- cult to get rid of by filtration and evaporation respective- ly. The fixed ratio of functionality means one cannot use more of the good solvent, or more of the poor one, to opti- mize the recrystallization. ‘\\ Chem 322/5b Period 3 JBE ANSWERS to QUESTIONS for CYCLOHEXANONE BY HYPOCHLORITE OXIDATION 1. If HOAc and cyclohexanol form the ester, it must first hydrolyze back to cyclohexanol before oxidation can occur. 2.(a) The system is buffered because HOAc is a weak acid and NaOCl is the salt of a weak acid (even weaker than HOAc). (b) Since acetic acid dissolves the organic SM and is miscible with aq NaOCl, the reactants are in the same phase. It turns NaOCl into HOCl but is not itself oxidized. One could separate the (neutral) product from HOAc by making the mixture basic giving NaOAc, which is not soluble in the product nor in organic extracting solvents. 3.(a) See the decomposition of a chromate ester, Solm p 524 top. (b) A hypochlorite ester of a tertiary alcohol, e.g. t—Bu-OCl, does not have an H on the C bound to O and therefore cannot undergo the E2-like elimination. 4. H20 + Hso3‘ + C12 --—-—> H804" + 2 c1‘ + 2 H+ 5.(a) MW is just MW sum of the product and NaHSO3, = 98 + 104. (b)(1) Organics can be washed away from a BAC, then the BAC can be decomposed back to the carbonyl compound. (2)(i) H303“ + H+ ——-—> "H2803" --——> H20 + so?¢ H303“ + "OH ----> so3= -k[110 (ii) :9 OWMH‘ .OJWW ' x-C>“F( c:) C£;{§I) @?:%”"C) (after [4-0 H o \\ CHEM 322bL Period 4. J.B. Ellern Answers to Egg-Write-Up Questions for ALDOL CONDENSATION 0F ISATIN AND ACETOPHENONE 1. (a) 30.....———-ketonic carbonyl Ofi ‘— amide function H <—————). / 0 (b) E Usual amide resonance ' c9 _ / 0 <—-——-> o ‘ N Note delocalization of charge to 3-C=O and g—quinoid structure; this correlates with color. OH H 2. (a) CHsz 2 CH_BZ ' .l 0‘ we —» Last step could involve carbocation formation t 3-C and then loss of . . (b) Overall driving force is formation of extended conjugated system. New C-C double bond is conjugated with an aromatic ring and two carbonyls (That's right aldol fans, not one but tw_o whole car—bonylsl). CHEM 322bL Period 4. J.B. Ellern Answers to Post~lnlrite~Up Questions for ALDOL CONDENSATION 0F ISATIN AND ACETOPHENONE 3. (a) Compound III is 3—phenacyloxindole, derived from 3-phenacylidene- oxindole (cpd II) by hydrogenation (addition of 2 H's) to the exocyclic C—-C double bond. Reasoning: The formula difference implies that exactly one double bond has hydrogenated to a single bond. The fact that III is white means that the extended conjugated system has been broken._ Since III still has an amide carbonyl (given) and a ketonic carbonyl (implied by reaction with 2,4—DNPH), reduction of the alkene double bond is the most probable reaction. s 046 + 21420 ——> 2Hso39 + 2[H] 2 cpd II + 2[H] ————) cpd III Sumning above gives 1:1 stoichiometry; equation is balanced. \ I (c) C=C OH H\ /c\ ————-> ,li—C=C—R + so2 R o—‘fs; H 0 I i enol of )CH-CH- —R $02 is other product; in aqueous medium, it dissolves (forms HSO3(D and/or 5036 depending on pH). 4. Orange-red color of isatin is associated with a (minor contributing) g-quinoid resonance structure of the neutral, like that of anion of dues 1(b) but with plus charge on N. de I, having no double bond to 6-3, has no chromophoric (color-producing) unit. de II (or-red) has a lengthy continuous conjugated system and Q-quinoid character. In cpd III (white), hydrogenation of the alkene double bond has broken the extended conjugated system. Chem 322bL/325bL Period 5 JBE ANSWERS to QUESTIONS for METHYL BENZOATE 1. Ac—— is acetyl, CH3-g-; Bz-— is benzgzl, Ph—g- (n9; benzyl, which is Ph—CH2—). 2. (a) A reaction can be driven by (l) removal of a product as it is formed (by, e.g., distillation or precipitation, or binding of water by a DA), and/or (2) by use of an excess of a reagent to maximize the conversion of the other reagent(s). (b) Use of an excess of a reagent is applicable here. (Since MeOH is the lowest boiling material here, continuous product distillation is not applicable. One might be able to find a three—component water/MeOH/hydrocarbon azeo— trope.) For the direct methyl formate prep, direct dis— tillation is usable since the product is the lowest boil- ing component in that case. (c) MeOH is the better reagent to use in excess here because it is also the reaction solvent; the benzoic acid will readily dissolve in it at its bp. (d) For [B10 = l M, let x = [C]. Then [A] = [B] = l — x. Solution to quadratic x/(l—x)2 = 3 is x ~ 0.57, and [CJ/[A] ~ 1.3. [B10 = 10 M.:$ x/(l—x)(lO-x) = 3. Can use approx here: x must be ~ 1:; lO—x ~ 9 :; x/(l-x) F 27, i.e., the ratio of product to hard-to—separate A is large. Page 2 - Answers to Period 5 (Methyl Benzoate) Questions (e) If one uses a reagent in excess, one had better have a plan to separate the product from the large amount of unreacted excess SM. (f) MeOH and EtOH are (l) are cheap (2) easily water—washed out of an organic solution and (3) low boiling point; what escapes washing can usually be readily separated by distillation. One treats the solution with a solution of aqueous base, one strong enough to convert BzOH quantitatively to its anion but not strong enough to hydrolyze the ester rapidly. Aq Na2C03 or bicarbonate is usually used; nearly all sodium carboxy~ lates are water soluble. Equations: 2 PhCOOH + CO3=' — - - - -—> 2 PhCOO_ + H20 + C02 PhCOOMe + aq CO3: — - - — -—> NR (a) The DA removes (at least) all the dispersed (i.e., sepa— rate phase) aqueous soln, thus cutting the water conc down to no more than 0.17%, the saturation conc. Although the distillate from this soln contains only 1.5% water, this distillate/vapor is about 9 times richer in water than is the pot liquid from which it comes. This ratio stays about constant throughout the distillation (a consequence of Henry's Law for this dilute system). Thus the water conc in the pot falls rapidly; cf ques 5 below. (b) Any aq droplets containing Na2C03 w0uld, if not removed before distillation, give hot, anhydrous Na2C03, which could decompose the product. 5. (a) (b) Page 3 — Answers to Methyl Benzoate (Pd 5) Questions Recall that [WJdist/[W] ~ 9 throughout the distilla— pot tion. As the [WJdiBt falls, the distillate will clear when its W conc falls to 0.17 wt %. At this conc in the distillate, the pot W conc is 1/9 as much, ~ 0.02 wt %. The differential equation can be written in the separable form dy/y = 9(dx/x). Integrating both sides gives ln y = 9 ln x + C. Requiring that y yo when x = x0 gives C = 1n yo — 9 ln x0. Putting y and yo on one side and the two terms in x and x0 on the other and recalling that 1n t — ln to = ln(t/t0) casts the equation into the form given. Even without (much) knowledge of calculus, one can derive the above result by realizing that the logarithm is the measure of relative change; cf Background handout p B—Zf. An infinitesimal relative change in t equals dt/t, which equals d(ln t); see discussion p App-fo. From this the equation follows immediately. The desired equation is gotten by moving 1n x and 1n x0 to the ln(y/y0) side and then dividing through by 8. The values 1/10, 1/100, 1/1000 are (y/x)/(y0/x0). Taking ln of each, dividing the result by 8, then taking the anti ln, i.e , exp, gives x/xo. One gets x/xo with fewer key strokes by simply taking the 1/8 (= 0.125) power, i e., the eighth root. Either way, the answers for the fraction distilled, l — (x/xo), are, to two significant figures, 0.25, 0.44, and 0.58, respectively. Thus distillation of, e.g.,<% the soln, cuts its [W] by a factor of 100. Page 4 — Answers to Methyl Benzoate (Pd 5) Questions Answer to ques 8 in Drying Agents handout: A proper— ly chosen DA (cf ques 4) removes all the dispersed, i e., separate phase, water plus some dissolved water. Distil— lation then does the thorough removal of the (typically small) amount of remaining dissolved water. The values of x/xo are (three significant figures): 0.990, 0.980, 0.951, 0.905, 0.819. Note that even the last value differs from 1.0 - 0.2 by <2%% of its own this reflects the fact that for t e—t value. Mathematically, close to zero, ln(1 — t) ~ -t, i.e., ~ 1 — t. Physi— cally, this means that if drying by distillation is very efficient, i.e., r is sufficiently large for the desired [WJ/[WJO that the whole right side of the equation of part (c) is only sl <0, then x/xo is only slightly <1, i e., But then y/yo is nearly x ~ x0. (y/y0)/(x/x0). Chem 322bL/325bL Period 6 J33 ANSWERS to QUESTIONS for DILANTIN BY VITAMIN Bl-CATALYZED SYNTHESIS l. (a) See structures: ..® (b) For tetrahedral IM and a-anion, see pG-l, last few lines of second paragraph, with R =- Ph. If Y" is ‘OI-l. there are no other resonance structures for the or-anion, but if ‘1‘ is the TICB of (a) one can draw the following non-charge separated (enamine) structure: "ox / (Jee- “(K6- IC '2 C.\ ’0 aye] R N The catalyst must have its nucleophilic atom multiply bonded to an atom which readily accepts (additional) electron density, i.e., (another) unshared pair.‘ (c) For the mechanism, see the next page top. The C-Y bond must be strong enough to give a reasonable cone of the Page 2 - Chem 322bL/325bL Benzoin (Pde) Answers adduct but not so strong that the catalyst is not readily lost at the end. (3 . t ’;\e +- (I! S Inn I@,/C=\__/,I‘H :2 [9)Cl—H c ' l u C3}! C3}! Pk-crf S / S <—-—— MN C H [@156 ...fi3;,}{ \. QJ )OAfEYHIrHiHELg )oL' ’CD‘ 5) :3 ’2 I? '\ / pL—cH—Cfl' I C?” 3 pk g ‘—-—-'> PA CH—C"I011 rem/)(WM 44mm I]. +77% 2. (a) In a—hydroxy carbdnyl compounds, the carbinol H is alpha to a carbonyl and therefore much more readily picked off by a base than is the carbinol H of a simple alcohol. (An oxidizing agent is of course necessary to remove the electrons left on carbon.) (b)(1) Note that benzoin is C14H1202; benzil has two fever H‘s. Rxn (1) 2Cu+2 + Benzoin -----> 2Cu+ + Benzil + 2H+ Rxn (2) 2Cu+ + 23* + NH4N03 ----> N2 + 3H20 + 2Cu+2 Sum Benzoin + NH4NO3 -----> Benzil + N2 + 3H20 Page 3 - Chem 322b/5b Benzoin (Pd 6) Answers (2) It may appear that the first step must be the rds since having more Cu(II) speeds up the overall reaction. But this is not certain: Suppose the oxidation of benzoin is fast, the (re)oxidation of Cu(I) by ammonium nitrate is slow, and its rate is proportional to [Cu(I)]. Then in the steady state situation, nearly all the copper is Cu(I) and the [Cu(I)] is nearly equal to the initial conc Of Cu(II). Then the overall rate is still proportional to the initial [Cu(II)]. Thus one cannot tell for sure which step is the rds solely from the facts given. (3) If rxn (1) is fast, the mixture should be Cu(I) colored since most of the copper is in the reduced state. If rxn (1) is slow, the mixture should be Cu(II) colored. (c)(1) There must be an intermediate with an odd number of elec— trons, i.e., a radical. (2) Likely its formation is slow and its further reaction (to give product) fast. Radicals are usually difficult formed but easily converted, by loss or gain of one elec- tron, to even-number-of-electron species. (d) The brown gas is N02, the brown haze of smog. Its conc gets high enough to make it visible here only after there is little or no Cu(I) around to reduce it to N02‘. Thisv occurs when (nearly) all the benzoin has been oxidized, and therefore the Cu(I)-producing rxn (l) is over. 3. (a) See Solutions Manual p (b)(1)-(4), see top of next page. (b)(5) The group is isoelectronic with -CO-CH2-CO-. Thus removal of H+ from the doubly a (to C=O's) central atom Page 4 - Chem 322b/5b Benzoin (Pd 6) Answers gives an anion resonance—stabilized by tag carbonyls. Mechanism for Dilantin formation: cam?» NSC: 0 _ C.HN|:¢N\C|_O __ CoHS\(T-¢N\cr-_o _ Ic‘zr + /- /C*°- N“: /CN'I Co"! 0 NH: Co": 0 C9“: 0 H 3- cm, I C.H, H cJ-l,\C,N\ (WC/ax" \C,N\ __ \C,N\ I c—o—- ’| c—o=‘3- ’l c-o— ’| c...o C / CoHsC / CoHsC\ / C‘H, \ \ \ CJh/5;N\H 0’ N\H 0’ "Q4’ 0’ N9 69 3. (c) The compound is Ph -CCE£% an.a—amino-acid (Solm ch 24), written in the dipolag form (p {HZ/FF). 4. (Post—Write Up Question) The fact that at reflux the mixture stays blue-green (getting somewhat greener as reaction proceeds since a white SM is giving a yellow product) means most of the copper is Cu(II) and therefore the rds is in rxn (1). If you watched carefully as the mixture was heating, you may have noticed that the color fades to beige and then the blue-green color returns. This implies that there is some lower-than-reflux temperature range over which rxn (2) is slow compared to rxn (1). Page 5 — Chem 322b/325b Benzoin (Period 6) Answers Answer to Chem 322/5b Period 6 Question 1(d)(e) (d) C02 reacts with aq NaOH to form carbonate, then bicarbonate, HCO3=, both much weaker bases than 'OH. Neither of these will de—protonate the thiazolium ion to the TICB. Air oxidation of benzaldehyde produces benzoic acid, which similarly neutralizes NaOH. Since we are using 0.1 equivalent of Thiamine and NaOH here, oxidation of just 10% the PhCHO will kill the reaction. If only 1 mol % were being used, even 99% aldehyde with only 1% BzOH will shut down the catalyst system. (e) At high pH the hydroxyfinamine (see structure in ansWer to 1(c) on p 2) is de—protonated to an englatg anion (draw its structure). This species is dead with respect to a route to benzoin via reasonably stable intermediates. Careful acidification of a too-basic reaction mixture to pH 5 10 permits the benzoin condensation to proceed. Chem 322b/325b period 46 ' JBE ANSWERS to QUESTIONS for PREPARATION OF GLUCOSE PHENYLOSAZONE 1. See below for a two—step version of the tautomerization in the box in Solm p [(DALI‘ Note the similarity to the tautomerization of an aldehyde or ketone to its enol form. (A = 0H,) 65 A‘EA v fi‘ fi‘ CI—I=N—NHC6H5 a—a CH=\_’.%—NHC6H5 (3}1—1\1—1\1———(36Hs l “‘4 RI —» u }{——(f——CH1 l“ I{-—fi}—‘Cni ¢———— C}__C)__}{ I 2. See below; the structure below left is just the one above right written with an internal H-bond This is an attractive route to the imino ketone (or ketoimine) in that it has very little movement of atoms from the SM to transition state, is one step, and requires no external acid or base. Note that moving the three electron pairs counterclockwise rather than as shown would give the same result. _ CH=NH /CH,, Nl—I I I — \fi Q,NHceH5 ——+ C=o + C6H5NH2 o—H’ ' 3. For the imino N of the MonoPhenylHydrazone (ques 1 answer, left structure) to end in the NH3, the MPH N-N bond must be getting hydrogenolyzed, giving the )C=NH group (and aniline, Ph-NHZ). Since the =N——NH—Ph is getting reduced (do the oxidation numbers), it is the oxidizing agent by definition. ...
View Full Document

Page1 / 18

lab_solutions-1-6 - Chem 322bL/325bL Period 1 JBE ANSWERS...

This preview shows document pages 1 - 18. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online