Answers To Chapter 7 Problems.

Transcription

1 Answers To Chapter Problems.. Most of the Chapter problems appear as end-of-chapter problems in later chapters.. The first reaction is an ene reaction. When light shines on in the presence of light and ose Bengal, singlet oxygen is obtained. This compound can do cycloadditions or ene reactions. If the reaction were a free-radical autoxidation, neither light nor ose Bengal would be required. hν ose Bengal C C C C C C C C econd reaction. Air is not required for formation of the keto-enol. The C C and bonds are broken, and a new C bond is made. It makes sense that the driving force for breaking the C C bond should be provided by migrating C from C to (note: a,-shift) and expelling. Then can add back to C to give a hemiketal which can open up to the ketone. C C C C Tf air C C C C C C C C C C + C C C C C C

2 Chapter C C ~ + C C + C C C C C C C C Air is required for conversion of the keto-enol to the endoperoxide. The most likely reaction is autoxidation. The makes bonds to C and C, neither of which has an atom attached for abstraction. But abstraction of from gives a radical, A, that is delocalized over and C. Addition of to C gives a hydroperoxy radical, which abstracts from of the starting material to give a hydroperoxide and A again. The hydroperoxide thus obtained can then add to the C ketone in a polar fashion to give the observed hemiketal. C C C C C C C C C Initiation: C C C C C C A C C Propagation: C C C C A C C C C

3 Chapter C C C C C C C C C C + A C C Polar reaction (acid is still present): + C C C C C C C C C C + C C C C C C C C The fourth reaction is transformation of the aldehyde into an acetal. This proceeds by acid-catalyzed addition of an alcohol to the carbonyl, loss of, and then addition of the acid to the carbocation. ther perfectly correct sequences of steps could be written here. C C + C C C C ~ + C C C

4 Chapter. (a) Compound is obviously made by a Diels Alder reaction between cyclopentadiene and methyl acrylate. Cyclopentadiene is made from the starting material by a retro-diels Alder reaction. The product is obtained stereoselectively because of endo selectivity in the Diels Alder reaction. C C The starting material is called dicyclopentadiene. Cyclopentadiene itself is not stable: it dimerizes to dicyclopentadiene slowly at room temperature by a Diels Alder reaction. It does this even though it is not an electron-deficient dienophile, demonstrating the enormous reactivity of cyclopentadiene as a diene in the Diels Alder reaction. (b) LDA is a strong base. Compound is obtained from the enolate of by a simple substitution reaction. (i-pr) I MM C C ow DM is treated with a, then with, then with Zn and a, to give overall substitution of C for C. The C group must come from DM, so we need to make a new bond between the DM C and the C= carbon. a is a good base; it deprotonates DM to give the dimsyl anion. This adds to the carbonyl C, and then loss of occurs to give the β-ketosulfoxide. This is a very good acid (like a,-dicarbonyl), so it is deprotonated under the reaction conditions to give the enolate. Workup gives back the β-ketosulfoxide. This part of the mechanism is directly analogous to a Claisen condensation.

5 C C a C C Chapter C C C C a C work-up C To get to, we need to cleave the C bond. Zn is an electron donor, like a or Li. Electron transfer to the ketone gives a ketyl, which undergoes fragmentation to give the enolate. The second electron from the Zn goes to the leaving group to give. Workup gives the methyl ketone. C e C C e C work-up (c) The conversion of to is a [+] cycloaddition, the Paterno Büchi reaction. This four-electron reaction proceeds photochemically. (d) The conversion of to is an E elimination. (i-pr) C The conversion of to is a wern oxidation. The of DM is nucleophilic, and it reacts with oxalyl chloride. Cl then comes back and displaces from to give a electrophile. The of is then deprotonated, whereupon it attacks, displacing Cl. Then deprotonation of a group and a retrohetero-ene reaction occur to give the ketone.

6 Chapter Cl C C Cl Cl C C Cl C C Cl Cl Cl Cl + C + C C C Et Cl C C Et C C C C The conversion of to is a dissolving metal reduction. umber the atoms. The key atoms are, C, C, C0, and C. Make: none. eak: C C. 0 Li C0 The first step is formation of the ketyl of. This species can undergo fragmentation to form the C C enolate and a radical at C. A second electron transfer gives a carbanion at C, which deprotonates. Upon workup, C0 is protonated to give. e e

7 Chapter work-up + C The conversion of to is a simple hydrolysis of an acetal. Acetals are functionally equivalent to alcohols + carbonyls and can be interconverted with them under acidic conditions. everal reasonable mechanisms can be drawn for this transformation, but all must proceed via substitutions. + The conversion of to uses PPh and I. The former is a nucleophile, the latter is an electrophile, so they react to give Ph + P I. The P is attacked by the alcohol to give an P bond, and the I then displaces Ph P from C to give the alkyl iodide. Ph P I I I PPh Ph P + Ph P I I The next reaction is obviously a free-radical chain reaction. Initiation: AIB C nbu C nbu Propagation: nbu + I nbu C I C

8 Chapter nbu 0 + nbu C C Finally, conversion of 0 to involves addition of the very nucleophilic Li to the ketone; workup gives the alcohol. Then E elimination promoted by the acid Ts gives the alkene. C C work-up + C C C C C C C C. (a) The transformation of to (not shown) is a simple deprotonation with LDA, followed by substitution on e, displacing eph. The conversion of to requires making C C and C C, and breaking C. The BuLi deprotonates C to give a sulfur ylide. This makes C nucleophilic. It adds to C, making an enolate and making C nucleophilic. The enolate at C then attacks C, displacing to give the product. C eph C I C C BuLi C eph C C C Bu C C C eph C

9 eph Chapter eph C C C C The conversion of to is a free-radical chain process. ote two equivalents of Bu n are required. Make: C C, n e. eak: e C, C C. Let s deal with the e first. After initiation, Bu n abstracts eph from C. The C radical then adds to C, giving a radical at C which abstracts from Bu n to regenerate nbu. The C C bond still needs to be broken, and C and C both need to have attached to them. We know that a cyclopropane ring cleaves very easily if a radical is generated at a C attached to it, e.g. at C. We can generate a radical at C by having Bu n add to. Then the C C bond cleaves, making a C radical and a tin enolate at C C. The C radical abstracts from Bu n to propagate the chain. The tin enolate is protonated upon workup to give. C eph 0 Bu n cat. AIB C 0 C C Initiation: AIB C nbu C nbu Propagation: eph nbu C C C C C nbu Bu n C C C

10 Chapter 0 Bu n C Bu n C C Bu n C C Bu n C C Bu n work-up C C C C (b) LiAl is a source of very nucleophilic. It must add to an electrophilic C. If you obey Grossman s ule, you will see that C and C in the product have extra s. f these two only C is electrophilic, because when adds to C, a very stable (aromatic) cyclopentadienyl anion is obtained. This anion is protonated at C upon workup to give the alcohol. (Actually, the anion can be protonated on C, C, or C, but all three isomers are in equilibrium with one another, and only the isomer protonated on C is able to undergo the subsequent Diels Alder reaction.) When the alcohol is oxidized to the ketone, the C=C0 π bond becomes electron-deficient and electronically suitable to undergo an intramolecular Diels Alder reaction with the cyclopentadiene to give. 0 C C ) LiAl ) Al(-i-Pr), acetone 0 C C Al C C C C work-up C C ppenauer C C

11 Chapter (c) Make: C C. eak: C C. The first step is electron transfer to form the ketyl. Fragmentation of the C C bond occurs to give a radical at C, which can add to C to make the C C bond and put the radical on C. A second electron transfer gives a carbanion at C. Upon workup it is protonated, as is C, to give. 0 LiDBB C TF C 0 e C e C work-up. First step. Make: C, C C. eak: C. C C C cat. h (Ac) C C C The product is a γ,δ-unsaturated carbonyl compound (a,-diene), hinting that the last step is a Claisen rearrangement. C C C C C C The diazo compound combined with the h(ii) salt tells you that a carbenoid is involved. The carbenoid can be drawn in the h=c form or as its synthetic equivalent, a singlet carbene. In either case, C can

12 Chapter undergo one of the typical reactions of carbenes, addition of a nucleophile, to form the C bond. After proton transfer to and loss of [h], a Claisen rearrangement can occur to give the product. C C h(ii) C C [h] C C [h] C ~ + C [h] C C C C C C C C C econd step. Make C C. eak C C. The reaction proceeds by a,-shift. C C C BF Et C C C C C C BF C C C BF C C C work-up C C C BF Third step. tandard ozonolysis with workup. C C C ; C C

13 Chapter The Criegee mechanism should be drawn. The initially formed,,-trioxolane can be split up in two ways, one of which gives the desired aldehyde, but the mechanism can t stop there. C C C C C C Fourth step. It is not clear whether the ring is or. If the ring is, then make: C, C, C, and break: C. If the ring is, then make: C, C, C, and break: C. C C cat. Ts or C C First step is protonation of one of the carbonyl s. An intramolecular addition is likely to occur faster than an intermolecular one. Because a better carbocation can be formed at C than at C, addition of to is more likely than addition of to C. C C + C C C C C C ~ + C C C C

14 C Chapter C C C It should be stressed that this mechanism is not the only reasonable one for this reaction. Any reasonable mechanism should avoid an substitution, however.. Make: C C, C C. eak: C, C. The light suggests a free-radical or pericyclic reaction is operative in at least part of the mechanism. aph C + C t-buk hν, liq. aph The base may deprotonate either C or C. Deprotonation of C makes it nucleophilic. We need to form a new bond from C to C via substitution. The mechanism of this aromatic substitution reaction could be addition elimination or. The requirement of light strongly suggests. ee Chap., section C., for the details of drawing an reaction mechanism. aph C + C hν, liq. C aph After the substitution is complete, all that is required is an aldol reaction, dehydration by Ecb, and deprotonation. Workup then gives the product.

15 Chapter C aph aldol aph Ecb work-up aph aph aph Alternatively, deprotonation of C makes it nucleophilic, and an aldol reaction and dehydration by Ecb gives an enone. aph C + C aldol Ecb C aph We still need to form C C. Deprotonation of C gives a dienolate. The more stable, (E) isomer will form. Light causes this isomer to isomerize to the (Z) isomer. An electrocyclic ring closing, which may also require light because it destroys aromaticity, gives the C C bond. Expulsion of and deprotonation gives the conjugate base of the product. C aph t-buk aph hν hν aph aph aph aph. Make: C C, C, C C. eak: C.

16 Bn C Chapter 0 C PhC Bn 0 C + The combination of an amine and an aldehyde under weakly acidic conditions almost always gives an iminium ion very rapidly. uch a reaction forms the C bond. ucleophilic C can then attack this iminium ion to give a new iminium ion. We still need to make C C. Deprotonation of C gives a neutral enamine and a,-diene. Cope rearrangement of the diene gives the C C bond, but it breaks the C C bond that was just formed! owever, C can be made electrophilic again by protonation of C0. Attack of nucleophilic C on C gives an iminium ion again, and deprotonation of C gives the product. Bn Bn ~ + C C Bn Bn C ~ + C Bn Bn + C C Bn 0 C + Bn C

17 Chapter Bn Bn C + C. First reaction. Make: C, C C. eak: C i, C. Dbs i Ph C ZnI, Et Dbs The combination of an amine and an aldehyde under weakly acidic conditions almost always gives an iminium ion very rapidly. uch a reaction forms the C bond. ucleophilic C can then attack this iminium ion to give a carbocation. Fragmentation of the C i bond gives the product. Dbs I Zn Dbs I Zn ~ + i Ph i Ph Dbs I Zn Dbs i Ph i Ph Dbs Dbs i Ph econd step. Make: C C0. eak: C.

18 Chapter Dbs 0 cat. (Ph P) Pd( CCF ) i-pr Et Dbs 0 The catalytic Pd complex and the aryl bromide together suggest the first step is oxidative addition of Pd(0) to the C bond. (The reduction of Pd(II) to Pd(0) can occur by coordination to the amine, β-hydride elimination to give a Pd(II) complex and an iminium ion, and deprotonation of Pd(II) to give Pd(0).) The C0 C π bond can then insert into the C Pd bond to give the C C0 bond. β-ydride elimination then gives the C C π bond and a Pd(II), which is deprotonated by the base to regenerate Pd(0). The overall reaction is a eck reaction. Dbs 0 Dbs L Pd 0 insertion II Pd L coordination, insertion Dbs II L Pd β-hydride elimination Dbs + II Pd L 0 L Pd

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