CHAPTER 16 HW: CONJUGATED SYSTEMS

Transcription

1 APTER 6 W: JUGATED SYSTEMS NAMING PLYENES. Giv th IUPA nam for ach compound, including cis/trans or E/Z dsignations whr ndd. ompound no E/Z trans or E Nam trans-2-mthyl-2,4-hxadin 2-mthoxy-,3-cyclohptadin ompound boxd = zam zid Nam (Z)-3-chloro-,3,5-hxatrin JUGATIN 2. Which of th following molculs or ions has a conjugatd π systm? 3. Explain why th alkns in compound A ar conjugatd whil th alkns in compound B ar not. Us orbital diagrams with your answr. A 3 3 Ths alkns ar conjugatd. Th p orbitals can align and ovrlap ( can dlocaliz). B 3 3 Ths alkns arn t conjugatd. Th cntral carbon has an alkn on ithr sid, and has to us a diffrnt p orbital for ach π bond. Th p orbitals usd thrfor hav to b 90 from on anothr, which prvnts thm from ovrlapping. 4. Which should b th major product of th following limination? ifly xplain. 3 Na DMF or or This product is th lowst nrgy: it has a trisubstitutd alkn that is conjugatd to th first alkn. Pag

2 5. Which bst dscribs th bond lngth of bond c? ifly xplain your answr. a).32 Å b) Å c) >.53 Å.53 A.32 A bond c but longr than a doubl (btwn Angstroms). Thr is som π orbital ovrlap on bond c, so it has partial doubl bond charactr. It should thrfor b somwhat shortr than a singl bond, 6. Estimat bond angls a and b. Thn us orbital diagrams and words to xplain why th bond angls ar not th sam. LPs in sp3 a b LP in sp 2 LP in p Bond angl b is 20. Th oxygn atom is nxt to an alkn, so its lon pair (LP) is in a p orbital. onjugation of th LP with th alkn lowrs its nrgy. Bond angl a is Th oxygn atom is not conjugatd (is nxt to an sp 3 cntr), so it is normal and has angls typical for an sp 3 hybridizd atom. 7. Napthaln and Anthracn ar whit solids, whil Ttracn is an orang solid. Us a gnral discussion of molcular orbitals (Ms/LUMs) to dscrib why ttracn is colord and th othr compounds ar not. Napthaln Anthracn Ttracn Aromatics and small polyns absorb UV light, but as thy bcom mor conjugatd (lik with ttracn), th M-LUM gap shrinks. This lads to lowr nrgy absorptions, which can b in th visibl rang. Ttracn must hav a M-LUM gap that corrsponds to a visibl wavlngth, lading to it bing colord (som λ absorbd). Th othr compounds don t absorb visibl wavlngths, so rflct all visibl λ and appar whit. 8. Th common acid-bas indicator phnolphthalin can xist in four forms, two of which ar rarly ncountrd (at p lss than 0 and gratr than 2). Th two common forms of phnolphthalin ar shown blow. n form is found in a p of and is colorlss. Th othr form is found in a p of and is a fuchsia (pink) color. Prdict which structur is colorlss and which is pink. ifly xplain your answr. Form 2 olor olorlss Fuchsia (pink) Th colord form is th highly conjugatd on (right). In th structur on th lft, th cntral carbon nar th oxygn is sp 3 and prvnts ovrlap of th p orbitals from th various rings. Also th fuchsia form is found in solutions with a high p, and anions ar common forms in basic solutions. Pag 2

3 MLEULAR RBITAL DIAGRAMS 9. Draw th molcular orbital diagram for ach molcul or ion, considring only π orbitals. Show th lctronic occupancy of ach orbital and idntify th M and LUM. Structur π 6 π 5 π 4 π M.. s LUM π M LUM M π 4 LUM M π π π 4 can also b Structur 3 LUM Et 3 LUM π M.. s 3 LUM M Et 3 M 3 π M π Et 3 π 0. Explain how th π M.. s for th allylic anion (drawn in th prvious problm) corrlat with th ion s rsonanc structurs, which dmonstrat that th ngativ charg is locatd on both nds of th ion, but not in th middl. δ δ Th rsonanc structur of th allylic anion has partial ngativ chargs on th nds of th structur, but not on th cntral carbon (as shown in th hybrid abov). This corrlats with th π M.. s bcaus th M (π2) rprsnts th lon pair. Th M has lctron dnsity on th nds of th structur, but not in th middl, just lik in th rsonanc hybrid! Pag 3

4 . Rgarding th radical blow: ow many total lctrons ar in th π systm? Draw th lowst nrgy π M.. Draw th highst nrgy π M.. 7 π DIENE REATIN WIT X 2. For th following raction, () (2) a. Draw th curvd arrow mchanism that accounts for th formation of both products () and (2). b. Which nrgy diagram would rprsnt th pathway to product (2)? c. Explain with structurs why vry littl (if any) of this product is obtaind: This product would form from th allylic carbocation abov. This carbocation is highr nrgy than th on that lads to th major products bcaus it has a δ on a and cntr, whil th lowr nrgy carbocation has a δ on a and cntr. Pag 4

5 3. Giv th two possibl monoaddition products of ach raction blow (assum quivalnt of X), procding through a mchanism whr only th lowst nrgy carbocation is formd. a. 3 lowst nrgy b. lowst nrgy 3 KINETI VS. TERMDYNAMI TRL 4. For th following addition raction, -78 A B a. Explain why th major product is A, not B undr ths raction conditions. δ δ Th chlorid ion is gnratd in clos proximity to th partial positiv at position #2. It taks lss nrgy to ract at #2 (highr ffctiv concntration, dosn t nd to disrupt solvnt). In cold conditions (-78 ), thr isn t nough availabl nrgy to ovrcom th activation barrir to ract at position #4 (raction is controlld by kintics). b. What should b th major product if th tmpratur is incrasd to 50? Explain. W ll assum that at 50 thr is nough nrgy to ovrcom any activation barrir (although this is rally xprimntally dtrmind), thrfor th chlorid could ract at position #2 or #4. Th raction is thn controlld by thrmodynamics, or th product stability. Product B should b th major product at 50 thn bcaus it is lowr nrgy than A: it is a mor substitutd alkn (disubstitutd compard to monosubstitutd). Pag 5

6 5. Explain why in th raction blow an incras in tmpratur dos not caus a diffrnt product to form. I -50 I,2 addition,4 addition I I or Thr is only on possibl product:,2 and,4 addition giv th sam thing. 6. Giv th major product of ach raction. Assum quivalnt of ractant in ach cas. a. c Kintic control (,2) Kintic control (,2) D D b. 50 Thrmo control (mor substitutd alkn) DIELS-ALDER REATIN d. 35 Thrmo control (mor substitutd alkn) 7. Giv th major product of ach raction. Show strochmistry whr ndd (cis/trans, ndo/xo). Indicat if a racmic mixtur is formd. 3 a. 60 o 3 (rac.) or 3 (rac.) b. (rac.) trans c. 3 3 (rac.) trans Pag 6

7 8. Giv th major product of ach raction. Show strochmistry whr ndd (cis/trans, ndo/xo), but don t worry about strochmistry for f. Indicat if a racmic mixtur is formd. a. c d b a a c b N 2 d 2 N (not rac: achiral) b. b c d a 3 f 3 f cis a b 2 c d 2 (rac) cis (rac) c. (rac) d. trans (rac) trans f Pag 7

8 9. Explain why th major product in this raction is, not D. Us diagrams with your xplanation. (60%) D (40%) Th ndo product is favord bcaus th transition stat is lowr in nrgy (product is formd fastr). With th ndo approach thr is a scondary orbital ovrlap (xtra bonding /stabilization) btwn th din and th group. Mor bonding in th transition stat lowrs th nrgy. 20. Explain why and how a racmic mixtur is formd in this Dils Aldr raction. Draw both nantiomric products. (50% ach is a racmic mixtur). Th din is flat, so th dinophil approachs with no prfrnc from th top and bottom. Th two approachs produc th diffrnt nantiomrs 2. Explain why a racmic mixtur is not formd in this raction. Th dinophil still approachs th din from th top and bottom, but th products from both approachs ar idntical. Th product has an intrnal mirror plan of symmtry (slic btwn th 2 s of th bridg) so it s achiral. Enantiomrs (or a racmic mixtur) ar not possibl with an achiral product. 22. Although compound E has four doubl bonds, only two of thm can ract with malic acid through a Dils-Aldr raction. Explain, thn draw th xpctd major product. 2 E 2 Malic acid To ract in a Dils-Aldr raction, th din must b in th s-cis conformation. In compound E, only on din is in th appropriat conformation (s-cis) to mak a ring. Pag 8