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by: Michelle Gulgowski DVM


Marketplace > University of Kentucky > Chemistry > CHE 230 > ORGANIC CHEMISTRY I
Michelle Gulgowski DVM
GPA 3.91


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This 31 page Class Notes was uploaded by Michelle Gulgowski DVM on Friday October 23, 2015. The Class Notes belongs to CHE 230 at University of Kentucky taught by Staff in Fall. Since its upload, it has received 28 views. For similar materials see /class/228298/che-230-university-of-kentucky in Chemistry at University of Kentucky.




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Date Created: 10/23/15
2 Alkanes 21 sp3 Hybridization Remember that C has four valence electrons in the con guration 2s22px 2py where pX and py are two of the three p orbitals in shell 2 Since C has two singly occupied orbitals we might expect that it would form two bonds In fact C almost always likes to form four bonds as in methane CH4 How does C accomplish this The rst thing we can do is promote one electron from 2s to 2pz Then we have the con guration 2s 2px 2py 2pzl with four singly occupied atomic orbitals Now we can say that each orbital is used to form one bond to H in CH There is still a problem though The angle between each of the p orbitals is 90quot so we might guess that three of the four C H bonds in methane are 90 apart In fact every bit of evidence shows that all four C H bonds are equivalent convince your self with a model w can we account for the true structure of methane We must mix the four valence orbitals 2s2px2py2pz of carbon to form new hybrid orbitals this process is termed hydbridization i C 89gtH H I We call the new orbitals sp3 hybrid orbitals because they consist of 1 part s orbital and 3 parts p orbital They point at 109 angles from each other that is to the four corners of a tetrahedron 22 Derivatives of methane We can replace one of the H atoms in methane with another atom or group These atoms or groups are called heteroaz oms Examples Methyl bromide methyl iodide methanol methylamine methanethiol Here the central C atom is not quite as perfectly tetrahedral but for our purposes it is close enough There are three ways we can remove a hydrogen from methane see Jones Fig 216219 We can remove just the H nucleus and leave behind the two electrons in the bond this gives CH3 with sp3 hybridization It has the same structure as ammonia just one fewer proton and neutron in the nucleus in the center We can take away the H39 This gives us CH3 The hybridization here We don39t want to waste valuable low energy s orbital in a hybrid orbital in which there are no electrons Instead the valuable s electron is divided equally among the six electrons the three bonds remaining Three bonds means three orbitals one s and two p That means the third p orbital remains unhybridized This situation is called sp2 hybridization Two p orbitals occupy a plane so in sp2 hybridization the three hybrid orbitals are coplanar They point 120 apart to the corners of a triangle The unhybridized p orbital is perpendicular to this plane What if we take away the H nucleus and one electron We have a situation in which CH3 methyl radical is between sp2 and sp3 hybridization It39s a shallow pyramid These species are high in energy and don39t exist for long The word quotmethylquot also refers to a CH3 group that is attached to the rest ofa molecule In general the root quotmeth quot means quota onecarbon groupquot The abbreviation Me is often used for CH3 80 Mel is the same as CH3I 23 Ethane If we replace one of the H atoms in methane with another CH3 group we get a compound called ethane ln ethane checHsr each c atom has tour orrtereht groups attached so each c rs splhyorrorzeo Each OH ooho rs tormeo oy oyerTap or a Csp orortaT wrth a Hms orortaT The cc ooho rs tormeo by the oyerTap or a CSp orortaT trom each c atom AH or these bonds are o ooh s o oh has cyhhorrcaT symmetry rt you Took down the am or the ooho rt ooks We a crrcTe As you H see rh a bTL hot aH oohos haye that property e way we represent s oohos rh orgamc compounds rs to Just use hhes Thus we represent the structure or ethahe as f1 GeneraHy we condense the structure to CHTCHL rh whrch omy the most rmportaht ooho the co ooho rs showh 24 Conformational stereoisomers of ethane Let s take a doser ook at ethahe The crc o oohd rh ethahe has cyhhdrrcaT symmet y so we can rotate the two CH3 gr ups wrth respect to one another wrthout chahgrhg the amourrt of oyertap between the two sp orbrtats makmg up the o oohd As a reamt the rotatroh rs yery tache ahd yery fast Thrs rs most ooyrous rt you make a mode of eth ahe n a The three structures aooye haye the same atoms attached to each omen out the shapes or the compounds are orttereht As a Tesu t they are stereorsomers These partrcutar stereorsomers can be rnterconverted srmptv bv rotatrng about o bonds so tnev are caHed contormatrona stereorsomers or cunfnrmersfor snort Because tne rnternat drmensrons drnedrat angtes HrH drstances or tne tnree contormers above are dtfferent tnev are contormatrona drastereomers Conrormers usuaHv rnterconvert so raprdtv tnattnev can t be separated trom eacn otner ans pnenomenon rs caHed free rotatron Tnere are tnree ommon wa s or draw above We caH rt sawnorse ecause t at s wnat rt ooks We Anotner wav or drawrng contormers rs caHed tne man pmecnm m tnrs projecuon we took drrecttv down tne aws ott e C7 o bon Wmch rs represented as a crrcte We can see tne prowmat c atom but tne drstat c atom rs obscured b tne crrcte or tne o bond Tne bonds to tne prowmat c atom are ruuv vrsrbte but tne bonds to tne drstat c atom are pamy obscured Tne tnree etnane contormers drawn above are redrawn betow as NeWman prdectrons We atso use er ectrve drawrngswrtn tne Wedged nasned nne tormansm to rndrcate drrterent contormatrons Tnese a atso snown betcw Ht n L u L t g Y f lxiniQ tn u 1 t at Ef a at t 1 a H n H quot N n r n gtriv gt t c m H H n u H n n Problem 1 Draw aH ecnpsed and staggered contormatrons Lbromor rcmoroet ane W rc o tnese stereotsomertc 0 one another WWCH are tdenttcat g Because the different conformers of ethane are diastereomeric we can expect that some of them are higher in energy less stable than others This is in fact the case The first and third conformers that I drew are called the staggered and eclipsed conformations of ethane The terms refer to the mutual arrangements of the 0 H bonds The eclipsed conformer is illustrated most dramatically by the Newman projection The eclipsed conformation is higher in energy than the staggered conformation The actual difference in energy is 29 kcalmol 12 kJmol which is a small but measurable amount At room temperature every 135 kcalmol difference in energy means a 10 fold difference in ratio 80 rotation about the 0 0 0 bond is not perfectly free but passes through an energy barrier of 29 kcalmol every time it passes through the eclipsed conformation We can draw a graph of energy versus the dihedral angle between two C H bonds See Jones Fig 224 The dihedral angle of W X Y Z is the angle that we see when we line X and Y up so that one is laying on top of the other as in a Newman projection More formally it is the angle between the line defined by W and X and the plane defined by X Y and Z A number of points can be made regarding this graph First of all note that the eclipsed conformers represent energy maxma A molecule in the eclipsed conformation is at unstable equilibrium and it will not remain there for more than one molecular vibration As a result we say that eclipsed conformers are transition states for the interconversion of the staggered conformers The reason traditionally given for the lower energy of the staggered conformer is that the 0 H bonds consist of electrons and electrons repel one another so the eclipsed conformation where the can bonds are an ahgned experiences a greater amount or eiectronic rebuision than the staggered contorrnation Recenti though this expianation has been shown The reason that the staggered conformation is lower in energy is that in this conform 39 lt03 deiocaiizie the electrons iowerin their energy interaction is better in the staggered conformer because the i iobe or the o orbitai oyeriaos With the big iobe of the 0 orbitai in the echosed conformer the big iobe of the o orbitai oyeriaos With the srnaii iobe of the 0 orbitai 2mm csenai new mime e u e e three eciiosing interactions ch eciiosed conformer We can diyide up the totai energy or the eciiosed conformer 2 kcaimoi into i o kcaimoi tor each Chceht destabiiizrng interaction The tack or stabiiization in the eciiosed contorrner or ethane is caiied torsionai strain We WiH see soon that there are other hnds or strain that can iead a oarticuiar rnoiecuie o contorrner to haye higher than expected energy 25 Derivatives of ethane We can make denyatiyes or ethane by reoiacing the H atorns in ethane itiust one or the H atorns in ethane is reoiaced by a different group then the CZH5 group is called an ethyl group So CzH5OH is called ethyl alcohol CzH5C is called ethyl chloride etc Another way of naming these compounds is to call them derivatives of ethane So CzH5Br in which one bromine replaces a H is called bromoethane By this logic we might call CzH5OH hydroxyethane However the OH group is so important that it is given its own suffix o as in alcohol 80 CZH5OH in which one H atom of ethane is replaced by the OH group is called ethanol The word quotethylquot also refers to a CH30H2 group that is attached to the rest of a molecule In general the root quotethquot means quota twocarbon groupquot with associated H atoms The abbreviation Et is often used for CH30H2 80 EtOH is the same as CHscHZOH 26 Propane amp its derivatives Skeletal isomerism Higher alkanes can be made by replacing more H atoms Replace one of the H atoms of ethane it doesn39t matter which they39re all equivalent with another CH3 group We now have a threecarbon compound propane If we look down one of the 0 0 bonds of propane we see that we can have different conformers There is one kind of staggered conformer and one kind of eclipsed conformer The barrier to rotation about the 0 0 bonds in propane is 34 kcalmol which is higher than the barrier for ethane 29 kcalmol see Jones Fig 230 The reason is that C is slightly more electronegative than H 25 vs 21 on the Pauling scale so the 0 0 0 orbital is slightly lower in energy than the 0 H o orbital so the hyperconjugative interaction between the 0 H 0 bond and the 0 0 0 orbital in staggered propane is slightly more favorable than that between the 0 H 0 bond and the 0 H o orbital 15 kcalmol vs 10 kcalmol Propane has a property that is not present in ethane or methane There are two different kinds of C atoms in propane and there are two different kinds of H atoms The terminal CH3 groups in propane which are indistinguishable or equivalent are called methyl groups and the middle CH2 group is called a methylene group The H atoms on the CH3 groups are called primary 1 and the H atoms on the methylene group are called secondary 2 If we want to make a derivative of propane by replacing a H atom with a heteroatom we now have two different H atoms we can choose from If we replace a 3 H atom with X we get a different compound than if we replace the middle H atom Replacing a 3 H atom gives what are called propyl compounds or oldfashioned npropyl compounds Replacing a 2 H atom gives isopropyl compounds 80 ifX OH we have isopropyl alcohol or isopropanol rubbing alcohol The abbreviations Pr and iPr or iPr or iPr are often used for propyl and isopropyl lsopropanol and propanol have the same chemical formula but the atomtoatom connections are different They are called skeletal isomers aka constitutional isomers Skeletal isomers have different properties taste smell melting point boiling point everything They are as different as if their elemental compositions were different This concept is incredibly important The different properties of skeletal isomers is one reason why there is such a diversity of organic compounds We have already met another kind of isomer Eclipsed and staggered ethane or propane have the same atom atomatom connections but different shapes They are stereoisomers Skeletal isomers and stereoisomers are the two different kinds of isomers Stereoisomers can be further subdivided into four types We have already met conformational diastereomers one of those four types We will soon meet the other three types The root quotpropquot means quota threecarbon group with the assoctateo hydrogerts 27 Butane 3 its derivatives Suppose we want to add one more carbon to propane We naye two otrrerent ktrtdS or H we can reptace a rnetnytene H or a rnetnyt H tr we reptace tne rnetnyt H we n buta e tr rt C trtvtat namet tsobutane More Systemattcatty we can htrtk or tt as propane n t roup on tne second carbort or 27 rnetnytpropane att one word tr we took down tne ctec2 o bond we get a tcture tnat ts tdenttcat to that of propane exceptmatme Me tn pro ane ts reptaceo by Et Tne grapn or energy ys otneorat angte about tne ctec2 bond tn butane doesn t took terrtbty otrrerent rrorn E two ecttpseo conrorrnattonat otastereorners 51 and 52 tn sz tne two Me groups are as rar apart as posstbte wttn a otneorat angte or 180 tt ts catteo tne antt conrorrner tn st tne two Me groups are ctose to one another wttn a otneorat angte or 60 tt ts catteo tne gaucne conrorrner S1 ts o 6 kcatmot ntgner tn energy tnan 52 The terms anti and quot a arm rrorn tne two Me groups trytng to occupy approxtmatety tne sarne regton or space mm Among the ecttpsed conformersy E1 has two CrH CrH interactions and one C MeC Me and E2 has one C H C H and two C H C Me Neither has good hyperconjugative interactions so the difference between them must be due to steric interactions It turns out that the steric cost of the eclipsed C MeC Me interaction in E1 is 10 kcalmol considerably worse than the steric interaction in E2 We can draw a graph of energy vs C C C C dihedral angle show see Jones Figure 235 Again the eclipsed conformers are transition states but the energy required to go through one of the eclipsed conformers is appreciably higher than the energy required to go through the other Also the gauche conformer is higher in energy than the anti conformer This doesn39t mean that all of the butane in any particular sample exists only in the anti conformation There is an equilibrium between the anti and gauche conformations and the equilibrium constant is calculated by K e AG RT R is the universal gas constant 1987 calmolK and T is the temperature at which the measurement is being made usually room temperature 295 K Using this formula we determine that for the anti gauche equilibrium K 46 in favor of the anti conformer So in a given sample of butane approximately 80 of the material is in the anti conformation and approximately 20 is in the gauche conformation Actually these numbers are not quite right because we didn t consider that there are two gauche conformations and only one anti but it gives us a ballpark estimate For larger alkanes we can conduct similar analyses We will see that the most stable conformation is the one in which all C C C C dihedral angles are 180 Problem 2 Of the different conformers of 1bromo2 chloroethane that you drew earlier which are anti and which are gauche Let s ook rnore ctosety at tne W0 gaucne contorrners ot outane tnose wrtn cececec orneorar angtes ot 00 and 300 Tnese W0 contorrners have tne sarne rnternar orrnensrons eg tne sarne orstances between C1 no tne sarne cececec orneorar angte etc Yet tne two tne corn ounos nrnoosaore One rs nonsuoenrnoosaore rmrror rrnage ot tne otner Tnese two structures are contormatronaenantrorners Even etnane nas enantrornenc contorrners Two ot tnese contorrners are shown oerow m eacn contorrner two H s are abeHed tor reterence Tne HtCrCrH orneorar angte ot tne contorrner on tne e rs somewhere between 0 and r 30 m tne co torrner on tne ngnt tne an e rs 7 30 fyou make rnooers ot thesetwo contorrners you Wm see tnat they are nonsuoenrnoosaore rmrror rrnages ot o e anotner AH tne rnternar orrnensrons atomratom orstances orneorar angtes oono angtes are denucaL out tne two structures are not mutuaHy supenmposame IL say H H H ortterent kmda ot orornooutanes Tne re caHed nebuM orornroe secrbum or sebutw orornroe rsooutyr orornroe n and tertputyt or Lbuty orornrde Tne rnore torrnat narnes are torornooutane Zebromobutane t rornoz rnetnytpro ane and Zebromorzrnemy propane However tne tnwat narnes tor tne tour outyt groups are very cornrnonw used Tne apprewatrons anu or nBu rrBu orrBut srBu or 58d and Bu or tBu are otten used tor tne tour outyt groups wnen mat Bu rs used rt reters to anu T e root out rneans a fourrcarbon group wrtn tne assocrated hydrogens 28 Pentane 3 its derivatives From outane and rsooutane we generated tour drtterent s so rt woutd seern we c e er te tour drtterent kmds of pentanes However reptacrng a 2 H m outane wrtn cHs gwes exac y tne sarne compound as reptacrng a 1 H n Zmemy propanet Botn reptaoernents gwe Zmemy butane or rsopentane eptaorng a 1 H n utane gwes pentane and reptacrng tne 3 H m Zmemy propane gwes Z ZHdtmemy propane or neopentane Two very trnportant pomts now 1 Orgamc onernrsts are tazy Once we getup to twe or rnore 3 g LO Q o o a o 3 U 2 o 3 lt o Latcuwnsm m There are many dlffererlt Ways to draw a compound such as pentane The mutual onentatlons of the dltterent llnes can change and the compound ls stlll pentane The same ls true of any other compound What matters ls the connecllons netween the atoms not the relallye poslllons orthe atomsl V y 7 1 L One more polnt Methane has thetormula cm ethane czhla propane CsHa outane and lso utene 0 my and the three entanes came The formula for any acyclic alkane is Problem 3 How many promopentanes are mere 29 Higher alkanes N menclature W earned aoout methane ethane propane outane and pentane Aooye pentanethe roots of the names d ye trom the Greek Word tor that number hexane alkarles name tells you how many c atoms there are the umoer of H atoms ln an alkarle ls glyen by the tormula cnhlm A oranched alkarle ls named as follows aFlrld the longest stralght chaln tn the molecule That proyldes the root of the name to Then descnoe the suostltuents on that long chaln lt the subsmuent has one carbon n 5 memw v we emw etc The posmon of me subsmuent 5 descnbed by a number w there 5 a meme the number of me rst subsmuent Shomd be as ow as posswb e w there 5 more than one subsutuent n ubsmuent of dwfferent kmds the are hsted a phabeucaHy and he rst Subsutuent m the name SHOU d havethe OWest number You can use IrPr srBu IVBU and 7 Bu as subsmuent names 1 mnemwue 2 new 3 3 dmh umgt Waugh 210 Cycloalkanes cycUpema e m fact we can make cychc a kanes of any swze A threermembered mg 5 cydobrobane a mun membered mg 5 cydo mane a swmembere mg 5 cycbbexane etc The number of H atoms m a cydoa kane 5 gwen by me formu a CHHZH When we have a cydoa kane w m a 5mg e subsutuent for exam e a uonne atom on a w membered Hng we caH n uorocydohexane No number 5 needed because the mg 14 nas no oegmmng and no end We can have atkyt substttuents on an atkane for exampte tsopropytcyctopropane n we n e more than one suosmuent the atphabettcatty rst substttuent 5 at posttton 1 an t e 0t er substttuent t5 numbered retatwe to the rst one and as tow as posstbte for exampte Muoro Srtodocyctooctane not 1r uoro wdocyctooctane Let s took at cyctopropane from the Stde Tne three c atoms 2 o o m a 3 m g 0 compounds have dtfferent snapes Tne onw Way to convert one of them mto the other 5 to break a pond Because they have dtfferent snapes they have dtfferent u u u u u u u u u propemes For exampte the one thh the W0 Br atoms o o m 3 o a o g a LG 3 m E Q Lo 9 g u 9 o 3 o 3 g than the otner Tne one thh the two ratoms on opposne StdeS t5 mucn thtcker than the other Here a am we have sterotsomers because the atomrtc atom connecttons are the same but the Structures are not denttcat tn ms pamcutar case the tntemat dtmenStonS of the compounds are dnrerent took anne BrrBr dtstance and the BrrCrCrBr dmedrat angte so the compounds are 15 diastereomers However these compounds are not conformational diastereomers because they can39t be interconverted by rotations without breaking covalent bonds As a result they are called con gurational diastereomers our third kind of stereoisomer Configurational diastereomers have different melting and boiling points different reactivities different spectral behavior in short they are completely different chemical entities If you could separate conformational diastereomers the same would be true of them however conformational diastereomers usually interconvert so rapidly on the time scale in which we operate that we can39t separate them and examine their properties As a result we don39t usually think of conformational diastereomers as being different compounds In fact if someone uses the word quotdiastereomerquot without specifying conformational or configurational you should assume that they mean configurational diastereomer Two groups that reside on the same side of a ring are called cis Two groups that reside on opposite faces of a ring are called trans The two compounds above would be called trans12dibromocyclopropane and cis12 dibromocyclopropane Any time you have at least two C atoms each of which has two nonidentical groups attached in a ring of any size you can have cis trans isomers We used the hashed wedged line convention to denote cis trans isomers Eg for 12dibromocyclopropane we write the following Some people use dashed lines instead of hashed lines but dashed lines can have several different meanings so hashed lines are better Note that the doubly wedged and doubly hashed structures that are drawn for the cis isomer are identical if we pick one up out of the plane of the paper flip it over and put it back down we get the other This is not true of the two trans structures that are drawn these represent examples of 16 the tourth dass of stereotsomersy conhgurahona enanhorners Enarmomers are nonsupenrnposaoe mrrror rrnages of one another One atr of enanuomers that ts ye fammar to you tsyour hands Your tett hand ts an enanuomer of your no t and The two teature the same shape the same otnmetothumo dtstance etc but they are h suoenmoosaote rmrror tmages m m to Fr n m to u dlemuw vVNw rm 2 uhomutdduxtvxu Dtastereomers have dtfferent mtema dtmenstonsy dmedra ang es and dtstances between nonbonded atoms 7 for exam e the W0 r Whereas dtastereomers dtrter m energy Some compounds don t haye enanttomers tor examp e 0157127 dtoromocyctoorooane ts rdentrca to tts mtrror tmage We Wm tath aoout enanuomers m much greater detatt soon t make models to understand the concept of stereoisomerism of cycloalkanes Note wellthat rtt use the term rsomersj t coutd be reternng to any W o tsorner constttuttonat tsomer Stereotsomer dtastereomer or enanuomer Problem 3 Whrch or the cychc sketetat rsomers or cam and ahh2 haye ms and trans mowers Draw them usrng the hashedwedged hne conyentron Many orgamc compounds haye more than one nng A compound can haye nngs that are tsotated Sptroy tused or 17 ondged dependmg on Whether they share 0 atorns Any athane wrth m nngs m rt has the to m t 2 or gt2 rrnuta chm 1 mary s far Two structures Wth the sarne e m r sm sum 0 formu a are tsom r5 f the ave dtfferent atomrtoratom eorsorners that have drtterent drtterent drhedrat angtes atomrtoratom drastereorners turther dryrded rnto tour types 1 ontorrnatronat drastereorners haye the sarne atomic atorn connectrons out drtterent rnternat drrnensrons and can be conyerted rnto one another easHy oy rotatrng around o oonds Eg echosed and staggered torrns of et ane Contorrnatronat enantrorners are nonrsupertmposabte can be conyerted rnto one another by rotatron around 0 bonds They aye the same atomrtoatom connecuons and the same mtema drrnensrons E9 the two gauche contorrners of outane Connguratronat drastereorners haye the s atomrtor atorn connectrons out drtterent rnternat drrnensrons and cam conye n rnternat drrnensrons drstances etc are N u at be A o 8 E Q g g g m 3 at one another by rotatron around 0 bonds They aye the same atomtoatom connections and the same internal dimensions Eg the two structures for trans 12dibromocyclopropane You need to make models to understand these points pKa is an Energy Term 0 A proton abstracted by a base is generalized in the following reaction AH B A BH 0 We will worry about the descriptive details of how this happens later 0 The half reaction in water is AH H20 A H30 0 has an equilibrium constant associated with it 39 Keq AHsOAHH20 o The terms in this equation are concentration The molarity of water H20 is 55 M or 55 molesLiter o The acidity constant is Ka KeqHZO AlleolAH Considerthis equation Ka KeqH20 0 Taking the negative log of both sides and multiplying by RT gives 0 RTogKa RTlogKeqH20 RTlogKeq RTogHZO R is the gas constant and T is the absolute temperature 230RTlogKa RTLnKeq RTLnH20 but ogKa pKa and RTLnKeq AG 230RTpKa RTLnH20 AG Result PKa increases linearly with energy conj bases AG species with lone Acids pairs Deprotonation Reaction Coordinate The species with lone pairs conjugate bases respond more to structural changes than the acids 111 Addition of x to Alkeuex and Alkynex to give 1 Dihalidex We have rheriticried that aikeries arid aikvries are rict su iciehtiv ac riucienphiies ta react With aikvi haiides HnwevEr thev are su icieritiv quad riuciecphiies tc react with eieriieritai haineeris X1 ta h th w g 33 E E a the in r ieht eiectrcrieeative scam in tead a riucienphiie r in art direcuv Sirice each ar aireadv has its actet the arc ar hand must break at the same time as the new band tn the riucienphiiercrriis crie crthe erminE atcriis acts as aieaving group L a Wee a Aikeries react with art arid ch ta 9W2 127dihainaikan25 cr 127 dihaiides This additmn reacticri is usuaHv carried nut in a HDH39 riucieuphiiic sciverit iike dichicrcriietharie R 114 Br ta Tp m m i if it in We can think cr the reactmri rciicwirie a mechanism siriiiiar tn the crie we discussed rcr additmn ci a strung acid tc ari aikerie the aikerie attacks crie ar in art Just the same as aikerie attack cri H in HBr orice attack cccurs a carbncaunn arid Br39 is generated Theri ar39 a riucieuphiie adds tn the carhccaticri tn ENE the ubserved Druduct 7 E39im R x t a et iricrriiaticri abuut maher rhech i t reactmri cr Evciupentme With art in the rst stED we wcuid get a carbncaunn With a ar atcrii ah the heiehhcririe carbun This carbocation could combine with Br39 to give two possible diastereomers The Br39 could attack from the same face of the ring as Br resides to give a cisproduct or it could attack from the opposite side to give a trans product Since a Br atom is larger than a H atom we would expect to get more trans than cis but not a whole lot more 1 Br 39 gBr Br gt S I quot H H H H Br Br H H H H b Br H In fact we get exclusively trans Our mechanism doesn39t explain this fact How can we modify the mechanism so that it explains it It turns out that we had an incorrect structure for the intermediate carbocation Imagine the Br atom in the carbocation reaching over and forming a bond to the electrondeficient carbon atom using one of its lone pairs This gives a threemembered ring called a bromonium ion H H See Jones Figs 10121013 Also note that the bromonium ion and the 2halocarbocation are structural isomers not resonance structures since they have different atomtoatom connections The Br atom has a formal positive charge The C atom is no longer electrondeficient but it is still electrophilic because the Br atom which is electronegative wants to leave and have its lone pair back to itself again The Br39 comes along and attacks the C atom as its electrons come in to the C atom other electrons must leave so that the C atom doesn39t gain more than an octet the ones that leave are in the C Br bond which go back to Br The Br39 attacks from opposite the bond that breaks We obtain trans12dibromocyclohexane as product This is called overall anti addition because the two Br atoms add m DDDDstte stdes at the duuhte band Ah acvette substrate wttt atsu uhdetgu amt addmnn In the examp e at brnmtne addmnn tn cvetupehtehe We can see that eneS the atagen atnms trans m Dne anmher and the Dru uct ts a 12sdtbmmnatkene A secnnd addttmn can quDW 92 at 9 WE 4 e tn Ewe the tetrahahde see lanes g 1B 74 102 mhal yemtion 2ilaloaltwholiV ZrHanetherx We39ve seen nudeDDhth Br39 attack the etectrDDhth brnmnntum b m d A attack the Um DH addttmn reacttun ts eatted mhaogehaztoh cuhatneehattuh ts WU eummuhtv canducted bv addtnu atl m an atkehe th water h ateuhut a eathuxvtte actd as suheht Fur exampte t We add ant ah atkehe th th ahut We a tath a pm uct th whteh a CHzO39 emup ts theatpurated thtu the Drnduct tnstead at a ar39 emuu Thts Druduct ts eatted a thmne her h a a x x one at 1 t The rst Dart at the meehahtsth ts the same as hatneehatturt a ermDntum tuh ts farmed Then mstead at at39 actth as a hueteephtte tnward the ermDmum tart the atmth sutveht acts as a hueteephtte bv usthe tts nxvgen tn attack c and dtsutaee at m we a eattuhtc etectmnssaturated thtetthedtate FtnaHV the umhtuth tuh trtehhedtate ts deptatuhated m we the Drnduct Meehahtsth I u I 14 BI R R3 3 3 t 5r 9 9 H293 39 o ESE Q H R i 1 R 31 39 VO R Br HFR H k 3 31 BI 62 530 BI 2Q 13 9 R RI R H3 39 FL is R R3 When we make 2haloethers and 2haloalcohols we are adding one group to one of the C atoms of the double bond and a different group to the other The question of regioselectivity now arises For example what happens if we use 1methylcyclohexene as a substrate Remember that we said that 3 carbocations were more stable than 2 ones We can imagine that in the bromonium ion the Cl Br bond is weaker than the CZBr bond because C1 is better able to bear a positive charge Since the Cl Br bond is weaker this bond is more prone to cleavage Therefore MeOH attacks C1 Markovnikov addition takes place and the product is 1bromo2methoxy2 methylcyclohexane the second one as drawn Note that both regioisomeric products that might be obtained have Br and OMe in a trans relationship 9 Me M Br 2 CH30 H NOT B attack on C1 attack on C2 OxymerucrationReduction similary yields alcohols from akenes via a mercurinium ion and gives the Markovinkov addition product eg see below read more about it in your textbook texomareo e 2 NaEH 0H ht Marmnrhcuaddccn Vamu 113 Catalytic Hydrogenation Alkynex to Alkenex to Alkanex can Hvdruuenate a kmes bv aHDng them react nun H1 m the presence cr a name metax catahst hke pd an charma pdc cr ptot Thrs rs caHed areduczron It rs an were addmon reamnn n u 3 F H quot1 14 22 x quot d c x x t a SuDDDSE we start wwm 127mm2 wv cvc uhexene Twu steremsumenc crsc Dr transclZcmme wwcvduhexane In fact 0 the t ned r the a ene surf e Bf the rneta EJEEWSE hhrch remams undrssahed and recewes path rts H atarnsrrarn the sameface Thrs rs caHed svn addman Ms M am ye u H vs catahtrc hvdradenatran rs a useruw reacnan because rt usuaHv reduces DnW unpoanzedx hands We cc 3 bands pdarrzed 3 hands We 20 3 bands are reduced wrth ddner reagents such as hthrurn a ummum hvdnde UA HA The reductan Bf CN n dwf cukv and depends Damv an the nature cr the draups attached ta c d Th s d c ze d dn dthat eraups can be se ectwa v transfurrned mm twu d deuendmu an the chmce cr r eadert a demure a mm A quot 01 L u Catahtrc hvdrcdenatrcn has crten been used tc decernhne haw manv cr the degrees cr unsaturatrcn cr a ccmpcun Wrth a dwen furmu a are due tc the presence cr cc x bands smce Evdna kanes dun t react Wereth Drnduc sr rh W CHXI the ceh hdhd Ts rather acidic pKa 15 almost the same as H20 A strong base like tBuOK can deprotonate it to give a carbanion a trivalent C species with a negative charge C is a somewhat electropositive element and it doesn39t like to bear a negative charge To relieve this charge one of the X groups attached to C can leave as X39 with the electrons in the C X bond The C species that is left behind is a divalent sixelectron species called a carbene In this case a dihalocarbene X e 3 c3 Quiet c x 9 16 3 X As you might imagine carbenes are extremely reactive species even more reactive than carbocations Since they are electrondeficient they chomp on anything with a pair of electrons They react with the 7 bond of alkenes in a pericyclic no intermediate reaction The pair of electrons in the 7 bond goes to form a bond between Ccarbene and C1 of the alkene while the lone pair on Ccarbene goes to form a bond between the Ccarbene and C2 The reaction is stereospecific with trans alkenes giving trans12disubstituted cyclopropanes xl A gt rquot I Mtg The SimmonsSmith reagent works similarly The Zn inserts into one of the C I bonds to give a compound with a C Zn bond The C Zn bond is heavily polarized with a 539 on C so you can imagine that I39 might leave from C just like in the formation of a dihalocarbene This isn t exactly what happens which is why we call the SimmonsSmith reagent carbenoid which is a metalcomplexed reagent with carbenelike reactivity Thus the SimmonsSmith reagent reacts with alkenes more or less in the same way as a true carbene does The product is a cyclopropane The reaction again is stereospecific with cis alkenes giving cis12disubstituted cyclopropanes IZn CH22 ZnCu gt r quotCH2quot an 4 i H H I39 quotCH2quot Zl ll an2 gt F M Me M Me 810 Oxidation of Alkenes Epoxides 12Diols Carbonyls When alkenes are treated with a peracid RC03H an epoxide is obtained The peracid used most often in the lab is called mCPBA but any peracid even peracetic acid AcOOH will work A peracid is related to a carboxylic acid but instead of an OH group attached to the carbonyl there is an OOH grgup 1 R RJKOIO H gtlt R 1 R4 3 R5 is Alkenes are nucleophiles so peracids should be electrophiles In fact as in Brz the two electronegative O atoms are fighting over the pair of electrons in the 0 bond between them The CO group attached to one 0 helps it win the battle so the O atom attached to H is even more electronpoor The alkene attacks it displacing a carboxylate anion and the 0 being attacked uses its lone pair to make a bond back to one of the alkene C s to give a protonated epoxide The carboxylate then deprotonates the O to give the products w u R Iquot 0 v n O R I A3 We will now learn two other oxidation reactions whose mechanisms are too complex to discuss You must simply memorize these reactions It is important to learn them because they accomplish useful functional group transformations When alkenes are treated with one equivalent of OsO4 osmium tetroxide and then aqueous NaH503 a 12diol is formed This reaction is called dihydroxylation The overall reaction is an addition reaction H OH 3 4 0H R2 g R3 la 39 3 1 R3 1 R3 R 1 R4 Basic permanganate similarly gives vicinal diols 12diols KM no4 OH CHGOHHZO NoOH 20 0 OH There are no regiochemical issues in dihydroxylations because the same group is added to each atom of the former 7 bond The stereuchermstrv ct the reactan S stereDSDem caHv syn bdth OH umups add tn the same stde ct the duub e band cm the ms 1 Dbta he b s a k t 5 Emma mastereumehc Druduct the eurrespuhmhe ms a kene a sn Ewes a smu e Drnduct dwerert frnm the DHE derwed fmm the trans a ehe k 241 The c duub e band m a kmes can be oxydatrveV cleaved and remeeed wtth twu co duume bands Thts can be dune m severa Wavs We Wm Eam unh DHE mum when an a kene 5 treated wtth DZDHE 0 an then 5 mm reducmg agent such as Mels reductwe Wurkruu twu ketuhes Dr emehvdes are Dbtamed other a e used 2 e z reduer reagents bestdes M21 can b H thp and Even hi eatet M t t at t e 4 2 Nu R 2mm t o t a 2 We Ngt Ngtquot mote meme L 2 v25 t N 2 meme See Jones Figure 1062 Oxidative workup with H202 leads either to ketone or carboxylic acids oxidative worku p R3 1 03 R1 Ft gt gto gto R2 R4 2 H202 H R2 R4 2 KETONES R1 Ft gtlt 1 03 R1 Ft 39gt gto gtO R2 H 2 H202 H R2 HO I I KETONE Carboxyllc aCId R1 R2 1 03 R1 R2 gt F0 H H 2 H202 H HO HO 2 Carboxylic acid R alkyl The mechanism of this reaction is wellknown and it is in the book if you are interested but I do not expect you to know it Oxidative cleavages of alkenes have neither regiochemical nor stereochemical issues


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