Structure of Organic Molecules
Structure of Organic Molecules CHEM 14C
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CFQ amp PP Carbohydrates Reading 39 Bruice sections 221 224 2210 2211 2214 2217 2221 39 Klein section 77 Suggested Text Exercises Bruice Chapter 22 1 5 7 8 20 23 27 29 35 45 Lecture Supplement Carbohydrates page 36 this Thinkbook Tutorial Link on course web site Carbohydrate Representations Must I Memorize All These Carbohydrate Structures You do not need to memorize the structures of all the carbohydrate examples given in the Carbohydrates Lecture Supplement OWLS and Thinkbook practice problems are written with the assumption that you will need to refer to lecture notes your textbook etc in order to find molecular structures Frequent reference to these structures will help you become more intimately familiar with their similarities and differences These molecules are the stuff of life as so life science majors should become very very familiar with their structures You may be asked to reproduce one or more of the more important structures on an exam More important structures are usually also more commonly encountered so use this as a guideline to decide the relative probability that you might be asked to reproduce a given structure on an exam Concept Focus Questions 1 Provide clear and concise definitions for the following terms a Aldose g Furanose m Pentose b Amylopectin h Glycoside n Polysaccharide c Amylose i Haworth projection o Pyranose d Anomeric carbon j Hemiacetal p Saccharide e Disaccharide k Hexose q Tetrose f Fischer projection l Ketose r Triose 2 What similarity in the chemical formulas of sucrose glucose fructose starch cellulose etc lead to this class of compounds being named quotcarbohydratesquot 3 What is general name for a five carbon saccharide that contains a ketone Draw the structure of an example of this class of compounds 206 CF Q amp PP Carbohydrates 4 List two different features that J39 quot 39 D cai from L cai 5 Describe how you locate the anomeric carbon of a saccharide 6 Describe how you can determine if a saccharide has an CC or 5 configuration 7 In aqueous solution D glucose exists as an equilibrating mixture of three structures Draw these three structures 8 Brie y explain why glucose and ribose might be considered the two most important carbohydrates in biological systems 9 Why is starch easily digestible by humans but cellulose is not Concept Focus Questions Solutions 1 a Aldose A carbohydrate containing an aldehyde b Amylopectin A type of starch having both L4 and 16 linkages between the oc D glucose polymer chains c Amylose A oc D glucose polymer that has only 14 linkages between the glucose molecules d Anomeric carbon In a pyranose or furanose the carbon atom that lies between the two oxygen atoms of the acetal or hemiacetal group e Disaccharide A carbohydrate that can be J into two ides f Fischer projection A method of representing an acyclic carbohydrate structure The meeting of two perpendicular lines indicates a stereocenter Vertical lines at this stereocenter indicate groups that are pointing away from the Viewer as if on a dashed wedge Horizontal lines indicate groups that are pointing toward from the Viewer as if on a solid wedge g Furanose The five membered cyclic form of a monosaccharide h Glycoside A carbohydrate in which the anomeric OH group has been replaced by some other C 0 0 glycoside C N N glycoside or C C bond C glycoside The text book definition 39an acetal at the anomeric carbon of a saccharide is too narrow i Haworth projection A method of representing a cyclic carbohydrate structure The ring is draw flat with substituents pointing straight up or straight down j Hemiacetal A functional group containing C O C O H commonly found in carbohydrates CF Q amp PP Carbohydrates 207 k Hexose A monosaccharide consisting of a six carbon chain l Ketose A monosaccharide containing a ketone m Pentose A monosaccharide consisting of a five carbon chain n Polysaccharide A carbohydrate that consisting of many monosaccharides o ranose The six membered cyclic form of a monosaccharide p Saccharide A carbohydrate that cannot be hydrolyzed into simpler carbohydrates Also called monosaccharide q Tetrose A carbohydrate consisting of a four carbon chain r Triose A carbohydrate consisting of a three carbon chain 2 All of these molecules have a formula of or very close to CnHZHOn or CnH20n Because this formula these compounds were originally considered to be quothydrates of carbonquot from which the term quotcarbohydratequot was derived 3 A saccharide with a five carbon chain that includes a ketone is termed a ketopentose Keto ketone pent 5 carbons ose carbohydrate Example CHZOH o HO H H OH CHZOH 4 Two features 39 In a Fischer projection with the primary alcohol CHZOH at the bottom the OH on the carbon immediately adjacent to the CHZOH is on the right as shown in the answer to the previous question 39 All natural carbohydrates are in the D configuration 5 In the pyranose or furanose ring locate the ring carbon mm mm attached to the primary alcohol unit Recall that a primary Q muse alcohol is a CHZOH group This carbon is also attached to H0 ag oxygen the ring oxygen The carbon on the other side of the ring H 011 oxygen is the anomeric carbon OH 6 An CC or 5 configuration refers to the stereochemistry at the anomeric carbon Locate the primary alcohol and decide if is pointing up or down In the example shown in the 208 CF Q amp PP Carbohydrates previous question the primary alcohol group is pointing up the axial hydrogen on the same carbon is pointing down Next locate the group attached to the anomeric carbon that is not part of the pyranose or furanose ring In the example shown in the previous question this group is an OH If the primary alcohol and anomeric attach ments are pointing in the same direction the carbohydrate has the 5 configuration If the primary alcohol and anomeric attachments are pointing in the opposite directions then the carbohydrate has the OL configuration In the example shown in the previous question the primary alcohol and anomeric OH groups both point up so the carbohydrate has the 5 configuration at the anomeric carbon HO HO HO o O OH o 7 Hgg OH H k H H o OH OH o OH OH S D Glucopyranose D Glucose oc D Glucopyranose 8 Glucose is important because it is the most common monosaccharide Ribose is important because it is an essential component of nucleic acids DNA and RNA 9 The anomeric linkage between two monosaccharides of starch has the OL configuration and this is easily digestible by humans The anomeric linkage of cellulose has the 5 configuration and is thus not easily digestible by humans Practice Problems 1 Using Fischer projections draw a an aldohexose and b a D ketohexose 2 Consider the absolute stereochemistry R or S of the aldoses a Is D glyceraldehyde the R or S enantiomer b Do all D aldoses have the same R or S configuration at the carbon that differentiates D from L 3 Draw the 0L and 5 anomers of D glucopyranose 4 How would you answer the following question which is frequently asked by Chem 14C students When discussing whether an anomeric carbon is in the CC or 5 configuration can we say is UL when the substituent is axial and 5 if equatorial 5 Why is D glucopyranose the single most common aldohexose in nature 6 Very briefly explain why glucose behaves as an aldehyde in some reactions but not in others 7 For the carbohydrate shown OH OH a Locate the anomeric carbon of the glucose unit Is it HO 0 CC or 5 OHHO OH H OH b What is the significance to humans of the anomeric carbon s stereochemistry CF Q amp PP Carbohydrates 209 8 Draw an oc disaccharide composed only of D glucose 9 Select the adjectives that describe part or all of the OH OH naturally occurring structure shown starch cellulose 3 ampO pyranose furanose fructose glucose nucleoside OHHO Kg aldohexose ketopentose D sugar and L sugar OH 10 Draw the structure of a disaccharide that is not easily digestible by humans 11 Refer to the hydrolysis of sucrose in the Carbohydrates Lecture Supplement Answer these questions concerning the fructose product of the hydrolysis reaction A model is useful a Draw the fructose product in its acyclic form using a Fischer projection b Is the product D fructose or L fructose c Is fructose an aldose or a ketose d Is fructose a triose tetrose pentose or hexose Practice Problems Solutions CH0 Aldo aldehyde hexose 2 six carbons By convention the 1 a H OH primary alcohol is shown at the bottom of the Fischer projection Ho H When drawn this way natural D carbohydrates have the OH H OH immediately above the CHZOH pointing to the right H OH CHon CHZOH The quotDquot means the carbohydrate has the most b 0 common natural configuration A ketohexose is a H OH carbohydrate whose backbone cons1sts of a Six Ho H carbon cha1n One of these carbons is part of a H OHCZIDconfrgumtion carbonyl CHon 2 a Cahn Ingold Prelog priorities OH gt CHO gt CHZOH gt H So D glyceraldehyde is the R enantiomer b In a generic sense the difference between glyceraldehyde and all of the other aldoses is changing the CH0 of glyceraldehyde 2 d highest priority to HCOH still 2nd highest priority Because the direction of decreasing priority does not change all of the D aldoses have the same R configuration at this carbon OH OH 3 H30 0 O 0 H51 OH HO OH HO oc D glucose S D glucose 210 CF Q amp PP Carbohydrates 4 The statement is not correct because the ocS designation refers to the relative position cis or trans of the CHZOH and anomeric substituents For example in the most stable conformation of oc glucopyranose all substituents on the pyranose ring are equatorial except the anomeric OH If the molecule experiences a chair chair ring ip just like cyclohexane then the anomeric OH becomes axial Verify this with a model Regardless of which chair conformation the molecule is in however the CHZOH and anomeric OH groups are still trans ie opposite faces of the ring and both chairs are oc glucopyranose CHZOH and anomeric re mans HO CHZOH and anomeric OH are tmn rmg lp 5 S D Glucopyranose is the most stable stereoisomer possible for any aldohexose in the pyranose form as all the ring substituents are equatorial Except for the anomeric carbon where the OH may be either equatorial or axial 0 In solution glucose is in equilibrium between the cyclic hemiacetal form and the acyclic aldehyde form The reactivity of the molecule depends upon the nature of the reaction conditions which control the position of the equilibrium 7 a The anomeric carbon of the glucose unit is labeled The other saccharide is not glucose because of the axial OH group It is the 5 stereochemistry because the anomeric OH shown in bold is on the same side of the glucose pyranose ring as the primary alcohol unit CHZOH primry alcohol OH 3 O OH pyranose HO O Oltl ring oxygen Ho OH OH OH 3 HO anomeric not glucose mun b Humans can readily digest carbohydrates that have 0L stereochemistry at the glycosidic linkage the ether connecting two monosaccharides at the anomeric carbon S Glycosides are much slower to undergo acidic hydrolysis and digestion H30 0 8 Maltose is a good example HO 0 Ho0 HO OH OH OH CF Q amp PP Carbohydrates 211 Elimination Reactions Reading from Brown and Foote 39 Chapter 9 Sections 95 98 Optional Reading 39 Klein Chapter 10 Suggested Text Exercises from Brown and Foote 39 Chapter 9 6 8 37 46 48 and 49 Lecture Supplement 39 Elimination Reactions Concept Focus Questions 1 Provide precise yet concise definitions for the following terms a Hydrogen h Elimination reaction 0 Tetrasubstituted alkene b Anti periplanar i Hofmann s rule p Trisubstituted alkene c Cis alkene j Internal alkene q Thermodynamic control d Disubstituted alkene k Kinetic control r Trans alkene e E alkene l Monosubstituted alkene s Vinyl hydrogen f E1 mechanism In Syn periplanar t Z alkene g E2 mechanism n Terminal alkene u Zaitsev s rule 2 Give an example of an E2 reaction including a curved arrow mechanism and the transition states for each step of the mechanism 3 Brie y explain the E2 transition state geometry requirement 4 What structural and reactivity factors are necessary for an E2 reaction to occur 5 Give an example of an E1 reaction including a curved arrow mechanism and the transition states for each step of the mechanism 6 What structural and reactivity factors are necessary for an E1 reaction to occur 7 What is the rate determining step of the E1 mechanism Brie y explain your choice 8 Why is acid necessary to eliminate water from an alcohol to form an alkene 9 Why can skeletal rearrangement occur in an E1 reaction but not an E2 reaction Give an example of an E1 reaction with mechanism in which rearrangement occurs H O If multiple alkene products are possible in an elimination reaction which alkene is the major product What factors control this Elimination Reactions 1 11 When deciding if an elimination reaction proceeds via the E2 or E1 mechanism why is the E2 mechanism considered before the E1 mechanism Concept Focus Questions Solutions 1 Illustrated definitions can be found at the Illustrated Glossary of Organic Chemistry available at the course web site a Hydrogen A hydrogen attached to a S carbon In an elimination reaction this is a hydrogen atom attached to the carbon atom that is attached to the carbon atom bonded to the leaving group H C C LG b Anti periplanar Two bonds or groups with a 180 dihedral angle ie lying in the same plane but pointing in opposite directions c Cis alkene An alkene in which the substituents lie on the same face of the molecule ie point in the same direction d Disubstituted alkene An alkene in which two vinyl hydrogens have been replaced by another atom or group e E alkene Describes a double bond in which the two groups of highest Cahn Ingold Prelog priority lie on opposite sides f E1 mechanism An elimination mechanism is which the carbonleaving group scission step and the carbocation deprotonation step do not occur simultaneously Always has a carbocation intermediate g E2 mechanism An elimination mechanism is which the carbonleaving group scission step and the carbocation deprotonation step occur simultaneously Never has a carbocation intermediate h Elimination reaction A reaction in which a molecule loses atoms or groups of atoms usually from adjacent atoms often resulting in a new pi bond i Hofmann s rule An elimination reaction occurs to give the less substituted alkenes as the major product j Internal alkene An alkene in which both ends of the carbon carbon double bond are directly bonded to at least one other carbon k Kinetic control A reaction in which the product ratio is determined by the rate at which the individual products are produced 1 Monosubstituted alkene An alkene in which one vinyl hydrogen has been replaced by another atom or group 2 Elimination Reactions m Syn periplanar Two bonds or groups with a 0 dihedral angle ie lying in the same plane and pointing in the same direction n Terminal alkene An alkene in which one end of the double bond is not directly bonded to any other carbons except the other end of the double bond 0 Tetrasubstituted alkene An alkene in which all vinyl hydrogens have been replaced by another atom or group p Trisubstituted alkene An alkene in which three vinyl hydrogens have been replaced by another atom or group q Thermodynamic control A reaction in which the product ratio is determined by the relative stability of the individual reaction products r Trans alkene An alkene in which the substituents lie on the opposite face of the molecule ie point in the opposite direction s Vinyl hydrogen A hydrogen atom bonded to the S172 carbon of an alkene t Z alkene Describes a double bond in which the two groups of highest Cahn Ingold Prelog priority lie on the same side u Zaitsev s rule An elimination reaction occurs to give the most substituted alkenes as the major product e r QCH3 t HquotOCH3 E I 2 N CH30H Cl g 3 The energy of the E2 transition state is lower if the carbon leaving group and carbon hydrogen bonds are parallel Incipientpibond v R n a CHgO R I 1 h39lm Note parallel pZ orbitals Elimination Reactions 3 4 U 0 This arrangement is termed antiperiplanar and is preferred over a syn periplanar arrangement to minimize steric and torsional strain As the bonds are breaking the carbons are changing from sp3 to sp2 hybridization and lobes of p orbitals are beginning to replace the bonds If the incipient p orbitals are parallel they can immediately begin to form the new pi bond The energy from the incipient pi bond helps replace the bonding energy lost from the carbon leaving group and carbon hydrogen bonds thus stabilizing the transition state There are three fundamental requirements for the E2 reaction a moderate or better leaving group a strong base and a hydrogen atom that is 5 and periplanar to the leaving group the H C C LG arrangement if a CC bond is formed The first two factors interact With a better leaving group a weaker base can be used With a stronger base a poorer leaving group can be eliminated o o E1 example OSCF3 H20 Mechanism i gt Ph 5 4 Ph ahxoso CF Ph oso2CF3 a 2 3 Ph K Ph HAOHZ 66H2 Some students include a strong base such as H039 or CH3O39 in an E1 mechanism If these species are not part of the given reactants or they are not generated in any reasonable concentration as part of the reaction then they are not present in the reaction and cannot be part of the mechanism 7 There are three fundamental requirements for the E1 reaction a moderate or better leaving group a stable carbocation and a polar solvent These three factors interact For example when a good leaving group and highly polar solvent are present a less stable carbocation can be formed A S hydrogen that is periplanar to the leaving group is not a requirement because carbocation deprotonation is not the rate determining step In addition the leaving group has already departed before the hydrogen is removed so the periplanar requirement is moot Every E1 mechanism includes these two steps ionization of the carbon leaving group bond to form a carbocation and deprotonation of the carbocation to form a pi bond There may be other steps as well such as protonation of an OH group These extra steps are not considered in this question because they are not present in every E1 Elimination Reactions 9 gt0 reaction Formation of a carbocation is energetically expensive because a bond is lost and no new bond is formed In the deprotonation step one sigma and one pi bond are gained while one sigma bond is lost for a net bonding energy increase Recall that activation energy controls rate and that to a reasonable approximation activation energy is controlled by bond energy changes Therefore the more energetically expensive step ionization is slower Acid is necessary to transform a poor leaving group HO into a better but still moderate leaving group water Hydroxide ion is a poor leaving group due to the small atomic radius of oxygen and the oxygen atom gains a charge as it departs Water is a better leaving group because the positive charge on oxygen is neutralized when it departs Rearrangement requires a carbocation intermediate Because carbocations are not formed in the E2 mechanism there can be no rearrangement KH OHZ OH Mm HquotOH2 3 CH3 gt gt gt CH3 CH3 CH3 CH3 CH3 CH3 CH3 CH3 gt CH3 10 Q When the base is sterically hindered or the leaving group is NR3 SR2 or F the 11 least substituted alkene is favored Hofmann orientation Otherwise the most highly substituted alkene is favored Zaitsev orientation 1 The most stable alkene is favored The E2 mechanism avoids the energetically expensive carbocation formation step of E1 Thus we consider E2 before E1 However there may be cases where E2 is not disallowed but E1 is faster because the reaction conditions and reactants favor E1 over E2 OWLS Problems 1 Suggest products and mechanisms for these elimination reactions Explain how you choose between E2 and E1 If more than one organic product is possible select the major product and explain your choice If no elimination occurs write no elimination and explain why Elimination Reactions 5 N 9 4 U 0 gt1 CH3 KOH CH3 CH30H DMF Cl I a CH33N H O b gt e MF 2 Br 0 NaOH 0 CH OH f CH OH o Isl CF3 3 OH 3 0 Which carbocation fates might be observed in an E2 reaction mechanism Provide examples Which carbocation fates might be observed in an E1 reaction mechanism Provide examples Explain why the most stable reaction product is usually the major reaction product as well A mechanism or structural effect that has both orbital electronic and geometric stereochemical factors is called a stereoelectronic effect Using both words and diagrams together briefly discuss the stereoelectronic effects that influence the E2 reaction transition state geometry Select the most stable and least stable alkene in each set Label your choices as cistrans and EZ as appropriate a gtlt gtlt gtlt W a d E 65 Complete the following reactions by writing the missing starting materials or reactants in each box Make sure the reaction gives mostly elimination and not substitution products Elimination Reactions Cl x i w w I KOH c CH3CHZOH 9 Mechanisms are the language of organic reactions Lewis structures formal charges and curved arrows are the grammar and punctuation Not only must grammar and punctuation be used properly and precisely for the meaning of the words to be clear they can also enrich the sentence beyond the simple implications of the naked words Here is a chance for you to practice your organic chemistry language skills The following mechanism steps were drawn from student answers on a midterm Each contains one or more errors Identify each error brie y explain why it is wrong and write the corrected mechanism step H20 COH c 0 d 0A HACI e r ch QH gt H3C OH OHAH HSO4 0H2 a OK gt b d Elimination Reactions 7 HAHsoI o o I X k H Hso4 mu quotU Practice Problems 1 Label each alkene as cis or trans as well as E or Z a c H 1 Cl Br JQ H F l 2 a Draw an alkene of molecular formula C4H8 that is both cis and Z b By changing just one atom redraw your part a molecule so that it is now cis and E 3 Select the most stable alkene 4 Select the most stable and least stable molecule 5 Does the carbocation mechanism step shown below proceed as written ie nothing else is more likely to occur or not ie something else is more likely to occur Brie y explain your answer H HACI 8 Elimination Reactions 6 For elimination reactions and many other reactions we will encounter later it is important to be able to quickly categorize an acid or base as strong moderate or weak For example HZSO4 pKE 9 and H3O pKE 18 are strong acids Their conjugate bases HOSO3 and H20 are weak bases HO and CH3O are strong bases Review your Chem 14C notes if necessary in order to understand the relationship between molecular structure and acidbase strength a Categorize each of the following acids as strong moderate or weak CH3OH2 H20 CH3OH and NH3 b Categorize each of the following bases as strong moderate or weak CH3CH20 CH33CO NH3 1 and F 7 State Zaitsev39s Rule Write an E2 reaction including mechanism that clearly illustrates Zaitsev39s Rule 8 Unlike E2 reactions E1 reactions always give the more stable alkene product regardless of the leaving group Explain why 9 For the elimination reaction shown below a Write an E2 mechanism for the formation of methylenecyclopentane b Write an E1 mechanism for the formation of 1 methylcyclopentene Br NaOCH gt cngoH cngoH l Methylcyclopentene Methylenecyclopentane 10 Examination of the elimination reactions of menthyl and neomenthyl chlorides was important in establishing the mechanism of ionic substitution and elimination reactions Give the E2 products formed in the following reactions along with a brief explanation CH3 CH3 CH33CO39K and gt elimination products E Cl 5 quotquot001 Menthyl chloride Neomenthyl chloride 11 Give the major alkene products that result when menthyl and neomenthyl chlorides are subjected to E1 elimination water heat Elimination Reactions 9 12 Provide mechanisms Brie y explain your choice E2 versus E1 Cl NaBr a I DMF 0 CH3 CH3 b V E V I DMIF H20 OX P 0 ethanol OH st04 d TZO t t 13 Very brie y explain why the reaction of question 12d does not occur in the absence of a strong acid such as HZSO4 or H3PO4 14 Provide a mechanism H3PO4 gt OH H20 Provide a mechanism Label the rate determining step with quotrds OH st04 1 H20 16 Which reaction is faster Write a very similar reaction that occurs by the same mechanism but is obviously faster NaO CH3 NaOCHg 39 or 39 CHgOH CHgOH OH Cl 10 Elimination Reactions H U 17 For the reaction shown below a write the transition state for the slowest step in the E2 mechanism and b determine the more likely mechanism E2 or E1 a xp NaOCH3 gt CH30H For the reaction shown below a Write an E1 mechanism b Write an E2 mechanism c What is the more likely mechanism Brie y explain 04 CH3CO239 Na gt CH30H For the reaction shown below a Select the major product b Write the mechanism for the major product of this reaction c Very brie y explain your choice for the reaction mechanism gtlt Y l Br a Select the major product b Write the mechanism for the major product of this reaction c Very brie y explain your choice for the reaction mechanism msmor NaOH gt CH30H For the reaction shown below For the reaction shown below a Write the major product of this reaction b Provide a curved arrow mechanism including all transition states showing how this major product is formed c Very brie y explain your choice of mechanism for this reaction CH3 quotquotquot c1 KOH CH30H Elimination Reactions 11 22 N W to 4 For the reaction shown below a Write all the products of this reaction b Provide a mechanism that clearly shows how all of your products are formed c Brie y explain why you eliminated other ionic substitution and elimination mechanisms in part a gt CH3CH20H The Hofmann degradation of an amine consists of three steps Conversion of the amine to a tetraalkylammonium salt R4N changing of the ammonium salt anion to hydroxide and finally E2 elimination The elimination step gives the less substituted alkene as the major product Consider the Hofmann degradation shown below Y Y AgZOY H20 gt NH2 CH31 gt large mess NCH33 139 NCH33OH Tetraalkylammonium salt K Y J 5 total heat gt M 95 Explain from the viewpoint of torsional and steric strain why E2 elimination of the tetraalkylammonium salt gives the less substituted alkene as the major product Hint Use models and Newman projections Consider the E2 reaction of the molecule shown below with NaOCH3 in CH3OH a Write all three possible E2 reaction products b Does the location of the phenyl group enhance or decrease the amount of each of these three products c Does the alignment anti periplanar versus syn periplanar of the carbon 5 hydrogen and carbon leaving groups bonds enhance or decrease the amount of each of these three products 12 Elimination Reactions d Select the major product of the reaction Carefully consider all assumptions that you make e What would you use in place of NaOCH3 to favor formation of the product with the least number of vinylic hydrogens N U Alkyl iodides A and B were both subjected to E2 elimination by treatment with KOCCH33 One alkyl iodide gave an alkene of molecular formula CBHIZ while the other failed to react Write the structure of the alkene product that was formed and explain why the other alkyl iodide failed to give an elimination product Hint Explore alkyl iodides A and B with models N 0 An important Chemistry 30A skill that you will need to develop is the ability to determine the most probable mechanism for a reaction even when the products are not given The mechanism choice is based on a series of logical decisions as well as gut instinct which you will develop by doing lots of problems A owchart is a convenient way to lay out the sequence of the logical choices Make such a owchart for the four reaction mechanisms we have studied so far assuming the reaction product is known You might want to create a owchart when the product is not known as well We will add to this owchart throughout the course so keep it handy 27 The following reactions may occur by the E2 8N2 E1 or SNl mechanisms Predict the major products in each case If no reaction occurs write NR Provide a brief explanation for your mechanism choices K4r 39CEN KI a CH3Br gt d CH3OH gt CHECHZOH acetone CH3 CH33CO39 K st04 b e gt at least5products Cl CH33COH H2O quot 0H Cl NaOH 0 I gt f H20 CH30H Elimination Reactions 13 28 For the reaction shown below a What features of the reactants favor each of the following mechanisms E2 8N2 E1 and 8N1 b The reaction produces four major SNl products and two major E1 products Write the products along with the mechanisms for their formation Ph CH30H gt quotquot0302CF3 29 Consider the dehydration of citric acid by the enzyme aconitase H COOH COOH HOOC aconitase HOOC H20 COOH COOH Citric acid Cisaconitic acid a What is the most probable mechanism for this reaction b Brie y explain your mechanism choice Clearly state any assumptions you make about the structure of aconitase c Draw a curved arrow mechanism for the dehydration of citrate by aconitase 30 Consider the 39 39 of J 39 a 391 building block of nucleic acids The final step in adenosine biosynthesis is an elimination reaction 1 HOOC HN coon NH2 HOOC N N enyzme N N lt J lt J gt7 Purlne N N N N COOH deoxyribose deoxyribose Adenosine Fumaric acid a What is the mechanism of this reaction b Briefly explain your mechanism choice You may make any assumptions that you want as long as they are logical and are clearly stated c Provide the detailed and complete curved arrow mechanism for this elimination reaction You may use quotpurinequot as an abbreviation for the fused aromatic ring system with the four nitrogen atoms 14 Elimination Reactions 31 As your knowledge and understanding of organic reactions increases so should your ability to work out mechanisms for reactions that have odd quirks or for reactions you have never seen before a The reaction of NaI with ICHZCHZCl in acetone produces 12 H2CCH2 and NaCl The observed rate law is rate k Nal ICHZCHZCl Write a mechanism that is consistent with these facts b Suggest a mechanism for the following reaction Na H sodium hydride is a strong base The mechanism does not involve any carbocations H o 0 Na H39 H3C So 0 E CH3 CH3 32 Why are E1 and SNl reactions faster in the Arctic and Antarctic than at the equator Hint This is a joke so think like Dr H Practice Problems Solutions 1 a This is a trans alkene because the carbon chains are on opposite faces of the carbon carbon double bond This is an E alkene because the highest priority group on each of the S172 carbons methyl and ethyl is on opposite faces of the double bond Highest prio 39 H CHZCH3 this end of 7C Highest priority on this end of cc 39 H3O H b This alkene cannot be labeled as cis or trans because the double bond has four carbon groups This is a Z alkene because the highest priority groups on each S172 carbon ethyl on the left and propyl on the right are on the same face of the double bond c This is a trans alkene because the carbon groups are on opposite faces of the double bond This is a Z alkene because the highest priority groups on each spz carbon iodine and ethyl are on the same face of the alkene Not all trans alkenes are Ealkenes and not all cis alkenes and Z alkenes d This alkene cannot be labeled as cis or trans because there are no carbon groups attached to the alkene This is an E alkene because the highest priority groups on each S172 carbon Cl and I are on opposite faces of the double bond Elimination Reactions 15 Note that among these structures we cannot always label the alkene as cis or trans but we can always use E or Z This is why the EZ system is preferred over the cistrans system a Only one answer is possible Z 2 butene This alkene is cis because the methyl groups are on the same side of the double bond This alkene is Z because the highest priority group on each sp2 carbon ie the methyl groups lie on the same side of the double bond H3C CH3 Z 2 butene CC 01 H H Cis 2 butene b The methyl groups must remain as they are in order to preserve the cis configuration E configuration requires that the two highest priority groups be on opposite side of the double bond Therefore we must transmute a vinyl hydrogen atom into an atom of higher priority Luckily any atom has a higher Cahn lngold Prelog priority than hydrogen H3C CH3 E 2 iodo butene CC 0 1 H Cis 2 iodo 2 butene The left and middle alkenes are both internal and trisubstituted The alkene on the right is terminal and disubstituted and therefore the least stable of the three We saw in lecture that large substituents can destabilize an alkene due to severe van der Waals eclipsing interactions Recall the case of tetra tert butylethylene discussed in lecture Tetra tert butylethylene A space filling model of tetra tert butylethylene showing severe torsional and steric strain among the four tert butyl groups In this problem the left alkene has a methyl group eclipsing with the alkene hydrogen whereas the middle alkene has a much larger tert butyl group eclipsing with the alkene hydrogen The most stable alkene is shown below Elimination Reactions 4 U The most stable alkene is 23 dimethyl 2 butene tetrasubstituted The least stable alkene is 2 propylcyclopropene disubstituted with ring strain HM 23 Dimethyl 2 butene 2 Propylcyclopropene 2Propylcyclopropene is a disubstituted alkene because each of the alkene carbons is bonded to a carbon It does not matter that both alkene carbons are bonded to the same carbon because substitution in uences alkene stability by the number of bonds and not the number of carbons This mechanism step does not occur as written This mechanism step is another one of the three carbocation fates lose a proton to form a pi bond In this case proton loss leads to a disubstituted alkene whereas loss of a different proton leads to a trisubstituted alkene A trisubstituted alkene is more stable than an isomeric disubstituted alkene isomer so this proton loss is unlikely to occur a CH3OH2 is a strong acid similar to H3O H20 is a weak acid CHSOH is a weak acid similar to H20 NH3 is a very weak acid b CH3CHZO is a strong base similar to H0 and CH3O CH33CO is a strong base similar to HO CH3O and CH3CH20 NH3 is a moderate base I is a very poor base due to large atomic radius and high polarizability F is a moderate base due to small atomic radius and low polarizability In a S elimination reaction the most highly substituted alkene is the major product The E2 example must have a strong base moderate or better leaving group and 5 hydrogen The mechanism is a single step To clearly illustrate Zaitsev39s Rule there must be at least two possible alkene products with different levels of substitution and the most highly substituted must be the major product QTY n y CH36AH Trisubstituted internal alkene Major product Disubstituted terminal alkene Minor product Elimination Reactions 17 8 In an E2 reaction the nature of the leaving group can in uence which alkene product is major because the leaving group is departing in the same mechanism step in which the alkene is formed In an E1 reaction the leaving group has departed before the alkene is formed so the leaving group has little if any in uence on which alkene product is major amp a gt HOCH3 Br H OCH3 PBr b CH30H2 Br H KEEH3 Carbocations are very strongly driven to lose a proton they have very low pKa that just about any species in the reaction with an electron pair is sufficient to deprotonate them In this case the carbocation can be deprotonated by CHSOH as shown or by Br CHSOH is the reaction solvent and therefore present in much higher concentration than Br Thus the carbocation is more likely to encounter and thus be deprotonated by CH3OH than by Br The mechanism shows this Recall that an E2 reaction requires the C H bond and C Cl bond to be anti to each other Review the E2 transition state discussion in the Elimination Reactions Lecture Supplement Recall also that cyclohexane rings prefer a chair conformation Thus we must see if there are any chair conformations that have an anti periplanar arrange ment of the C H and C Cl bonds Explore a model of cyclohexane to convince your self that this can only occur when the C Cl and C H bonds are both axial CH3 Cl Menthyl chloride gt same as HA39ocCH33 Thus we see only one alkene product from this reaction because there is only one hydrogen atom that is anti periplanar to the leaving group The more stable trisub stituted alkene cannot form because the C H bond next to the isopropyl group cannot become anti periplanar to the C Cl bond 18 Elimination Reactions H OcCH33 Neomenthyl chloride Wm gt wow same as CD CH3 H ltocCH33 Also sameas CD Thus we expect to see two alkene products from this E2 reaction But which is the major one Because the base is sterically hindered Hofmann elimination dominates ie the least substituted alkene is the major product 11 E1 elimination proceeds Via a carbocation Once the carbocation is formed think about the three carbocation fates Since we are considering only elimination and not substitution in this case we need only think about lose a proton to make a pi bond and rearrange we can ignore capture a nucleophile CH3 CH3 4 c1 E A either diastereomer Lose a proton to form a pi bond Rearrange 12 to 3 in this case 1 CH3 CH3 CH3 Hquot390H2 gt gt A We must now consider the three fates of this new tertiary carbocation A Once again we ignore capture a nucleophile A more stable carbocation 2 with resonance or 3 Elimination Reactions 19 with resonance is not available via rearrangement This leaves only lose a proton to form a pi bond Deprotonation to form the more stable alkene is favored CH3 CH3 CH3 CH3 gt versus gt H 39OH2 HA390H2 Pathway B Pathway C A tetrasubstituted alkene is more stable than an isomeric trisubstituted alkene so deprotonation pathway B occurs more readily than pathway C 12 a This is obviously an elimination reaction so the mechanism choices are E2 or E1 We consider E2 before E1 because E2 avoids the energetically expensive step of carbocation formation The three E2 requirements are 39 Strong base None present 39 Moderate or better leaving group Cl is a moderate leaving group and 39 H C C LG Present The absence of a strong base prevents this elimination from occurring via the E2 mechanism With weak bases such as bromide ion the leaving group must be truly exceptional for the E2 mechanism to occur Because the E2 mechanism cannot occur the E1 mechanism is the only other possibility To verify that E1 is a reasonable choice consider the three E1 requirements 39 Moderate or better leaving group Cl is a reasonable leaving group 39 Stable carbocation A secondary carbocation is formed and 39 Polar solvent DMF has a high dielectric constant Therefore E1 is reasonable although sluggish under these reaction conditions Carbocations are generally so strongly driven to fill the octet of the carbon bearing the positive charge that even a weak base can deprotonate it forming a new pi bond In this case either Cl or Br is adequate for the job 20 Elimination Reactions Cl 4 gt gt HBr CH3 CH3 CH3 H ABr Carbocation rearrangement is possible but its presence is invisible because it leads to the same major product HABr39 gt HBr CH3 CH3 CH3 H b When considering elimination reaction mechanisms we consider E2 before E1 The absence of strong base indicates this cannot be an E2 reaction The E1 requirements are 39 Moderate or better leaving group iodide ion is among the best 39 Stable carbocation the carbocation intermediate is 3 with resonance and 39 Polar solvent the solvent is a mixture of water a 80 and DMF s 37 so its a is somewhere between the two The exact value of 8 depends upon the water DMF ratio The reaction therefore occurs by the E1 mechanism V gt lt7Hnj W H30 13 Any base that was present at the start of the reaction or generated during the reaction can be used for the deprotonation step Carbocations are very reactive so even weak bases iodide ion water or even DMF can perform the deprotonation that produces a full octet on carbon c This is an E1 reaction because there is no strong base present i HO gt K gt 3 d The mechanism cannot be E2 because no strong base is present Elimination Reactions 21 In water HZSO4 pKa 9 is completely ionized to form H3O pKa 178 Only in instances in which there is no water or in which the number of moles of water is less than the number of moles of HZSO4 are there any non ionized HZSO4 in solution Thus in aqueous H2804 the material that protonates the alcohol is H3O not H2804 l H OH2 OH 0H2 Ph 39 Ph f P11 Ph H OH2 HAOHZ Ph 39 Ph Ph HA390H2 gt Ph Ph 13 The reaction involves elimination of water from an alcohol to form an alkene In the absence of acid the leaving group is hydroxide ion This is a very poor leaving group due to the small atomic radius of oxygen and the change from neutral to negative charge and in conjunction with the energetically expensive process of carbocation formation does not allow the reaction to proceed Protonation of the alcohol by acid converts the leaving group into water Water is a moderate leaving group because the charge of the oxygen atom changes from positive to neutral when it leaves That water is a better leaving group than hydroxide ion helps to overcome the difficulty in forming the carbocation so the reaction can proceed H 390H2 1 4 gt gt gt OH 5H 2 HbaH 0 q C J v Hquot 0H2 H 0H2 H 7 22 Elimination Reactions HAOHZ EgtMmeaxfgt 16 These are obviously elimination reactions When analyzing elimination mechanisms we consider E2 before E1 The E2 requirements are c m am 0H2 HAOHz EH EH4 3 OR Egti Ox 39 Moderate or better leaving group H0 is very poor Cl is moderate 39 Good base CH3O is a strong base and 39 H C C LG arrangement present Thus we predict that cyclohexyl chloride reaction can occur by the E2 mechanism and cyclohexanol reaction cannot The cyclohexanol reaction cannot occur by the E1 mechanism under these conditions because hydroxide ion is a poor leaving group Thus the cyclohexyl chloride reaction is faster because the cyclohexanol reaction cannot occur at all under the given reaction conditions regardless of mechanism We can accelerate this E2 reaction by using a better leaving group illustrated below NaOCH Faster reaction gt cngoH 1 7 i CH30 a a 17 a w o i 03 7 b The E1 reaction mechanism involves the energetically expensive step of carbo cation formation E2 avoids this step and so is energetically cheaper Thus if other reaction conditions allow an elimination reaction usually proceeds by the E2 mechanism in preference to the E1 mechanism Under conditions where both are allowed both mechanisms may operate simultaneously For an elimination to proceed via the E2 mechanism there are three requirements strong base methox ide ion is present moderate or better leaving group Br and the leaving group Elimination Reactions 23 must be beta to the hydrogen being removed H C C LG arrangement All of these requirements are met so the E2 mechanism predominates in this case 1 18 a 04 O CH35H2 H IREH3 gt1 b gt 0 CH3COOH 1 N H3 HAOTC O c An E2 elimination requires a moderate or better leaving group iodide ion H C C LG arrangement and a good base Acetate ion is a modest base but iodide is an excellent leaving group Therefore the reaction can occur by the E2 mechanism The E1 mechanism requires a moderate or better leaving group iodide ion polar solvent methanol and stable carbocation tertiary These requirements are met so the reaction can occur by the E1 mechanism Given a choice we predict the reaction occurs by the E2 mechanism because this avoids the energetically expensive step of carbocation formation However E2 is slowed in this case because acetate is a modest base Thus the reaction might occur by either mechanism 19 a The first alkene is more substituted and is therefore the major product Hquot0H b I gt Y H20 Br c This is an elimination reaction The common elimination mechanisms are E2 and E1 We consider E2 before E1 because E2 is less energetically expensive no carbocation formation E2 requires a strong base HO a moderate or better leaving group Br and H C C LG arrangement These are all present so the reaction proceeds via the E2 mechanism 20 a The more substituted alkene the alkene on the right is the major product CH30H2 24 Elimination Reactions The carbocation can also be deprotonated by iodide ion but it is less prevalent than CHSOH the solvent The carbocation most likely encounters a molecule of methanol before an iodide ion c The reaction mechanism cannot be SNl or 8N2 because this is an elimination not a substitution reaction When deciding between E2 and E1 we examine E2 first because it is less energetically demanding no carbocation is formed E2 requires a strong base moderate or better leaving group and H C C LG arrangement There is no strong base present so E2 is ruled out CH3OH is a weak base CH3O is a strong base but is not present E1 requires moderate or better leaving group stable carbocation and polar solvent all of which are present in this case CH3 CH3 KOH 21 a gt CH OH quot c1 3 b In an E2 reaction the C H and C LG bonds being broken must be periplanar In a cyclohexane ring this can only be achieved when both bonds are axial The C H next to the methyl cannot be axial at the same time as the C Cl bond thus the trisubstituted alkene 1 methylcyclohexene cannot form in this case CH3 H CH3 H CH3 H same as H Cl quotWIC1 H H Cl No CiH antiperiplanar Only one CiH is to CiCl E2 not possible antiperiplanar to CiCl Ho 5 1 quot CH3 CH3 CH3 H N H H same as H H i H l H 61 5 c This is an elimination reaction so the mechanism choices are E2 and E1 We consider E2 before E1 because E2 avoids the energetically expensive step of carbocation formation E2 requires strong base H0 is strong H C C LG present and moderate or better leaving group Cl is a moderate leaving group The E2 requirements are satisfied so we predict the E2 mechanism dominates Elimination Reactions 25 22 a Mechanism analysis suggests this to be an E1 SNl reaction Br gt OCH2CH3 CH3 CH2 OH SNl product E1 product H ECHZCH3 bx l Ph Gr P11 Ph HOCH2CH3 P11 Ph AOCHgCH Ph OCH2CH3 HP HOCHch3 c EH2 Ethanol CH3CH20H is a poor nucleophile Bromide is not a good enough leaving group to overcome this poor nucleophilicity so 8N2 is ruled out More obviously the carbon bearing the leaving group is 3 Ethanol is a poor base Bromide is not a good enough leaving group to overcome this poor basicity so E2 is eliminated if you will pardon the pun N W Recall that E2 elimination requires a periplanar alignment dihedral angle of 0 or 180 for the C H and C LG bonds that are being broken A Newman projection viewed along the carbon carbon bond bearing both the hydrogen removed and the leaving group looks like this gauche conformation H I m H and LG are anti periplanar LG Any Csp3 Csp3 bond has just three gauche conformations like this Here are the three staggered conformations along the C2 C3 bond of 2 butanamine that might lead to 2 butene H CH3 H H CH3 H CH3 H CH3 H3C H H H H C N CH N CH N CH 3 3 3 A B C 26 Elimination Reactions The NCH33 group is larger than a methyl group so conformation B has less torsional strain than conformations A and C The molecule spends most of its time in conformation B and only rarely is found in conformations A or C Conformation B does not lead to elimination because there are no S hydrogen atoms that are periplanar to the leaving group Therefore elimination to form 2 butene is slow because the molecule does not spend much time in a conformation that can lead to this product All three of staggered conformations of the C1 C2 bond of 2 butanamine look alike I H CH2CH3 H H N CH 3 All of these conformations have a S hydrogen that is periplanar to the leaving group so regardless of conformation elimination to form 1 butene can happen anytime the base approaches Formation of 1 butene is therefore the major reaction pathway 3h 3h Ph Eh I NaOCH3 24 a gt CH3 CH3 CH3 CH30H CH3 CH3 CH3 CH3 A B C b Conjugation of the alkene with the phenyl group favors enhances the amount of B produced but not the amounts of A or C c Alkenes A and C result from an anti periplanar alignment of the carbon leaving group and carbon S hydrogen bonds Formation of alkene B requires a syn periplanar arrangement of these bonds An anti periplanar alignment has less torsional strain than a syn periplanar alignment so the bond alignment favors A and C and disfavors B d Product A is not conjugated but it comes from an anti periplanar transition state less torsional strain Product B is conjugated but comes from a syn periplanar transition state more torsional strain Therefore our prediction of major product depends on which factor we assume has a greater amount of influence If torsional strain carries more weight than conjugation then product A is major If conjugation carries more weight than torsional strain then product B is major Product C is formed in the least amount because it is a terminal disubstituted and nonconjugated alkene Elimination Reactions 27 e The product with the most number of vinylic hydrogens hydrogens attached to the alkene carbons is C This is the major product under Hofmann conditions so it will be major when the base highly hindered such as K OCCH33 25 The leaving group of alkyl iodide A is not periplanar to any S hydrogen so E2 elimination is not possible Therefore alkyl iodide A does not react CH3 K 39OCCH33 gt 1 I H H No E2 reaction In alkyl iodide B the S hydrogen next to the methyl group is syn periplanar to the leaving group so E2 is possible CH3 Kt 39OCCH33 gt I H H H 1 Aynperiplanar 26 The owchart format shown here is just one of many possible formats Yes S 1 N subsmuuon Yes Nucleophile is poor X Y gt X Z orX is 3 carbon No sN2 No Yes E1 Ellmlnauon Yes Base is poor or WXYZ i xY hydrogen not periplanar N0 E2 1 No Other mechanisms 27 Recall that we consider the substitutionelimination mechanism possibilities in the order listed below In each case the requirements for that mechanism to occur are noted If one of the requirements is not met then the reaction cannot occur by that mechanism The owchart developed in the previous question may be useful here 39 E2 strong base H C C LG moderate or better leaving group 39 8N2 good nucleophile moderate or better leaving group carbon undergoing subs titution not 3 An aprotic solvent is better than a protic solvent but a protic solvent does not necessarily prevent 8N2 Moderate solvent polarity is best 8 20 to 40 when the reaction involves a negatively charged nucleophile and neutral electrophile Other charge combinations may call for high solvent polarity 28 Elimination Reactions 39 ElSNl stable carbocation moderate or better leaving group polar solvent a Q lacks the H C C LG arrangement s g has good nucleophile cyanide ion has little steric hindrance CH3 and a moderate leaving group Br The solvent is protic and has modest polarity s 28 which is acceptable for an 8N2 reaction with Br Therefore this is an 8N2 reaction A CEN Br CH3 gt H3C CEN Br b strong base tert butoxide moderate or better leaving group Cl H C C LG arrangement present Hydrogen bonding by a protic solvent reduces basicity in the same way it reduces nucleophilicity but R0 in ROH retains enough basicity for the E2 mechanism Therefore this is an E2 elimination Tert butoxide is a large base so the Hofmann elimination product less substituted alkene is favored H OCCH33 f lt gt HOCCH33 Cl 01 sag no good nucleophile c Q no strong base ElS l The carbocation is easily formed 3 with resonance moderate or better leaving group I and polar solvent water Thus the reaction proceeds via the ElSNl mechanisms woe Remember the three carbocation fates This carbocation cannot rearrange to become more stable Iodide capture returns it to starting material So we need to consider water capture FOH2 gt ZCH gt OH H 0H2 Elimination Reactions 29 Or losing a proton to form a pi bond H 0H2 d E no strong base sag no moderate or better leaving group ElSNl the methyl carbocation is too unstable to form None of these three mechanisms can operate so this is a case of no reaction e No strong base 72 no strong nucleophile ElS l protonation of the OH affords water a moderate leaving group loss of which leads to a secondary carbocation Water is a polar solvent So we conclude ElSNl to be reasonable in this case Because SNl and E1 compete we cannot easily predict the major product Recall however the among the E1 products the more highly substituted alkene is more stable and therefore is produced in a greater amount than other alkenes CH3 CH3 gt 01 3 OH H2 HDOHZ CH3 CH3 CH3 gt gt L OH OH 0H2 4 HA 0H2 HAOHZ I7 OH 0H2 OH CH3 CH3 CH3 Elimination Reactions CH3 CH3 CH3 CH3 gt gt J HI 0H2 H OH2 Hh 0H2 CH2 CH2 gt f E Strong base HO H C C LG present and moderate or better leaving group Cl is moderate The E2 requirements are met so we predict E2 is the major mechanism c W M Trisubstituted internal alkene J Major product HNOH 35 xx Disubstituted terminal alkene pg Minor product HKOH 28 a Q Good leaving group and periplanar S hydrogen sag Good leaving group attached to a carbon that is not tertiary E1 and 8amp1 Good leaving group polar solvent and resonance stabilized secondary carbocation intermediate A tertiary carbocation with resonance stabilization is also an intermediate but because this carbocation is not formed in the rate determining step its presence does not have a significant impact on the reaction rate These mechanisms have the same ratedetermining step so the reactant features that favor their operation are also the same b SNl products A H HOCH3 HOCH3 I OCH3 OCH3 Ph Ph Ph Ph gt gt gt quotdsozcrg Two enantiomers Elimination Reactions 31 H HOCH HOCH3 l7 3 Ph Ph CHQ39o Ph cngo Ph Two enantiomers The carbocation rearrangement 2 gt 3 with resonance is very highly favored so the 2 carbocation may not survive long enough to capture methanol E1 products Ph Ph A CH30H H 29 a The more probable mechanism is E2 The answer to this question depends upon your assumptions b Given a choice between E1 and E2 an elimination reaction generally proceeds via the E2 mechanism because this avoids the energetically expensive step of carbocation formation Thus we can make some reasonable assumptions that allow the E2 mechanism to proceed E2 requires 39 Moderate or better leaving group Hydroxide is a poor leaving group but protonation by aconitase converts hydroxide to water a much better leaving group Strong base No strong base is present However we might assume the enzyme assists its departure probably by hydrogen bonding H C C LG arrangement This is present in citrate c H enz acidic group in the enzyme active site B enz 2 basic group in the enzyme active site fH Enz H0 H20 coon COOH COOH HOOC HOOC I HOOC COOH COOH EnzB H COOH 30 a This more probable mechanism for this elimination reaction is E2 32 Elimination Reactions
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