Once familiar with chiral centers, models, drawings and mental images NOW: Final representation of chiral centers: Fischer Projections

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1 Once familiar with chiral centers, models, drawings and mental images NOW: Final representation of chiral centers: Fischer Projections Fischer Projections are 2-dimensional representations of 3-dimensional chiral centers (Ch. 25). -Developed by Emil Fischer during his work on carbohydrates -Must always be viewed so the horizontal lines are coming TOWARDS you and the vertical lines are going AWAY from you. A B View from Top D D C D A C B A C B (R) Lactic Acid O CO 2 C 3 O 2 C C 3 O Everyone sees things differently how about TIS angle (between the C 3 line and O wedge)? O CO 2 C 3 CO 2 O C 3 Or TIS angle (between O wedge and CO 2 line)? O CO 2 C 3 O C 3 CO 2 There are lots of correct answers for drawing a Fischer projection. Unless you can see these comfortably and confidently, I would suggest that you view them the same way every time. Remember: If the lines form a cup, then DAS is UP but if lines are facing down, then DAS is DOWN

2 The lines are already placed on the left and right sides as you view the Fischer Projection. It s sorting out the wedge and dash that can be tricky sometimes Draw a Fischer Projection for the following: A B D C Draw Fischer Projections: 3 C 3 C 2 C O O 3 C 2 C 3 C C 3 O Can use Fischer Projections to determine if two chiral centers are the same or opposite (i.e. mirror images) by moving the Fischer Projections and comparing: If all four groups line up the same, the chiral carbons are the same. If two match and two do not, the chiral carbons are mirror images. SUGGESTION: When comparing two molecules, just rotate/move one of the Fischer Projections. If you start moving both of them, in different ways, trying to get them to match, you may just confuse yourself Rules for moving Fischer Projections: 1. Can rotate a F.P. 180º but not 90º or 270º. C 3 O O C 3

3 2. Can hold any one group constant and rotate the other three groups, CW or CCW. O 3 C O C 3 Inadvertently violating either of these rules will result in the accidental inversion of configuration! What should be R will appear to be S Single Chiral Center: Same or Enantiomers? C 3 Cl O O C 3 Cl O Cl O 3 C C 3 Cl Multiple Chiral Centers: Same, Enantiomers or Diastereomers? C 3 O C 3 O O O C 3 C 3 Simplest way to compare: eak the multiple centers into individual chiral carbons. Compare C 1 with C 1 and then C 2 with C 2. O C 3 1 C 3 O O 1' O C 3 2 2' C 3 If all chiral centers match exactly the same compound C 1 = C 1 and C 2 = C 2 If ALL centers do not match (no match means mirror image, so all are mirror images) enantiomers

4 C 1 C 1 and C 2 C 2 If some match and some do not diastereomers C 1 = C 1 and C 2 C 2 or C 1 C 1 and C 2 = C 2 C 3 O 1 C 3 O 1' R R R O 2 C 3 R O C 3 2' The More Evil Version: What makes this more confusing? 3 C 2 N N Cl 3 C C 3 C 3 Cl 3 C Cl 1' 2' C 3 Determine the relationships: 3 C C 3 O O C 3 C 3 C 3 O C 3 N Can Assign R/S configuration to Fischer Projections: 1. Prioritize according to CIP Rules. 2. Move/manipulate the F.P. so the lowest priority (#4) is at the top OR bottom of the F.P. drawing. This places it in a position that is FACING AWAY from you. 3. Assign R/S: If is CCW, the configuration is S.

5 If is CW, the configuration is R. Assign R/S configurations: Cl O C 3 C 3 O Cl Final Section: Stereochemistry of Reactions What stereochemical outcome is expected in a reaction? Will products will the reaction produce? o Enantiomers? A Racemic Mixture? Diastereomers? Meso Compound? Will it be optically active or not? Sometimes reactions do not generate chiral carbons when the reagents add to an alkene: This IS a possible outcome that can occur. The product is not chiral - it has no chiral center and it is not optically active. For those reactions that do generate chiral centers though, there are two options for what might occur: Option I: Achiral + Achiral = not optically active product(s) (Enantiomers or a Racemic Mixture) Ex 1: Addition of to an alkene The alkene shown below has no chiral centers in it ( achiral ) and the reagent,, is not chiral either ( achiral ). Draw the product: ow many chiral centers formed in the reaction? Draw the product(s) WIT STEREOCEMISTRY:

6 + This is an example of a reaction where an achiral alkene is reacting with an achiral reagent. The product produced is NOT OPTICALLY ACTIVE (i.e. either a racemic mixture or a meso compound). The addition of to this alkene generates ONE new chiral center for a molecule that had no chiral centers. The total number of stereoisomers is 2 N, where N=1 so 2 1 = 2 possible stereoisomers, which are shown above. If Achiral + Achiral = Not Optically Active and the only options are a Racemic Mix or Meso Compound, which did you form? With only one chiral center, this must be a not optically active racemic mixture that formed (since meso compounds must have at least two chiral centers). Recall that the mechanism for the addition of to an alkene involves the addition of - to a carbocation, which allows for both possible enantiomers to form. Ex 2: Addition of OsO 4 to an alkene 1. OsO 4 2. NaSO 3, 2 O Achiral + Achiral = not optically active product(s) ow many new chiral centers are formed? ow many possible stereoisomers are formed? This reaction generates two new chiral centers. 2 2 = 4 but because we know this reaction only occurs as a SYN addition, so only two of the 4 possible stereoisomers can form: 1. OsO 4 2. NaSO 3, 2 O O O + O O What is the relationship between these two products? Did this reaction form a racemic mixture (non-superimposable mirror images) or a meso compound (possess at least two chiral centers and a plane of symmetry)?

7 With two chiral carbons and a plane of symmetry: Meso Compound! Option II: Chiral + Achiral = Optically Active Diastereomers What about additions to chiral alkenes? The previous examples were reactions done on achiral alkenes. What is the difference when an alkene is reacting that has a pre-existing chiral center? C 3 (S) Chiral alkenes inherently do not have a plane of symmetry. Because chiral alkenes are asymmetric, one face of the alkenes would be more accessible by the reagent than the other face, so the product mixture that forms will NOT be 50:50. C 3 (S) C 3 (S) (S) C 3 (R) (S) Chiral + Achiral = Optically active product(s) What is the relationship between these two products? The old chiral center remains unchanged in the process. The new chiral center is a mixture of R and S (albeit not 50:50). The result is unequal amounts of (2R, 3S) and (2S, 3S) which are a mixture of optically active diastereomers! Determine the stereochemical outcome in the following reactions: Ex: Addition of 2 to an alkene 2, Pd/C Chiral or Achiral Alkene? achiral + achiral = not optically active product(s) Like the reaction with OsO 4, this reaction generates two new chiral centers. 2 2 = 4 possible stereoisomers that could be drawn but there will not be 4 products because this reaction can only occur as a SYN addition. With stereochemistry:

8 2, Pd/C + What is the relationship between these two products? Did this reaction form a racemic mixture or a meso compound? With two chiral carbons and a plane of symmetry: Meso Compound! What about: 2, Pd/C Chiral or Achiral Alkene? achiral + achiral = not optically active product(s) Again, this reaction generates two new chiral centers but only 2 possible stereoisomer products, due to SYN addition. With stereochemistry: 2, Pd/C + Note that there is no central plane of symmetry. The hydrogen atoms are aligned but not the methyl and ethyl groups. If you rotate to match the methyl and ethyl groups, the hydrogen atoms are no longer aligned: What is the relationship between these two products? Did this reaction form a racemic mixture or a meso compound? Although there are two chiral carbons, there isn t a plane of symmetry: Racemic mixture!

9 Ex: Oxymercuration 1. g(oac) 2, 2 O 2. NaB 4 Chiral or achiral alkene? Chiral + achiral = optically active diastereomers Products: (S) (R) (S) (S) * * O O One old chiral center. ow many new chiral centers are created? What is the relationship between these two products? One chiral center the same, one a mirror image Diastereomers Ex: Addition of 2 to an alkene 2 achiral + achiral = not optically active product(s) This reaction generates two new chiral centers. 2 2 = 4 but there will not be 4 possible stereoisomers, because this reaction can only occur as an ANTI addition. With stereochemistry: 2 + There is no plane of symmetry because the alkene has two different groups attached at each end (left: methyl, right: ethyl). What is the relationship between these two products? Racemic mixture!

10 Ex: Addition of 2 to an alkene 2 achiral + achiral = not optically active product(s) Again, this reaction generates two new chiral centers but there will not be 4 possible stereoisomer products, because this reaction can only occur as an ANTI addition. With stereochemistry: 2 * * + * * Notice how the alkene was symmetrical with the methyl groups side by side. Notice also that the methyl groups in the product are still aligned side by side, but one bromine atom faces up and the other faces down. Is there a plane of symmetry? No What is the relationship between these two products? Racemic mixture! Ex: One more time: Addition of 2 to an alkene 2 achiral + achiral = not optically active product(s) Only two possible products, because this reaction can only occur as an ANTI addition. With stereochemistry: 2 + What s the relationship between these two products? Is there a plane of symmetry? No groups are lined up to show any mirror plane. Rotate that second molecule around on its center C-C bond and see how the methyl groups line up: 3 C C C C 3 3 C C 3

11 Is there a plane of symmetry? Yes Two or more chiral centers? Yes Meso Compound! Other reactions could include hydroboration, acid-catalyzed hydration or halohydrin formation. Chapter 9: Alkynes Structure: R C C terminal alkyne R C C R' internal alkyne The alkyne functional group contains two sp hybridized carbons, with a 180º bond angle, linear geometry There are two sets of perpendicular p orbitals to form the two pi bonds 6 e- between the two carbon nuclei o shortest bond shortest hybrid orbitals (sp) o strongest sigma bond - best overlap with roundest hybrid orbital (sp) Nomenclature: Suffix: -yne 1. Find the longest chain containing the C C triple bond 2. Number from the end closest to the triple bond (use branches if equidistant from both ends). If more than one alkyne is present, number from end closest to first triple bond. 3. Identify all substituents and their position numbers 4. Write the full name: Prefixes (alphabetized) - # - Parent YNE OR Prefixes (alphabetized)-parent-#-yne As seen with alkenes, use the lower of the two numbers to indicate the alkyne position.

12 Can have adiynes,-atriynes, etc, just include numbers for each triple bond. Molecules with both alkenes and alkynes: Two Suffixes: -en yne [IN TAT ORDER!!] 1. Find the longest chain containing both alkene and alkyne 2. Number from the end closest to a MULTIPLE bond. If equidistant, the ALKENE gets the priority and thus numbering occurs from the end closest to the alkene. So: Which end do we start from? 3. Identify all substituents and their position numbers 4. Write the full name: Prefixes - # Parent, followed by en - # yne or Parent -# en - # yne 3-decen-7-yne OR dec-3-en-7-yne Preparation of Alkynes: Double dehydrohalogenation: (Need a dihalide) 2 equiv KOtBu Step-Wise:

13 2 equiv KOtBu 1 equiv KOtBu 1 equiv KOtBu Mechanism: tbuo 2 equiv KOtBu 1 equiv KOtBu 1 equiv KOtBu OtBu Where does the dihalide come from to make an alkyne? X 2 Addition to an alkene: 2 So, how would you do the following conversion (synthesis, more than one step)?

14 NOTE: Cannot do double dehydrations to form alkynes. Reactions of Alkynes: 1. Addition of X (Addn of, X, where X = Cl, ) Markovnikov 1x or 2x Types of Alkynes Terminal one end is substituted with R and one has an. The end with the R is more substituted, and therefore forms a more stable carbocation intermediate, thus Markovnikov can be easily distinguished: 1 equiv 2 equiv Note that when two equivalents of X are added to a terminal alkyne, both X s wind up on the same carbon! Internal Symmetrical Alkynes Both ends are equally substituted with the same R group so it doesn t matter which end the or X adds to, TE FIRST TIME This is not a regioselective reaction both ends are the same: 1 equiv. Both carbocations that could form are exactly the same: is the same as

15 The SECOND addition is problematic. Once the first addition has occurred, the two ends of the (now) alkene are not symmetrical. Each end still has one alkyl group so neither is more substituted but the mere presence of the first halide throws off the symmetry. 2 equiv 1st equiv 2nd equiv??? 1st equiv 2nd equiv and That s a bad reaction makes two different products! Internal Asymmetrical Alkynes In this version, the alkyne is equally substituted with a single R group on each end but the R groups are different so Markovnikov cannot be distinguished mixture of at least two products always results, even with only 1x addition of X. Bad reaction!! 1 equiv. and through two different carbocations: and A second addition would just make even more products (four total!) Examples: 1 equiv. Cl

16 2 equiv. 1 equiv Cl 2. Addition of X 2 (X = Cl, ) ANTI 1x or 2x -Can only see ANTI for a single addition (1x) -ANTI is shown by the relative positions of the halides in the alkene product, NOT by the use of wedges/dashes!! 1 equiv Cl 2 2 equiv Cl 2 More Examples: 1 equiv. Cl 2 1 equiv. 2 2 equiv. 2 An Oldie: Cl 2 3. ydrogenation of alkynes (2X Addn of, ) SYN (can t see it but it IS happening!) 2x ALWAYS

17 2, Pd/C 2, Pd/C 2, Pd/C 2, Pd/C 2, Pd/C 4. Formation of CIS alkenes (1X Addn of, ) SYN 1X ALWAYS -Catalyst is poisoned with lead (Pb) and thus reacts slower than Pd/C. This reaction gives us some insight into the relative reactivity of alkenes and alkynes. Since only the starting alkynes continue to react without any of the product alkenes undergoing hydrogenation, we can see that alkynes ARE more reactive than alkenes in a hydrogenation. 2 Pd/CaCO 3 /Pb 2 Pd/CaCO 3 /Pb

18 2, Pd/CaCO 3 /Pb 5. Formation of TRANS alkenes - (1X Addn of, ) -ANTI 1X ALWAYS Li or Na, N 3 Li, N 3 Li, N 3 The mechanism for this reaction is completely different than those in the past this reaction involves an electron transfer process. Both Na and Li are Group I elements that want to give up one electron in order to become Nobel Gas configurations. They dump an electron into the pi system, creating a radical anion, which immediately removes a proton ( + ) from the solvent, N 3. Then the process repeats itself: Li Li + + 1e- N 2 Li Li + + 1e- N 2 The stereochemistry is determined when the second electron is transferred. Which is more stable? And why?

19 The first one is more stable because it places the alkyl groups around the newly forming alkene farther away from each other (less steric interactions!).