Unit 3 Packet: Stereochemistry PDF

Name: _______________________________ Unit 3 Packet: Stereochemistry 129  Key Terms For Unit 3 General Plane Polarized Light – light of only one particular orientation...

Name: _______________________________ Unit 3 Packet: Stereochemistry 129  Key Terms For Unit 3 General Plane Polarized Light – light of only one particular orientation Optical Activity – the ability of certain compounds to rotate the plane of polarized light Isomers – compounds that have the same chemical formula, but different properties Structural Isomers – isomers that have different atom-to-atom bonding Stereoisomers – isomers that have the same atom-to-atom bonding, but different spatial orientation Diastereomers – stereoisomers that are not mirror images of one another Enantiomers – stereoisomers that are non-superimposable mirror images Geometric Isomers – diastereomers that originate from hindered rotation about a double bond or a ring Cis-trans (geometric) isomers – see “geometric isomers” – specific to the parent chain’s orientation Chirality – “handedness” – a principle that is true of anything that is non-superimposable on it’s mirror image Chiral Center – a carbon with 4 different groups attached (often called a stereocenter) Meso Compound – a compound that contains stereocenters but has an internal mirror plane – NOT chiral Fischer Projection – a projection where all horizontal groups point towards you, all vertical groups point away Cahn-Ingold-Prelog Rules – priorities based on atomic number and point of 1st difference used for R/S and E/Z R – when CIP priorities are assigned to a stereocenter and #4 is pointing away, 123 is clockwise S - when CIP priorities are assigned to a stereocenter and #4 is pointing away, 123 is counterclockwise D – a sugar whose last stereocenter is “R” L - a sugar whose last stereocenter is “S” d (+) – dextrorotatory – a substance that rotates PPL clockwise (to the right) from the POV of the observer l (-) – levorotatory – a substance that rotates PPL counterclockwise (to the left) from the POV of the observer (±) – racemic mixture – a 50%/50% mixture of a pair of enantiomers – not optically active ee – enantiomeric excess – the amount of excess of a given enantiomer in a non-racemic mixture E (entgegen) – German for “opposite” – when two higher CIP groups on a double bond are on opposite sides Z (zusammen) – German for “together” – when two higher CIP groups on a double bond are on the same side Scientists Jean Baptiste Biot Emil Fischer Lewis Carrol Louis Pasteur Jacobus van’t Hoff 130  Isomers Vocabulary Flowchart (5.1) Isomers same chemical formula, different properties Structural Isomers Stereoisomers Different atom-to-atom bonding same atom-to-atom bonding, different spatial arrangement Organic Inorganic Enantiomers Ex: butane vs. isobutane Diastereomers Non-superimposable mirror image. Usually Not mirror-images has a chiral center. Coordination Isomers Cis-trans Isomers -Ligand and counter ion switch in coordination (Geometric Isomers) compounds Cis-trans about a Other double bond axis, Diastereomers or on different Linkage Isomers sides of a fixed -Ligand’s point of attachment to cation differs Organic Inorganic 1. Hindered rotation Positions about central about double bond metal cation in 2. Different sides of a coordination compounds rigid ring varies 131  Optical Rotation of Plane Polarized Light (5.4) In the early 1800s, Jean Baptiste Biot, among other scientists, began to study the behavior of what is called ________ _______________ ________. When light is emitted from a source, the waves travel in all possible orientations. When any light source is passed through polarized film however, only certain orientations are allowed to pass through the film. These orientations are parallel with the orientation of the film’s gradient. Chemists discovered that certain chemicals had the ability to rotate the plane of polarized light. As polarized light passes through a solution of a chemical like this, the initial plane of polarized light is rotated either clockwise or counterclockwise, and the degree of rotation is dependent of the concentration of the solution and the path-length of travel. A chemical that exhibits this property is described as being ______________ ____________. One such chemical is tartaric acid, a chemical that is isolated from _______ _______, a waste product of the fermentation of grapes by yeast. Tartaric Acid that was isolated in this manner exhibits optical activity, but artificially synthesized tartaric acid does not. This discrepancy baffled scientists of the day. 132 In 1849, as he was studying tartaric acid, __________ ______________ observed that a pile of white tartaric acid crystals that had been synthesized in the lab was actually composed of two different kinds of crystals – crystals that appeared to be ____________ ___________ of each other. Upon noticing this fact, Pasteur went through the painstaking process of hand-separating the crystals into two separate piles of asymmetric mirror image crystals. When each pile was dissolved in solution, it was determined that both piles were optically active, and that separate solutions of equal concentration of both crystals rotated the plane of polarized light in _________________ _________________ by the same ______________, while an equal mix of the two piles was not optically active. From this result, Pasteur deduced that the molecules themselves must be asymmetric, and that they exhibit a “handedness”, or ______________, as a result. Much like your right and left hands are non-superimposable mirror images, the two piles of tartaric acid crystals exhibited much the same quality. For 25 years, the molecular secret to what makes this possible was unknown. In 1874, Jacobus van’t Hoff and Joseph Le Bel both independently proposed that the ability for molecules to exhibit chirality must be due to a carbon structure where the four groups attached at a carbon are pointing towards the four corners of a ________________. When this tetrahedral structure is considered, any carbon with 4 different groups attached exhibits the property of handedness. In many ways, our understanding of the three-dimensional structure of molecules owes its roots to the study of optical activity. 133  Identifying Stereocenters R and S – Nomenclature (5.3) Although drawings can provide a pictorial representation of stereochemistry, they are difficult to translate into words. Thus, a verbal method of specifying the exact three-dimensional arrangement (configuration) at a chiral center is also necessary. The method that is used employs the Cahn-Ingold-Prelog Rules for determining the priority of each group attached to a carbon. The method for determining CIP priority is as follows: 1. Look at the four atoms directly attached to the chiral center and assign priorities in order of decreasing atomic number. The atom with the highest atomic number is ranked first; the atom with the lowest atomic number is ranked fourth. 2. If two or more of the atoms attached to the chiral center are tied for atomic number priority, compare the atomic numbers of the atoms attached to each of the tied atoms, continuing on as necessary through the third or fourth atoms outward until the first point of difference is reached. 3. Multiple bonded atoms are considered as if they were an equivalent number of single- bonded atoms. For example, - CH = O substituent is equivalent to – CH – O | O 134 Example: Consider the following molecules that each have a chiral center. The carbon in the molecule shown to the left is attached to four distinct groups, H thus making this atom chiral. When trying to determine the CIP priorities of the | groups attached to this carbon, we look first at atomic mass. I – C – Br | All 4 groups attached to this carbon have different atomic numbers, so we can Cl make the following priority assignments. #1 = I #2 = Br #3 = Cl #4 = H The carbon that is circled in the drawing to the left is attached to four distinct groups: a chlorine, a hydrogen, an aldehyde and an alcohol. In assigning priority by atomic number, we can see that chlorine wins out for priority #1 and that hydrogen is clearly #4. Our two carbon atoms appear to be tied for 2/3. H H | | #1 = Cl #2 = ? #3 = ? #4 = H Cl – C – C = O | To make the distinction in priority between these two atoms, we look at the HO – CH2 atoms directly attached to those carbons. One (carbon “A”) has two bonds to H and one bond to O, while the other (carbon “B”) has two bonds to O and one bond to H. Since the atoms attached to carbon B have higher atomic numbers, carbon B wins out as priority #2. #1 = Cl #2 = CHO #3 = CH2OH #4 = H Assigning R/S 135 Stereochemical Relationships: Enantiomers and Diastereomers (5.5) Stereoisomers have the same constitution (connectivity of atoms) but are ___________________ (they differ in their spatial arrangement of atoms). Stereoisomers can be subdivided into two categories, as shown below: __________________ are stereoisomers that are mirror images of one another, while __________________ are stereoisomers that are not mirror images of one another. According to these definitions, we can understand why cis-trans isomers (discussed at the beginning of this chapter) are said to be diastereomers, rather than enantiomers. Consider, once again, the structures of cis-2-butene and trans-2-butene: They are stereoisomers, but they are not mirror images of one another and are therefore diastereomers. An important difference between enantiomers and diastereomers is that enantiomers have the same physical properties, while diastereomers have different physical properties. The difference between enantiomers and diastereomers becomes especially relevant when we consider compounds containing more than one chirality center. As an example, consider the following structure: This compound has two chirality centers. Each one can have either the R configuration or the S configuration, giving rise to four possible stereoisomers (two pairs of enantiomers): 136 To describe the relationship between these four stereoisomers, we will look at one of them and describe its relationship to the other three stereoisomers. The first stereoisomer listed above has the configuration (1R, 2S). This stereoisomer, like anything in the world, only has only one ________ ________, or enantiomer, which has the configuration (1S, 2R). The third stereoisomer is not a mirror image of the first stereoisomer and is therefore its diastereomer. In a similar way, there is a diastereomeric relationship between the first stereoisomer and the fourth stereoisomer above because they are not mirror images of one another. To help better visualize the relationship between the four compounds above we will use an analogy. Imagine a family with four children (two sets of twins). The first pair of twins are identical to each other in almost every way, except for the placement of one birthmark. One child has the birthmark on the right cheek, while the other child has the birthmark on the left cheek. These twins can be distinguished from each other based on the position of the birthmark. They are nonsuperimposable mirror images of each other. The second pair of twins look very different from the first pair. But the twins in the second pair are once again identical to each other in every way, except the position of the birthmark on the cheek. They are nonsuperimposable mirror images of each other. In this family of four children, each child has one twin and two other siblings. The same relationship exists for the four stereoisomers shown above. In this molecular family, each stereoisomer has exactly one enantiomer (mirror-image twin) and two diastereomers (siblings). Now consider a case with three chirality centers: Once again, each chirality center can have either the R configuration or the S configuration, giving rise to a family of eight possible stereoisomers: 137 These eight stereoisomers are arranged above as four pairs of enantiomers. To help visualize this, imagine a family with eight children (four sets of twins). Each pair of twins are identical to each other with the exception of the birthmark, allowing them to be distinguished from one another. In this family, each child will have one twin and six other siblings. Similarly, in the molecular family, each stereoisomer has exactly one enantiomer (mirror-image twin) and six diastereomers (siblings). Note that the presence of three chirality centers produces a family of four pairs of enantiomers. A compound with four chirality centers will generate a family of eight pairs of enantiomers. This begs the obvious question: What is the relationship between the number of chirality centers and the number of stereoisomers in the family? This relationship can be summarized as follows: where n refers to the number of chirality centers. A compound with four chirality centers can have a maximum of , or eight pairs of enantiomers. As another example, consider the structure of cholesterol: Cholesterol has eight chirality centers, giving rise to a family of stereoisomers (256 stereoisomers). The specific stereoisomer shown has only one enantiomer and 254 diastereomers, although the structure shown is the only stereoisomer produced by nature. 138  Symmetry and Chirality (5.6) In this section we will learn how to determine whether a compound is chiral or achiral. It is true that any compound with a single chirality center must be chiral. But the same statement is not necessarily true for compounds with two chirality centers. Consider the cis and trans isomers of 1,2-dimethylcyclohexane: Each of these compounds has two chirality centers, but the trans isomer is chiral, while the cis isomer is not chiral. To understand why, we must explore the relationship between symmetry and chirality. Chirality is entirely dependent on ________________________ symmetry. Let’s consider the structure of the cis isomer of 1,2-dimethycyclohexane: This compound exhibits reflectional symmetry, with the ________ of ___________ shown in the image below. Any molecules that possess this kind of internal plane of symmetry cannot be chiral. Trans-1,2-dimethylcyclohexane (shown below) does not possess an internal plane of symmetry. As a result, any structure of trans-1,2-dimethylcyclohexane would have to be non- superimposable on its mirror image, and thus it would demonstrate enantiomerism. Compounds like this are chiral; all molecules that are chiral always have an enantiomer. 139 Chirality is only dependent on the presence or absence of reflectional symmetry. Any compound that possesses an internal plane of symmetry in any conformation will be achiral (not chiral). The cis isomer of 1,2-dimethylcyclohexane exhibits a plane of symmetry, and therefore, the compound is achiral. It is identical to its mirror image. It does not have an enantiomer. Because cis-1,2-dimethylcyclohexane has stereocenters as well as an internal plane of symmetry, it and other structures like it are called _______ structures, which refers to a stereoisomer that does not have an enantiomer. Although the presence of a plane of symmetry renders a compound achiral, the converse is not always true—that is, the absence of a plane of symmetry does not necessarily mean that the compound is chiral. There are also other types of reflectional symmetry, but they are beyond the scope of our discussion. For our purposes, it will be sufficient to look for a plane of symmetry. We can summarize the relationship between symmetry and chirality with the following two statements: A compound that has a plane of symmetry will be ____________. A compound that lacks a plane of symmetry will most likely be ____________ (although there are rare exceptions, which can mostly be ignored for our purposes). Thalidomide Carvone Spearmint Oil Essence of Caraway 140  Drawing Enantiomers and Diastereomers __________________________________________________________ 141  Worksheet #1 I. For each of the following pairs of molecules, indicate if they are diastereomers, a pair of enantiomers, the same molecule drawn twice, or totally different molecules. Star all chiral centers. 1) CH3 CH3 6) Cl H H Cl HO CH2CH3 CH3CH2 OH H H H Cl Cl H 7) 2) Cl CH3 Cl Cl H H Br CH3 Cl H H H H H H Br 8) Cl Cl 3) Cl OH CH3 H H CH3 HO CH3 Cl H H H CH3 H CH3 CH3 Cl Cl 4) H H 9) CH3 HO H H Cl Cl CH3 NH2 H2N CH3 H H Cl Cl 5) H OH CH3 H HO OH HO | O=C CH2CH3 On a separate sheet, try these: CH3 CH3 1-bromobutane and its mirror image CH3CH2 C=O | 2-pentanol and its mirror image OH Remember: 1. If a compound has 1 chiral center, it is chiral 3-chloropentane and its mirror image 2. If a compound has no chiral center, it is usually not chiral 3. If a compound has more than one chiral center, it may or may not be chiral. 4. Any molecule w/ an internal mirror plane is not chiral 142  Enantiomers, Diastereomers, or Same Molecule? 1. A + B? 2. B + C? 3. A + C? 4. A + D? 5. B + D? 6. C + D? 7. E + F? 8. E + G? 9. E + H? 10. F + G? 11. F + H? 12. G + H? Are any of the structures to the right meso? 143  Worksheet #2 Draw each molecule in the section below with 3D perspective if needed Star all chirality centers Determine whether each molecule is chiral or not Draw each molecule in the section below with 3D perspective if needed Draw an internal plane of symmetry where applicable Star each chirality center Assign R or S to each chirality center 144  Stereoisomerism in Alkenes – cis/trans and E/Z (8.4) Using cis and trans Designations Recall that a double bond is comprised of a bond and a bond (Figure 8.4). The bond is the result of overlapping sp2-hybridized orbitals, while the bond is the result of overlapping p orbitals. We have seen that double bonds do not exhibit free rotation at room temperature, giving rise to stereoisomerism: Cycloalkenes comprised of fewer than seven carbon atoms cannot accommodate a trans bond. These rings can only accommodate a bond in a cis configuration: In these examples, there is no need to identify the stereoisomerism of the double bond when naming each of these compounds, because the stereoisomerism is inferred. For example, the last compound is called cyclohexene, rather than cis-cyclohexene. A seven-membered ring containing a trans double bond has been prepared, but this compound (trans-cycloheptene) is not stable at room temperature. An eight-membered ring is the smallest ring that can accommodate a trans double bond and be stable at room temperature: 145 Using E and Z Designations The stereodescriptors cis and trans can only be used to indicate the relative arrangement of the parent chain in an alkene or ring. When the parent chain does not continue beyond a double bond, the use of cis-trans terminology would be ambiguous. Consider, for example, the following two compounds: These two compounds are not the same; they are stereoisomers. But which compound should be called cis and which should be called trans? In situations like this, IUPAC rules provide us with a method for assigning different, unambiguous stereodescriptors. Specifically, we look at the two groups on each vinylic position and choose which of the two groups is assigned the higher priority: In each case, a priority is assigned by using the Cahn-Ingold-Prelog priorities used to determine R vs. S. Specifically; priority is given to the element with the higher atomic number. In this case, F has priority over C, while N has priority over H. We then compare the position of the higher priority groups. If they are on the same side (as shown above), the configuration is designated with the letter Z (for the German word ______________, meaning “together”); if they are on opposite sides, the configuration is designated with the letter E (for the German word _____________, meaning “opposite”): These examples are fairly straightforward, because the atoms connected directly to the vinylic positions all have different atomic numbers. Other examples may require the comparison of two carbon atoms. In those cases, we will use the same tie-breaking rules that we used when assigning the configurations of chirality centers. 146  Worksheet #3 I. Draw the structure for the following: a. trans-3,4-dimethylhex-3-ene c. (E)-1-deuterio-2-chloropropene b. (Z)-3-chloro-4-methylhex-3- d. cis-hex-3-ene ene II. Indicate which of the following compounds show geometric isomerism; draw the isomeric structures of those that do and specify each as Z or E. a. But-1-ene f. Pent-1-ene b. But-2-ene g. Pent-2-ene c. 1,1-dichloroethene h. 1-chloropropene d. 1,2-dichloroethene i. 1-chloro-2-methylbut-2-ene e. 2-methylbut-2-ene j. 3-methyl-4-ethylhex-3-ene 147  Fischer Projections – D-aldoses (5.7) Shown above: The D family of aldoses. All of these sugars occur naturally except threose, lyxose, allose and gulose. Notice the similarity at carbon #5 for all of these “D” aldoses. To qualify as D, carbon 5 must be “R”. While all of these sugars are D, they vary as to whether they are (+) or (-). 148  Interpreting Fischer Projections 1. A + B? 2. A + C? 3. A + F? 4. B + J? 5. B + H? 6. C + H? 7. D + J? 8. D + K? 9. E + L? 10. G + I? 149  Worksheet #4 Consider the following structures: A B C D E F G H I J K L 150 Worksheet #5 151 Worksheet #6 152 153 154 Reading #1 – Mirror Image Molecules 155 156 157 Reading #2 – Sinister Molecules 158  Reading #3 – Chirality in Nature Reading #4 – Pasteur’s Discovery of Enantiomers 159 Reading #5 – The Importance of Being Cis 160 Stereochemistry Practice Questions 161 162 163 164 165 166 167 168 169 170 171 SKILLBUILDER REVIEW 5.1 IDENTIFYING CIS-TRANS STEREOISOMERISM STEP 1 Identify and name all four groups attached to STEP 2 Look for two identical groups on different the bond. vinylic positions and assign the configuration as cis or trans. Try Problems 5.1, 5.2, 5.3, 5.35 5.2 LOCATING CHIRALITY CENTERS STEP 1 Ignore sp2- and sp-hybridized STEP 2 Ignore CH2 and CH3 groups. STEP 3 Identify any carbon centers. atoms bearing four different groups. Try Problems 5.4, 5.5, 5.6, 5.34b, 5.48 5.3 DRAWING AN ENANTIOMER Either place the mirror behind the or place the mirror on the side of or place the mirror below the compound… the compound… compound. Try Problems 5.7, 5.8, 5.33, 5.34a, 5.38a-f,i-l 172 5.4 ASSIGNING CONFIGURATION STEP 1 Identify the STEP 2 If two (or more) atoms are STEP STEP STEP 5 Identify four atoms attached identical, make a list of substituents 3 Redraw the 4 Rotate direction of 1-2-3 to the chirality center and look for the first point of chirality molecule so sequence: clockwise and prioritize by difference. center, that the fourth is R, and atomic number. showing only priority is on a counterclockwise the priorities. dash. is S. Try Problems 5.8, 5.9, 5.10, 5.32, 5.39a-g,i, 5.45 5.7 DETERMINING THE STEREOISOMERIC RELATIONSHIP BETWEEN TWO COMPOUNDS STEP 1 Compare the configuration STEP 2 If all chirality centers have opposite configuration, the compounds of each chirality center. are enantiomers. If only some of the chirality centers have opposite configuration, then the compounds are diastereomers. Try Problems 5.21, 5.22, 5.36a-j,l, 5.41c, 5.49a,c,e-j 5.8 IDENTIFYING MESO COMPOUNDS Draw all possible stereoisomers and then look for a plane of symmetry in any of the drawings. The presence of a plane of symmetry indicates a meso compound. Try Problems 5.26, 5.27, 5.28, 5.37 5.46, 5.47, 5.51, 5.53a,b,d,f-h 173 5.9 ASSIGNING CONFIGURATION FROM A FISCHER PROJECTION EXAMPLE Assign STEP 1 Choose STEP 2 Prioritize. STEP 3 Rotate so STEP 4 Assign the configuration of one horizontal line that the fourth configuration. this chirality center. and draw it as a priority is on a dash. wedge. Choose one vertical line and draw it as a dash. Try Problems 5.29, 5.30, 5.31, 5.39h, 5.41a,b, 5.43c, 5.53c,e, 5.54, 5.55, 5.56 8.2 ASSIGNING THE CONFIGURATION OF A DOUBLE BOND STEP 1 Identify the two groups STEP 2 Repeat step 1 for the STEP 3 Determine whether the priorities are connected to one vinylic other vinylic position, moving on the same side (Z) or opposite sides (E) of position and then determine away from the double bond and the double bond. which group has priority. looking for the first point of difference. Try Problems 8.5, 8.6, 8.51 Material Covered on the Unit 3 Exam 1. Be able to identify stereocenters and assign R and S designations 2. Be able to assign E and Z designations to double bonds 3. Be able to identify pairs of molecules as enantiomers, diastereomers, geometric isomers, identical molecules, meso structures, structural isomers or totally different molecules 4. Be able to interpret and draw Fischer projections 5. Be familiar with the Cahn-Ingold-Prelog rules for assigning priorities 6. Be able to identify chirality in nature 174