(1) Check to see if the two compounds are identical. (2) Recall the definitions of stereoisomers, conformational isomers, and constitutional isomers.

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1 MCAT Organic Chemistry Problem Drill 04: Stereochemistry Question No. 1 of 10 Question 1. Determine the relationship of the molecules shown: O O Question #01 (A) Identical (B) Constitutional isomers (C) Conformational isomers (D) Stereoisomers One molecule is an ether and the other is an alcohol so they can not be identical. The connectivity of the atoms is different in each compound. B. Correct! While each molecule has the same molecular formula, the atoms are connected differently. The ether and the alcohol are constitutional isomers. C. Incorrect! Conformational isomers have the same molecular formula and atom connectivity but differ in the rotation about a single bond or bonds to give different conformations. These two compounds do not have the same atom connectivity. Stereoisomers have the same molecular formula and atom connectivity but differ in their spatial orientation. These two compounds do not have the same atom connectivity. (1) Check to see if the two compounds are identical. Identical compounds would have the same molecular formula, atom connectivity, and spatial orientation. These compounds do share the same molecular formula: C 8 18 O. owever, their atom connectivity differs. One contains an ether and the other an alcohol. They are not identical. (2) Recall the definitions of stereoisomers, conformational isomers, and constitutional isomers. Stereoisomers have the same molecular formula and atom connectivity but differ in their spatial orientation. Conformational isomers have the same molecular formula and atom connectivity but differ in the rotation about a single bond or bonds to give different conformations. We have already established that the two given compounds share the same molecular formula but do not have the same atom connectivity. This fact eliminates both stereoisomers and conformational isomers as possible answers. Constitutional isomers are compounds that have the same molecular formula but have different atom connectivities. Therefore, the correct answer is (B).

2 Question No. 2 of 10 Question 2. List these three conformers of pentane in terms of increasing stability. 3 C 2 C C 2 C 3 3 C 2 C Question #02 C 3 C 3 3 C I II III (A) I, II, III (B) I, III, II (C) III, II, I (D) II, III, I The ordered list I, II, and III is incorrect. Go back and determine the type of each conformer (anti, gauche, eclipsed) based on the relationship between the two largest groups the ethyl and methyl groups of the molecule. The ordered list I, III, II is partially correct. Review each conformer and make sure you have them listed in order of increasing stability. C. Incorrect! The ordered list III, II, I is partially correct. Go back and determine which conformer would have the most steric repulsions of the ethyl and methyl groups and would therefore be the least stable. D. Correct! II, the totally eclipsed conformer, is the least stable due to steric repulsions of the ethyl and methyl groups, and I, the anti conformer, is the most stable. (1) Look at each conformer and decide whether it is anti, gauche, eclipsed or totally eclipsed. I shows the two largest groups, the ethyl and methyl groups, 180 degrees apart in a staggered conformation. This means I is the anti conformer. II shows the ethyl and methyl groups are lined up one in front of the other in an eclipsed conformation. This means II is the totally eclipsed conformer. III shows a staggered conformation with the two largest groups 60 degrees apart. This means III is a gauche conformer. (2) Determine the relative stability of each and rank them in order of increasing stability. The further the two largest groups are from each other, the more stable the conformer will be since there are less steric interactions. Since the anti conformer has the two largest groups as far apart as possible (180 degrees), it should be the most stable. The totally eclipsed conformer, with a dihedral angle of 0 degrees, is the least stable. The ranking of the conformers in order of increasing stability should be II, III, I. Therefore, the correct answer is (D).

3 Question No. 3 of 10 Question 3. Which object below is not chiral? Question #03 (A) An ear (B) A foot (C) A circular table (D) A molecule of DNA Your ears, like your hands, are chiral. Look for the item with a mirror plane of symmetry. Your feet, like your hands, are chiral. Look for the item with a mirror plane of symmetry. A circular table is superimposable with its mirror image and therefore is not chiral. A molecule of DNA has the form of an α-helix which is chiral. Look for the item that is superimposable on its mirror image. (1) Picture each item (it may help to make a quick drawing of each choice) and ask the following questions: Is there a mirror plane of symmetry? Can you take the object, place it on top of its mirror image and have everything line up? If you can answer yes to either of these questions, then the object is probably chiral. In the list of possible answers, the answers of an ear, a foot, and a molecule of DNA can not be superimposed on their mirror images and are therefore chiral. A circular table can be superimposed on its mirror image and there are numerous planes of symmetry so the circular table is not chiral. Therefore, the correct answer is (C).

4 Question No. 4 of 10 Question 4. Which of the molecules below is chiral? Question #04 O C 3 I II III IV S (A) I, II, III, IV (B) I, III (C) II, IV (D) II, III, IV Not all the molecules in this answer are chiral. Do any of them have a mirror plane of symmetry? Neither of these molecules are chiral. Look closely at their structures. Can you find a carbon in either that has four different groups attached? Both molecules contain a stereogenic center and as a result are chiral. Not all the molecules in this answer are chiral. Do any of them have a mirror plane of symmetry? (1) Look at each molecule and determine whether it is superimposable on its mirror image. One of the easier ways to do this is to look for a stereogenic center in the molecule. A stereogenic center (sometimes called a chiral center) is a carbon that has four different things attached to it. If a molecule contains a stereogenic center, then it must be chiral. In compound I, carbon 3 would appear to be a stereogenic center. owever, on closer inspection, one would notice that carbon 3 has two ethyl groups attached to it in addition to the alcohol and methyl group. It can not be chiral because it has two of the same groups attached to it. In compound II, the center carbon has a hydrogen, chlorine, bromine and methyl group attached to it. Since it has 4 different groups attached to it, it is a stereogenic center and the molecule must be chiral. In compound III, there appears to be three possible chiral centers. owever, on closer inspection, there is a mirror plane of symmetry (indicated by dashed line) that runs through the methyl group and ring so the molecule can not be chiral: In compound IV, the carbon attached to the sulfide (-S) has four different groups attached so the molecule is chiral. Therefore, the correct answer is (C).

5 Question No. 5 of 10 Question 5. What is the relationship between these two molecules? Question #05 (A) Identical (B) Enantiomers (C) Diastereomers (D) Conformational isomers Identical molecules must have the same molecular formula, the same atom connectivity and the same spatial orientation. These molecules have a different orientation in space. Enantiomers are molecules that are nonsuperimposable mirror images of each other. The molecules above contain two chiral centers. Only one set of the centers are mirror images of each other. What about the other set of centers? These molecules are diastereomers or non-mirror image stereoisomers. If you assign R/S configurations to each center you will find the first molecule is S,R while the second molecule is R,R. Conformational isomers have the same molecular formula and atom connectivity but differ in the rotation about a single bond or bonds to give different conformations. These molecules do not differ by a bond rotation but by their spatial orientation. (1) Determine if the molecules are identical. Identical compounds would have the same molecular formula, atom connectivity, and spatial orientation. These molecules do not have the same spatial orientation and therefore can not be identical. (2) Review the definitions of enantiomer and diastereomer. Enantiomers are molecules that are non-superimposable mirror images of each other. Diastereomers are molecules that are non-mirror image stereoisomers. One of the easiest ways to determine if a set of molecules are enantiomers or diastereomers, is to assign R/S configurations to every chiral center present in the molecules. For example, if there are two molecules that contain two chiral centers each, assign the R/S configurations of each center. If the set of configurations are opposite each other (R,R versus S,S or R,S versus S,R) then the molecules are enantiomers. If all the centers do not have the exact opposite configuration (R,S versus S,S or R,R versus S,R), than the molecules are diastereomers. In the problem above, after assigning R/S configurations, you find that the first molecule is S,R while the second molecule is R,R. These molecules are diastereomers. Since these molecules do not differ by a bond rotation but by their spatial orientation, they can not be conformational isomers. Therefore, the correct answer is (C).

6 Question No. 6 of 10 Question 6. Which stereoisomer shown is meso? Question #06 C 3 C 3 C 3 C 3 C 3 C 3 I II III IV (A) I (B) II (C) III (D) IV None are meso. A. Correct! Compound I, while containing two chiral centers, also contains a mirror plane of symmetry and is meso. A meso compound must have a mirror plane of symmetry and are superimposable on its mirror image. This compound does not have a plane of symmetry and can not be superimposed on its mirror image. C. Incorrect! A meso compound must have a mirror plane of symmetry and are superimposable on its mirror image. This compound does not have a plane of symmetry and can not be superimposed on its mirror image. There is a meso compound present. Look for a molecule that contains a mirror plane. (1) Recall the definition of meso. A meso compound is one that contains more than one stereogenic center but is achiral due to the presence of a mirror plane of symmetry in the molecule. (2) Look at the possible answers and determine if any contain more than one stereogenic center. All three molecules contain two stereogenic centers. (3) Determine if any of the molecules contain a plane of symmetry. For compound I, there is a plane of symmetry present. The top chiral center reflects into the bottom chiral center so the compound is meso. C 3 C 3 The other molecules do not contain a plane of symmetry and are not meso. Therefore, the correct answer is (A).

7 Question No. 7 of 10 Question 7. Which statement concerning these four compounds is incorrect? Question #07 3 CO C 3 O O 3 CO C 2 C 3 C 3 C 2 C 3 C 3 O OC 3 C 2 C 3 O I II III IV C 3 OC 3 C 2 C 3 (A) I and IV are enantiomers. (B) I and II are diastereomers. (C) II and IV are enantiomers. (D) III and IV are diastereomers. I and IV are non-superimposable mirror images of each other and are enantiomers. Assign R/S configurations to each center to help determine sets of enantiomers and diastereomers. I and II are diastereomers or non-mirror image stereoisomers. Look at I and II. Can you see that the top chiral centers are mirror images of each other but the lower chiral centers are not? This makes the two compounds non-mirror image stereoisomers. II and IV are diastereomers not enantiomers. Only the bottom stereocenter in each is a mirror image of the other. The top stereocenter is identical in each. III and IV are diastereomers. The top chiral centers in each are mirror images but the bottom chiral center are identical to each other. (1) Recall the definitions of enantiomer and diastereomer. Enantiomers are molecules that are non-superimposable mirror images of each other. Diastereomers are molecules that are non-mirror image stereoisomers. (2) Assign R/S configurations to each chiral center. One of the easiest ways to determine if a set of molecules are enantiomers or diastereomers, is to assign R/S configurations to every chiral center present in the molecules. The configurations for I are R,R. The configurations for II are S,R. The configurations for III are R,S. The configurations for IV are S,S. There are two sets of enantiomers: I and IV; II and III. Any other pairings must be diastereomers. Therefore, the correct answer is (C).

8 Question No. 8 of 10 Question 8. If a molecule has 5 chiral centers, how many possible stereoisomers can exist? Question #08 (A) 10 (B) 16 (C) 32 (D) 36 There are more than 10 possible stereoisomers for this molecule. Recall the formula for calculating possible stereoisomers: 2 n where n is the number of chiral centers. There are more than 16 possible stereoisomers for this molecule. Recall the formula for calculating possible stereoisomers: 2 n where n is the number of chiral centers. There are 5 chiral centers present. Therefore, in this case, the formula for calculating chiral centers is 2 n where n is equal to 5. Two raised to the fifth power is 32. There are not 36 possible stereoisomers for this molecule. Recall the formula for calculating possible stereoisomers: 2 n where n is the number of chiral centers. (1) Recall the formula for calculating possible stereoisomers. The formula is 2 n where n is the number of stereogenic centers in the molecule. (2) Plug in a number for n and calculate the number of possible stereoisomers. In this case, we know there are 5 stereogenic centers in the molecule. 2 n = number of possible stereoisomers 2 5 = 32 Therefore, the correct answer is (C).

9 Question No. 9 of 10 Question 9. Determine which statement regarding the stereoisomers of alkenes is correct. Question #09 (A) The relative configuration of alkenes is determined by the orientation of the parent chain through the alkene though this process is difficult for tri and tetra-substituted alkenes. (B) An alkene is cis if the parent chain is on opposite sides of the double bond. (C) An alkene is trans if the parent chain is on the same side of the double bond. (D) To assign the configuration of tri- and tetra-substituted alkenes, one uses the same R/S configuration for chiral centers. A. Correct! The relative configuration of an alkene (cis versus trans) is determined by the orientation of the parent chain through the alkene. Relative configurations are typically used for disubstituted alkenes since the process is more complicated with a tri- or tetra-substituted alkene. An alkene is cis if the parent chain is on the same side of the double bond. Recall the definitions of cis and trans in relation to alkene geometry. C. Incorrect! An alkene is trans if the parent chain is on opposite sides of the double bond. Recall the definitions of cis and trans in relation to alkene geometry. To assign the configuration of tri- and tetra-substituted alkenes, one uses the E/Z system for establishing priority of groups coming off the double bond. The R/S system is used only for determining the configuration of chiral carbons. (1) Recall the conventions for determining the relative and absolute The relative configuration of an alkene (cis versus trans) is determined by the orientation of the parent chain through the alkene. Relative configurations are typically used for disubstituted alkenes since the process is more complicated with a tri- or tetra-substituted alkene. An alkene is cis if the parent chain is on the same side of the double bond. An alkene is trans if the parent chain is on opposite sides of the double bond. The absolute configuration of an alkene is determined by the E/Z system. The E/Z system uses priority rules for assigning R/S to chiral centers. If the two groups with highest priority are on the same side of the double bond, then the alkene is Z. If the two groups with highest priority are on opposite sides of the double bond, then the alkene is E. This system can be used for di-, tri-, and tetra-substituted alkenes. (2) Read each statement and determine which one is correct. Based on our knowledge of relative and absolute configurations of alkenes, we know statements (B), (C), and (D) are incorrect in their details. Statement (A) however contains no incorrect information. Therefore, the correct answer is (A).

10 Question No. 10 of 10 Question 10. Using the E/Z system, determine (1) which two groups on the alkene below are of highest priority and (2) the absolute configuration of the alkene. I O Question #10 O (A) I and ; Z (B) C 2 O and (C=O); E (C) I and (C=O); Z (D) and C 2 O; E When determining absolute configuration, divide the alkene in half and determine the highest priority group on each carbon of the double bond. The two highest priority groups will never occur on the same carbon. Because of this fact, we can eliminate (A) as a possible answer since the iodide and bromine atoms are on the same carbon. When determining absolute configuration, divide the alkene in half and determine the highest priority group on each carbon of the double bond. The two highest priority groups will never occur on the same carbon. Because of this fact, we can eliminate (B) as a possible answer since the C 2 O and the aldehyde groups are on the same carbon. Based on the priority rules of the E/Z system, we can determine that the iodide and the aldehyde groups will have the highest priority. Since both groups are on the same side of the double bond, the alkene is assigned a Z configuration. The E/Z priority rules are based on giving the groups with the higher atomic mass/number the higher priority. Go back and check to see if you determined the correct atomic masses for each atom directly attached to the double bond. (1) Divide the alkene down the center. I O O (2) Look at the left end of the alkene. Determine which group has the highest priority. On the left end of this alkene, we have an iodide atom and a bromine atom. Iodine has an atomic number of 53 while bromine has an atomic number of 35. Since iodine has the higher atomic number, it will be given the higher priority. Circle the atom you determine to have higher priority. I O O (3) Look at the right end of the alkene. Determine which group has the highest priority. ere is a situation where the same atom, in this case it is carbon, is bonded to the same end of the double bond. The priority of the groups can not be determined using the atomic number of the atoms directly attached to the double bond since they are the same. In cases like this, it is customary to compare the next group of atoms, with any double bonds being duplicated (and triple bonds triplicated) with imaginary atoms. I O C O O This carbon is attached to a, O, O., O, O is given the higher priority over,, O. This carbon is attached to a,, O. The aldehyde group is therefore given the higher priority. I O O Since both groups of higher priority are on the same side of the bond, the alkene s configuration is Z. Therefore, the correct answer is (C).