Midterm 2 Review Packet · Organic Chemistry Peer Tutoring Department CHEM 51A University of California, Irvine Professor Sim Wesley ... by the volume of solvent: (0.5-0.05 g)/(100

Organic Chemistry Peer Tutoring Department CHEM 51A University of California, Irvine Professor Sim Wesley Vu ([email protected]) https://sites.uci.edu/ochemtutors/ Emily Nguyen ([email protected])

Midterm 2 Review Packet

1. Given the following molecule, assign R/S configuration to any stereocenters that are present. From the left view, draw the Newman projection for the exact conformation of the molecule and for the highest energy conformation for bond C1-C2.

Answer:

Remember that if the lowest priority group is in the front, the assigned configuration is inverted! Left view = C1 is in the front! Exact conformation: Highest energy conformation:

The highest energy conformation should always have the most steric interactions. Therefore, the largest substituents should be eclipsed! Notice that both this conformation has two steric interactions between OH & I and methyl & Cl. Note: the exact conformation is the staggered, anti conformation and is the lowest energy conformation.

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Organic Chemistry Peer Tutoring Department CHEM 51A University of California, Irvine Professor Sim Wesley Vu ([email protected]) https://sites.uci.edu/ochemtutors/ Emily Nguyen ([email protected]) 2. The following molecule is tartaric acid.

(a) Determine how many possible stereoisomers there are for this molecule (b) Draw all possible stereoisomers and label each stereogenic center with R/S

Answers:

(a) In order to find out the number of possible stereoisomers for a molecule, use the equation X= 2n where n is the number of stereogenic centers and X is the number of possible stereoisomers. 22= 4 possible stereoisomers

(b) Remember that hydrogen atoms in skeletal structure are often implied, so they may not always be explicitly drawn in! If this is confusing to you, draw out the hydrogen atoms yourself before solving the questions. Also remember that if the lowest ranking substituent is not in the back, you must switch the configuration you get at the end.


Organic Chemistry Peer Tutoring Department CHEM 51A University of California, Irvine Professor Sim Wesley Vu ([email protected]) https://sites.uci.edu/ochemtutors/ Emily Nguyen ([email protected]) 3. Given the following molecules, identify any enantiomers, diastereomers, or identical compounds. Also identify which compounds are achiral and which compounds are meso. Tips for determining molecule relationships:

● Assign R/S configuration to check relationships! ○ Enantiomers = all stereocenters inverted ○ Diastereomers = some (NOT ALL) stereocenters inverted

● If the substituents on the stereocenters are the same, rotate & flip the molecules to double check if they are identical!

● Remember that meso compounds refer to one molecule → looking for an internal plane of symmetry

● ALL meso compounds must be achiral!

Answer: All these molecules only have two stereocenters. The R/S configuration is as follow:

Identical: A & B — this can be visualized by flipping & rotating the molecule!

● While it may seem like it, there is no inversion of stereocenters here. After you rotate and flip the molecule, the stereocenters align and have the same configuration



Enantiomers: C & D Diastereomers: A/B & C, A/B & D Meso: A & B

● If you draw a line down the middle of A/B, you see that there is an internal plane of symmetry! The middle hydroxyl group is split in half!

Achiral: A & B 4. A solution of 0.50 g (+)-Atropine and 0.05 g (-)-Atropine in 100 mL methanol is placed into a tube with length 1 dm. What is the observed rotation that will be measured from this solution if (-)-Atropine has a specific rotation of -22°? Answer: Use the formula where [ ] is specific rotation, is observed rotation, c isα] [ = αc×l α α concentration (g/mL) , and l is the path length (dm) There is an excess of the (+) enantiomer, so the (+) enantiomer will produce the observed rotation. Since the specific rotation of the (-) enantiomer is -22°, the specific rotation of the (+) enantiomer is 22°.


Organic Chemistry Peer Tutoring Department CHEM 51A University of California, Irvine Professor Sim Wesley Vu ([email protected]) https://sites.uci.edu/ochemtutors/ Emily Nguyen ([email protected]) To find c, subtract the mass of the (-) enantiomer from the mass of the (+) enantiomer and divide by the volume of solvent: (0.5-0.05 g)/(100 mL) = 0.0045 g/mL Plug in values to the equation and solve. =0.099°α 5. Draw both chair conformations for the following compounds and determine which one is more stable. Tips for chair conformations:

● Substituents are axial in one conformation → equatorial in the flipped conformation ● Most stable chair = largest substituents in equatorial position

○ Equatorial is the more energetically favorable (and stable) position!

(a)

Answer: Remember that there are multiple ways to draw chair conformations—it all depends on carbon numbering! That means that your structure may not match mine depending on where you started counting carbons. As long as your substituents are on the correct carbon numbers and are in the correct orientation, the answer is correct!

The second chair conformation is the most stable because it has two substituents in the equatorial position vs. one equatorial substituent in the first conformation.

(b)


Organic Chemistry Peer Tutoring Department CHEM 51A University of California, Irvine Professor Sim Wesley Vu ([email protected]) https://sites.uci.edu/ochemtutors/ Emily Nguyen ([email protected]) Answer:

Both chair conformations are equally stable; both have one substituent in equatorial and one substituent in axial. Because the substituents are identical, they have the same energy values.

(c)

Answer:

The second chair conformation is the most stable because both of the substituents are in the equatorial position. 6. Determine the relationship between the two compounds in each pair.

(a)

(b)



(c) Answers:

(a) Enantiomers: the configurations at both stereocenters are different between the two molecules.

(b) Constitutional Isomers: The first ring has substituents located on the 1 and 2 positions, while the second ring has substituents located on the 1 and 3 positions.

(c) Enantiomers: Redraw the structure into a cyclohexane with dash and wedged bonds to make it easier to determine.

7. Name the following compounds using IUPAC nomenclature and indicate how many 1°, 2°, 3°, and 4° carbons are present in each compound. Concept review:

● 1° (primary carbon) = bonded to 1 other carbon ● 2° (secondary carbon) = bonded to 2 other carbons ● 3° (tertiary carbon) = bonded to 3 other carbons ● 4° (quaternary carbon) = bonded to 4 other carbons



(a)

Answer: (4S,5S)-4,5-dibromo-3-ethyl-6-methyloctane

When naming IUPAC structures with stereocenters that can be assigned configuration, you must list the configurations separately in the front. All the stereocenters are listed in parentheses with their configurations next to them (as seen above).

(b)

Answer: trans-1-chloro-4-iodocyclohexane


Organic Chemistry Peer Tutoring Department CHEM 51A University of California, Irvine Professor Sim Wesley Vu ([email protected]) https://sites.uci.edu/ochemtutors/ Emily Nguyen ([email protected]) In this case, there are two substituents that are in opposite directions (up vs. down). Therefore, this molecule can be classified as trans. Note: The stereochemistry of the molecule is denoted by the trans relationship between the two substituents. In addition, this molecule has no stereocenters so R/S configuration cannot be assigned!

(c)

Answer: 7-tert-butyl-6-ethyl-2-fluoro-5-methyldecane

8. Explain the difference between enantiomers and diastereomers. Answer: Enantiomers have different configurations at all stereogenic centers, while diastereomers have a different configuration at one stereogenic center and the same configuration at another stereogenic center. Also note that molecules with only one stereogenic center cannot have a diastereomer since they cannot meet the requirement of having one stereogenic center with opposite configuration and one stereogenic center with the same configuration.


Organic Chemistry Peer Tutoring Department CHEM 51A University of California, Irvine Professor Sim Wesley Vu ([email protected]) https://sites.uci.edu/ochemtutors/ Emily Nguyen ([email protected]) 9. Draw the curved arrow mechanisms for the following reaction and identify the rate-determining step. Given that the first step is rate-determining and the reaction is thermodynamically favored, write the rate equation for this reaction. Then draw the energy diagram for the reaction, including transition states and intermediates. Label the activation energy of the rate-determining step, ΔH, and ΔG in the energy diagram. Key things to remember:

● Rate-determining step = slowest step—limits how fast the reaction can occur ● Rate equation is made up of reactants in the rate-determining step only!

Answer:

reaction rate = k[CH3CH2COCH3][-OH]


Organic Chemistry Peer Tutoring Department CHEM 51A University of California, Irvine Professor Sim Wesley Vu ([email protected]) https://sites.uci.edu/ochemtutors/ Emily Nguyen ([email protected]) 10. The following molecule is Lisinopril, an ACE inhibitor used to treat high blood pressure. Label any functional groups present in this molecule.

Answer:


Organic Chemistry Peer Tutoring Department CHEM 51A University of California, Irvine Professor Sim Wesley Vu ([email protected]) https://sites.uci.edu/ochemtutors/ Emily Nguyen ([email protected]) 11. The following molecule is R-(+)-Limonene. It’s pure specific rotation is +11.5°.

(a) Calculate the %ee of a solution with a specific rotation of +6.3°.

ee 00% = [α]pure[a]mixture × 1

ee 00 4.8% excess R% = +6.3+11.5 × 1 = 5 (b) What is the pure specific rotation of S-(-)-Limonene?

α] − 1.5°[ pure = 1 Remember that enantiomers have the same specific rotation but in opposite directions!

(c) What is the percent of the S-(-)-Limonene in the solution of part a? R-enantiomer in solution 4.8= x + 5 S-enantiomer in solution asa sf= x

x 4.8 002 + 5 = 1 2.6x = 2

Therefore, there is 22.6% of S-(-)-Limonene in the enantiomeric solution.

12. Do the following sets of molecules have different conformations or different configurations?

(a)

(b)


Organic Chemistry Peer Tutoring Department CHEM 51A University of California, Irvine Professor Sim Wesley Vu ([email protected]) https://sites.uci.edu/ochemtutors/ Emily Nguyen ([email protected]) Answers:

(a) These two molecules have different conformations since conformational changes happen through rotation about a single bond. These cannot have different configurations since all of the bonding is the same. Newman projections can be used to visualize this rotation about the single bond.

(b) These two molecules have different configurations. Configurational changes occur when the position of two substituents is switched; this requires the breaking of bonds, unlike in conformational changes where no bonds are broken. In the second molecule, the methyl and ethyl groups are switched.

13. Calculate the ΔH values of each reaction using Table 6.2 from your lecture notes. Also determine the signs of ΔG and ΔS. Then state whether the starting material or product is favored at equilibrium. Classify each reaction as addition, substitution, or elimination.

H nergy of bonds broken nergy of bonds formedΔ = e − e (a) Note: C=C bond energy = 614

Answer: Bonds broken: C-C, H-Br Bonds formed: C-H, C-Br H 614 68) 435 85) 62 kJ /molΔ = ( + 3 − ( + 2 = 2

Because ΔH & ΔG are similar in value, ΔG is also positive. Because entropy in the system is going down (two molecules → one), ΔS is negative.

(b)

Answer: Note: NaI and NaCl are ionic bonds! Therefore, no energy is input/released. Bonds broken: C-Cl Bonds formed: C-I H 335) 222) 13 kJ /molΔ = ( − ( = 1

Because ΔH & ΔG are similar in value, ΔG is also positive. Because entropy in the system is neither going up or down, ΔS=0.

(c)



Answer: Note: C=C bond energy = 614 Bonds broken: C-H, C-Br Bonds formed: C-C, H-OH H 397 85) 614 98) − 30 kJ /molΔ = ( + 2 − ( + 4 = 4

Because ΔH & ΔG are similar in value, ΔG is also negative. Because entropy in the system is going up, ΔS is positive. 14. Rank the following compounds from least acidic to most acidic.

CH3CH2COOH, BrCH2COOH, CH3CH2CH2OH

Answer: When determining acidity, look at the conjugate bases:

A more stable conjugate base will be a stronger acid. Thus, the order from weakest acid to strongest acid is CH3CH2CH2OH, CH3CH2COOH, BrCH2COOH.


Organic Chemistry Peer Tutoring Department CHEM 51A University of California, Irvine Professor Sim Wesley Vu ([email protected]) https://sites.uci.edu/ochemtutors/ Emily Nguyen ([email protected]) 15. Given the following energy diagram:

(a) Determine the rate-determining step.

Because the second step in the reaction has the highest activation energy, it is the rate-determining step.

(b) Identify what each variable represents in the reaction. Write the values of activation energy for both reactions, ΔG, and ΔH for the overall reaction. A: starting materials B: transition state (first step) C: intermediate D: transition state (second step) E: products

(c) Write the values of activation energy for both reactions, ΔG, and ΔH for the overall reaction. Write these same values if the reaction was reversed.

Normal Reaction Reversed Reaction

Ea (first step): B-A Ea (second step): D-C

ΔG≈ΔH=C-E

Ea (first step): D-E Ea (second step): B-C

ΔG≈ΔH=C-A


Organic Chemistry Peer Tutoring Department CHEM 51A University of California, Irvine Professor Sim Wesley Vu ([email protected]) https://sites.uci.edu/ochemtutors/ Emily Nguyen ([email protected]) 16. Classify the following reactions as an oxidation or reduction.

(a)

(b)

(c) Answers:

(a) In this reaction, a C=O bond is converted to a C—O bond, so this is a reduction. (b) In this reaction, a C—O bond is converted to a C=O bond, so this is an oxidation. (c) Remember that a C=C double bond counts as two C—C bonds. In this reaction, two

C—C bonds are being converted into two C—H bonds, so this is a reduction.

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Midterm 2 Review Packet · Organic Chemistry Peer Tutoring Department CHEM 51A University of California, Irvine Professor Sim Wesley ... by the volume of solvent: (0.5-0.05 g)/(100