Chemistry

  1. How many chiral (asymmetric) C atoms are in the molecule below?
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6

 

Version II, III, same choices

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  1. What is the correct formula of the molecule below?
  2. C15H23BrClNO
  3. C14H23BrClNO2
  4. C14H22ClBrNO
  5. C15H24BrClNO
  6. C14H22ClBrNO

Version II

  1. C14H22ClBrN2O
  2. C14H25BrClN2O
  3. C15H22ClBrN2O2
  4. C15H23BrClN2O
  5. C15H24ClBrN2O

Version III:

  1. C14H22BrClN2O
  2. C14H20BrClN2O
  3. C15H22ClBrN2O
  4. C14H18BrClNO2
  5. C15H20ClBrN2O
  6. How many π bonds are there in the molecule (E)-5-bromo-6-methylocta-1,3-dien-7-yne?
  7. 2
  8. 3
  9. 4
  10. 5
  11. 7

Version II:  (Z)-4-chloro-3,8-dimethylnona-5,7-dien-1-yne;

Version III: (E)-3-ethyl-4,8-dimethylnona-4,7-dien-1-yne;

same choices, same key

 

 

  1. Which pair of the structures are isomers?

II, III: change some of the molecules from right to left, change -OH to -Br in A, change -OH to -Cl in B, change toluene to aniline in D, change -Br to -Cl or NH2 in E. Combine these changes to make two different versions.

 

 

  1. Which one of the following statements is correct?
  2. Molecules with a π bond adjacent to a nitrogen atom will undergo fast interconversion between resonance forms at room temperature.
  3. The molecule (E)-4,6-dimethylhept-2-ene has 3 methyl groups in total.
  4. All pure compounds that contain chiral carbon atoms are optically active.
  5. The formula of spiro[4.5]decane is C10H18.
  6. The molecules of tert-butylcyclohexane do NOT undergo ring-flip because the A-value of tert-butyl group is very large.

Version II:

  1. Molecules with a π bond adjacent to a nitrogen atom will undergo fast interconversion between resonance forms at room temperature.
  2. The molecule (E)-4-ethyl-6-methylhept-2-ene has 5 methyl groups in total.
  3. All pure compounds that contain asymmetric carbon atoms are optically active.
  4. The formula of spiro[4.5]decane is C10H16.
  5. The molecules of tert-butylcyclohexane undergo ring-flip despite that the A-value of tert-butyl group is very large.

Version III:

  1. Molecules with a π bond adjacent to a nitrogen atom will undergo fast interconversion between resonance forms at room temperature.
  2. The molecule (Z)-5-ethyl-6-methylhept-2-ene has 4 methyl groups in total.
  3. All pure compounds that contain asymmetric carbon atoms are optically active.
  4. The formula of spiro[5.5]undecane is C11H18.
  5. The molecules of tert-butylcyclohexane do not undergo ring-flip because the A-value of tert-butyl group is very large.

 

 

  1. Assuming all the atoms follow the octet rule, add formal charges to the species below, and select the total number of charges you added. Note each structure may have 0, 1 or multiple formal charges.
  2. 1 positive charge, 3 negative charges
  3. 2 positive charges, 3 negative charges
  4. 3 positive charges, 2 negative charges
  5. 3 positive charges, 3 negative charges
  6. 2 positive charges, 2 negative charges

Version II:

Same answer

Version III:

  1. 1 positive charge, 3 negative charges
  2. 2 positive charges, 3 negative charges
  3. 3 positive charges, 2 negative charges
  4. 3 positive charges, 3 negative charges
  5. 2 positive charges, 2 negative charges

 

 

  1. What is the relationship between the two molecules below?
  2. They are a pair of enantiomers.
  3. They are two diastereomers.
  4. They are conformers of the same molecule.
  5. They are the same conformation of the same molecule, viewed from different angles.
  6. They are cis-trans stereoisomers.

 

Version II:

Version III:

Answer is the same for all versions.

 

 

  1. This molecular structure below is shared 8 stereoisomers because it has 3 asymmetric carbon atoms. This is based on a rule proposed by which scientist?
  2. Jacobos van’t Hoff
  3. Gilbert Lewis
  4. Friedrich Wöhler
  5. Emil Fisher
  6. Dmitri Mendeleev

Add one more methylene to the chain to make Version II, change -OH to -Cl to make Version III.

 

 

  1. What is the degree of unsaturation of the molecule below:
  2. 4
  3. 5
  4. 6
  5. 7
  6. 8

Version II

  1. 4
  2. 5
  3. 6
  4. 7
  5. 8

Version III:

  1. 4
  2. 5
  3. 6
  4. 7
  5. 8

 

  1. Which one of the following Newman projections correctly depicts the lowest conformation of the meso compound (2R,3S)-butane-2,3-diol? (hint: 3 of the choices are wrong conformers, 1 of the choice is a wrong isomer)

Version II: change -OH to -Cl; Version III: change -OH to -Br. Don’t forget to change the compound name in the problem text.

 

 

  1. In their lowest energy conformations, are all the non-hydrogen atoms in the following molecules in the same plane? Write “Yes” or “No” for each molecule [Do not add any other text, including spaces, to your answer or it will be marked as incorrect.]. (1 pt each, 4 pts total)
  2. a) . b) . c)          . d)            . (YYNN)

Version II:

  1. a) . b) . c)          . d)            . (NYYN)

Version III:

  1. a) . b) . c)          . d)            . (YNYN)

 

 

  1. Assign the configuration for the following carbon atoms where wedge and dash bonds are drawn. Write R, S, or achiral for each molecule. [Do not add any other text, including spaces, to your answer or it will be marked as incorrect.] (1 pt each, 4 pts total)
  2. a) . b) . c)          . d)            . (R, S, Achiral, S)

Version II:

  1. a) . b) . c)          . d)            . (S, Achiral, S, R)

Version III:

  1. a) . b) . c)          . d)            . (R, Achiral, S, R)

 

 

  1. Which one of the following molecules is aromatic? (hint: consider their resonance structures) [For carbonyl-containing molecules consider only the resonance forms that have a formal charge on oxygen.]

Version II:

Version III:

 

 

  1. Which one of the following molecule’s boiling point is irrelevant to hydrogen-bonding?
  2. CH3CH2CH2OH
  3. HOCH2CH2CH2OCH3
  4. H2O

Version II:

Change A from propanol to ethanol; add a -Br to the para position of C.

Version III:

Rewrite B from right to left; add a -Me to the para position of C; rotate E.

 

 

  1. For the molecule below, write the number of sp3, sp2, and sp hybridized carbon atoms, respectively. (note: this molecule has 19 carbon atoms in total) (1.5 pts each if one or two correct answers, 4 pts total for three correct answers) [This 4 point scheme might be difficult to realize in Canvas.]

       sp3,         sp2,        sp hybridized carbon atoms. (7, 10, 2)

 

Version II: For the molecule below, write the number of sp3, sp2, and sp hybridized carbon atoms, respectively. (note: this molecule has 20 carbon atoms in total)

       sp3,         sp2,        sp hybridized carbon atoms. (8, 10, 2)

 

Version III: For the molecule below, write the number of sp3, sp2, and sp hybridized carbon atoms, respectively. (note: this molecule has 17 carbon atoms in total)

       sp3,         sp2,        sp hybridized carbon atoms. (6, 9, 2)

 

 

  1. Which one of the following molecules does NOT have a molecular dipole?

 

Version II: change all -Cl to -Br, rotate all molecules, then change -OH in C to -Cl

Version III: keep all -Cl, rotate all molecules, change -OH in C to methyl

 

 

  1. Like carbon atoms, sp3 hybridized nitrogen atoms can be chiral. However, when nitrogen atoms that have a pair of unshared electrons are attached to three different groups, the pyramidal structure can undergo fast inversions at 1,000-100,000 times per second even at low temperatures, as shown below. Therefore, these nitrogen atoms are not chirally stable stereocenters and do not lead to optical activity of the molecules. In most context, a nitrogen atom is considered a chiral center only if it is attached to four different groups as an ammonium ion (like the case of an asymmetric carbon), or such inversion is prevented by conformational strain (so the inversion does not happen or one structure has energetic advantage).

With the information above in mind, which one of the nitrogen atoms in the following structures is a chiral center that can lead to optical activity?

Version II:

Version III:

 

  1. With the A-value table at 298 K below, what is the free energy difference between the two chair conformations of the molecule below at 298 K? (hint: absolute value for the “difference”, do not worry about the positive/negative sign) (4 pts) [Consider making this multiple choice. Students will want partial credit if they are close … you will be doing yourself a favor.]

 

Version I:

ΔG =                    .  0.7 kcal/mol

Version II:

ΔG =                    .  2.0 kcal/mol

Version III:

ΔG =                    .  2.2 kcal/mol

 

-1 pt for not having the unit.

 

 

  1. Which one of the bonds indicate by arrow in the molecule below have the longest bond length? (hint: consider resonance structures when you determine bond orders)
  2. 1
  3. 2
  4. 3
  5. 4
  6. 5

Version II:

 

Version III:

Key is B. 2 for both Versions II and III.

  1. Assuming the atoms follow the octet rule, how many unshared electron pairs need to be added to the hypothetical structure shown to make it complete? How many positive formal charges need to be added? How many negative formal charges need to be added?
  2. 3 pairs, one positive charge, one negative charge
  3. 4 pairs, one positive charge, no negative charge
  4. 4 pairs, no positive charge, one negative charge
  5. 3 pairs, one positive charge, no negative charge
  6. 4 pairs, one positive charge, one negative charge

Version II:

 

Version III:

(Same choices, same key)

 

 

  1. How many resonance contributors does this structure have? Include in your total only those forms that have the same number of π bonds as the structure shown below. Also, do not include the below structure itself, in your total. (Non-bonded/unshared valence electrons are not shown.)

 

 

  1. 1
  2. 2
  3. 3
  4. 4
  5. 5

Version II:

Version III:

(Same choices, same key)

 

 

  1. How are the compounds shown related?
  2. They are enantiomers.
  3. They are diastereomers.
  4. They are constitutional isomers.
  5. They are identical.
  6. They are meso structures.

Version II:

  1. They are enantiomers.
  2. They are diastereomers.
  3. They are constitutional isomers.
  4. They are identical.
  5. They are meso structures.

 

Version III:

  1. They are enantiomers.
  2. They are diastereomers.
  3. They are constitutional isomers.
  4. They are identical.
  5. They are meso structures.

 

 

  1. How should the compound shown be depicted as a Fischer projection? Select the option that correctly indicates which functional groups (OH or H) correspond to the letters in the template provided.
  2. R1 = R3 = H; R2 = R4 = OH
  3. R1 = R3 = OH; R2 = R4 = H
  4. R1 = R4 = OH; R2 = R3 = H
  5. R1 = R4 = H; R2 = R3 = OH
  6. R1 = R2 = R3 = R4 = OH

Version II:

  1. R1 = R3 = H; R2 = R4 = OH
  2. R1 = R3 = OH; R2 = R4 = H
  3. R1 = R4 = OH; R2 = R3 = H
  4. R1 = R4 = H; R2 = R3 = OH
  5. R1 = R2 = R3 = R4 = OH

Version III:

  1. R1 = R3 = H; R2 = R4 = OH
  2. R1 = R3 = OH; R2 = R4 = H
  3. R1 = R4 = OH; R2 = R3 = H
  4. R1 = R4 = H; R2 = R3 = OH
  5. R1 = R2 = R3 = R4 = OH
  6. What is the lowest energy chair conformer for the compound shown? Select the option that correctly indicates which functional groups would correspond to the letters in the template provided.

 

  1. R1 = R4 = H; R2 = tBu; R3 = Me
  2. R1 = R3 = H; R2 = tBu; R4 = Me
  3. R2 = R3 = H; R1 = tBu; R4 = Me
  4. R2 = R4 = H; R1 = tBu; R3 = Me
  5. R1 = R2 = H; R3 = tBu; R4 = Me

Version II:

  1. R1 = R4 = H; R2 = tBu R3 = Me
  2. R1 = R3 = H; R2 = tBu; R4 = Me
  3. R2 = R3 = H; R1 = tBu; R4 = Me
  4. R2 = R4 = H; R1 = tBu; R3 = Me
  5. R1 = R2 = H; R3 = tBu; R4 = Me

Version III:

  1. R1 = R4 = H; R2 = tBu; R3 = Me
  2. R1 = R3 = H; R2 = tBu; R4 = Me
  3. R2 = R3 = H; R1 = tBu; R4 = Me
  4. R2 = R4 = H; R1 = tBu; R3 = Me
  5. R1 = R2 = H; R3 = tBu; R4 = Me

 

 

  1. Convert the structure shown (right) to a Newman Projection. Select the option that correctly indicates which functional groups would correspond to the letters in the template provided (left). (hint: pick the correct depiction on the template. The structure shown with the template may not correspond to the lowest energy conformer.)
  2. R1 = R2 = H; R3 = SPh; R4 = tBu
  3. R1 = SPh; R2 = R3 = H; R4 = tBu
  4. R1 = SPh; R3 = tBu; R2 = R4 = H
  5. R1 = R4 = H; R2 = tBu; R3 = SPh
  6. R1 = SPh; R2 = tBu; R3 = R4 = H

Version II:

  1. R1 = R2 = H; R3 = SPh; R4 = tBu
  2. R1 = SPh; R2 = R3 = H; R4 = tBu
  3. R1 = SPh; R3 = tBu; R2 = R4 = H
  4. R1 = R4 = H; R2 = tBu; R3 = SPh
  5. R1 = SPh; R2 = tBu; R3 = R4 = H
  6. R1 = R2 = H; R3 = SPh; R4 = tBu
  7. R1 = SPh; R2 = R3 = H; R4 = tBu
  8. R1 = SPh; R3 = tBu; R2 = R4 = H
  9. R1 = R4 = H; R2 = tBu; R3 = SPh
  10. R1 = SPh; R2 = tBu; R3 = R4 = H

 

 

 

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