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Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 1 of 20 PART A: Drawing Conventions and Molecular Geometry Review Overview for Part A: By completing these problems, you will learn four different ways to represent organic molecules. Names of organic molecules are provided in these questions, but you will learn about naming organic molecules later in this module. For Part A of this packet, your focus should be on drawing and interpreting organic structures. To complete these problems, refer to the CHEM 104 reference sheet, “Molecular Representations and Drawing Conventions,” which you read as part of your pre-discussion activity. The information and skills used in the problems in Part A will only be covered in your discussion class. This material will not be repeated in your whole class session, so it is important that you complete all of these problems (even if you need to finish them outside of discussion class). 1. Draw 3-methyl-2-butanol (C 5 H 12 O) using the given conventions. (To be demonstrated by TA) a. Lewis Structure c. Bond Line Structure b. Condensed Formula d. Dash-Wedge Structure CH 3 CH(OH)CH(CH 3 )CH 3 or CH 3 CH(OH)CH(CH 3 ) 2 or (CH 3 ) 2 CHCH(OH)CH 3 (Note: Condensed formulas may be written in multiple ways) The CH 3 and OH groups shown coming out of the plane of the paper are not shown with all bonds. Trying to draw these CH 3 and OH groups accurately and completely highlights the drawbacks of using this convention.
Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 2 of 20 2. Draw the hydrocarbon 2,3-dimethylbutane (C 6 H 14 ) using the following conventions: a. Lewis Structure c. Bond Line Structure b. Condensed Formula d. Dash-Wedge Structure CH 3 CH(CH 3 )CH(CH 3 )CH 3 or (CH 3 ) 2 CHCH(CH 3 ) 2 (Note: Condensed formulas may be written in multiple ways) 3. Draw trans-2-butene (C 4 H 8 ) using the following conventions: a. Lewis Structure c. Bond Line Structure b. Condensed Formula d. Dash-Wedge Structure CH 3 CH=CHCH 3 or CH 3 CHCHCH 3 (Note: Condensed formulas may be written in multiple ways) The CH 3 groups shown coming out of the plane of the paper are not shown with all bonds. Trying to draw these CH 3 groups accurately and completely highlights the drawbacks of using this convention.
Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 3 of 20 4. The compounds shown below are the active ingredients in over-the-counter drugs. Determine the molecular formula for each. Molecular formula: C 9 H 8 O 4 Molecular formula: C 8 H 9 O 2 N 5. Following convention, unbonded electron pairs are not shown in the line drawings for Question #4. Add them to the drawings above. See drawings above. 6. Condensed structural formulas for some organic functional groups (common bonding arrangements) may not always be obvious at first. Below are some examples of types of molecules you will study in this organic section. Match the condensed formula to the molecule: a. CH 3 CH 2 COOCH 2 CH 2 CH 3 b. CH 3 CH 2 CH 2 OH c. CH 2 CHCH 3 d. CH 3 CH 2 CHO e. CH 3 CH 2 CH 3 f. CH 3 COCH 3 g. CHCCH 3 h. CH 3 CH 2 CH 2 OCH 2 CH 2 CH 3 i. CH 3 CH 2 COOH c. a. b. d. e. f. g. h. i.
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Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 4 of 20 7. There are many ways to draw structures incorrectly. Look over the following drawings with your group to spot any errors in the way the molecule is represented. If you find an error(s), then redraw the molecule correctly by either changing bond angles or adding/removing bonds or atoms. There may be multiple ways to “fix” the molecule. a. b. h. d. g. f. e. c.
Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 5 of 20 8. The Lewis Structure for Demerol, an opioid analgesic, is shown to the right. a. How many lone pairs of electrons are in Demerol? ___ 5 _____ b. What is the hybridization of carbon a ? ____ sp 2 ______ c. What is the molecular geometry for carbon a ? _ trigonal planar _ d. What is the bond angle indicated by label c ? ___ 120 ____ e. What is the electron domain geometry of nitrogen d ? _ Tetrahedral _ f. What is the approximate bond order of bond b ? What orbitals are involved in forming the bond between the carbon atoms at label b ? What types of bonds are formed? Bond order is 1.5 (aromatic ring). A σ bond is formed between sp 2 (C)- sp 2 (C) and a bond is formed between 2p(C) and 2p(C). (Think about resonance and pi delocalization in benzene!)
Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 6 of 20 PART B: Hydrocarbons, Isomers, and Intermolecular Forces 9. Structural (constitutional) isomers are molecules with the same molecular formula but with different arrangements of the atoms and bonds. There are 18 possible arrangements for C 8 H 18 (one of which is octane). Using bond line drawings, identify all 18 of these possible arrangements. As you are working, check to make sure you are drawing unique structures and adhering to the original chemical formula. 10. A discussion class of CHEM 104 students was asked to draw all possible structural isomers for C 3 H 8 O. One group came up with the list below. Several students raised objections to some of these. Describe how the group may have erred. There are only 3 possible structural isomers for C 3 H 8 O Identical… Only one of these counts. Identical… AND wrong – too many carbons! This is one isomer. This is one isomer.
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Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 7 of 20 11. What are the primary intermolecular forces between alkanes? What affects the strengths of these forces? London dispersion forces (temporary attraction between induced dipoles (nonpermanent) in adjacent molecules) are the primary intermolecular forces between alkanes. Molecules with larger molar masses and/or greater surface area will have greater London dispersion forces. 12. A. Arrange the following alkanes in order from lowest to highest boiling points: 3 (36°C) 2 (28°C) 1 (lowest, 10°C) 4 (highest, 108°C) B. List all the intermolecular forces present in pure samples of each of the compounds in part A. Use the intermolecular forces to justify your relative order of boiling points. n-pentane, 2-methylbutane, and 1,2-dimethylpropane contain only hydrogen and carbon, and they all have the same molar mass. They each interact with other molecules identical to themselves via dispersion forces. The greater surface area of n-pentane molecules means that there will be more interactions between particles and greater dispersion forces, compared to the weaker dispersion forces between compact 1,2-dimethylpropane molecules. 1-chloropentane is a polar molecule, and it will interact with other identical molecules via dipole-dipole forces and also through dispersion forces. The molar mass of 1-chloropentane is significantly greater than the other three molecules, so 1-chloropentane will have greater dispersion forces than the other molecules. In addition, dipole-dipole interactions are stronger than dispersion forces. The greater the intermolecular forces, the more energy required to separate particles at boiling, and the higher the boiling point temperature. 13. A. Circle the molecule that you predict is a gas at room temperature. *Gas
Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Date of last revision: 7/14/23 Page 8 of 20 B. List all the intermolecular forces present in pure samples of each of the compounds in part A. Use the intermolecular forces to justify which one you predict to be a gas. All three molecules contain only hydrogen and carbon, and they all have different molar masses. All three molecules interact with molecules identical to themselves via dispersion forces. Larger molecules are more polarizable and will have stronger dispersion forces. The greater the intermolecular forces, the more energy required to separate particles at boiling, and the higher the boiling point temperature. Therefore the smallest molecule (propane) is a gas, while the other two molecules (octane and hexane) are liquids at room temperature. 14. Match the name and structure for the following alkanes: a. 4-propylnonane b. 2,2,4-trimethylpentane c. heptane d. 2,2,4-trimethyloctane e. 3-ethylpentane Information for Questions 15-17: When asked for the “general rule for the molecular formula” in these questions, you should use the subscript “n” to represent the number of carbon atoms (C n ) and use an algebraic term including “n” as the subscript for the number of hydrogen atoms. 15. Draw a structure for a noncyclic alkane with five carbons and write its chemical formula. Draw the structure for hexane and write its chemical formula. What is the general rule for the molecular formula for a noncyclic alkane? Note: There are many isomers that could be drawn! C 5 H 12 C 6 H 14 The general molecular formula for a noncyclic alkane is C n H (2n+2) . Note that any branching does not change this general formula: C 5 H 12 Organic, Module 1 2 C 6 H 14 e. d. a. c. b.
Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 9 of 20 16. Draw at least three different cycloalkanes with a total of 8 carbons; try using different numbers of rings and alkyl branches. Next, draw at least three different cycloalkanes with a total of 9 carbons, again using different numbers of rings and alkyl branches. For each cycloalkane structure, write its molecular formula. Determine a general rule for the molecular formula of a cycloalkane. Note: There are many isomers that could be drawn! C 8 H 16 C 8 H 16 C 8 H 16 Note: There are many isomers that could be drawn! C 9 H 18 C 9 H 18 C 9 H 18 The general formula for a cycloalkane is C n H (2n) . Stated another way: each ring takes 2 hydrogens out of the general molecular formula 17. Draw the structures for two alkane isomers with a total of 7 carbons and write their chemical formulas. Next, redraw your two isomers of the 7-carbon molecules, but this time draw in one double bond within the structure to represent an alkene; write the chemical formulas for these structures. Lastly, redraw your two 7-carbon alkenes, but this time add a second double bond within the structure to represent; write the chemical formulas for these structures. Determine a general rule for how adding a double bond changes the molecular formulas for hydrocarbons. Note: There are many isomers that could be drawn! C 7 H 16 C 7 H 16
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Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 10 of 20 Note: There are many isomers that could be drawn! C 7 H 14 C 7 H 14 Note: There are many isomers that could be drawn! C 7 H 12 C 7 H 12 Each π bond takes 2H from the general alkane formula. 18. Which of these two molecules would you expect to have a higher boiling point? Why? The molecule on the right is polar, while the molecule on the left is nonpolar (since the dipoles from the polar C-Cl bonds cancel each other out). The molecule on the right, therefore, would have dipole-dipole intermolecular forces that the other molecule would not have. The greater the intermolecular forces, the more energy required to separate particles at boiling, and the higher the boiling point temperature. Thus, we would expect the molecule on the right to have the higher boiling point.
Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 11 of 20 CH 3 CH 2 CHCHCH 3 CH 3 CH 2 CH 3 CH 3 CH 2 CHCH 2 CHCH 3 Cl Cl HC CCH 2 CH 3 CH 3 CH 2 CH 2 CH 2 CH 3 CHCHCH 2 CH 2 CH 3 CH 2 CH 2 CH 3 19. Match the name and structure for the following molecules: a. 3-ethyl-2methylpentane b. cyclohexane c. 4-methyl-5-propylnonane d. cis-2-butene e. 2,4-dichlorohexane f. 1-butyne g. cyclohexene h. trans-2-butene 20. Which of the following molecules are chiral? Place an asterisk (*) next to each chiral carbon center. Ibuprofen (Advil) – chiral Acetaminophen (Tylenol) – achiral Azidothymidine (AZT) – chiral Aspirin – achiral a. g. e. f. d. h. c. b.
Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 12 of 20 Folic Acid (Vitamin B 9 ) – chiral Naproxen (Aleve) – chiral Vitamin E – chiral Nandrolone – chiral
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Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 13 of 20 PART C: Functional Groups 21. Identify (by circling and naming) all functional groups in each of the molecules below: 22. You are not required to be able to name the molecules in Question #21. However, you should be able to identify functional groups in molecules using naming “clues”, such which suffix corresponds to which functional group. Assign the given names below to each molecule in #21. See names written in / matched with compounds above. phenylmethanal phenylmethanol 3-methylpentanoic acid hexyl ethanoate 2-pentanone cyclohexyl ethanoate hexyl ethanoate (hexyl acetate) Ester Aromatic Aldehyde Carboxylic Acid Ester Ketone cyclohexyl ethanoate phenylmethanol phenylmethanal 3-methylpentanoic acid hexyl ethanoate 2-pentanone Alcohol Aromatic
Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 14 of 20 23. Suppose CHEM 104 students were instructed to circle and identify all functional groups in the molecule below, and one student marked the molecule as shown. Determine how correct the student’s answer is. For any incorrect circling and/or identifications, specify why the answer is incorrect. 24. In the space below, draw a single molecule with the following functional groups: one ether , one aldehyde , one alkene , one aromatic , and one amine. Your molecule MUST obey the octet rule! (Hint: You may find it useful to separate all of your functional groups by CH 2 groups.) This question has no single correct answer. Any molecule with all functional groups is correct. An example is shown below. aromatic alkene ketone amide ester Amide should not include groups off of nitrogen Alkene should include both carbons of double bond Ketone should include 3 total carbons, including one on each side of ketone carbon
Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 15 of 20 25. List the intermolecular forces you’ve learned about in CHEM 103/104. Then, identify all intermolecular forces that would be present in pure samples of the molecules drawn below. Types of intermolecular forces Characteristics of molecules that exhibit these forces London dispersion | van der Waals | exists between all molecules instantaneous dipole-induced dipole | dipole-dipole exists between polar molecules hydrogen bonding exists between a molecule that has a hydrogen attached to an oxygen, nitrogen, or fluorine atom and another molecule that has a lone pair of electrons on an oxygen, nitrogen, or fluorine atom All of these molecules exhibit London dispersion forces. In addition to this, they also exhibit the forces listed below: 26. Use line notation to draw each of the following molecules. Which one of them would you expect to have highest boiling point? Explain. butanone 1-butanol n-butane butanal 2-methylpropane We would expect 1-butanol to have the highest boiling point due to hydrogen bonding and its linear structure (stronger London dispersion forces). dipole-dipole hydrogen bonding, dipole-dipole dipole-dipole dipole-dipole dipole-dipole hydrogen bonding, dipole-dipole
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Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 16 of 20 27. A student vigorously shakes blue tinted ethanol (d = 0.87 g/mL, tinted with a water-soluble dye), octane (d = 0.703 g/mL), and water together and then lets it sit a few days until their next laboratory session. What will this solution look like right after shaking? What will this solution look like after a few days? Use pictures to explain what the solution will look like right after shaking and after sitting for a few days. Right after shaking, the solution will look like it is completely mixed and will appear to have blue color throughout. After a few days, the solution will have two layers. The upper layer will appear colorless and will contain octane. The lower layer will appear blue and will contain a thoroughly-mixed solution of water and ethanol. 28. CH 3 CH 2 CH 2 CH 2 OH and CH 3 CH 2 OCH 2 CH 3 have the same molar mass but are structural isomers. What would expect in terms of melting point, boiling point, and solubility in water? (Hint: Draw structures to aid in your explanation.) CH 3 CH 2 CH 2 CH 2 OH (alcohol) CH 3 CH 2 OCH 2 CH 3 (ether) Boiling Point Higher (117.2 °C) Lower (34.5 °C) Melting Point Higher (-89.8 °C) Lower (-116.3 °C) Solubility Predicted to be more soluble (73 g/L at 25 °C) Predicted to be less soluble (61 g/L at 25 °C) CH 3 CH 2 CH 2 CH 2 OH is capable of hydrogen bonding with other molecules identical to itself, and CH 3 CH 2 OCH 2 CH 3 is not. Therefore, the alcohol will have much stronger intermolecular forces in pure form, giving it a higher melting and boiling point than the ether. When considering solubility of these substances in water, we need to consider not only the intermolecular forces between the molecules of the substances themselves, but also the intermolecular forces between the molecules of the substances and the solvent water molecules. CH 3 CH 2 CH 2 CH 2 OH is capable of hydrogen bonding both with itself and with water molecules. The hydrogens on the hydroxyl group of the alcohol can hydrogen bond with the oxygens in water, and the hydrogens on water can hydrogen bond with the oxygens in the alcohol. This results in a good number of hydrogen bonds between the water and alcohol. CH 3 CH 2 OCH 2 CH 3 is capable of some hydrogen bonding when in water (but not as a pure substance). The hydrogen bonding of ether in water occurs solely between the hydrogen atoms from water molecules and the oxygens of the ether. Due to the lesser extent of hydrogen bonding between the ether and water molecules, the ether is less soluble than the alcohol.
Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 17 of 20 29. Complete the table below, determining whether each molecule is an is a primary, secondary, or tertiary alcohol or amine. Molecule Alcohol or Amine? 1°, 2°, or 3°? Alcohol Alcohol Alcohol Amine Alcohol Amine Alcohol 1° and 2° Alcohol
Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 18 of 20 PART D: Hydrolysis and Condensation Reactions For each of the reactions below, provide the missing reactants or products. Then, identify the type of reaction taking place (such as hydrolysis, condensation, etc.). 30. 31. 32. 33. 34. 35. Condensation Condensation Condensation Condensation Hydrolysis Hydrolysis
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Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 19 of 20 36. 37. Questions 38-39: Directions for group work – Work in a small group to determine answers to these questions. Your TA will ask selected groups to report out their answers to the class. 38. A student is unable to make acetic acid (CH 3 COOH) and N(CH 3 ) 3 undergo a condensation reaction to form an amide, even after an acid catalyst is added. Why not? Draw structures to use in your explanation. In amide-producing condensation reactions between a carboxylic acid and an amine, the OH from the carboxylic acid and a hydrogen atom from the amine bond together to form a water molecule. The amine in this case does not have a hydrogen atom to donate since no hydrogens are bonded directly to the nitrogen atom. Condensation Hydrolysis
Chemistry 104 Discussion Activity Packet ANSWER KEY Fundamental Organic Chemistry Organic, Module 1 2 Date of last revision: 7/14/23 Page 20 of 20 39. In the laboratory, a student was given three bottles labeled X, Y, and Z. One of the bottles contained acetic acid, the second contained acetaldehyde, and the third ethanol. The student ran a series of experiments to determine the contents of each bottle. Substance X reacted with substance Y to form an ester under certain conditions and substance Y formed an acidic solution when dissolved in water. a. Identify which compound is in which bottle and write the Lewis structural formula for each compound. Substance X is ethanol. Substance Y is acetic acid. Substance Z is acetaldehyde. b. Write a chemical equation for the ester formation reaction and for the ionization of the acid. Ester formation: Ionization of acid: CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) End of Module Reflection: Discuss with the members of your group which Learning Objectives you each find most difficult and what resources are available for support. **If you are unsure where to find LO, go to Canvas site and click on “Exam Resources” in week 4.