Hour Exam 1 - Chemistry 204 PDF

# Hour Exam 1 - Chemistry 204 **Name:** Oluwatamilore (Tami) Adetula **Date:** Sept. 9, 2024 **Instructor:** Dr. Hellwig **Total Points:** 110 (graded on the basis of 100) **Score:** 33/100 ## Question 1: Infrared Spectroscopy (8 points) Circle the compound that corresponds most closely to th...

# Hour Exam 1 - Chemistry 204 **Name:** Oluwatamilore (Tami) Adetula **Date:** Sept. 9, 2024 **Instructor:** Dr. Hellwig **Total Points:** 110 (graded on the basis of 100) **Score:** 33/100 ## Question 1: Infrared Spectroscopy (8 points) Circle the compound that corresponds most closely to the infrared spectrum. Point out the presence or absence of the relevant peaks. **Image Description:** The image contains an infrared spectrum with the following peaks: - A strong broad peak around 3200-3600 cm-1. - A sharp peak around 1700 cm-1. - A peak around 1600 cm-1. - A peak around 1500 cm-1. - A peak around 1450 cm-1. **Compound Options:** - Two structures of cyclic compounds with different functional groups (one with a hydroxyl group and the other with an amine group) - A linear alkane with a triple bond **Possible Answer:** The compound with the hydroxyl group (-OH) corresponds most closely to the infrared spectrum. **Justification:** - The strong broad peak around 3200-3600 cm-1 indicates the presence of an -OH stretch. - The peak around 1700 cm-1 suggests the presence of a carbonyl group (C=O). ## Question 2: IR Spectroscopy (6 points) How would you use IR spectroscopy to distinguish between the two compounds, A and B? What absorbance(s) would you see in one that you would not see in the other? You only need to list the absorptions that would be present in one and not in the other. **Compound Options:** - Compound A: An aromatic ring with a carbonyl group. - Compound B: A linear alkane with a hydroxyl group. **Answer:** | Compound | Absorbance(s) | Justification | |---|---|---| | A | 6-8.5 shift (1600, 1500, 1450 peak) | Aromatic C=C bonds | | B | 3200-3600 peak | -OH stretch (alcohol) | ## Question 3: Mass Spectrometry (7 points) Consider this mass spectrum. The compound that gives rise to this spectrum most likely contains: - a) bromine - b) chlorine - c) neither bromine nor chlorine - d) only carbon and hydrogen **Answer:** (c) neither bromine nor chlorine. **Explanation:** The mass spectrum shows no significant peaks at M+2 and M+4, which would indicate the presence of bromine and chlorine, respectively. ## Question 4: Infrared Spectroscopy (5 points) In infrared spectroscopy, the incoming electromagnetic radiation: - a) causes the nuclei to flip and oppose the magnetic field - b) causes bonds to bend and stretch - c) causes other electrons to jump to a higher energy level - d) knocks an electron out of the sample molecule. **Answer:** (b) causes bonds to bend and stretch. **Explanation:** Infrared spectroscopy works by detecting the absorption of infrared radiation by molecules, causing their bonds to vibrate (bend or stretch). ## Question 5: 13C NMR (5 points) How many different signals will there be for this molecule in the 13C NMR? **Image Description:** The molecule shown is a linear chain with 5 carbons, a chlorine atom, and a bromine atom. **Answer:** 5 **Explanation:** Each carbon in the molecule is in a unique chemical environment and will produce a separate signal in the 13C NMR spectrum. ## Question 6: 13C NMR (5 points) How many different signals will there be for this molecule in the 13C NMR? **Image Description:** The molecule shown is a branched alkane with 5 carbons, a bromine atom, and a methyl group. **Answer:** 4 **Explanation:** There are 4 unique carbon environments in the molecule, so there will be 4 signals. ## Question 7: 13C NMR (5 points) How many different signals will there be for this molecule in the 13C NMR? **Image Description:** The molecule shown is a branched alkane with 3 carbons, a methyl group, and a chlorine atom. **Answer:** 3 **Explanation:** There are 3 unique carbon environments in the molecule, so there will be 3 signals. ## Question 8: 13C NMR (5 points) How many different signals will there be for this molecule in the 13C NMR? **Image Description:** The molecule is a cyclic alkane with 5 carbon atoms. **Answer:** 2 **Explanation:** The molecule has two distinct types of carbon atoms: those in the ring and those in the methyl group. ## Question 9: 13C NMR (5 points) How could you distinguish between these two compounds in the 13C NMR spectrum? What signals would you see where? **Compound Options (with approximate chemical shift ranges):** - A: Aromatic ring (120-160 ppm) with hydroxyl group (-OH, 50-100 ppm) - B: Aromatic ring (120-160 ppm) with alcohol group (-OH, 50-100 ppm) **Answer:** | Compound | Signal (ppm) | Justification | |---|---|---| | A | 120-160 | Aromatic ring | | A | 50-100 | Hydroxyl group (-OH) attached to an sp3 hybridized carbon | | B | 120-160 | Aromatic ring | | B | 50-100 | Alcohol group (-OH) attached to an sp2 hybridized carbon | ## Question 10: 13C NMR (5 points) Which one of these four compounds would give a 13C NMR spectrum with only these three peaks: one at 210 ppm, one at 30 ppm, and one at 20 ppm? **Compound Options:** - A: Cyclic ketone with a methyl group. - B: Cyclic ketone with two methyl groups on the same carbon. - C: Cyclic ether with three carbons in the cycle. - D: Acyclic ketone with two methyl groups on the same carbon. **Answer:** **A** **Explanation:** - **210 ppm:** indicates a carbonyl group (C=O). - **30 ppm:** indicates a methyl group (CH3) attached to a carbon with a single bond. - **20 ppm:** indicates a methylene group (CH2) attached to a carbon with a single bond. - **Compound A** is the only one that has both a carbonyl group and a methyl group attached to a carbon with a single bond. ## Question 11: 1H NMR (12 Points) Consider the 1H NMR of this molecule: **Image Description:** The molecule contains: - 5 carbon atoms - 2 bromine atoms - 2 methyl groups (CH3) **Answer:** 1. What is the splitting pattern of the H's marked a? - (c) Triplet 2. What is the splitting pattern of the H's marked b? - (d) Quartet 3. What is the splitting pattern of the H's marked c? - (d) Quartet 4. What is the splitting pattern of the H marked d? - (b) Doublet **Explanation:** The splitting pattern of a proton signal is determined by the number of neighboring protons. - Protons marked 'a' have two neighboring protons, so they will be split into a triplet. - Protons marked 'b' have three neighboring protons, so they will be split into a quartet. - Protons marked 'c' have three neighboring protons, so they will be split into a quartet. - Proton marked 'd' has one neighboring proton, so it's split into a doublet. ## Question 12: 1H NMR (5 points)? Which one of these compounds would give exactly two signals between 80 and 85 in the proton (1H) NMR? - a) 1,2-dichloropropane - b) 1,1-dichloropropane - c) 2,2-dichloropropane - d) 1,3-dichloropropane **Answer:** (c) 2,2-dichloropropane **Explanation:** - 2,2-dichloropropane has two chemically distinct protons. - One is a triplet due to the two neighboring methyl groups, which will be split into a triplet, usually appearing in the region 80-85 ppm. - The other is a singlet since it has no neighboring protons, appearing typically in the region 4-5 PPM. ## Question 13: 1H NMR (5 points) How could you distinguish between these two compounds in the 1H NMR spectrum? What would you see for one that you would not see for the other? **Compound Options:** - Aromatic ring with a methyl group attached to it. - Aromatic ring with a hydroxyl (-OH) group attached to it. **Answer:** | Compound | Signal (ppm) | Justification | |---|---|---| | Aromatic with methyl group | 2.1-2.5 | Signal corresponding to the methyl group (CH3) | | Aromatic with hydroxyl group | 1.0-3.0 | No signal corresponding to the methyl group (CH3); instead, you might see a broad peak in the range 1.0-3.0 ppm for the -OH group, depending on the solvent and concentration. | ## Question 14: 1H NMR (16 points) Consider these two isomeric compounds: - 1, 4-dichlorobutane - 1, 2-dichloro-2-methylpropane a) Draw each compound and label the H's on each with letters a, b, c, etc. b) Then fill in this table for the 1H NMR spectrum for each compound. ### 1,4-dichlorobutane | Letter of hydrogen (a, b, etc.) | # of H's | shift (δ) | splitting | |---|---|---|---| | A | 3 | 3.5-3.8 | triplet | | B | 2 | 1.8-2.1 | multiplet | | C | 2 | 1.5-1.8 | quintet | | D | 3 | 3.5-3.8 | triplet | ### 1,2-dichloro-2-methylpropane | Letter of hydrogen (a, b, etc.) | # of H's | shift (δ) | splitting | |---|---|---|---| | A | 2 | 3.3-3.6 | doublet | | B | 6 | 1.4-1.7 | singlet | ## Question 15: Unsaturation (16 points) a) How many unsaturations does a C4H8O2 compound have? Show your work. **Answer:** The formula for calculating the degree of unsaturation is: **Unsaturation = [(2 * C) + 2 - H + N - X] / 2** Where: - C = Number of carbon atoms - H = Number of hydrogen atoms - N = Number of nitrogen atoms - X = Number of halogen atoms In this case, we have: - C = 4 - H = 8 - O = 2 Substituting the values: **Unsaturation = [(2 * 4) + 2 - 8] / 2 = 1** Therefore, the C4H8O2 compound has **one unit of unsaturation**. b) What is the structure of the C4H8O2 compound that gives this 1H NMR spectrum? **Image Description:** The 1H NMR spectrum shows 3 peaks: - A singlet at 1.3 ppm with an integral value of 3. - A singlet at 3.7 ppm with an integral value of 2. - A singlet at 4.1 ppm with an integral value of 3. **Answer:** The structure of the C4H8O2 compound is: **CH3CH2OCH2CH3** **Explanation:** - The singlet at 1.3 ppm with an integral value of 3 corresponds to the methyl group (CH3) of the ethyl group. - The singlet at 3.7 ppm with an integral value of 2 corresponds to the methylene group (CH2) directly attached to the oxygen atom. - The singlet at 4.1 ppm with an integral value of 3 corresponds to the methylene group (CH2) attached to the oxygen atom and the other ethyl group. This is a simple ether with one carbon, one oxygen and that oxygen is double bonded to the carbon, and attached to the central carbon are two ethyl groups.