CHEM 219 Module 1,2,3,4,5,6,7 and 8 Principles of Organic Chemistry with Lab Latest 2024 Versions 100% Verified; Portage Learning

CHEM 219 Modules 1–8 Exams & Final Exam – Principles of Organic Chemistry (2025) | Portage Chemistry Geneva College 164 pagesCHEM 219 PRINCIPLES OF ORGANIC CHEMISTRY MODULE 1 - 8 EXAM & FINAL EXAM Portage Learning Inside you will get: #### 1. Multiple-Choice Questions #### 2. True/False Questions #### 3. Short-Answer Questions #### 4. Drawings and Mechanism Problems #### 5. Long-Answer or Essay Questions (Occasional) #### 6. Expert-Level Rationales (Portage CHEM 219 – 2025 Ready)Table of Contents CHEM 219 Module 1 Exam..............................................................................3 CHEM 219 Module2Exam...............................................................................13 CHEM 219 Module 3 Exam............................................................................24 CHEM 219 Module 4 Exam............................................................................22 CHEM 219 Module 5 Exam............................................................................42 CHEM 219 Module 6 Exam............................................................................65 CHEM 219 Module 7 Exam............................................................................87 CHEM 219 Module 8 Exam..........................................................................100 CHEM 219 FINAL EXAM .................................................................................117CHEM 219 Module 1 Exam Question 1 On a piece of scratch paper, write out the Lewis Dot Diagram for each of the indicated elements and then complete the following table for each: 1. Bromine 2. Silicon 3. Carbon 4. Sodium 5. Selenium *Write number using number (1, 2, 3 etc.) rather than typing out the number (one, two, three). Answer Element Total Number of Valence Electrons The number of lone pairs of electrons The number of Unpaired Electrons The number of bonds the atom will form Br 7 3 1 1 Si 4 0 4 4 C 4 0 4 4 Na 1 0 1 1 Se 6 2 2 2 **Expert Rationale:** The Lewis Dot Diagram for each atom visualizes their valence electrons based on their group number in the periodic table. Bromine (Br, Group 17) has 7 valence electrons, typically arranged as three lone pairs and one unpaired electron, allowing it to form one bond. Silicon (Si, Group 14) and carbon (C, Group 14) each have 4 valence electrons, all unpaired, signifying the ability to form 4 bonds. Sodium (Na, Group 1) has one valence electron, which is unpaired, so it generally forms one ionicbond. Selenium (Se, Group 16) possesses 6 valence electrons, commonly displayed as two lone pairs and two unpaired electrons, allowing it to form two bonds. Question 2 Classify the bonding between the given pairs of atoms as ionic, covalent, polar covalent, or purely covalent. Use the table of electronegativities shown below to help with the classification. Answer Bond X Classification O-H 1.4 Polar Covalent Cl-Br 0.2 Covalent N-P 0.9 Polar Covalent N-N 0 Purely Covalent K-S 1.7 Polar Covalent **Expert Rationale:** The classification hinges on the difference in electronegativity (ΔEN) between the two atoms. Generally, ΔEN 0.4 corresponds to pure (nonpolar) covalent; 0.4 ≤ ΔEN ≤ 1.7 denotes polar covalent bonds, and ΔEN 1.7 typically indicates ionic character. However, bonds with ΔEN near the cutoff (like K-S) may display significant covalent character even if ionic bonding is possible. In these cases, the bonding is described as polar covalent unless the difference greatly favors electron transfer. For example, the O-H bond is strongly polar covalent due to the substantial electronegativity difference, whereas N-N is nonpolar (purely covalent) because the atoms are identical.Question 3 What is the relationship between the compounds shown? Are they the same compound, constitutional isomers, or two different compounds that are not related to one another? Explain. Answer: Two different compounds. They have different molecular formulae. **Expert Rationale:** Compounds with differing molecular formulae are fundamentally distinct chemical entities. Constitutional isomers have the same molecular formula but different connectivity of atoms. Here, since the formulae are not the same, the compounds cannot be isomers and share no direct chemical relationship, signifying that they are unrelated compounds. Question 4 What is the molecular formula for the indicated structural formula?Answer: C8H14 **Expert Rationale:** Determining the molecular formula from a structural formula requires accounting for every atom depicted in the structure. The count of carbon (C) and hydrogen (H) atoms present leads to the concise formula, C₈H₁₄, which expresses the precise atomic composition of the molecule. Question 5 What are the formal charges on each of the indicated atoms in the nitric acid molecule? Answer: (a)=0, (b) = -1, (c)= +1, (d) = 0**Expert Rationale:** Formal charge is calculated as: Formal charge = (Valence electrons) - (Non-bonding electrons) - ½(Bonding electrons) Each atom’s formal charge is assigned based on the electron distribution in the Lewis structure. Charges summing to the overall molecular charge indicate proper assignment, confirming the electronic distribution within the molecule is accurately represented for nitric acid. **Question 6** True or False: It is okay for resonance structures to have different numbers of electrons from one another if the formal charges do not balance out the same overall total. **Answer:** False **Expert Rationale:** Resonance structures are alternative Lewis structures for the same molecule or ion, reflecting the delocalization of electrons. Critically, all resonance structures must depict the same number of electrons; only electron arrangements, not the actual count or connectivity of atoms, may differ. Changing the total number of electrons invalidates the resonance relationship and often results in representations of different species altogether. Therefore, proper resonance structures maintain consistent atom connectivity, overall electron count, and total formal charge. --- **Question 7** True or False: Moving single bonds in electron resonance is allowed, as that would change the connectivity of the atoms in the molecule. **Answer:** False **Expert Rationale:**In the context of resonance, only the distribution of π electrons, lone pair electrons, or formal charges is altered via the use of curved arrows. The σ (single) bonds signify the fundamental connectivity of atoms within a molecule, which must not be changed during the depiction of resonance structures. Modification of single bonds would alter the molecular framework, generating isomers rather than different resonance forms of the same species. **Question 8** True or False: A single-barbed arrow is used when drawing resonance structures to show that a pair of electrons is moving. **Answer:** False **Expert Rationale:** In resonance structures, electron movement is represented exclusively by doublebarbed (full-headed) arrows, which indicate the transfer of pairs of electrons. Singlebarbed (half-headed or fishhook) arrows are specific to the movement of individual electrons, as seen in the mechanisms of free radicals, not in the context of resonance, where electron pairs move. --- **Question 9** True or False: A resonance double-headed arrow points to where the electrons being redistributed begin. **Answer:** False **Expert Rationale:** The double-headed arrow used in resonance does not indicate electron movement but rather signifies that the two structures are resonance forms of a single molecule. The redistribution of electrons is indicated by curved arrows, where the tail shows the origin of electrons and the head shows their destination. The double-headed arrow (⇌) simply connects resonance structures and communicates that the true electronic structure is a hybrid of those forms.Question 10 Using the VSEPR method, predict the molecular geometry (shape and bond angle) around the indicated atoms for the following structure Answer i) Trigonal planar, 3 bonding pairs, 120° ii) Trigonal pyramidal, 3 bonding pairs, 1 lone pair, 107° iii) Trigonal planar, 3 bonding pairs, 120° iv) Tetrahedral, 4 bonding pairs, 109.5° **Expert Rationale:** The Valence Shell Electron Pair Repulsion (VSEPR) model predicts molecular geometry by minimizing electron pair repulsions around the central atom. For a central atom with 3 bonding pairs and no lone pairs, the arrangement is trigonal planar with bond angles of approximately 120°. If there are 3 bonding pairs and one lone pair (such as in ammonia), the geometry shifts to trigonal pyramidal, with bond angles slightly less than 109.5°, typically about 107°, due to lone pair–bonding pair repulsion. A central atom with 4 bonding pairs adopts a tetrahedral geometry with bond angles close to 109.5°.**Question 11** Briefly explain the theory of atomic orbital hybridization. **Answer:** Hybridization theory postulates that atomic orbitals within an atom, specifically those in the valence shell, blend or combine to produce new hybrid orbitals of equivalent energy and character. These hybrid orbitals are then utilized to form chemical bonds in molecules. The process of hybridization allows for the observed geometries and bond angles in molecules—such as the linear arrangement in BeCl₂ (sp hybridization), trigonal planar geometry in BF₃ (sp² hybridization), and tetrahedral arrangement in CH₄ (sp³ hybridization)—which could not be readily rationalized using only unmodified (‘pure’) atomic orbitals. Thus, hybridization is a crucial concept for explaining molecular shape and bonding, as predicted by both VSEPR theory and experimental observation. Question 12 Classify each of the following structural formulae as bond-line, condensed, or dash representations:Question 13 On a piece of scrap paper, draw the resonance structure (“B”) obtained by redistributing the electrons in the azide ion (N3-) as shown by the curved arrows, then answer the questions which follow. a. What are the formal charges on N(I), N(II), and N(III) in resonance structure “B”? b. What is the overall charge on the ion for structure “B”? c. Are the two resonance forms equivalent contributors to the hybrid structure? Briefly explain. Answera. N(I) = -2, N(II) = +2, N(III) = -1 b. (-2) + (+2) + (-1) = - 1 overall c. The two forms have different charges on different atoms and different patterns of bonding between the atoms. This makes them not equivalent.CHEM 219 Module2Exam Question 1 Classify each of the following molecules as being: I. Aliphatic or Aromatic II. Carbocyclic (cyclic), acrylic, or heterocyclic Question 2 Generate the IUPAC systematic name for each of the following compounds (identify your answer for a, b, c, and d). Be sure to include the designation of cis- and trans- stereoisomerism where appropriate.Answer a. 6-bromo-3,4 dimethyloctane b. 2,2,3- trimethylpentane c. 2-ethyl-1-methyl-3-propylcyclohexane d. cis- 1,2-dichlorocyclobutane Question 3 Assuming that only monosubstituted products are generated, how many different monobrominated products (constitutional isomers) can be formed in the following radical halogenation reaction? Generate the IUPAC names of each of the products that would be produced. Answer Two different products can be formed. a. 1-bromopropaneb. 2-bromopropane Question 4 Consider the Newman projection shown below. Answer the questions which follow, a. What is the IUPAC name of the molecule shown? b. Does the Newman projection depict a staggered or eclipsed conformation? Answer: a. The molecule shown is 1,1-dichloropropane. b. The molecule is shown in a staggered conformation Question 5 Generate the IUPAC systematic name for each of the following compounds. Be sure to include the designation of cis- and trans- stereoisomerism where appropriate.Answer a. 6-ethyl 1,3-dimethylcyclohexene b. Cis-6 methyl 3 heptene c. 3-chloro- 4-methyl-1,5-hexadiene (or 3-chloro-4methyl-hexa-1,5-diene) d. 6-ethyl-1-nonyne Question 6 A chemist finds two bottles in the chemical storeroom. The old labels have come loose and fallen from the bottles and are lying on the shelf. One label reads “cyclohexene” and the other label reads “cyclohexane”. The chemist needs to determine which bottle contains which compounds. She remembers that bromine can be used in qualitative test to distinguish between alkene and alkanes. Describe how an experiment could be designed that would allow the chemist to distinguish the two clear, colorless liquids from one another. Answer The chemist could put some of each substance into its own test tube. She could then add a few drops of bromine solution to each tube and mix the tubes. After a few minutes, the alkene would have reacted with the red/orange bromine to produce a colorless product, while the alkane will still retain its color.Question 7 Generate the IUPAC systematic name for each of the following aromatic compounds Answer a. Toluene (methylbenzene) b. Ethylbenzene c. O-dichlorobenzene (ortho-dichlorobenzene or 1,2 dichlorobenzene) d. 1,2,3-tribromobenzeneQuestion 8 Describe the following statements about benzene as either true or false. The carbons of the benzene ring are all trigonal pyramidal and the bond angles are 107° Answer False Question 9 Describe the following statements about benzene as either true or false. The repeating pattern of π – σ – π bonding in the ring is known as catenation. Answer False Question 10 Describe the following statements about benzene as either true or false The typical mechanism by which reactions occur on the benzene ring is substitution. Answer True Question 11 Describe the following statements about benzene as either true or false. Electrophilic reagents seek the benzene ring because of its low electron density. Answer False Question 12Describe the following statements about benzene as either true or false. Answer EAS stands for Electrophilic Aromatic Substitution Question 13 For each of the following, identify the functional group(s) present in each molecule. There may be more than one group present in a single molecule – you should identify them all. Note that you are not being asked to generate the IUPAC name of the compound, only to state which functional groups are present. Answer a. Aldehyde b. Alkyl Halide c. Ether d. Alkyne, alkene e. Ester Question 14 The structure shown below is Aspartame. Aspartame is an artificial non-saccharide sweetener 200 times sweeter than sucrose and is commonly used as a sugar substitute in food and beverages. Identify by name the functional groups (A-E) that have been circled in the aspartame molecule.Answer a. Carboxylic Acid b. Amine c. Amide d. Ester e. Aromatic (Benzene) RingCHEM 21G Module 3 Exam 1. Characterize each of the following objects as being chiral or achiral A. Drill bit B. Screw C. ScrewdriverD. Boot Carrot: Answer A. Chiral B. Chiral C. Achiral D. Chiral E. Achiral Expert Rationale: - A drill bit and a screw are classic examples of objects that display handedness (i.e., chirality) because their helical threads can exist in two non-superimposable mirror image forms (right- or left-handed). - A screwdriver (with a simple cylindrical shaft and symmetric handle) displays bilateral symmetry and is achiral. - A standard boot (not a symmetric galosh) fits only one foot, so it is chiral—left and right boots are non-superimposable. - A carrot, if symmetric and not twisted, is typically achiral, though some biological carrots may exhibit subtle twists (rarely relevant in general chemistry context). 2. Characterize each of the following objects as being chiral or achiral: a. Pencil b. Wine Glass c. Baseball Glove d. Orchestra Conductors' Baton Scissors: Answer a. achiralb. achiral c. chiral d. achiral e. chiral Expert Rationale: - A pencil, lacking features like a grip or label, is a uniform cylinder and thus achiral—its mirror image is superimposable. - A wine glass (without decorations or asymmetry), due to its radial symmetry about the stem, is achiral. - A baseball glove is purposely designed to fit either the left or right hand only, making it chiral. - An orchestra conductor's baton (symmetrical stick) is achiral as it has no distinguishing asymmetry. - Scissors (not ambidextrous) are chiral; a right-handed pair cannot be superimposed on a left-handed pair. 3. Draweachofthefollowingstructuresonapieceofpaper.Identifyanystereo centers in each by indicating the carbon number corresponding to a stereo center.If there are no stereo centers in a compound,state "none". A. 2,3-dimethylpentane B. 3-methylcyclohexene C. 1-bromo-2,2-dichloropropane d. 2,2 dichloropropane E. 1,2 dichlorobutane 2 methylhexane:Answer A. C3 is a Stereocenter B. C3 is a stereocenter C. None D. none E. One stereo center at C2 F. One stereocenter at C2 Expert Rationale and Explanation: - A stereocenter (often a tetrahedral carbon atom bonded to four different groups) is the basis for chirality in organic molecules. - *2,3-dimethylpentane*: Carbon 3 is attached to -CH3, -H, -CH2CH3, and -CH(CH3)2, qualifying it as a stereocenter. - *3-methylcyclohexene*: C3 is bonded to four distinct groups due to the ring system and the double bond. - *1-bromo-2,2-dichloropropane* and *2,2-dichloropropane*: No carbon is attached to four different groups, hence none. - *1,2-dichlorobutane*: C2 is attached to -Cl, -H, -CH2Cl, and -CH3, making it a stereocenter. - *2-methylhexane*: C2 is bonded to -CH3, -H, -C4H9, and -CH2CH3, qualifying as a stereocenter. 4. Assignapriorityorder(1-4;is greatestpriority) tothefollowingfourgroups for the purposes of determining R and S configurations at stereo center. Group A: -CH3Group B: -SH Group C: -CH=CH2 Group D: -H: Answer 1. B 2. C 3. A 4. D Expert Rationale: Priorities are assigned using the Cahn–Ingold–Prelog (CIP) rules. The higher atomic number (Z) receives higher priority: - Sulfur (–SH) Carbon (–CH=CH₂, due to double bond) Carbon (–CH₃) Hydrogen. - Double bonds increase priority for –CH=CH₂ over –CH₃. 5. Assigna priorityorder (1-4;is greatest priority) to the following fourgroups for the purposes of determining R and S configurations at stereo center. Group A: -H Group B: -CH3 Group C: - C6H5 Group D: -I: Answer 1. D 2. C 3. B 4. AExpert Rationale: Iodine has the highest atomic number (Z=53), thus top priority; benzene ring (aromatic carbon system) methyl; hydrogen is always last. 6. Assign a priority order (1-4;is greatest priority)to the following four groups for the purposes of determining R and S configurations at stereo center. Group A: -NH2 Group B: -CL Group C: -CH3 Group D: -H: Answer 1. B 2. A 3. C 4. D Expert Rationale: Chlorine (Z=17) has the highest atomic number, followed by nitrogen (Z=7), carbon (Z=6), and hydrogen. 7. Assign a priority order (1-4;is greatest priority)to the following four groups for the purposes of determining R and S configurations at stereo center. Group A: -C(CH3)3 Group B: -CH(CH3)2Group C: -CH3 Group D: - CH2CH3: Answer 1. D 2. C 3. B 4. A Expert Rationale: When comparing alkyl groups, priority is based on the atomic number of directly attached atoms; longer chain (–CH₂CH₃) methyl; secondary, tertiary alkyls have lower priority unless branching creates more significant differences at the first point of difference. 8. Based on the direction of twist of the following groups, assign an R or S orientation: Answer A.R orientation B. S orientationC. S orientation 9. Based on the direction of twist of the following groups, assign an R or S orientation: Answer A.S orientation B. R orientation C. R orientation 10. Based on the direction of twist of the following groups, assign an R or S orientation:Answer A.S orientation B. R orientation C. S orientation 11. The Fischer projection formula of the molecule fructose is shown below. A. How many stereocenters are present in the fructose molecule? B. What is the maximum number of stereoisomers that could exist for this sugar?: Answer A. 3 B. 8 12. Assign an R/S configuration for the following chiral molecule . Explain:Answer R configuration because this is a clockwise rotation 13. Assign an R/S configuration for the following chiral molecule . Explain: Answer S configuration because it moves counter clockwise way 14. For each of the following molecules, determine if the C=C is in the E or Z configuration:Answer A. E B. Z 15. For each of the following molecules, determine if the C=C is in the E or Z configuration: Answer A. Z B. E16. For each of the following molecules, determine if the C=C is in the E or Z configuration: Answer A. Z B. E 17. The molecule 2,3-dibromobutane has two stereocenters (C2 and C3) and thus can exist as a maximum of four different stereoisomers.The Fischer projections of these stereoisomers are shown below. Answer the questions which followa. Which stereoisomers (if any) represents pairs of enantiomers? (list all the pairings that apply) b. Which stereoisomers (if any) represents pairs of diastereomers? (list all the pairings that apply) c. Which stereoisomers (if any) represent a mess compound?: Answer A.C/D are enantiomers b. the pairs of A/C, A/D, B/C, and B/D are diastereomers c. memo compound is A 18. Determine if the statements below are true of false regarding pair of molecules shown.They have different boiling points: Answer False 19. Determine if the statements below are true of false regarding pair of molecules shown. One rotatesplane-polarizedlight in theoppositedirectionfromtheother: Answer True 20. Determine if the statements below are true of false regarding pair of molecules shown.They have the same density: Answer True 21. Determine if the statements below are true of false regarding pair of molecules shown. One rotates plane-polarized light by a different number of degrees than the other: Answer False 22. Determine if the statements below are true of false regarding pair ofmolecules shown. They are non-superimposable mirror images of each other: Answer True 23. Determine if the statements below are true of false regarding pair of molecules shown. They are constitutional isomers of one another: Answer False24. Briefly explain why it is important for chemists, biologist,doctors and nurses to understand the principle of chirality in organic molecules.: Answer It is important for people dealing in medicine because it is very important to know that chiralcompounds can changegreatlyintheircharacteristics and ifnot administered carefully can lead to abnormalities. 25. Why is the concept of chirality an important topic to learn to learn in the study of organic chemistry?: Answer Chirality is relevant to organic chemistry because substances like carbohydrates and amino acids are chiral.Chiral molecules have different properties then their isomers and typically in many biological systems only one enantiomer is important or can be metabolized.Therefore, the concept of chirality is important to understand the difference between stereoisomers.CHEM 21G Module 4 Exam 1. Name each of the following using the IUPAC systemic nomenclature methodology: A. Trans-1-bromo-3-chloro-cyclopentane B. 5-bromo-3-chloro-3-ethylheptane C. 1,2,4-trichlorobenzene2. Classify each alkyl halide as primary, secondary, or tertiary: A. Secondary B. primary C. tertiary D. primary 3. Classify each alkyl halide as primary, secondary, or tertiary: a) secondary b) tertiary c) secondary d) primary4. In the following reaction, identify the nucleophile, substrate, and leaving group. 1. Water 2. 2-chloro-2-methylpropane 3. Chlorine: 1. Nucleophile 2. Substrate 3. Leaving group 5. In the following reaction, identify the nucleophile, substrate, and leaving group. 1. Acetate Ion 2. 1-bromopropane 3. Bromine:1. Nucleophile 2. Substrate 3. Leaving group 6. For the following reaction. predict what functional group the product of the reaction would contain. Note that you are not being asked to generate the IUPAC name of the product, simply to state what functional group it would containAnswer Alcohol 7. For the following reaction. predict what functional group the product of the reaction would contain. Note that you are not being asked to generate the IUPAC name of the product, simply to state what functional group it would contain Answer Ester 8. For the following reaction. predict what functional group the product of the reaction would contain. Note that you are not being asked to generate the IUPAC name of the product, simply to state what functional group it would containAnswer Amine 9. For the following reaction. predict what functional group the product of the reaction would contain. Note that you are not being asked to generate the IUPAC name of the product, simply to state what functional group it would contain Answer Nitrile 10. For the following reaction. predict what functional group the product of the reaction would contain. Note that you are not being asked to generate the IUPAC name of the product, simply to state what functional group it would containAnswer Alcohol 11. For the following reaction. predict what functional group the product of the reaction would contain. Note that you are not being asked to generate the IUPAC name of the product, simply to state what functional group it would contain Answer Amine 12. How many different alkene products could be formed if elimination oc- curredon each of the following alkyl halide substrates? Do not take E/Z stereoisomers into account Only consider the different con- stitutional isomers which can form: A. 2 B. 3 13. Determine if the following statement regarding substitution and elimination mechanisms are true or false: The "2" in SN2 means that this mechanism occurs in a series of steps: Answer False Expert Rationale: The "2" in SN2 refers to the bimolecular nature of the rate-determining step, not the number of steps in the mechanism. The SN2 reaction is a concerted, single-step mechanism in which bond formation and bond breaking occur simultaneously. Thus, it does not proceed through a series of steps. ---14. Determine if the following statement regarding substitution and elimination mechanisms are true or false: The "2" in SN2 stands for bimolecular or second order: Answer True Expert Rationale: The "2" in SN2 designates the reaction as bimolecular, meaning the rate law depends on the concentration of two reactants: the substrate and the nucleophile. This corresponds to a second-order kinetic process. --- 15. Determine if the following statement regarding substitution and elimination mechanisms are true or false: The "1" in SN1 means that this mechanism occurs in a single step: Answer False Expert Rationale: Though the "1" in SN1 stands for unimolecular, indicating a first-order rate law that depends only on the substrate concentration, the mechanism itself is multistep, typically involving formation of a carbocation intermediate before nucleophilic attack. --- 16. Determine if the following statement regarding substitution and elimination mechanisms are true or false: SN1 mechanisms are unimolecular, meaning they only depend on the concentration of the substrate: Answer True Expert Rationale:In an SN1 mechanism, the rate-determining step is the loss of the leaving group to form a carbocation, which depends only on the concentration of the substrate. Thus, the reaction is first order, or unimolecular, with respect to the substrate. --- 17. Determine if the following statement regarding substitution and elimination mechanisms are true or false: The "2" in SN2 indicates that the rate of the reaction depends on the concentration of two molecules: Answer True Expert Rationale: The SN2 mechanism exhibits bimolecular kinetics, with its rate law depending on both the substrate and the nucleophile. Therefore, the "2" indicates reaction order and the involvement of two different species in the rate-determining step. --- 18. Determine if the following statement regarding substitution and elimination mechanisms are true or false: SN1 mechanisms occur in a series of steps: Answer True Expert Rationale: The SN1 mechanism involves at least two discrete steps: the dissociation of the leaving group to form a carbocation intermediate, followed by the attack of the nucleophile. Some SN1 reactions may also involve additional steps, such as solvent mediation or rearrangement. ---19. Determine if the following statement regarding substitution and elimination mechanisms are true or false: The 2 in SN2 means that this mechanism occurs in a single step: Answer True Expert Rationale: The SN2 mechanism is a concerted, single-step process wherein the nucleophile attacks the substrate simultaneously as the leaving group departs. This one-step nature is a hallmark of the SN2 pathway. --- 20. Determine if the following statement regarding substitution and elimination mechanisms are true or false: "Inversion" of geometry is typically associated with the SN2 mechanism: Answer True Expert Rationale: The SN2 reaction occurs via a backside attack, resulting in inversion of configuration at the carbon center (Walden inversion). This stereochemical outcome is diagnostic for the SN2 pathway. --- 21. Determine if the following statement regarding substitution and elimination mechanisms are true or false: "Racemization" of geometry is typically associated with the SN1 mechanism: Answer True Expert Rationale:Because the SN1 mechanism proceeds through a planar carbocation intermediate, nucleophilic attack can occur from either side, often producing a racemic mixture of enantiomers when the substrate is chiral. --- 22. Determine if the following statement regarding substitution and elimination mechanisms are true or false: The use of weak bases as nucleophiles promotes elimination over substitution in most cases: Answer False Expert Rationale: Weak bases (often weak nucleophiles as well) generally favor substitution reactions (especially SN1) over elimination, as elimination typically requires strong bases (E2) to effectively abstract a proton and form an alkene. --- 23. Determine if the following statement regarding substitution and elimination mechanisms are true or false: The use of strong bases as nucleophiles promotes elimination over substitution in most cases: Answer True Expert Rationale: Strong bases are most effective at promoting elimination reactions, especially E2, as they can efficiently abstract protons. In cases where a strong base is also a good nucleophile and substrates are bulky, elimination often predominates over substitution. ---24. Determine if the following statement regarding substitution and elimination mechanisms are true or false: Bulky substrates prefer to react by SN1 or E1 mechanism is the departure of the leaving group Answer True Expert Rationale: Bulky substrates hinder backside attack required for SN2 or E2 reactions, making the SN1 or E1 mechanisms—both of which proceed via carbocation intermediates after the leaving group departs—more favorable. --- 25. T/F The use of strong bases nucleophiles promotes substitution over elimination in most cases: Answer False Expert Rationale: As previously explained, strong bases typically favor elimination (E2), especially when steric hindrance precludes substitution. --- 26. Determine if the following statement regarding substitution and elimination mechanisms are true or false: The first step in SN1 or E1 mechanism is the departure of the leaving group: Answer True Expert Rationale: The initial step in both SN1 and E1 mechanisms is the loss of the leaving group, resulting in the formation of a carbocation intermediate, which is essential for subsequent reaction steps.--- 27. Determine if the following statement regarding substitution and elimination mechanisms are true or false: The first step in SN1 or E1 mechanism is the departure of the substrate: Answer False Expert Rationale: The proper terminology is the "departure of the leaving group," not the substrate. The substrate remains the molecules undergoing change; only the leaving group exits as the first mechanistic event. --- 28. Determine if the following statement regarding substitution and elimination mechanisms are true or false: The first step in SN2 or E2 mechanism is the departure of the leaving group: Answer False Expert Rationale: In both SN2 and E2 mechanisms, the departure of the leaving group is concerted with nucleophilic attack (SN2) or proton abstraction (E2); these are single-step mechanisms, not stepwise processes where the leaving group departs first. Identify the stronger nucleophile in each pair below. Briefly explain your choice A. Hydroxide (-OH) vs Water (H2O) B. Hydroxide (-OH) vs bisulfide (-SH):A. Hydroxide ion is stronger because of the negative charge it holds compared to the water B.Bisulfide is stronger because Sulfur is larger and less electronegative than oxygen Classify the following solvents as protic or aprotic.Would this solvent enhance the rate of an Sn1 or Sn2 reaction Answer A. Protic, Sn1 B. Aprotic, Sn2 C. Aprotic, Sn2 D, Protic, Sn1 Rank the followingsubstrates according to their increasing rate of reaction in an Sn2 subsitution reaction. Briefly explain your rankingAnswer CDAB C is the bulkiest therefore there is greater hinderance so it reacts the slowest. B is a methyl halide and has less hinderance therefore it reacts the fastest. Classify the following solvents as protic or aprotic.Would this solvent enhance the rate of an Sn1 or Sn2 reaction Answer a) Aprotic, Sn2b) protic, Sn1 c) aprotic, Sn2 d) protic, Sn1 13. Consider the substitution reaction of 1-bromobutane with hydroxide ion Assuming no other changes, what effect on the rate would result from simul- taneously doubling the concentrations of both butyl bromide and hydroxide ion? Explain A) No effect B) It would double the rate C) It would triple the rate D) It would increase the rate four times E) It would increase the rate six times: D. The substrate proceeds with an SN2 mechanism, therefore the rate relies on the concentration of both the nucleophile and substrate. If the concentrations of each are increased by 2 the rate would increase by 4 Rank the followingsubstrates according to their increasing rate of reaction in an Sn2 subsitution reaction. Briefly explain your rankingAnswer B,A,C,D Expert Rationale: • SN2 reactions proceed best at carbons with little steric hindrance because the nucleophile must attack from the back side. • Methyl halides react the fastest due to total lack of steric crowding. • Tertiary halides do not undergo SN2 reactions appreciably; they are reserved for SN1 or E1 mechanisms. • Cyclohexyl chloride, though cyclic, reacts slower than a primary alkyl halide due to its secondary position and ring strain. Consider the substitution reactionAssuming no other changes, what effect on the rate would result from simulta- neously doubling the concentrations of 2-chloro-2-methylpentane and iodide ion? Explain A) No effect B) It would double the rate C) It would triple the rate D) It would increase the rate four times E) It would increase the rate six times: B The substrate proceeds with the Sn1 mechanism, and therefore only relies on the concentration of the substrate. Changing the effect of iodine will have no effect on the rate, therefore doubling 2-chloro-2-methylpentane will double the rateCHEM 21G Module 5 Exam 1. Name the following alcohol using IUPAC systematic nomenclature Answer 3-pentanol2. Name the following alcohol using IUPAC systematic nomenclature Answer 2-pentanol 3. Name the following alcohol using IUPAC systematic nomenclature Answer cyclopropanol4. Name the following alcohol using IUPAC systematic nomenclature Answer cyclobutanol 5. Name the following alcohol using IUPAC systematic nomenclatureAnswer 2,6-dichlorophenol 6. Name the following alcohol using IUPAC systematic nomenclature Answer 3,3-di- methyl-2-butanol 7. Name the following alcohol using IUPAC systematic nomenclatureAnswer 2-methyl-1-propanol 8. Name the following alcohol using IUPAC systematic nomenclature Answer Cyclopentanol 9. Name the following alcohol using IUPAC systematic nomenclatureAnswer 2-butene-1-ol 10. Name the following alcohol using IUPAC systematic nomenclature Answer 3,5-dinitrophenol 11. Generatethename forall thealkeneproducts possible when the following acid-catalyzed dehydration reaction occurs. You do not need to include cis/trans isomerism of the C=C in answering this question. If there is more than one alkene possible, predict which would be formed in the largest amountAnswer 3-methylcyclopentene 1-methylcyclopentene 12. Generatethename forall thealkeneproducts possible when the following acid-catalyzed dehydration reaction occurs. You do not need to include cis/trans isomerism of the C=C in answering this question. If there is more than one alkene possible, predict which would be formed in the largest amountAnswer 2-methyl2-butene (Major product) 2-methyl-1-butene 13. Name the following ether using IUPAC systemic nomenclature Answer 1,4-dioxa- cyclohexane14. Namethefollowing ether usingIUPACsystemic nomenclature Answer 2-methoxyhexane 15. Name the following ether using IUPAC systematic nomenclature Answer Oxacylohexane 50 /16. Name the following ether using IUPAC systematic nomenclature Answer 3-methoxypentane 17. Name the following ether using IUPAC systematic nomenclature Answer oxacyclobutane 51 /18. Name the following ether using IUPAC systematic nomenclature Answer anisole (methoxybenzene) 19. The ionic radius of the potassium ion (K+) is 2.66A(angstroms). (1 angstrom= 1x10^-10m). Using the table,select the best crown ether to "host" the potassium ion as a "guest" 52 /Answer 18-crown-6 20. Would she get the desired product in a good yield? State yes or no,briefly explain why is happening in the synthesis. Answer No,since the chemist did not design this experiment correct.For theWilliamson synthesis to work, the R group on the substrate must not be bulky.Since R is bulky, then elimination would occur which would result in an alkene to form. 21. If this nucleophile is added to an aldehyde or ketone an imine results Answer Amine 53 /22. If this nucleophile is added to an aldehyde or ketone an Cyanohydrin results Answer Cyanide ion 23. Complete the following table describing the various types of nucleophilic additions to the C=O of aldehydes and ketones. Answer 1. Carbanion 2. Acetylide ion 3. Water 4. Alcohol **24. Determine if the following statement regarding reactions of carbonyl compounds is true or false** The more electron rich the C=O of a molecule, the greater the reactivity of that compound with nucleophiles.: **False** 54 /**Expert Rationale:** The reactivity of a carbonyl compound toward nucleophilic attack depends upon the electrophilicity of the carbonyl carbon. An electron-rich carbonyl group is less electrophilic because the additional electron density reduces the partial positive charge on the carbon atom, making it less attractive to nucleophiles. Thus, the more electron-rich the C=O group, the lower its reactivity towards nucleophiles. --- **25. Determine if the following statement regarding reactions of carbonyl compounds is true or false** The more electron poor the C=O of a molecule, the greater the reactivity of that compound with nucleophiles.: **True** **Expert Rationale:** A carbonyl group that is electron-poor (more electrophilic) has an increased partial positive charge on the carbon atom, which enhances its ability to react with nucleophiles. Electronwithdrawing groups adjacent to the carbonyl, or reduced resonance donation, results in a more reactive carbonyl carbon due to increased susceptibility to nucleophilic attack. --- **26. Determine if the following statement regarding reactions of carbonyl compounds is true or false** Acid halides are the most reactive derivatives of carboxylic acids, due to the high electronegativity of the halogens: **True** **Expert Rationale:** 55 /Acid halides exhibit high reactivity because the halide substituent, which is highly electronegative, withdraws electron density from the carbonyl carbon, making it more electrophilic. Furthermore, halide ions are good leaving groups, facilitating nucleophilic acyl substitution reactions. This combination accounts for the exceptional reactivity of acid halides compared to other carboxylic acid derivatives. --- **27. Determine if the following statement regarding reactions of carbonyl compounds is true or false** Acid halides are the least reactive derivatives of carboxylic acids, due to the high electronegativity of the halogens: **False** **Expert Rationale:** This statement is incorrect. The high electronegativity of the halogen in acid halides increases, rather than decreases, the reactivity of these compounds. It does so by rendering the carbonyl carbon more electrophilic and by providing an excellent leaving group, thus making acid halides the most, not the least, reactive carboxylic acid derivatives. --- **28. Determine if the following statement regarding reactions of carbonyl compounds is true or false** The greater reactivity of acid anhydrides compared to esters is due to the presence of the additional carbonyl group in the anhydride: **True** **Expert Rationale:** Acid anhydrides are more reactive than esters because the adjacent carbonyl group in an anhydride exerts a strong electron-withdrawing effect through both inductive and resonance interactions. This increases the electrophilicity of the acyl carbon, enhancing its susceptibility to 56 /nucleophilic attack. In contrast, esters do not possess this additional carbonyl group, which accounts for their comparative unreactivity. --- **29. Determine if the following statement regarding reactions of carbonyl compounds is true or false** The greater reactivity of acid anhydrides compared to esters is due to the absence of the additional carbonyl group in the anhydride: **False** **Expert Rationale:** This statement is the inverse of the correct explanation. The increased reactivity of acid anhydrides actually results from the presence, not the absence, of a second carbonyl group. The adjacent carbonyl withdraws electron density, strengthening the electrophilicity of the acyl carbon, and thus facilitating nucleophilic substitution. --- **30. Determine if the following statement regarding reactions of carbonyl compounds is true or false** Aldehydes and ketones tend to react by nucleophilic acyl addition because their C=O is not bonded to a good leaving group: **True** **Expert Rationale:** Aldehydes and ketones undergo nucleophilic addition reactions rather than acyl substitution because the carbonyl carbon is not attached to a group capable of leaving as a stable anion during the reaction. The absence of a good leaving group (such as a halide or carboxylate) precludes substitution, resulting instead in addition to the carbonyl carbon. --- 57 /**31. Determine if the following statement regarding reactions of carbonyl compounds is true or false** Carboxylic acids and derivatives (esters, anhydrides, halides) tend to react by nucleophilic acyl substitution since the C=O is bonded to a good nucleophile: **False** **Expert Rationale:** This statement is incorrect because the driving force for nucleophilic acyl substitution is the presence of a good leaving group, not a nucleophile, attached to the carbonyl carbon. Reactive carboxylic acid derivatives have substituents that are easily displaced during substitution, such as halides or carboxylate groups, rather than nucleophilic groups. --- **32. Determine if the following statement regarding reactions of carbonyl compounds is true or false** Carboxylic acids and derivatives (esters, anhydrides, halides) tend to react by nucleophilic acyl substitution since the C=O is bonded to a good leaving group: **True** **Expert Rationale:** Carboxylic acids and their derivatives, such as esters, anhydrides, and acid halides, generally react through nucleophilic acyl substitution mechanisms. This is made possible because the group bonded to the carbonyl carbon in these compounds (e.g., halides, alkoxides, or carboxylates) acts as a good leaving group, facilitating the substitution process following nucleophilic attack. 58 /33. Complete the following table describing the various types of nucleophilic additions to the C=O of aldehydes and ketones. Answer a) a diol b) a hemiacetal c) an imine d) a cyanohydrin 59 /Nucleophile Product Reaction Description a) Diol Acetal Two equivalents of an alcohol add to the carbonyl group, forming a compound with two hydroxyl groups (diol) attached to the same carbon. b) Hemiacetal Hemiacetal One equivalent of an alcohol adds to the carbonyl group, forming a compound with one hydroxyl group and one alkoxy group attached to the same carbon. c) Imine Imine An amine adds to the carbonyl group, forming a compound with a carbon-nitrogen double bond. d) Cyanohydrin Cyanohydrin Cyanide ion (CN-) adds to the carbonyl group, forming a compound with a cyano group and a hydroxyl group attached to the same carbon. Explanation: • Diol formation: This reaction requires an acid catalyst and involves two steps. The first step is the nucleophilic attack of one alcohol molecule on the carbonyl carbon, forming a hemiacetal. The second step involves another alcohol molecule attacking the hemiacetal, resulting in the formation of a diol (acetal). • Hemiacetal formation: This is the first step in the formation of a diol. A single alcohol molecule adds to the carbonyl group, forming a compound with a hemiacetal linkage. • Imine formation: An amine acts as the nucleophile, attacking the carbonyl carbon and forming a carbon-nitrogen double bond (imine). Water is eliminated as a byproduct in this reaction. • Cyanohydrin formation: Cyanide ion (CN-) is a strong nucleophile that readily adds to the carbonyl group. The resulting compound is called a cyanohydrin. 60 /34. If carbanion is added to a aldehyde or ketone this product results Answer alcohol Acetylide ion is added to a aldehyde or ketone this product results nswer Alkynol Match the IUPAC ester name to the structural formula Answer Methyl benzoate Match the IUPAC ester name to the structural formula 61 /Answer Butyl Pentanoate Match the IUPAC ester name to the structural formula Answer Pentyl Butanoate Match the IUPAC ester name to the structural formula Answer propyl cyclopropanecarboxylate Describe the bonding interactions that occur in carboxylic acids and the effect that these interaction have on the boiling points of carboxylic acids Answer - Carboxylic acids form strong intermolecular hydrogen bonding to eachother.Due to these hydrogen bonds, they exist in the form of dimers.This strong hydrogen 62 /bonding are the reason for the increase of boiling points of carboxylic acids Complete the following table (provide answers for a-e) to predict the out- come of the indicated reactions of carboxylic acids and their derivatives Answer A. acid halide B. Ester C. Carboxylic acid D. Carboxylic acid (or ester) E. Carboxylic acid 63 /Complete the following table (provide answers for a-e) to predict the out- come of the indicated reactions of carboxylic acids and their derivatives Answer a) ester b) primary alcohol c) acid halide d) primary alcohol e) ester 64 /Explain why carboxylic acids and their derivatives tend toreactby substi- tution at the C=O,while aldehydes and ketones tend to react by addition to the C=O Answer Carboxylic acids and their derivatives tend to react by substitution at the C=O becausethey have a leaving group at the carbonyl carbon where aldehydes and ketones do not. CHEM 219 Module 6 Exam 1. Name the following amine using IUPAC systematic nomenclature method- ology 65 /Answer N,N-dimethylethylamine 2. Name the following amine using IUPAC systematic nomenclature method- ology Answer N,N – dimethylpropylamine 3. Name the following amine using IUPAC systematic nomenclature method- ology 66 /Answer N-ethylethylamine 4. Name the following amine using IUPAC systematic nomenclature method- ology Answer 1-amino-3-methylcyclohexane (3-methylcyclohexlamine) 5. Name the following amine using IUPAC systematic nomenclature method ology 67 /Answer 1-aminopentane 6. Name the following amine using IUPAC systematic nomenclature method- ology Answer 1-methylcyclohexane-1-amine 7. Name the following amine using IUPAC systematic nomenclature method- ology 68 /Answer triethlylamine 8. Classify each of the following amines as primary, secondary, or tertiary Answer A. primary B. tertiary C. primary D.tertiary 69 /9. Classify each of the following amines as primary, secondary, or tertiary Answer A. primary B. secondary C. primary D.tertiary 10. Complete the following table describing the pattern of reactions in the synthesis of amines via alkylation by filling in all of the blanks. Use (') notation to distinguish between the "R" groups when writing the formula of the product that is produced. Classify the amine product as 1,2,3, or quaternary. 70 /Answer A.primary B. R-NH2 C. R2NH D.R'-X d. Quaternary 11. Complete the following table describing the pattern of reactions in the synthesis of amines via alkylation by filling in all of the blanks. Use (') notation to distinguish between the "R" groups when writing the formula of the prod- uct that is produced. Classify the amine product as 1,2,3, or quaternary. 71 /Answer 1) Nucleophile: R3N 2) Substrate:R'-X 3) Product:R-NR'H 4) A) primary C) tertiary 12. Provide the IUPAC systemic name and subclassification (1,2,3) for each of the following amides 72 /Answer A. N-ethyl-N-methylcyclohexanecarboxamide; tertiary B. N,N-dimethylpropanamide; tertiary C. Propanamide; primary 13. Provide the IUPAC systemic name and subclassification (1,2,3) for each of the following amides Answer a) N-methyl-N-propylcyclopentanecarboxamide; Tertiary b) N-ethylpropanamide; Secondary 73 /c) N-methylethanamide; Secondary 14. Provide the IUPAC systemic name and subclassification (1,2,3) for each of the following amides Answer a) N,N-diethylcyclohexanecarboxamide Tertiary b) N-methylpropionamide secondary c) pentanamide primary 15. Based on the reactivity trend for carboxylic acid derivatives, state whether the following conversions are likely to occur (or not) 74 /1) Ester to Amide 2) Acid Anhydride to Acid Halide 3) Acid Anhydride to Acid Halide 4) Amide to Ester Answer 1) Likely to occur 2) Unlikely to occur 3) Likely to occur 4) Unlikely to occur 16. Based on the reactivity trend for carboxylic acid derivatives, state whether the following conversions are likely to occur (or not) 1) Acid Halide to Amide 2) Acid Anhydride to Amide 3) Amide to Acid halide 4) Ester to Acid Anhydride Answer 1) Likely to occur 2) Likely 3) unlikely 4) unlikely 75 /17. Generate the IUPAC name of the final product "B" of the following reaction Sequence Answer Phenol 18. Generate the IUPAC name of the final product "B" of the following reaction sequence 76 /Answer Nitrile 19. Generate the IUPAC name of the final product "B" of the following reaction sequence Answer Aryl halide 20. Determine if the following statement regarding reduction of nitrogen compounds is true or false Since the nitrogen of an amine is in a reduced state, compounds containing nitrogen in more oxidized forms can be reduced to produce amines Answer: True Expert Rationale: Amines possess nitrogen atoms in their most reduced oxidation state, typically with three bonds to hydrogen or carbon and a lone pair. Many nitrogen-containing functional groups, such as nitro, nitrile, or amide groups, have nitrogen atoms in higher oxidation states. These oxidized nitrogen compounds can be chemically reduced through processes that add hydrogen or remove oxygen, ultimately leading to the formation of amines. This reduction is a fundamental transformation in organic 77 /synthesis, utilized extensively to access various amine derivatives from more oxidized precursors. --- 21. Determine if the following statement regarding reduction of nitrogen compounds is true or false The nitrogen of an amine is in an oxidized state Answer: False Expert Rationale: Amines contain nitrogen in its most reduced state within organic compounds, where the nitrogen atom is bonded primarily to carbon and/or hydrogen atoms. This reduced state is evidenced by the lack of multiple bonds to oxygen or lower electronegativity atoms. In contrast, other functional groups, such as nitro or nitroso compounds, DO exhibit nitrogen in oxidized states. Thus, the statement is inaccurate; nitrogen in an amine is not oxidized, but rather reduced. --- 22. Determine if the following statement regarding reduction of nitrogen compounds is true or false Amides can be reduced to produce primary, secondary, or tertiary amines, depending on the structure of the catalyst used Answer: False Expert Rationale: The ability to obtain primary, secondary, or tertiary amines from amide reductions principally depends on the structure of the amide substrate (i.e., whether the nitrogen is bonded to hydrogens, alkyl groups, or aryl groups), not the catalyst used. For example, reduction of a primary amide yields a primary amine; reduction of a secondary or tertiary amide yields a secondary or tertiary amine, respectively. The catalyst (e.g., LiAlH₄) is necessary for the reduction but does not determine the classification of the resulting amine—the nature of the substituents on the amide nitrogen does. --- 23. Determine if the following statement regarding reduction of 78 /nitrogen compounds is true or false The conversion of aromatic nitro compounds to aromatic amines is an example of oxidation Answer: False Expert Rationale: The chemical transformation of an aromatic nitro group (–NO₂) into an aromatic amine (–NH₂) involves the gain of hydrogen atoms and/or the loss of oxygen atoms by the nitrogen, which is the hallmark of a reduction process, not an oxidation. Reduction decreases the oxidation state of nitrogen by providing electrons (i.e., through hydrogenation or chemical reduction with agents such as iron/HCl or tin/HCl). Therefore, classifying this transformation as oxidation is incorrect. --- 24. Determine if the following statement regarding reduction of nitrogen compounds is true or false The conversion of aromatic nitro compounds to aromatic amines is an example of reduction Answer: True Expert Rationale: Reduction is defined as a process involving the gain of electrons, addition of hydrogen, or removal of oxygen. Converting a nitro group (which is highly oxidized) to an amine (which is reduced) clearly exemplifies a reduction reaction. Such reductions are fundamental in organic synthesis and are often performed using agents like catalytic hydrogenation or metals in acid. --- 25. Determine if the following statement regarding reduction of nitrogen compounds is true or false The effect of the reduction of an amide with LiAlH₄ is to turn the carbonyl group of the amide into a methylene group (–CH₂–) Answer: True Expert Rationale: Lithium aluminum hydride (LiAlH₄) is a potent reducing agent capable of 79 /reducing amides by cleaving the carbonyl group via hydride addition, which ultimately removes the oxygen atom entirely. This reaction converts the original carboxamide carbon into a methylene (–CH₂–) group adjacent to the nitrogen, yielding an amine. Specifically, a primary amide becomes a primary amine (–CONH₂ → –CH₂NH₂), with the carbonyl oxygen completely replaced by two hydrogens. --- 26. Determine if the following statement regarding reduction of nitrogen compounds is true or false Amides can be reduced to produce primary, secondary, or tertiary amines, depending on the structure of the amide substrate Answer: True Expert Rationale: The amide substrate’s nitrogen substituents determine the class of the resulting amine upon reduction. A primary amide (RCONH₂) reduces to a primary amine (RCH₂NH₂), a secondary amide (RCONHR') to a secondary amine (RCH₂NHR'), and a tertiary amide (RCONR'R'') to a tertiary amine (RCH₂NR'R''). The type of amine produced thus directly depends on the substituents originally present on the amide nitrogen, not the reducing agent. 27. Determine if the following statement regarding reduction of nitrogen compounds is true or false Amides cannot be reduced to produce primary, secondary, or tertiary amines, even by changing the structure of the amide substrate Answer: False Expert Rationale: This statement is incorrect because it overlooks the well-documented ability to obtain amines of different degrees (primary, secondary, tertiary) from the reduction of appropriately substituted amides. The nature of the resulting amine is determined by the substituents present on the amide nitrogen; by modifying these, primary, secondary, or tertiary amines can be systematically produced through reduction. 80 /--- 28. Describe what happens in a reductive amination reaction Answer In the reductive animation reaction, the amine is added to the C=O of the aldehyde/ketone to produce an imine. In the next step, Then the reducing agent converts the imine to an amine. 29. During a reductive lamination reaction, what type of reactant is converted into what type of product? How many steps are involved in the process? Answer A ketone is converted into an amine and it takes two steps. 30. Activation of the nitrogen of an amine converts it into what functional group? Answer Imine 31. By what process are amines converted to amides? Answer Acylation of the amine nitrogen 32. Name the following nitrile using IUPAC systemic methodology 81 /Answer Heptanenitrile 33. Name the following nitrile using IUPAC systemic methodology Answer Hexanenitrile 34. Name the following nitrile using IUPAC systemic methodology 82 /Answer 3-methylhexanenitrile 35. Name the following nitrile using IUPAC systemic methodology Answer Cyclobutanecarbonitrile 36. Name the following nitrile using IUPAC systemic methodology 83 /Answer 2-methylcy- clopentanecarbonitrile 37. Name the following nitrile using IUPAC systemic methodology Answer Propanenitrile 84 /38. Name the following nitrile using IUPAC systemic methodology Answer 2,2-di- methylcyclopentane-1-carbonitrile 39. Generate the IUPAC name of the final product "B" of the following reaction sequence 85 /Answer Benzonitrile 86 /CHEM 219 Module 7 Exam 1. Complete the following description of heterocyclic molecules by filling in each of the blanks: Heterocyclic (heterocyclic compounds) are organic molecules in which one or more atoms are replaced by heteroatomes- elements other than or hydrogen. The most common heteroatom are , nitrogen, and sulfur, but heterocyclic with other elements are known. More than heteroatom can be present, and the heteroatom can be the same or different. Answer: Cyclic Carbon Carbon Oxygen One Expert Rationale: The definition specifies that heterocycles must be cyclic and organic. Replacing carbon atoms, not hydrogens, with heteroatoms (commonly oxygen, nitrogen, or sulfur) gives rise to heterocyclic chemistry. The inclusion of “one” allows for mono- or polynuclear heterocyclic rings with varying numbers and types of heteroatoms. --- 2. Determine if the following statement regarding the structure of heterocyclic molecules is true or false: 87 /*Heterocyclic compounds by definition do not contain any multiple bonds.* Answer: False Expert Rationale: The statement is incorrect; heterocyclic compounds may contain single, double, or even triple bonds within their ring structures. Multiple bonds are especially common in aromatic heterocycles, such as pyridine or furan. --- 3. Determine if the following statement regarding the structure of heterocyclic molecules is true or false: *Heterocyclic compounds may contain multiple bonds.* Answer: True Expert Rationale: Many heterocyclic compounds, both aromatic and non-aromatic, have pi bonds within their rings. The presence of multiple bonds is essential for aromaticity and influences the reactivity and electronic properties of the ring system. 88 /--- 4. Determine if the following statement regarding the structure of heterocyclic molecules is true or false: *The main two subclassifications of heterocyclic compounds are aromatic and non-aromatic.* Answer: True Expert Rationale: Heterocyclic compounds are indeed categorized broadly into aromatic and non-aromatic classes. The classification distinguishes between compounds with delocalized pi electron clouds (aromatic) and those without (nonaromatic), affecting their chemical behavior dramatically. --- 5. Determine if the following statement regarding the structure of heterocyclic molecules is true or false: *Many natural products and drug molecules contain heterocyclic rings.* Answer: True 89 /Expert Rationale: A significant proportion of pharmaceuticals and bioactive molecules, such as alkaloids, nucleic acids, and vitamins, have heterocyclic ring systems that contribute crucially to their biological function. --- 6. Determine if the following statement regarding the structure of heterocyclic molecules is true or false: *Non-aromatic heterocyclic compounds have chemical behavior that is very similar to their acyclic counterparts.* Answer: True Expert Rationale: Non-aromatic heterocycles generally lack special stabilization from delocalization and thus resemble structurally related open-chain analogues in reactivity and properties. --- 7. T/F Non-aromatic heterocyclic compounds have chemical behavior that is very different from their acyclic counterparts: Answer: 90 /False Expert Rationale: Without aromatic conjugation, non-aromatic heterocycles exhibit reaction profiles similar to their straight-chain analogues, lacking the enhanced stability or distinct chemistry associated with aromatic systems. --- 8. Explain the difference between nitrogen heterocycles where one molecule has a C=N bond but the other has only C-N bonds. Answer: A nitrogen heterocycle containing a C=N bond (such as pyridine) has a nitrogen atom whose lone pair does not participate in aromatic delocalization and is instead available for basicity and nucleophilicity, akin to pyridine nitrogen. When the nitrogen is involved only in C–N bonds and is part of the aromatic system (as in pyrrole), its lone pair is delocalized, making it less available for bonding and less basic. Expert Rationale: The distinction reflects differing electronic environments and reactivity: a pyridine-like nitrogen's lone pair lies orthogonal to the aromatic cloud, retaining basicity, while in pyrrole-like systems, nitrogen's lone pair is part of the aromatic sextet, reducing its availability for external bonding. --- 91 /9. Which is more basic in a heterocyclic compound, a "pyridine-like" nitrogen or a "pyrrole-like" nitrogen? Answer: A "pyridine-like" nitrogen is more basic, since its lone pair is not involved in aromatic delocalization and thus is available for protonation. Expert Rationale: Pyridine’s nitrogen lone pair is localized in an sp2 orbital and not delocalized into the aromatic ring, making it significantly more basic than the pyrrole-type nitrogen, whose lone pair contributes to aromatic resonance. 10. Identify each of the following 5-membered aromatic heterocyclic compounds: A. Furan B. Pyrrole 92 /C. Thiophene Expert Rationale: Furan contains an oxygen heteroatom, pyrrole contains a nitrogen, and thiophene contains a sulfur. All three are classic 5-membered aromatic heterocycles. --- 11. Explain the effect of the presence of the nitrogen atom in pyridine when that molecule undergoes EAS reactions (as compared t