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INSTANT PDF DOWNLOAD. Complete CHEM 219 Principles of Organic Chemistry Modules 1–8 Exams & Final Exam Questions & Answers (2026/2027) for Portage Learning. Includes multiple-choice questions, true/false items, short-answer problems, mechanism drawings, and detailed explanations covering all modules to help students review effectively and succeed in the full course and final exam. chem 219 exams, chem 219 answers, organic chemistry exam, portage chem 219, chem 219 module exams, chem 219 final exam, chem 219 test bank, chem 219 study guide, organic chemistry questions, portage learning exam, chem 219 practice test, organic chemistry answers, chem 219 modules 1-8, chemistry exam prep, organic chemistry notes, chem exam pdf
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Group 1) has one valence electron, which is unpaired, so it generally forms one ionic bond. 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.
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.
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.
What is the molecular formula for the indicated structural formula?
****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 double- barbed (full-headed) arrows, which indicate the transfer of pairs of electrons. Single- barbed (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 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.
CHEM 219 Module 2 Exam
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-bromopropane b. 2-bromopropane
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-tribromobenzene
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) Ring
