Organic Chemistry Quiz: Curved Arrow Formalism And Mechanistic Bookkeeping

What this quiz covers

This quiz focuses on Curved Arrow Formalism And Mechanistic Bookkeeping, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The mechanism should show a carbocation intermediate, requiring the C-Cl bond to break before the C-Br bond forms completely
2. The first curved arrow should point to the chlorine atom instead of the carbon atom to show direct displacement of the leaving group
3. Both arrows should be drawn in opposite directions to show the reversible nature of the $S_N2$ mechanism under these conditions
4. The second curved arrow should originate from the C-Cl bond, not from the carbon atom itself, since bonds contain the electron pairs that move (correct answer)

Explanation: When analyzing curved arrow mechanisms, remember that arrows represent the movement of electron pairs, not atoms or charges. In $S_N2$ reactions, you must show exactly where electrons come from and where they go. The correct mechanism requires the second curved arrow to originate from the C-Cl bond itself, not from the carbon atom. This is because the electron pair in the C-Cl bond is what actually moves to form the chloride ion as it leaves. Carbon atoms don't contain free electron pairs that can move independently—only bonds and lone pairs contain the electrons that participate in reactions. When you draw an arrow from the C-Cl bond to chlorine, you're showing that the bonding electrons become a lone pair on the departing chloride ion. Option A is incorrect because $S_N2$ reactions are concerted, meaning bond breaking and forming occur simultaneously without carbocation intermediates. Option B misunderstands the mechanism—the nucleophile attacks the carbon center, not the leaving group directly. This backside attack is what causes the characteristic inversion of stereochemistry in $S_N2$ reactions. Option C incorrectly suggests the arrows should show reversibility, but curved arrows in a single mechanistic step show electron movement in one direction, not equilibrium. Study tip: Always trace electron pairs, not atoms. In mechanisms, curved arrows must originate from bonds (lines) or lone pairs (dots), never from atom symbols themselves. This fundamental rule will help you draw and evaluate any organic mechanism correctly.

Question 2

1. $\pi$ electrons attack both bromine atoms simultaneously, the Br-Br bond breaks with electrons staying on the first bromine, and the second bromine forms the bridge
2. $\pi$ electrons attack one bromine atom, both bromine atoms simultaneously attack different carbons, and the Br-Br bond breaks symmetrically
3. $\pi$ electrons attack one bromine atom, the Br-Br bond breaks with electrons going to the other bromine, and a lone pair from the first bromine forms the three-membered ring (correct answer)
4. A lone pair from one bromine attacks the alkene, the $\pi$ bond breaks and attacks the other bromine, and the Br-Br bond breaks with electrons going to the leaving bromide

Explanation: When you encounter alkene bromination mechanisms, focus on the sequential nature of electron movements and the formation of the distinctive three-membered bromonium ion intermediate. The correct mechanism involves three distinct curved arrows showing electron flow. First, the $\pi$ electrons of the alkene act as a nucleophile and attack one of the bromine atoms in $\text{Br}_2$. Second, this attack causes the Br-Br bond to break heterolytically, with the bonding electrons moving to the bromine that wasn't attacked, forming a bromide ion ($\text{Br}^-$). Third, a lone pair of electrons on the bromine atom that was attacked forms a bond to the other carbon of the alkene, creating the cyclic bromonium ion intermediate. Option A incorrectly suggests simultaneous attack on both bromines, which isn't how this mechanism proceeds. The process is sequential, not simultaneous. Option B also incorrectly describes simultaneous processes and doesn't properly account for the lone pair involvement in ring formation. Option D reverses the roles, incorrectly showing bromine as the initial nucleophile attacking the alkene, which contradicts the electrophilic nature of this reaction. The key study tip here is to remember that bromonium ion formation is a coordinated but sequential process: nucleophilic attack by alkene → Br-Br bond breaking → intramolecular cyclization by lone pair. This creates the distinctive three-membered ring that explains the stereochemical outcome of anti-addition in alkene bromination reactions.

Question 3

1. The arrow shows a lone pair from the central carbon moving to form a new double bond; students commonly use $\pi$ electrons instead of lone pairs
2. The arrow shows the positive charge moving directly to the other end of the molecule; students commonly forget that charges don't move, only electrons do
3. The arrow shows $\pi$ electrons moving from the C=C double bond toward the positively charged carbon; students commonly draw the arrow backwards from the positive charge (correct answer)
4. The arrow shows simultaneous movement of both $\pi$ electrons and the positive charge; students commonly show only one of these movements

Explanation: When analyzing resonance structures, you're tracking the movement of electrons—never charges—to show how electron density is distributed across a molecule. In the allyl cation, you have a positively charged carbon adjacent to a C=C double bond, creating a perfect setup for electron delocalization. The correct electron movement shows the $\pi$ electrons from the C=C double bond moving toward the positively charged carbon. This movement accomplishes two things: it creates a new C=C double bond where the positive charge originally was, and it places the positive charge on the carbon that donated its $\pi$ electrons. The curved arrow starts at the center of the existing double bond and points toward the positively charged carbon. Choice A is incorrect because the central carbon in the allyl system has no lone pairs—it's $sp^2$ hybridized and fully bonded. Choice B represents a fundamental misconception: charges don't "move" in resonance structures. The positive charge appears in different locations because electrons redistribute, but you never draw arrows showing charge movement. Choice D is wrong because you only show electron movement—the charge redistribution is simply the consequence of where electrons end up. The most common student error is drawing the arrow backwards (from the positive charge toward the double bond), which incorrectly suggests that the positive charge is donating electrons. Remember: arrows always show electron movement from areas of high electron density to areas of low electron density. In resonance, always start your arrows from bonds or lone pairs, never from positive charges.

Question 4

1. A single arrow from the C-Br bond to the bromine atom, followed by a second step where the base removes a proton.
2. Three simultaneous arrows: one from the base to a C1 proton, one from the C1-H bond to form the C1-C2 pi bond, and one from the C2-Br bond to the bromine atom. (correct answer)
3. Two simultaneous arrows: one from the base to the C2 carbon, and one from the C2-Br bond to the bromine atom, forming a carbanion intermediate.
4. Three simultaneous arrows: one from the base to a C3 proton, one from the C3-H bond to form the C2-C3 pi bond, and one from the C2-Br bond to the bromine atom.

Explanation: An E2 mechanism is a concerted, one-step process. For the formation of but-1-ene (the Hofmann product, favored by a bulky base), a proton must be removed from C1. This requires three electron movements to happen at once: (1) the base attacks a proton on C1, (2) the electrons from the C1-H bond swing down to form a double bond between C1 and C2, and (3) the leaving group (bromide) on C2 departs, taking the C-Br bonding electrons with it. Option B correctly describes this concerted process. Option A describes an E1 mechanism. Option C describes an impossible mechanism. Option D describes the arrows to form but-2-ene, which is the minor (Zaitsev) product with a bulky base.

Question 5

1. A single arrow from the alkene pi bond attacks one bromine atom, breaking the Br-Br bond and forming a secondary carbocation.
2. An arrow from the alkene pi bond attacks one bromine atom, while a second arrow from a lone pair on the same bromine atom attacks the other carbon of the double bond, and a third arrow breaks the Br-Br bond. (correct answer)
3. An arrow from a lone pair on a bromine atom attacks the more substituted carbon of the alkene, and a second arrow breaks the alkene pi bond, moving the electrons to the less substituted carbon.
4. Two arrows are shown: one from the alkene pi bond to one bromine atom, and a second from the other bromine atom to the other carbon of the double bond, breaking the Br-Br bond.

Explanation: The formation of the cyclic bromonium ion is a concerted process involving three electron movements. (1) The electron-rich pi bond of the alkene acts as a nucleophile, attacking one of the bromine atoms. (2) To avoid forming a high-energy carbocation, a lone pair on that same bromine atom immediately attacks the other carbon of the original double bond, forming a three-membered ring. (3) Simultaneously, the Br-Br bond breaks, with the electrons moving onto the other bromine atom, which becomes a bromide ion. Option B correctly describes this entire concerted process. Option A describes an incorrect mechanism that would lead to a discrete carbocation, which does not explain the observed anti-stereochemistry of addition. Options C and D depict incorrect electron flow.

Question 6

1. The arrow should point to one of the carbon atoms, not the center of the bond.
2. The arrow indicates the movement of a proton, but curved arrows must show the movement of electrons.
3. The arrow should be double-barbed to indicate the movement of two electrons.
4. The arrow is pointing in the wrong direction; it should originate from the electron source (pi bond). (correct answer)

Explanation: The most fundamental rule of curved-arrow formalism is that arrows must show the movement of electrons, originating from an electron source (lone pair, pi bond, sigma bond) and pointing to an electron sink (an atom or a location where a bond will form). A proton (H⁺) is an electron sink with no valence electrons. Therefore, an arrow cannot originate from it. The arrow must originate from the electron-rich pi bond and point towards the electron-deficient proton. While it's true the arrow should point to the proton (as described in the stem), the critical error is its origin. Thus, the arrow is pointing in the reverse direction of electron flow.

Question 7

1. The substrate is a tertiary alkyl halide, which cannot undergo a concerted SN2 reaction due to severe steric hindrance. (correct answer)
2. The arrows violate the octet rule by creating a pentavalent carbon intermediate.
3. Water is a strong nucleophile, so the reaction should proceed via a single concerted step as drawn.
4. The arrow from the C-Br bond to the bromine atom is unnecessary; the bromide ion simply falls off.

Explanation: When analyzing substitution reactions, you need to identify the substrate type and match it to the appropriate mechanism. Tertiary alkyl halides like tert-butyl bromide follow predictable reaction pathways based on their structure. The correct answer is A because tertiary substrates cannot undergo SN2 reactions due to severe steric hindrance. The bulky tert-butyl group blocks the backside approach that SN2 mechanisms require. Instead, tertiary alkyl halides react with nucleophiles like water through an SN1 mechanism: first the C-Br bond breaks to form a carbocation intermediate, then water attacks the positively charged carbon in a separate step. Let's examine why the other options are incorrect. Option B misunderstands what creates octet rule violations - the student's arrows actually avoid creating pentavalent carbon by showing bond breaking concurrent with bond formation, which would be fine if the mechanism were plausible. Option C incorrectly characterizes water as a strong nucleophile; water is actually a weak nucleophile that typically requires carbocation intermediates (SN1 conditions) to react efficiently. Option D oversimplifies bond breaking - in any valid mechanism, you must show electron movement explicitly with curved arrows; bonds don't simply "fall off" without proper electron accounting. Remember this key pattern: primary substrates favor SN2, tertiary substrates favor SN1, and secondary substrates depend on conditions. When you see a tertiary substrate reacting with a weak nucleophile like water, immediately think SN1 mechanism with a two-step process involving carbocation formation.

Question 8

1. One curved arrow from the lone pair on nitrogen in $\text{NH}_2^-$ to the oxygen atom in ethanol, and another arrow from the O-H bond to the hydrogen atom
2. One curved arrow from the O-H bond in ethanol to the nitrogen atom in $\text{NH}_2^-$, and another arrow from the lone pair on nitrogen to oxygen in ethanol
3. One curved arrow from the hydrogen atom in ethanol to the lone pair on nitrogen in $\text{NH}_2^-$, and another arrow from oxygen in ethanol to the O-H bond
4. One curved arrow from the lone pair on nitrogen in $\text{NH}_2^-$ to the hydrogen atom in ethanol, and another arrow from the O-H bond in ethanol to oxygen, creating the alkoxide ion (correct answer)

Explanation: When you encounter acid-base reactions, curved arrow formalism shows the actual movement of electrons during bond breaking and forming. The key is to trace how electrons flow from the base (electron donor) to the acid (electron acceptor). In this reaction, $\text{NH}_2^-$ acts as the base and ethanol as the acid. The amide ion has a lone pair of electrons that will attack the acidic hydrogen, while the O-H bond must break to form the products. You need exactly two curved arrows: one showing the base attacking the hydrogen, and another showing the bond breaking to accommodate the new bond formation. Answer D correctly shows this mechanism: the first curved arrow goes from nitrogen's lone pair to the hydrogen atom in ethanol (the base attacks the acidic proton), and the second arrow shows the O-H bond electrons moving to oxygen, forming the alkoxide ion $\text{CH}_3\text{CH}_2\text{O}^-$. Answer A incorrectly shows the lone pair attacking the oxygen atom rather than the hydrogen, which wouldn't result in proton transfer. Answer B has the arrows going in impossible directions - bonds can't transfer to atoms, and the sequence doesn't make chemical sense. Answer C shows hydrogen moving as if it carries electrons, but hydrogen transfers as $\text{H}^+$ (without electrons) in acid-base reactions. Remember: in proton transfer reactions, always start your curved arrows from the base's lone pair attacking the acidic hydrogen, then show the breaking bond's electrons staying with the more electronegative atom.

Question 9

1. Ensuring that each arrow originates from a site with available electrons and that the total number of bonds and lone pairs remains constant throughout all steps
2. Verifying that formal charges are conserved in each individual step and that the sum of all formal charges equals the total charge of all species present (correct answer)
3. Confirming that the number of curved arrows in each step matches the number of bonds broken plus the number of bonds formed in that step exactly
4. Checking that all intermediates have complete octets and that no atom ever exceeds or falls below the octet rule during any mechanistic step

Explanation: The most critical aspect of mechanistic bookkeeping is charge conservation. In each mechanistic step, the sum of all formal charges must remain constant - electrons can move between atoms, but the total charge of the system cannot change unless species are added or removed. This requires careful tracking of where electrons go when bonds break and form. Choice B correctly emphasizes both individual step charge conservation and overall charge balance. Choice A describes electron movement correctly but doesn't address the critical issue of charge tracking. Choice C incorrectly suggests a 1:1 relationship between arrows and bond changes (some arrows show lone pair movements, others show bond breaking/forming). Choice D is incorrect because many stable intermediates (like carbocations) violate the octet rule, and this is often necessary in organic mechanisms.

Question 10

1. All four electron movements occur simultaneously: base abstracts proton, C-H bond breaks, C-C $\pi$ bond forms, and C-Br bond breaks, with all arrows drawn in the same mechanistic step (correct answer)
2. The arrows must be drawn in sequence: first base attacks hydrogen, then C-H bond breaks, then $\pi$ bond forms, finally C-Br bond breaks, requiring four separate mechanistic steps
3. Three arrows are simultaneous (base attack, C-H break, C-Br break) while the fourth arrow ($\pi$ bond formation) occurs in a subsequent fast step after the others are complete
4. Two pairs of arrows operate simultaneously: base attack and C-H bond breaking occur together, followed immediately by $\pi$ bond formation and C-Br bond breaking in the second step

Explanation: E2 elimination is a concerted mechanism, meaning all bond breaking and forming occurs simultaneously in a single step. The four electron movements are: (1) base lone pair attacks the $\beta$-hydrogen, (2) C-H bond breaks with electrons moving toward the $\alpha$-carbon, (3) these electrons form the new C-C $\pi$ bond, and (4) C-Br bond breaks with electrons going to bromide. All arrows are drawn in one mechanistic step because all changes occur at the same time. Choice A correctly describes this concerted nature. Choice B incorrectly describes a stepwise process, which would be E1, not E2. Choices C and D incorrectly break the concerted mechanism into multiple steps, contradicting the definition of E2.

Question 11

1. A curved arrow from the C-H bond at C-3 to the positive charge at C-2, with the positive charge moving to C-3 (correct answer)
2. A curved arrow from the positive charge at C-2 to the C-H bond at C-3, with the positive charge moving to C-3
3. A curved arrow from the C-H bond at C-1 to the positive charge at C-2, with the positive charge moving to C-1
4. A curved arrow from the C-C bond between C-2 and C-3 to the positive charge at C-2, with the positive charge moving to C-3

Explanation: In a 1,2-hydride shift, the C-H bond at the adjacent carbon (C-3) donates its electron pair to fill the empty p-orbital at the carbocation center (C-2). The curved arrow must originate from the electron-rich C-H bond and point toward the electron-deficient carbocation center. This results in the hydrogen moving to C-2 and the positive charge relocating to C-3. Choice A correctly shows the arrow from the C-H bond to the positive charge. Choice B incorrectly shows the arrow originating from the positive charge (which has no electrons to donate). Choice C describes a shift from C-1, which would be a different rearrangement. Choice D incorrectly shows a C-C bond breaking instead of a C-H bond.

Question 12

1. The student drew the nucleophile attack and leaving group departure as separate steps, causing carbon to temporarily have only six electrons around it
2. The student drew both arrows pointing toward carbon simultaneously, giving carbon ten electrons in the transition state representation (correct answer)
3. The student drew the leaving group arrow originating from the iodine atom instead of from the C-I bond, creating an impossible electron count
4. The student forgot to show lone pairs on the cyanide nucleophile, making it appear that carbon forms five bonds during the reaction

Explanation: The most common octet rule violation in student mechanisms occurs when both the nucleophile attack arrow and leaving group departure arrow point toward the same carbon atom. This incorrectly suggests that carbon simultaneously accepts electrons from the nucleophile while retaining the leaving group, giving carbon 10 electrons (hypervalent carbon). The correct $S_N2$ mechanism shows the nucleophile arrow pointing to carbon while the leaving group arrow points away from carbon (from the C-I bond to iodine). Choice B describes this common error. Choice A describes an $S_N1$ mechanism, which would be incorrect for methyl iodide but wouldn't violate the octet rule. Choice C describes an arrow origin error but doesn't create octet violations. Choice D describes a drawing convention issue, not an octet rule violation.

Question 13

1. Arrow (1) should originate from the alcohol oxygen lone pair attacking the hydronium hydrogen, not from hydronium attacking the alcohol (correct answer)
2. Arrow (2) is unnecessary because the original O-H bond remains intact when the alcohol is protonated by the acid
3. Arrow (3) should show the C-O bond electrons going to oxygen, but the student likely showed them going to carbon or nowhere
4. The mechanism requires an additional arrow showing water molecule formation from the protonated alcohol before C-O bond cleavage occurs

Explanation: In acid-base chemistry, the base (alcohol oxygen with lone pairs) must attack the acid (hydronium hydrogen), not vice versa. The student's error in arrow (1) violates the fundamental principle that curved arrows originate from electron-rich sites (lone pairs or bonds) and point toward electron-poor sites. Hydronium has no lone pairs to donate to the alcohol. The correct mechanism shows: (1) alcohol oxygen lone pair attacks hydronium hydrogen, (2) O-H bond in hydronium breaks with electrons going to oxygen (forming water), and (3) C-O bond breaks with electrons going to oxygen. Choice A identifies the fundamental error. Choice B is wrong because protonation does break the original alcohol O-H bond. Choice C incorrectly assumes arrow (3) is wrong. Choice D describes an unnecessary additional step.

Question 14

1. The terminal carbon has formal charge -1, the internal carbon has formal charge 0; drawing the arrow from the C-H bond to the amide nitrogen instead of to the nitrogen lone pair
2. The terminal carbon has formal charge -1, the internal carbon has formal charge 0; drawing the arrow from the amide nitrogen to the terminal carbon instead of to the acidic hydrogen (correct answer)
3. Both carbons have formal charge -1/2 due to resonance; failing to show the resonance arrow between the two carbons in the acetylide anion
4. The terminal carbon has formal charge 0, the internal carbon has formal charge -1; drawing both arrows in the same direction instead of showing the electron pair movement correctly

Explanation: In the acetylide anion, the terminal carbon (which lost its hydrogen) bears the -1 formal charge, while the internal carbon remains neutral with formal charge 0. The most common mechanistic error is drawing the nucleophilic attack incorrectly - students often show the amide attacking the carbon atom instead of abstracting the hydrogen. This error would suggest electrons are being added to carbon rather than hydrogen being removed, leading to incorrect formal charge assignments. Choice B identifies both the correct formal charges and the most likely arrow error. Choice A has the right formal charges but describes an impossible arrow (bonds don't attack atoms). Choice C incorrectly suggests resonance delocalization of charge between sp carbons. Choice D has incorrect formal charges and describes a vague error.

Question 15

1. A nucleophile attacking an electrophilic carbon center.
2. The delocalization of electrons in a conjugated pi system to draw a resonance structure.
3. A base abstracting a proton in an E2 elimination reaction. (correct answer)
4. The departure of a leaving group in the first step of an SN1 reaction.

Explanation: Curved arrows show electron movement. In an E2 elimination, a base removes a proton, and the electrons from the C-H sigma bond are used to form the new pi bond. This is explicitly shown with a curved arrow originating from the C-H σ bond and pointing to the space between the two carbons to form the double bond. In contrast, nucleophilic attack (A) arrows start from lone pairs or pi bonds. Resonance in conjugated systems (B) involves pi bonds and lone pairs. The departure of a leaving group in an SN1 reaction (D) involves an arrow starting from the C-LG sigma bond and pointing to the leaving group, representing bond cleavage, but in an E2 reaction, the C-H sigma bond arrow represents bond formation.