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Organic Chemistry I · Mechanisms

Arrow pushing practice problems, with answers

A curved arrow shows a pair of electrons moving from where they are (a lone pair or a bond) to where they end up (a new bond or a new lone pair). Almost every mechanism in a college organic course is built from the same five arrow patterns. Learn to spot them and the mechanisms stop looking like a wall of arrows.

Already know the basics? Jump to the 8 practice questions or the worked examples.

The rules every curved arrow follows

These are the checks to run on each arrow before you move on. If you want the basics first, start with the curved arrows guide.

  1. Tail on electrons. The tail starts at a lone pair or at a bond (σ or π). Never start at a bare atom, a hydrogen with no lone pair, or a + sign.
  2. Source to sink. Electrons flow from an electron-rich source (a nucleophile or base) to an electron-poor sink (an electrophile, an acidic H, or a carbocation). Arrows never point from an electrophile toward a nucleophile.
  3. Head on the destination. Point at an atom if the electrons become a lone pair there. Point at the space between two atoms if they become a new bond.
  4. Respect the octet. Carbon, nitrogen, and oxygen can't go past eight electrons, and hydrogen can't hold more than one bond. If an arrow adds a bond to a full atom, another arrow must take a pair away from it in the same step.
  5. Recount charges. Total charge is the same before and after every step. A negative nucleophile becomes neutral when it bonds; a neutral one (water, an alkene) becomes positive.

Five patterns behind most mechanisms

Common curved-arrow patterns
PatternArrowsExample
Nucleophilic attackLone pair (or π bond) to an electrophilic atomWater adds to a carbocation
Loss of a leaving groupThe bond to the leaving group moves onto the leaving groupC–Br breaks to give a carbocation and Br⁻
Proton transferTwo arrows: base lone pair to H, and the old H–X bond onto XWater removes H⁺ from a protonated alcohol
Carbocation rearrangementA C–H or C–C bond next to C⁺ moves over to C⁺Hydride shift, secondary to tertiary carbocation
Addition to C=OTwo arrows: nucleophile to carbonyl carbon, and the C=O π bond onto oxygenCyanide or hydride adds to a ketone

SN2 and E2 are combinations done in a single step. SN2 is nucleophilic attack plus loss of a leaving group. E2 is a proton transfer, a C–H bond becoming the new π bond, and loss of a leaving group, all at once (three arrows).

Curved arrows for a nucleophile adding to a ketone carbonylA negatively charged nucleophile uses a lone pair to attack the carbonyl carbon of a ketone with two R groups. A second arrow moves the electrons of the carbon-oxygen pi bond onto oxygen. The product is a tetrahedral carbon bonded to the nucleophile, both R groups, and an oxygen with a negative charge.One step, two arrowsNu−CORRCO−NuRRArrow 1: lone pair to carbon (new bond). Arrow 2: π bond to oxygen (new lone pair).
Nucleophilic addition to a ketone. Only the attacking lone pair is drawn; other lone pairs are left off. Oxygen ends with one bond and three lone pairs, so it carries the −1 charge. The total charge is −1 on both sides.

Mistakes that cost points

  • Arrow starts at a charge or an atom. Draw lone pairs on the nucleophile so you have something to start from. A − sign is a clue, not a source.
  • Arrow goes backward. If it points from H⁺ to a base, or from a carbocation to water, it's describing atoms moving, not electrons. Flip it.
  • Five bonds to carbon. If a nucleophile attacks a carbon that already has four bonds, a bond to that carbon has to break in the same step. Hydrogen is the same: a base can bond to H only if the old bond to H breaks.
  • Arrows that skip steps. In an SN1 reaction, the leaving group leaves first, and the nucleophile attacks in a separate step. Don't draw both arrows on one structure.
  • Forgetting charges in the product. After every step, recount: three bonds and one lone pair on oxygen means +1; one bond and three lone pairs means −1.

Worked examples

Example 1: HBr adds to propene

CH₃CH=CH₂ + HBr gives 2-bromopropane. Draw the arrows for each step.

  1. Step 1, alkene takes a proton. Arrow 1 starts at the C=C π bond and points at the H of HBr. Arrow 2 starts at the H–Br bond and points at Br, because hydrogen can't hold two bonds.
  2. Where does H go? It goes to the CH₂ end (C1), so the positive charge lands on C2, a secondary carbocation: CH₃C⁺HCH₃. Putting H on C2 would leave a less stable primary carbocation.
  3. Charge check: neutral reactants, products CH₃C⁺HCH₃ and Br⁻. Net zero.
  4. Step 2, bromide attacks. One arrow from a lone pair on Br⁻ to the positive carbon.

Answer: three arrows in two steps. π bond to H and H–Br bond to Br, then Br⁻ lone pair to C⁺. Product: CH₃CHBrCH₃.

Example 2: acid turns tert-butyl alcohol into 2-methylpropene

(CH₃)₃COH with H₃O⁺ (E1 dehydration). Draw every arrow.

  1. Protonate the OH (proton transfer). Arrow from a lone pair on the alcohol O to an H of H₃O⁺. Arrow from that H–O bond onto the H₃O⁺ oxygen. Result: (CH₃)₃C–OH₂⁺ and H₂O.
  2. Loss of the leaving group. One arrow from the C–O bond onto oxygen. Water leaves, neutral, and carbon is left with three bonds and a + charge: (CH₃)₃C⁺.
  3. Remove a β-hydrogen (proton transfer). Arrow from a lone pair on water to an H on a CH₃ next to C⁺. Arrow from that C–H bond to the C–C⁺ bond, where it becomes the new π bond.
  4. Charge check: +1 at the start (H₃O⁺), +1 at the end (H₃O⁺ regenerated). Acid is a catalyst.

Answer: five arrows in three steps: two, one, two. Product (CH₃)₂C=CH₂ plus water.

Example 3: a hydride shift during HCl addition

3-Methyl-1-butene, CH₂=CHCH(CH₃)₂, reacts with HCl and gives a lot of 2-chloro-2-methylbutane. Show the arrow that explains it.

  1. Protonation: π bond to H, H–Cl bond to Cl. H goes on C1, leaving a secondary carbocation at C2: CH₃C⁺HCH(CH₃)₂.
  2. Look next door: C3 has one H and two CH₃ groups. If the C3–H bond's electrons move to C2, the positive charge ends up on C3, which is tertiary.
  3. The arrow: tail on the C3–H bond, head on C2⁺. The hydrogen travels with the pair as a hydride. New cation: CH₃CH₂C⁺(CH₃)₂.
  4. Capture: arrow from a Cl⁻ lone pair to the tertiary C⁺, giving CH₃CH₂CCl(CH₃)₂.

Answer: the shift arrow runs from the C3–H bond to C2⁺. OpenStax reports roughly equal amounts of the rearranged product (2-chloro-2-methylbutane) and the unrearranged one (2-chloro-3-methylbutane).

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Practice questions

Pick an answer, then check it. A wrong answer tells you what went wrong, so you can try again. Every question also has a full solution.

Question 1Hydroxide, HO⁻, attacks CH₃Br in an SN2 reaction. Where does the first curved arrow start?

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Arrows start on electrons. Hydroxide's oxygen has three lone pairs, and one of them becomes the new C–O bond. So arrow 1 goes from an oxygen lone pair to the carbon. Arrow 2 goes from the C–Br bond onto Br, in the same step.

Question 2Protonated tert-butyl alcohol, (CH₃)₃C–OH₂⁺, loses water to form a carbocation. Which arrow is correct?

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Leaving-group departure is a single arrow: tail on the C–O bond, head on oxygen. Oxygen goes from three bonds and one lone pair (+1) to two bonds and two lone pairs (neutral water). Carbon loses a bond and becomes (CH₃)₃C⁺.

Question 3Ammonia reacts with HCl to form NH₄⁺ and Cl⁻. Which set of arrows is correct?

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Proton transfer is a two-arrow pattern. Nitrogen's lone pair forms the N–H bond, and at the same time the H–Cl bond pair moves onto Cl. Nitrogen ends with four bonds and no lone pair (+1). Chlorine ends with four lone pairs (−1). Net charge stays zero.

Question 4Cyanide, ⁻C≡N, adds to the C=O carbon of acetaldehyde, CH₃CHO. Which arrows describe this step?

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Addition to C=O is the two-arrow pattern from the figure above. Arrow 1: the cyanide carbon's lone pair to the carbonyl carbon. Arrow 2: the C=O π bond onto oxygen. The product is CH₃CH(CN)O⁻, an alkoxide. Charge is −1 before and after.

Question 5In the carbocation CH₃C⁺HCH(CH₃)₂, a 1,2-hydride shift makes a more stable cation. Where does the arrow go?

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C2 is the positive carbon. Its neighbor C3 carries one H and two CH₃ groups. Moving the C3–H bond pair to C2 makes C2 neutral (it now has the H) and leaves C3 with three carbons and a positive charge: CH₃CH₂C⁺(CH₃)₂, a tertiary carbocation. The arrow's tail is on the C3–H bond and its head is on C2.

Question 6A student draws one arrow for HO⁻ + CH₃Br: from an oxygen lone pair to the carbon, and nothing else. What's wrong?

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The methyl carbon already has four bonds. Adding a fifth would give it ten electrons, which breaks the octet rule. So in the same step the C–Br bond pair must move onto bromine. Two arrows, one step: lone pair to C, C–Br bond to Br. Products: CH₃OH and Br⁻.

Question 7Water's lone pair attacks (CH₃)₃C⁺. Before any proton is removed, what is the formal charge on oxygen in the new species?

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Formal charge = valence electrons − nonbonding electrons − bonds. Oxygen: 6 − 2 − 3 = +1. The carbocation's + charge has moved to oxygen: (CH₃)₃C–OH₂⁺. A second water molecule then removes an H⁺ to give tert-butyl alcohol.

Question 8Propene reacts with H₃O⁺. Which description of the first step is correct?

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Two arrows: the C=C π bond to an H on H₃O⁺, and the H–O bond onto oxygen, which leaves as neutral water. The H bonds to C1 (CH₂), so C2 loses its share of the π electrons and becomes positive: CH₃C⁺HCH₃, a secondary carbocation. This is the Markovnikov orientation.

Answer key

  1. C, an oxygen lone pair: arrows start at electrons, not at charges or atoms.
  2. A, C–O bond to O: the leaving group takes the bonding pair.
  3. D, two arrows: N lone pair to H, H–Cl bond onto Cl.
  4. A: Nu lone pair to carbonyl C, π bond onto O.
  5. B, C3–H bond to C2⁺: a hydride shift moves a bond, not an atom.
  6. D, add the C–Br arrow: no five-bond carbon.
  7. B, +1: three bonds and one lone pair on O.
  8. C: π bond to H, H–O bond to O, secondary cation at C2.

Frequently asked questions

What is the difference between arrow pushing and curved arrows?

They are the same thing. Arrow pushing is the informal name for drawing curved arrows to show electron pairs moving during a reaction mechanism.

How many arrows does a proton transfer need?

Two. One arrow from the base's lone pair to the hydrogen, and one from the old bond to that hydrogen onto the atom it was attached to.

Is arrow pushing on the MCAT?

The MCAT tests organic reactions and their mechanisms at a basic level, so you need to read nucleophiles, electrophiles, and leaving groups. You rarely draw full mechanisms, but the same source-to-sink logic helps you predict products quickly. Check the current AAMC content outline for the exact topic list.

When do I use a half-headed arrow?

A half-headed, fishhook arrow moves one electron and is used for radical reactions. The polar mechanisms on this page all use full-headed arrows, each moving two electrons.

Sources

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