Source: Chem 237 Marathon Worksheet Key (University of Washington)
Difficulty: Intermediate | Prerequisites: Decision Framework study notes (Part 1)
Tags: SN1 examples, SN2 examples, E1 examples, E2 examples, radical halogenation examples, worked problems, mechanism prediction, organic chemistry practice
This document walks through 19 worked examples from the Chem 237 Chapter 9 marathon worksheet, organised by mechanism type rather than problem number. Each example shows the reasoning chain: check the leaving group, assess the reagent, consider the substrate, then predict the product(s) and stereochemistry. Working through these in order, grouped by mechanism, builds the pattern recognition you need for the exam.
Nineteen worked problems from the Chem 237 marathon worksheet, sorted by mechanism. Five result in no reaction (bad leaving group or incompatible substrate/reagent pairing). The rest split across SN2, E2, mixed SN2/E2, SN1/E1, and radical halogenation. Each entry shows the full reasoning chain so you can practise the decision framework on real exam-style problems.
These are the problems students lose marks on by overthinking. If the leaving group is bad or the substrate/reagent combination cannot support any mechanism, the answer is simply "no reaction."
Problem 2: Alcohol + NaOH
Substrate has an OH group, not a halide. OH⁻ is a terrible leaving group (it is a strong base, not a weak one). All four ionic mechanisms (SN1, SN2, E1, E2) require a good leaving group. No reaction occurs.
Problem 9: Tertiary alcohol + NaOH
Same principle as Problem 2. Even though the substrate is tertiary (which would favour SN1/E1 if there were a good leaving group), the OH group simply will not leave. No reaction.
Problem 15: Alcohol + NaOH
Another bad-leaving-group problem. The worksheet repeats this pattern deliberately. On the exam, if you see an OH on the substrate and no acid to protonate it, write "no reaction" and move on.
Problem 18: Alkane + I₂ / hv
This is a radical halogenation setup, but iodine is generally unreactive for radical reactions. The H–I bond that would form is too weak relative to the C–H bond being broken, making the propagation step thermodynamically unfavourable. No reaction.
Problem 19: Primary alkyl bromide + H₂O
Good leaving group (Br⁻), but H₂O is both a weak base and a weak nucleophile. That rules out SN2 and E2. SN1 and E1 require a carbocation intermediate, but a primary carbocation is too unstable to form. No mechanism is viable. No reaction.
These problems share the same pattern: good leaving group, strong nucleophile (with or without strong base character), and an unhindered (primary or secondary) electrophilic carbon.
Problem 1: Primary alkyl bromide + NaOH → alcohol (SN2)
Reasoning chain:
Good leaving group? Yes, Br⁻.
Reagent? NaOH is both a strong base (E2 possible) and a strong nucleophile (SN2 possible).
Substrate? Uncrowded primary alkyl halide.
Verdict: SN2 dominates. Primary substrates strongly favour SN2 over E2. SN1/E1 are not required (strong base/nucleophile present, and primary carbocation would be too unstable anyway).
Product: the bromide is replaced by hydroxide. No stereochemistry issue at a primary carbon (no chiral centre created).
Problem 3: Secondary alkyl bromide + NaCN → nitrile (SN2)
Reasoning chain:
Good leaving group? Yes, Br⁻.
Reagent? CN⁻ is a strong nucleophile but a weak base. E2 is off the table.
Substrate? Secondary halide. SN2 is feasible (though slower than at primary).
Verdict: SN2. No SN1/E1 needed (strong nucleophile present). No carbocation intermediate, so no rearrangements.
Stereochemistry: inversion of configuration at the electrophilic carbon (Walden inversion).
Problem 11: Primary alkyl chloride + NH₃ → amine (SN2)
Reasoning chain:
Good leaving group? Yes, Cl⁻.
Reagent? NH₃ is a weak base (E2 is not an option) but a strong nucleophile (lone pair on nitrogen).
Substrate? Primary halide. SN2 is strongly favoured.
Verdict: SN2. No SN1/E1 needed. No carbocation, so no shifts.
Product: chloride replaced by NH₂ group.
These problems feature a strong base (often bulky) with a substrate where SN2 is blocked or disfavoured.
Problem 7: Tertiary alkyl bromide + NaNH₂ → alkene (E2)
Reasoning chain:
Good leaving group? Yes, Br⁻.
Reagent? NaNH₂ is both a strong base and a strong nucleophile.
Substrate? Tertiary halide. SN2 is completely blocked by steric hindrance.
Verdict: E2. SN1/E1 are not needed when a strong base is present.
Stereochemistry: E2 requires anti-coplanar H and leaving group. If Br is on a wedge, the H removed must be on a dash. The chirality of other groups (e.g. the ethyl group) is unchanged.
Product: the alkene forms according to E2 geometry.
Problem 8: Secondary alkyl chloride + KOtBu → alkene (E2)
Reasoning chain:
Good leaving group? Yes, Cl⁻.
Reagent? KOtBu is a strong base but a very bulky, poor nucleophile. SN2 is off the table.
Substrate? Secondary halide (not that it matters, since SN2 was already ruled out by the bulky base).
Verdict: E2. SN1/E1 not needed.
Product: the most stable (internal, more substituted) alkene is the major product (Zaitsev's rule).
Note: tert-butoxide can sometimes give the less substituted (Hofmann) product for steric reasons, but this nuance is deferred to Chem 241.
Problem 13: Tertiary alkyl chloride + KOtBu → alkene (E2)
Reasoning chain:
Good leaving group? Yes, Cl⁻.
Reagent? KOtBu. Strong base, poor nucleophile.
Substrate? Tertiary. SN2 impossible.
Verdict: E2. Same anti-periplanar geometry requirement as Problem 7.
The most stable alkene is the internal alkene. The H removed must be anti-coplanar to the leaving group. Chirality at other centres is unchanged.
Problem 16: Primary alkyl chloride + KOtBu → alkene (E2)
Reasoning chain:
Good leaving group? Yes, Cl⁻.
Reagent? KOtBu. Strong base, bulky, poor nucleophile.
Substrate? Primary. Normally SN2 territory, but the bulky base blocks nucleophilic attack.
Verdict: E2.
Only one possible alkene can form from this substrate. No carbocations, so no hydride shifts.
This is the classic borderline case that exams love to test.
Problem 17: Secondary alkyl fluoride + NaOCH₃ → substitution and elimination products
Reasoning chain:
Good leaving group? Yes, F⁻ (fluoride is borderline but workable here).
Reagent? NaOCH₃ (sodium methoxide) is both a strong base and a strong nucleophile. Both E2 and SN2 are on the table.
Substrate? Secondary alkyl halide. This is the zone where SN2 and E2 compete roughly evenly.
Verdict: both SN2 and E2 occur.
E2 product: the more substituted, more stable alkene (Zaitsev).
SN2 product: methoxide replaces fluoride with inversion of stereochemistry.
Neither SN1 nor E1 is required because a strong base/nucleophile is present.
This type of problem requires you to draw both products and state that both mechanisms operate. On exams, showing only one mechanism here would cost marks.
These problems all involve weak base/weak nucleophile reagents with substrates that can form stable carbocations. SN1 and E1 always compete because they share the same rate-determining step (formation of the carbocation).
Problem 4: Tertiary alkyl bromide + H₂O → alcohol + alkenes (SN1 + E1)
Reasoning chain:
Good leaving group? Yes, Br⁻.
Reagent? H₂O is a weak base and a weak nucleophile. SN2 and E2 are both off the table.
Substrate? Tertiary halide. SN1 and E1 are feasible because a tertiary carbocation is stable.
SN1 product: water attacks the carbocation to give an alcohol. The carbocation is planar (sp²), so the alcohol chiral centre is stereorandom (racemic). Other chiral centres in the molecule are unchanged.
E1 products: two different alkenes form, both equally substituted, so both are major products. No anti-periplanar requirement because the carbocation is planar.
Rearrangement: not needed. The carbocation is already tertiary.
Problem 5: Secondary alkyl iodide + CH₃OH → ether + alkene (SN1 + E1 with hydride shift)
Reasoning chain:
Good leaving group? Yes, I⁻.
Reagent? CH₃OH is a weak base and a weak nucleophile.
Substrate? Secondary halide.
Key twist: after the leaving group departs and a secondary carbocation forms, a hydride shift occurs to give a more stable tertiary carbocation.
SN1 product: methanol attacks the rearranged (tertiary) carbocation. Product has no chiral centres.
E1 product: the more substituted alkene forms from the rearranged carbocation.
This problem tests whether you remember to check for rearrangements whenever SN1/E1 is the mechanism.
Problem 10: Tertiary alkyl chloride + CH₃CO₂⁻ → acetate ester + alkenes (SN1 + E1)
Reasoning chain:
Good leaving group? Yes, Cl⁻.
Reagent? Acetate (CH₃CO₂⁻) is a weak base but a strong nucleophile. However, at a tertiary carbon, SN2 is impossible regardless of nucleophile strength.
Substrate? Tertiary. SN1 and E1 are the viable pathways.
E1 products: two equally substituted alkenes, both major.
SN1 product: acetate attacks the tertiary carbocation. The new chiral centre is stereorandom. Other chiral centres remain.
No rearrangement needed (already tertiary).
These problems all involve X₂ + hv (light). There is no leaving group on the substrate; instead, a C–H bond is being replaced by a C–X bond through a radical chain mechanism.
Problem 6: Alkane + Br₂ / hv → tertiary alkyl bromide
Bromine is highly selective for the most stable radical. Radical stability follows the same order as carbocation stability: 3° > 2° > 1° > methyl. The major product is the most substituted (most crowded) alkyl bromide, regardless of how many other hydrogens are available.
Problem 12: Alkane + Cl₂ / hv → mixture of all possible monochlorinated products
Chlorine is largely unselective. Its modest preference for more stable radicals is roughly cancelled by the statistical advantage of more abundant hydrogen positions. All possible monochlorinated products form in significant amounts. This makes chlorination generally impractical when a single product is needed.
Problem 14: Alkane + F₂ / hv → primary alkyl fluoride (major)
Fluorine is extremely reactive and essentially non-selective. It reacts based on statistics: whichever type of hydrogen is most plentiful determines the major product. In this substrate, there are many equivalent methyl (primary) hydrogens and fewer secondary or tertiary positions, so the primary alkyl fluoride is the major product.
Q: A tertiary alkyl bromide is treated with NaOH. What mechanism(s) operate and what is the major product?
A: E2 only. NaOH is a strong base and strong nucleophile, but SN2 is impossible at a tertiary carbon (steric hindrance). SN1/E1 are not required when a strong base is present. The major product is the most substituted alkene (Zaitsev).
Q: A primary alkyl chloride is treated with H₂O. What happens?
A: No reaction. H₂O is a weak base and weak nucleophile, ruling out SN2 and E2. SN1 and E1 are not viable because a primary carbocation is too unstable to form.
Q: A secondary alkyl bromide is treated with NaCN. What is the mechanism and what stereochemical outcome do you expect?
A: SN2. CN⁻ is a strong nucleophile but a weak base (E2 is off). Secondary substrate allows SN2. The product is a nitrile with inverted configuration at the electrophilic carbon.
Q: An alkane is treated with Br₂ and light. The alkane has six equivalent primary hydrogens and one tertiary hydrogen. Which product predominates?
A: The tertiary alkyl bromide. Bromine is highly selective for the most stable (tertiary) radical, regardless of the statistical advantage of the six primary hydrogens.
Q: A secondary alkyl iodide is treated with CH₃OH. The carbon adjacent to the electrophilic carbon bears three methyl groups. What products form?
A: SN1 + E1 with a hydride shift. The initial secondary carbocation rearranges to a more stable tertiary carbocation. SN1 gives an ether (stereorandom at the new centre); E1 gives the most substituted alkene.
Q: Why does KOtBu almost always give E2 rather than SN2?
A: The tert-butyl group makes the oxygen extremely sterically hindered. It remains a strong base (it can abstract a small, accessible proton), but it is far too bulky to act as a nucleophile in a backside attack.
Q: A tertiary alkyl chloride is treated with CH₃CO₂⁻ (acetate). What mechanism(s) operate?
A: SN1 + E1. Acetate is a strong nucleophile but at a tertiary carbon, SN2 is impossible. Acetate is a weak base, so E2 is off. SN1 and E1 proceed through the tertiary carbocation.
Q: What is the key geometric requirement for E2, and how does it affect product prediction in a cyclohexane ring?
A: E2 requires anti-periplanar (180°) geometry between the C–H bond and the C–leaving group bond. In a cyclohexane, both the H and the leaving group must be axial and on opposite faces of the ring. This may limit which hydrogen can be removed and therefore which regioisomeric alkene forms.
Q: An alkane is treated with I₂ and light. What happens?
A: No reaction. Iodine is generally unreactive for radical halogenation because the overall reaction is thermodynamically unfavourable.
Q: A primary alkyl bromide is treated with NaOH. Both SN2 and E2 are possible in principle. Which dominates?
A: SN2 dominates. At an uncrowded primary carbon, the nucleophile has easy access for backside attack. E2 is possible in principle but is a minor pathway at primary substrates with non-bulky bases.
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