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Organic Chemistry I · Substitution and Elimination

SN1, SN2, E1, and E2 practice problems, with answers

On an exam, an alkyl halide and a reagent show up together and you have to say which of four mechanisms wins and what the major product is. The good news: three questions settle almost every problem. What kind of carbon holds the leaving group? Is the reagent mainly a nucleophile or mainly a base? And is anything (a weak reagent, heat, a bulky base) tipping the balance?

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

A quick decision method

Run these four checks in order. They are a summary of the trends in the college Organic Chemistry I chapter on alkyl halide reactions, so treat them as strong defaults, not laws.

  1. Classify the substrate. Look at the carbon bonded to the leaving group: methyl, primary, secondary, or tertiary. Note whether it is allylic or benzylic, since resonance can stabilize a carbocation there.
  2. Classify the reagent. Put it in one of four bins: a weak nucleophile and weak base (H₂O, ROH); a good nucleophile that is a weak base (I⁻, Br⁻, CN⁻, N₃⁻, RS⁻, HS⁻); a strong, unhindered base (HO⁻, CH₃O⁻, CH₃CH₂O⁻); or a strong, bulky base (potassium tert-butoxide, (CH₃)₃CO⁻K⁺).
  3. Look up the likely mechanism in the table below. Polar aprotic solvents (DMSO, DMF, acetone) help SN2. Protic solvents with no added base point to SN1 and E1. Heat pushes any competition toward elimination.
  4. Draw the product, then check the details. SN2 inverts a stereocenter. SN1 goes through a flat carbocation, so it scrambles configuration and may rearrange first. E2 and E1 usually give the more substituted (Zaitsev) alkene, but a bulky base gives more of the less substituted (Hofmann) alkene.
Usual major pathway for simple alkyl halides
SubstrateWeak Nu, weak base (H₂O, ROH)Good Nu, weak base (I⁻, CN⁻, N₃⁻, RS⁻)Strong base (HO⁻, RO⁻)Bulky base (t-BuO⁻)
MethylVery slowSN2SN2SN2 (no β-hydrogen to remove)
PrimaryVery slowSN2Mostly SN2E2
SecondarySlow SN1/E1 mix; faster if allylic or benzylicSN2, best in a polar aprotic solventMostly E2E2
TertiarySN1 and E1 together; heat favors E1SN1 (some E1); never SN2E2E2

Two traps. First, tertiary substrates never do SN2: the backside of the carbon is blocked. Second, a strong base present in solution beats the slow SN1/E1 route, so tertiary plus alkoxide is E2, not SN1.

Turning the mechanism into a major product

Naming the mechanism is half the problem. The rest is drawing the right product.

  • SN2: the nucleophile ends up exactly where the leaving group was, attached from the back. A stereocenter inverts. No rearrangements.
  • SN1: the nucleophile adds to the carbocation. Before it does, check whether a neighboring hydrogen or alkyl group can shift (with its electron pair) to make a more stable carbocation. A stereocenter at the reacting carbon comes out as a mix of both configurations.
  • E2: the base pulls off a hydrogen on a carbon next door (a β-hydrogen) while the leaving group leaves. The new C=C forms between those two carbons. With small bases, the more substituted alkene wins (Zaitsev's rule).
  • E1: same carbocation as SN1, but a weak base (often the solvent) removes a β-hydrogen. The more substituted alkene usually wins, and rearrangements can happen here too.

If you only need to compare SN1 and SN2, see the SN1 vs SN2 guide. For the substitution-or-elimination call by itself, see substitution vs elimination.

Worked examples

Example 1: 2-bromo-2-methylbutane with sodium ethoxide

(CH₃)₂CBrCH₂CH₃ + NaOCH₂CH₃ in ethanol. Name the mechanism and the major product.

  1. Substrate: the C–Br carbon has three carbons on it, so it is tertiary. SN2 is out.
  2. Reagent: ethoxide is a strong, unhindered base.
  3. Mechanism: tertiary plus strong base is E2. The base doesn't wait for a carbocation.
  4. Which alkene? There are β-hydrogens on two different kinds of carbon. Taking one from a CH₃ group gives CH₂=C(CH₃)CH₂CH₃, 2-methyl-1-butene (two alkyl groups on the C=C). Taking one from the CH₂ gives (CH₃)₂C=CHCH₃, 2-methyl-2-butene (three alkyl groups on the C=C).
  5. Zaitsev: ethoxide is small, so the more substituted alkene wins. OpenStax reports about 70% 2-methyl-2-butene and 30% 2-methyl-1-butene for this reaction.

Answer: E2. Major product 2-methyl-2-butene, (CH₃)₂C=CHCH₃; minor product 2-methyl-1-butene.

Example 2: (S)-2-bromobutane with sodium azide in DMF

(S)-CH₃CHBrCH₂CH₃ + NaN₃ in DMF. Name the mechanism and the product, including configuration.

  1. Substrate: secondary, so the reagent and solvent decide it.
  2. Reagent: azide, N₃⁻, is a very good nucleophile but a weak base. That rules out E2 as the main path.
  3. Solvent: DMF is polar aprotic, which keeps azide reactive. That favors SN2.
  4. Stereochemistry: SN2 attacks from the back, so the reacting carbon inverts.
  5. Relabel carefully: in the starting material the priorities are Br > ethyl > methyl > H. In the product they are N₃ > ethyl > methyl > H. The new group takes the top priority, just as Br did, so the geometric inversion also flips the label from S to R.

Answer: SN2. Product (R)-2-azidobutane, CH₃CH(N₃)CH₂CH₃, plus NaBr.

Example 3: 2-chloro-2-methylpropane warmed in aqueous ethanol

(CH₃)₃CCl in 80% aqueous ethanol at 65 °C, with no added base. Which mechanisms run, and what forms?

  1. Substrate: tertiary. SN2 is out.
  2. Reagent: only water and ethanol, both weak nucleophiles and weak bases. Nothing is strong enough for E2.
  3. Mechanism: the C–Cl bond ionizes to give the tert-butyl carbocation, (CH₃)₃C⁺. That one intermediate can go two ways.
  4. SN1 path: water adds to the carbocation, and loss of a proton gives (CH₃)₃COH, 2-methyl-2-propanol (tert-butyl alcohol).
  5. E1 path: a solvent molecule removes a β-hydrogen from one of the methyl groups, giving (CH₃)₂C=CH₂, 2-methylpropene.
  6. Ratio: OpenStax reports a 64:36 mix of the alcohol and the alkene under these conditions. Warmer conditions shift the ratio toward the alkene.

Answer: SN1 and E1 together. Main products: 2-methyl-2-propanol (SN1) and 2-methylpropene (E1).

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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 11-Bromopropane, CH₃CH₂CH₂Br, reacts with NaCN in DMSO. What is the mechanism and the major product?

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Primary substrate, so SN1 and E1 are out. Cyanide is a strong nucleophile but a weak base, and DMSO is polar aprotic, which favors SN2. Cyanide attacks the CH₂ that holds bromine and bromide leaves. The product, CH₃CH₂CH₂C≡N, has four carbons including the nitrile carbon, so it is butanenitrile.

Question 22-Bromobutane reacts with sodium methanethiolate, NaSCH₃, in DMF. What is the main pathway?

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Secondary substrates are decided by the reagent. Thiolates (RS⁻) are excellent nucleophiles but weak bases, and DMF is polar aprotic. That combination is the SN2 case. Sulfur bonds to C2 from the back and bromide leaves, giving CH₃CH(SCH₃)CH₂CH₃. If the starting material were a single enantiomer, the product would be inverted.

Question 3tert-Butyl bromide, (CH₃)₃CBr, is treated with sodium methoxide in methanol. What is the major product?

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Tertiary: no SN2. Methoxide is a strong base, and OpenStax's summary for tertiary halides is that a base gives E2. Every β-hydrogen sits on an equivalent methyl group, so there is only one alkene: (CH₃)₂C=CH₂, 2-methylpropene. A little SN1 ether can form, but E2 is the main path.

Question 42-Bromo-2-methylbutane, (CH₃)₂CBrCH₂CH₃, is heated with potassium tert-butoxide in tert-butanol. What is the major product?

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Tertiary substrate plus strong base means E2. The question is which β-hydrogen leaves. Small bases favor the more substituted alkene, 2-methyl-2-butene. Potassium tert-butoxide is bulky, so it reaches the hydrogens on the outer CH₃ groups more easily than the ones on the inside CH₂. That gives more of the less substituted alkene, CH₂=C(CH₃)CH₂CH₃ (2-methyl-1-butene), the Hofmann product.

Question 5(S)-2-Bromopentane reacts with NaCN in DMSO. What is the product?

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Secondary substrate, strong and weakly basic nucleophile, polar aprotic solvent: SN2. Cyanide attacks C2 from the back, so the carbon inverts. The product CH₃CH(CN)CH₂CH₂CH₃ is named from the nitrile carbon (C1), so it is 2-methylpentanenitrile. Priorities before: Br > propyl > methyl > H. After: CN > propyl > methyl > H (the nitrile carbon has three bonds to N). The order is unchanged, so geometric inversion turns S into R.

Question 62-Bromo-3-methylbutane, CH₃CHBrCH(CH₃)₂, is left in warm water with no added base. What is the major substitution product?

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Water is a weak nucleophile and weak base, so the reaction goes through a carbocation (SN1, with some E1). Ionization gives a secondary carbocation at C2. The C3 hydrogen moves over to C2 with its pair of electrons (a 1,2-hydride shift), leaving the positive charge on C3, which is tertiary. Water adds to that carbon and loses a proton, giving (CH₃)₂C(OH)CH₂CH₃, 2-methyl-2-butanol (newer style: 2-methylbutan-2-ol).

Question 7Bromocyclohexane is heated with sodium ethoxide in ethanol. What is the major product?

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Secondary substrate. Ethoxide is a strong base, and OpenStax's summary says E2 predominates for secondary halides with strong bases; heat pushes it further. Removing a hydrogen from either neighboring CH₂ gives the same alkene, cyclohexene. Ethoxycyclohexane (SN2) is a minor product.

Question 82-Chloro-2-methylbutane is warmed in ethanol with no added base. Which description fits best?

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Tertiary substrate, weak nucleophile and weak base, protic solvent: the carbocation route. Ionization gives (CH₃)₂C⁺CH₂CH₃. Ethanol can add to it (SN1, giving (CH₃)₂C(OCH₂CH₃)CH₂CH₃, 2-ethoxy-2-methylbutane) or remove a β-hydrogen (E1). E1 gives mostly the more substituted alkene, (CH₃)₂C=CHCH₃. Warming increases the share of alkene.

Answer key

  1. A, SN2, butanenitrile: primary substrate, good nucleophile, polar aprotic solvent.
  2. C, SN2: secondary carbon, weakly basic good nucleophile, polar aprotic solvent.
  3. B, 2-methylpropene by E2: tertiary substrate plus strong base.
  4. D, 2-methyl-1-butene: E2 with a bulky base gives the Hofmann (less substituted) alkene.
  5. A, (R)-2-methylpentanenitrile: SN2 inversion, same priority order, so S becomes R.
  6. B, 2-methyl-2-butanol: SN1 with a hydride shift to a tertiary carbocation.
  7. C, cyclohexene by E2: secondary substrate plus strong base and heat.
  8. B, SN1 + E1: tertiary, weak base, protic solvent; E1 favors the Zaitsev alkene.

Frequently asked questions

Is SN1, SN2, E1, E2 on the MCAT?

The AAMC content outline for the MCAT includes nucleophilic substitution, and SN1 versus SN2 reasoning shows up in passages about alcohols and other functional groups. Elimination gets less attention than in a college organic course. Check the current AAMC outline for exact wording, but the substrate and nucleophile logic on this page is useful either way.

Does heat always mean elimination?

Heat shifts a competition toward elimination, because elimination makes more molecules and is favored by entropy. It does not override everything. A primary halide with cyanide still gives mostly SN2 when warmed.

Why does a primary halide with hydroxide give SN2, but a secondary one gives E2?

On a primary carbon the backside is open, so substitution is fast. On a secondary carbon the extra alkyl group slows backside attack, while the β-hydrogens are still easy to reach, so a strong base like hydroxide or an alkoxide mostly eliminates.

How do I spot a bulky base?

The classic one is potassium tert-butoxide, (CH₃)₃CO⁻K⁺. Its oxygen is surrounded by three methyl groups, so it reacts as a base and takes the most accessible β-hydrogen. That gives more of the less substituted alkene, called the Hofmann product.

Sources

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