Identifying Three C4H9Br Isomers from Kinetic Data and Product Distributions
Three unlabelled bottles A, B and C each contain one isomer of C4H9Br. A student measures initial rates of reaction with hydroxide in aqueous ethanol at 25 °C, keeping the haloalkane concentration fixed at 0.100 mol L⁻¹:
- Bottle A: with 0.100 mol L⁻¹ OH⁻, rate = 2.0×10⁻⁴ mol L⁻¹ s⁻¹; with 0.200 mol L⁻¹ OH⁻, rate = 4.0×10⁻⁴ mol L⁻¹ s⁻¹.
- Bottle B: with 0.100 mol L⁻¹ OH⁻, rate = 2.6×10⁻³ mol L⁻¹ s⁻¹; with 0.200 mol L⁻¹ OH⁻, rate = 2.6×10⁻³ mol L⁻¹ s⁻¹.
- Bottle C: with 0.100 mol L⁻¹ OH⁻, rate = 5.5×10⁻⁵ mol L⁻¹ s⁻¹; with 0.200 mol L⁻¹ OH⁻, rate = 7.2×10⁻⁵ mol L⁻¹ s⁻¹.
Further tests: warming B with KOH in ethanol gives a single alkene; warming A the same way gives a single, different alkene; warming C gives two alkenes. Adding AgNO3 in ethanol gives a pale yellow precipitate with B within seconds, no precipitate with A after 10 min, and a slow precipitate with C.
1. Assign a structure to each of A, B and C, and justify every assignment using the rate data and the alkene products.
Read the worked answer
Read the order in hydroxide first - it identifies the mechanism.
- A: doubling [OH⁻] doubles the rate (4.0×10⁻⁴ / 2.0×10⁻⁴ = 2.0), i.e. rate = k[RBr][OH⁻]. Second-order, SN2. SN2 is fast only for primary halides, so A = 1-bromobutane (butan-1-yl bromide). It is not chiral, and the only alkene it can give is but-1-ene, matching the single alkene observed.
- B: the rate is unchanged when [OH⁻] doubles (2.6×10⁻³ / 2.6×10⁻³ = 1.0), i.e. rate = k[RBr]. First order, SN1 (solvolysis), which needs a stable carbocation - tertiary. So B = 2-bromo-2-methylpropane (tert-butyl bromide). Its only possible alkene is 2-methylpropene, again a single product, and the tertiary carbocation explains the instant AgBr precipitate.
- C: doubling [OH⁻] raises the rate only 1.3-fold, between the two limiting cases - mixed SN1/SN2, characteristic of a secondary halide. So C = 2-bromobutane. Elimination can remove a hydrogen from C1 or C3, giving but-1-ene and but-2-ene, which is the two-alkene result.
The internal consistency is the check: the mechanism order (SN2 fast for primary, SN1 fast for tertiary), the rate orders, the number of alkenes and the AgNO3 test all point to the same three structures.
2. Write the mechanism by which B gives a precipitate with AgNO3 in ethanol, and explain why A gives none. Why is B's solvolysis about ten times faster than C's?
Read the worked answer
B with AgNO3/ethanol (SN1):
Step 1 (slow, rate-determining): (CH3)3C–Br → (CH3)3C⁺ + Br⁻ Step 2 (fast): (CH3)3C⁺ + C2H5OH → (CH3)3C–O(H)C2H5⁺ Step 3 (fast): loss of H⁺ → (CH3)3C–OC2H5 (tert-butyl ethyl ether)
The Br⁻ released is precipitated as AgBr(s), driving the equilibrium and giving the pale yellow solid. The rate depends only on [RBr], exactly as the kinetics show, because the nucleophile enters after the rate-determining step.
Why A gives nothing: ionising a primary C–Br to a primary carbocation costs far too much energy; the primary cation is not a viable intermediate. A only reacts when a nucleophile attacks the carbon directly (SN2), and ethanol is a weak nucleophile and water/ethanol a poor ionising medium, so no Br⁻ is liberated at a detectable rate in 10 min.
B versus C: both ionise by SN1, but B gives a tertiary carbocation stabilised by three electron-donating alkyl groups (plus hyperconjugation), while C gives a secondary carbocation stabilised by only two. The activation energy for ionisation is therefore much lower for B, and the rate constant is larger - here roughly 2.6×10⁻³ / 5.5×10⁻⁵ ≈ 50 at the lower hydroxide concentration, and the ratio grows as the SN1 pathway dominates for B.
3. The student wants to convert C into butan-2-ol, and separately into but-2-ene. Predict the stereochemical outcome of each route, and say which product mixture would be optically active.
Read the worked answer
C is 2-bromobutane, a chiral secondary halide (C2 carries H, CH3, C2H5 and Br), so stereochemistry is the whole point.
Route 1 - butan-2-ol.
- In water/ethanol with no strong nucleophile, the reaction goes by SN1: ionisation gives a planar secondary carbocation, which the nucleophile attacks from either face with equal probability. The product is a 50:50 racemic mixture of (R)- and (S)-butan-2-ol - optically inactive, even though both enantiomers are formed.
- With concentrated NaOH, SN2 competes and becomes dominant; backside attack gives inversion of configuration at the stereocentre, so the product is largely one enantiomer (partially racemised if some SN1 still operates). Here the rate data (partial dependence on [OH⁻]) warn that both pathways run, so expect inversion with a significant racemic component.
Route 2 - but-2-ene. E2 with KOH in ethanol removes a β-hydrogen. Loss of H from C3 gives but-2-ene (Zaitsev, the more substituted and therefore major alkene); loss from C1 gives but-1-ene as the minor product. But-2-ene is formed as cis/trans (E/Z) isomers, not enantiomers, and neither the alkene nor the cis/trans mixture is optically active.
Judgement: before answering, check for a stereocentre. Racemisation (SN1) destroys optical activity; inversion (SN2) does not fully destroy it but reduces it; an E2 product has no stereocentre at all. Reporting simply 'butan-2-ol is formed' without noting the racemate would miss the point of using a chiral substrate.