AQA A-Level Chemistry Paper 3, 2019: Question 33

AQA A-Level Chemistry • Organic Chemistry • Paper 1 / Paper 2 Multiple Choice

This multiple-choice question assesses your ability to:

Apply the mechanism of electrophilic addition of hydrogen halides (HBr) to unsymmetrical alkenes.
Predict the major product using Markovnikov's rule based on relative carbocation stabilities (tertiary > secondary > primary).
Convert between condensed structural formulas and systematic IUPAC nomenclature for branched haloalkanes.

Option D: (CH₃)₂CHCH=CH₂

1 Mark awarded: Correct letter selected.

The starting alkene is 3-methylbut-1-ene:

Electrophile H⁺ adds to carbon-1 to form the more stable secondary carbocation intermediate: (CH₃)₂CH-CH⁺-CH₃ rather than the primary carbocation (CH₃)₂CH-CH₂-CH₂⁺ .
The bromide ion (:Br⁻) attacks the secondary carbocation at carbon-2, forming 2-bromo-3-methylbutane as the major product.

Major vs Minor Products: When H-X adds across an asymmetrical double bond, H adds to the carbon atom already bonded to more hydrogens (Markovnikov's rule).
Carbocation Stability:
3° > 2° > 1° > methyl
Alkyl groups release electron density toward the positive carbon via the positive inductive effect (+I), stabilising the positive charge.
Naming Check: Number the parent chain from the end that gives substituents the lowest locants:
C4H₃-C3H(CH₃)-C2H(Br)-C1H₃ → 2-bromo-3-methylbutane.

A: (CH₃)₂C=CHCH₃ (2-methylbut-2-ene)
H⁺ adds to C3 forming a tertiary carbocation: (CH₃)₂C⁺-CH₂CH₃ .
Major product = 2-bromo-2-methylbutane (Incorrect).
B: CH₃CH₂CH=CHCH₃ (pent-2-ene)
Unbranched chain of 5 carbons. Addition produces a mixture of 2-bromopentane and 3-bromopentane. Contains no methyl branch (Incorrect).
C: CH₃CH₂C(CH₃)=CH₂ (2-methylbut-1-ene)
H⁺ adds to C1 (=CH₂) forming a tertiary carbocation: CH₃CH₂C⁺(CH₃)₂ .
Major product = 2-bromo-2-methylbutane (Incorrect).
D: (CH₃)₂CHCH=CH₂ (3-methylbut-1-ene)
H⁺ adds to C1 (=CH₂) forming a secondary carbocation: (CH₃)₂CH-CH⁺-CH₃ .
Attack by Br⁻ yields 2-bromo-3-methylbutane (Correct!).

Confusing C and D: Students frequently confuse 2-methylbut-1-ene and 3-methylbut-1-ene when drawing them from condensed formulas. Drawing out full display structures on rough paper avoids this mistake.
Selecting Tertiary Precursors: Students often jump to options A or C thinking that a tertiary product is desired. Notice that the target molecule 2-bromo-3-methylbutane has bromine attached to a secondary carbon ( -CH(Br)- ), not a tertiary carbon.
Numbering from the Wrong End: Remember that numbering gives bromine (alphabetically first substituent) priority for numbering when locants are tied (2,3 vs 3,2).

For "Which alkene gives product X?" questions, work backwards from the target product:

Draw the target: (CH₃)₂CH-CH(Br)-CH₃ .
Remove H and Br from adjacent carbons to find potential alkene precursors:
- Removing H from C3 and Br from C2 gives: (CH₃)₂C=CHCH₃ (Alkene A). But electrophilic addition to A would recreate a tertiary carbocation, giving 2-bromo-2-methylbutane as the major product, not our target!
- Removing H from C1 and Br from C2 gives: (CH₃)₂CH-CH=CH₂ (Alkene D). Electrophilic addition to D forms a secondary carbocation as the most stable intermediate, successfully yielding our target as the major product!
This eliminates the trap in less than 30 seconds.

Question