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Halogenoalkanes: Question 4

Syllabus 15.1

Multiple choice AS 1 mark

Four bromoalkanes are:

A: 1-bromobutane, CH3CH2CH2CH2Br\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} B: 2-bromobutane, CH3CH2CH(Br)CH3\text{CH}_3\text{CH}_2\text{CH}(\text{Br})\text{CH}_3 C: 2-bromo-2-methylpropane, (CH3)3CBr(\text{CH}_3)_3\text{CBr} D: 1-bromo-2-methylpropane, (CH3)2CHCH2Br(\text{CH}_3)_2\text{CHCH}_2\text{Br}

Which bromoalkane is classified as tertiary, and therefore reacts with aqueous sodium hydroxide almost entirely by the SN1 mechanism?

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Worked solution

Classifying by the carbon bonded to the halogen

A halogenoalkane is classified as primary, secondary or tertiary according to how many other carbon atoms are directly bonded to the carbon that carries the halogen, not by how branched the rest of the molecule looks.

  • A, 1-bromobutane: the C–Br carbon (C1) is bonded to just one other carbon (C2). Primary.
  • B, 2-bromobutane: the C–Br carbon (C2) is bonded to two other carbons (C1 and C3). Secondary.
  • C, 2-bromo-2-methylpropane: the C–Br carbon is bonded to three other carbons (three methyl groups). Tertiary.
  • D, 1-bromo-2-methylpropane: the C–Br carbon (C1) is bonded to only one other carbon (C2), even though C2 itself carries two methyl branches. The branching is on the neighbouring carbon, not the one bearing the halogen, so D is primary, not tertiary.

Linking classification to mechanism

Compound C’s C–Br carbon has three alkyl groups attached. These alkyl groups are electron-donating by the inductive effect, so if the C–Br bond breaks heterolytically the resulting carbocation is a relatively stable tertiary carbocation. This carbocation is stable enough to form as a genuine (if short-lived) intermediate, so compound C reacts with OH−\text{OH}^- almost entirely by the two-step SN1 mechanism: the C–Br bond breaks first (rate-determining), then OH−\text{OH}^- attacks the resulting carbocation.

By contrast, a primary carbon (as in A and D) cannot support a stable-enough carbocation, so these react instead by the single-step, concerted SN2 mechanism, in which OH−\text{OH}^- attacks the C–Br carbon directly as Br−\text{Br}^- leaves.

Why the other options are wrong

  • A is correctly described as primary/SN2, but the question asks for the tertiary compound, so A is not the answer.
  • B is correctly described as secondary, reacting by a mixture of mechanisms, but it is not tertiary.
  • D wrongly classifies the compound based on the overall shape of the molecule rather than on the carbon actually bonded to bromine, which has only one carbon neighbour.

Final answer

C. 2-bromo-2-methylpropane is tertiary (the C–Br carbon is bonded to three other carbons) and, because it can form a relatively stable tertiary carbocation, it reacts with aqueous NaOH almost entirely by the SN1 mechanism.