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Halogenoalkanes: Question 5

Syllabus 15.1

Structured AS 8 marks

2-Bromo-2-methylpropane, (CH3)3CBr(\text{CH}_3)_3\text{CBr}, is heated under reflux with aqueous sodium hydroxide.

(a) Give the structural formula and name of the organic product. [1]

(b) Describe, in words, the two-step SN1 mechanism by which this reaction proceeds. [4]

(c) Explain, in terms of the inductive effect of alkyl groups, why the intermediate formed in step 1 is stable enough to exist, and why this makes SN1 the favoured mechanism for a tertiary substrate. [2]

(d) State whether the rate of the rate-determining step depends on the concentration of OH−(aq)\text{OH}^-(aq), and explain your answer. [1]

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Worked solution

Part (a): Product

Hydroxide ion substitutes for bromine: (CH3)3CBr+OH−→(CH3)3COH+Br−(\text{CH}_3)_3\text{CBr} + \text{OH}^- \rightarrow (\text{CH}_3)_3\text{COH} + \text{Br}^-

The organic product is 2-methylpropan-2-ol, (CH3)3COH(\text{CH}_3)_3\text{COH}.

Part (b): The SN1 mechanism

Step 1 (slow, rate-determining): the C–Br bond breaks heterolytically, both electrons in the bond go to the more electronegative bromine atom, without any involvement of OH−\text{OH}^-. This forms a planar, positively charged tertiary carbocation, (CH3)3C+(\text{CH}_3)_3\text{C}^+, and a free bromide ion, Br−\text{Br}^-.

Step 2 (fast): the nucleophile, OH−\text{OH}^-, uses a lone pair on its oxygen atom to attack the empty p-orbital of the planar carbocation. Because the carbocation is planar, OH−\text{OH}^- can attack from either face with roughly equal probability, forming the new C–O bond and giving the alcohol product.

Because the mechanism involves one species (the halogenoalkane) in the slow, rate-determining step, it is called SN1 (substitution, nucleophilic, unimolecular).

Part (c): Why the tertiary carbocation is stable enough to form

The carbocation intermediate in step 1 carries three methyl groups directly attached to the positively charged carbon. Alkyl groups are electron-donating by the inductive effect: they push electron density towards the electron-deficient carbon, partially offsetting (delocalising) its positive charge. With three such groups donating electron density, the tertiary carbocation is considerably more stable than a primary or secondary carbocation would be (which have fewer alkyl groups to share the charge).

Because this intermediate is stable enough to exist for a measurable, if brief, lifetime, the two-step SN1 pathway through it becomes energetically favourable, which is why tertiary halogenoalkanes react almost entirely by SN1 rather than SN2 (a bulky tertiary carbon is also too crowded for the direct backside attack that SN2 requires).

Part (d): Rate dependence on [OH⁻]

No, the rate of the rate-determining step does not depend on [OH−(aq)][\text{OH}^-(aq)]. The rate-determining step (step 1) is the ionisation of (CH3)3CBr(\text{CH}_3)_3\text{CBr} into the carbocation and Br−\text{Br}^-, which involves only the halogenoalkane breaking apart, OH−\text{OH}^- plays no role until the fast second step. So the overall rate equation is rate =k[(CH3)3CBr]= k[(\text{CH}_3)_3\text{CBr}], first order overall and independent of [OH−][\text{OH}^-].

Final answers

  • (a) 2-Methylpropan-2-ol, (CH3)3COH(\text{CH}_3)_3\text{COH}
  • (b) Step 1 (slow): heterolytic fission of C–Br gives a tertiary carbocation +Br−+\text{Br}^-. Step 2 (fast): OH−\text{OH}^- attacks the planar carbocation from either face.
  • (c) Three electron-donating alkyl groups delocalise the positive charge, stabilising the tertiary carbocation enough for the SN1 pathway to be favoured.
  • (d) No. The rate-determining step (ionisation) does not involve OH−\text{OH}^-, so rate =k[(CH3)3CBr]= k[(\text{CH}_3)_3\text{CBr}] only.