Rate Data and Mechanism for Two Halogenoalkane Hydrolyses
A student investigates the hydrolysis of two halogenoalkanes in aqueous ethanol at 40.0°C by following the release of bromide ions with silver nitrate. The substrates are 1-bromobutane and 2-bromo-2-methylpropane. Initial-rate experiments give: for 1-bromobutane, when [RBr] is doubled at constant [OH⁻] the rate doubles, and when [OH⁻] is doubled at constant [RBr] the rate doubles; for 2-bromo-2-methylpropane, doubling [RBr] doubles the rate, but doubling [OH⁻] has no measurable effect. At 40.0°C the tertiary substrate reacts about ten times faster than the primary substrate at the same concentration. The student also finds that 2-bromo-2-methylpropane gives a small amount of an alkene, and that adding more OH⁻ increases the alkene yield only slightly. The student's report states: 'Both reactions are SN2 because hydroxide is a strong nucleophile; the tertiary compound is faster because it has more hydrogen atoms.' The supervisor asks for a mechanism-based correction using the rate data, stereochemistry, and the minor product.
1. Use the rate data to determine the rate law and mechanism for each substrate. Explain the structural factors that account for the difference.
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1-Bromobutane
Doubling [RBr] doubles the rate, and doubling [OH⁻] also doubles the rate. The reaction is first order in each reactant and second order overall:
rate = k[RBr][OH⁻].
This is the signature of a bimolecular SN2 mechanism. The rate-determining step involves both the halogenoalkane and the hydroxide ion. Hydroxide attacks the electrophilic carbon from the side opposite the leaving group in a single concerted step, and the C–Br bond breaks as the C–O bond forms. A primary carbon is relatively unhindered, so backside attack is easy. The transition state contains both species, which is why both concentrations appear in the rate law.
2-Bromo-2-methylpropane
Doubling [RBr] doubles the rate, but doubling [OH⁻] has no effect. The reaction is first order overall:
rate = k[RBr].
This is the signature of an SN1 mechanism. The rate-determining step is the unimolecular heterolysis of the C–Br bond to give a carbocation and Br⁻. Hydroxide is not involved in that step; it attacks the carbocation only after the rate-determining step. The tertiary carbocation is strongly stabilised by three electron-donating alkyl groups, so it forms much more readily than a primary carbocation. The tertiary substrate is also sterically crowded, which hinders SN2 backside attack. These two factors explain why it reacts about ten times faster than 1-bromobutane under the same conditions.
The student's statement that both are SN2 is therefore wrong, and the explanation based on 'more hydrogen atoms' is also wrong. The tertiary compound has fewer hydrogens on the reacting carbon, not more, and the key factor is carbocation stability plus steric hindrance.
2. Predict the products and stereochemistry for each reaction. Explain the minor alkene product and why adding more OH⁻ has only a small effect on it.
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1-Bromobutane with OH⁻
SN2 substitution gives butan-1-ol:
CH₃CH₂CH₂CH₂Br + OH⁻ → CH₃CH₂CH₂CH₂OH + Br⁻.
The reaction proceeds with inversion of configuration at the carbon bearing the leaving group. Butan-1-ol itself is achiral, so the inversion is not observable by optical rotation in this case. If the substrate were a chiral secondary halide, the product would show inversion, which is strong evidence for SN2.
2-Bromo-2-methylpropane
The major substitution product is 2-methylpropan-2-ol:
(CH₃)₃CBr + OH⁻ → (CH₃)₃COH + Br⁻.
The mechanism is SN1: ionisation gives the planar tertiary carbocation (CH₃)₃C⁺, which is then attacked by water or hydroxide. Because the carbocation is planar, attack can occur from either face. Here the product is achiral, but for a chiral tertiary substrate the substitution product would be racemic.
The minor alkene is 2-methylpropene, formed by E1 elimination:
(CH₃)₃C⁺ → CH₂=C(CH₃)₂ + H⁺.
A base removes a β-hydrogen from the carbocation. Hydroxide can act as that base, but the rate-determining step is still the formation of the carbocation. Adding more OH⁻ therefore has only a small effect on the overall rate and only a modest effect on the alkene yield. The alkene yield is increased more effectively by heating, because elimination has a higher activation energy than substitution, or by using a stronger base in a less polar solvent.
3. The supervisor suggests increasing the water content of the aqueous ethanol solvent. Predict the effect on each reaction rate and explain your reasoning in terms of solvation and mechanism.
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Increasing the water content makes the solvent more polar and more strongly hydrogen-bonding (more protic). The two mechanisms respond differently.
SN2 of 1-bromobutane: the rate is likely to decrease. Water is a protic solvent that hydrogen-bonds strongly to the hydroxide nucleophile, forming a bulky solvation shell and reducing its nucleophilicity. The SN2 transition state also develops charge separation, but the dominant effect for a small, strongly basic nucleophile such as OH⁻ is nucleophile solvation. A polar aprotic solvent would speed up SN2 by leaving the nucleophile relatively unsolvated. There may also be a solubility problem: 1-bromobutane is poorly soluble in water, so adding too much water could make the organic substrate less available to the aqueous hydroxide.
SN1 of 2-bromo-2-methylpropane: the rate is likely to increase. The rate-determining step is ionisation to a carbocation and Br⁻. A more polar protic solvent stabilises the developing carbocation and the separating bromide ion through solvation, lowering the activation energy for ionisation. Water is particularly good at solvating ions. This is the classic solvent effect that supports an SN1 mechanism.
The student could test this by measuring initial rates at several water:ethanol ratios. If the primary reaction slows while the tertiary reaction speeds up, that is further evidence for SN2 versus SN1. The result would also explain why the practical uses aqueous ethanol rather than pure water: ethanol helps dissolve the halogenoalkanes, while water provides the ionising medium needed for the tertiary substrate.