HardChemistryHigh School (11–12)Organic Compounds with Functional Groups › Haloalkanes and HaloarenesNumerical

A secondary haloalkane is hydrolysed in a solution of sodium hydroxide, and it reacts by SN1 and…

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A secondary haloalkane is hydrolysed in a solution of sodium hydroxide, and it reacts by $S_{\text{N}}1$ and $S_{\text{N}}2$ at the same time. The rate law is $\text{rate} = k_1[\text{RX}] + k_2[\text{RX}][\text{OH}^-]$ . The constants are $k_1 = 3.0 \times 10^{-5}\ \text{s}^{-1}$ and $k_2 = 4.0 \times 10^{-5}\ \text{L/(mol s)}$ (values chosen for this problem). The first term is the $S_{\text{N}}1$ route and the second term is the $S_{\text{N}}2$ route. Find the percentage of the reaction that goes by $S_{\text{N}}2$ when $[\text{OH}^-] = 0.50\ \text{mol/L}$ , and when $[\text{OH}^-] = 0.10\ \text{mol/L}$ . Find also the hydroxide concentration at which the two routes contribute equally.

Need a reminder? The key ideaShow

$S_{\text{N}}1$ is first order, $\text{rate} = k[\text{RX}]$ , with half-life $t_{1/2} = \dfrac{0.693}{k}$ . A planar carbocation gives a racemic or partly racemic product. The enantiomeric excess is the difference of the two shares, and the observed rotation is the specific rotation times tube length times concentration times the excess. A carbocation can rearrange to a more stable one by a hydride shift.

The worked solution

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