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

Sodium ethoxide is made by dissolving 2.30 g of sodium in an excess of dry ethanol…

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Sodium ethoxide is made by dissolving $2.30\ \text{g}$ of sodium in an excess of dry ethanol. Then $13.7\ \text{g}$ of 1-bromobutane is added and the mixture is warmed. The ether 1-ethoxybutane forms by $S_{\text{N}}2$ , together with sodium bromide. Use the molar masses $23.0\ \text{g/mol}$ for sodium, $137.0\ \text{g/mol}$ for 1-bromobutane, $102.0\ \text{g/mol}$ for 1-ethoxybutane and $103.0\ \text{g/mol}$ for sodium bromide. One mole of sodium gives one mole of ethoxide. The yield of the ether is $65\ \%$ . Find the amount of sodium ethoxide, the theoretical mass of the ether, the mass obtained and the mass of sodium bromide formed.

Need a reminder? The key ideaShow

$S_{\text{N}}2$ is one step with the rate law $\text{rate} = k[\text{RX}][\text{Nu}]$ , and it inverts the shape of the carbon. The speed falls from methyl to primary, secondary and tertiary. Rates of $S_{\text{N}}2$ reactions of different halides are compared through their rate constants. Time for the same conversion is inversely proportional to the rate constant. Yield calculations use the ratio of the balanced equation.

The worked solution

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