A student warms 9.25 g of 1-chlorobutane with 22.5 g of sodium iodide in dry acetone…
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A student warms of 1-chlorobutane with of sodium iodide in dry acetone. A white solid separates. Take the molar masses as for 1-chlorobutane, for sodium iodide, for sodium chloride and for 1-iodobutane. Assume the reaction goes to completion. Find the mass of the white solid, the theoretical mass of 1-iodobutane and the amount of sodium iodide left.
Need a reminder? The key ideaShow
In free radical halogenation the share of a product is proportional to the number of hydrogen atoms times their reactivity. Bond enthalpies give the energy change of a step: bonds broken minus bonds formed. Hydrogen halides add to alkenes by Markovnikov's rule, and an alcohol with NaBr and sulphuric acid gives a bromoalkane.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. This is the Finkelstein reaction. Find the amounts, then use the 1 : 1 ratio.
Step 1. Amounts. 1-Chlorobutane: . Sodium iodide: .
Step 2. Limiting reagent. The 1-chlorobutane limits the reaction. reacts.
Step 3. Sodium chloride. It does not dissolve in acetone. of white solid.
Step 4. 1-Iodobutane. .
Step 5. Sodium iodide left. .
Answer: of sodium chloride, of 1-iodobutane and of sodium iodide left over.
Check. The white solid shows the exchange has gone: sodium chloride is not soluble in acetone, and it leaves the equilibrium.
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