Daniel mixes 0.200 mol of bromobenzene and 0.200 mol of 1-bromobutane in dry ether and adds excess…
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Daniel mixes of bromobenzene and of 1-bromobutane in dry ether and adds excess sodium. Each coupling joins two halide molecules and uses two sodium atoms, one for each halide molecule. Assume that the two halides couple with equal ease, and that every molecule couples. Use the molar masses for biphenyl, for butylbenzene, for octane and for sodium. Find the mass of the wanted product, butylbenzene, the mass of biphenyl, the mass of octane and the mass of sodium used.
Need a reminder? The key ideaShow
A halogen deactivates the ring but directs to the ortho and para positions. There are two ortho positions, two meta positions and one para position, so the reactivity of one position is its share divided by the number of positions of that kind. The Wurtz-Fittig and Fittig reactions couple halides at random with sodium. A Grignard reagent takes a proton from any O-H or N-H bond and gives an alkane.
The worked solution
Have you tried it?
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. Count the couplings. Then split them into the three kinds by chance.
Step 1. Couplings. of couplings.
Step 2. Shares. Half of the halide is bromobenzene. A random pair is aryl-aryl with chance , aryl-alkyl with chance and alkyl-alkyl with chance .
Step 3. Amounts. Biphenyl: . Butylbenzene: . Octane: .
Step 4. Masses. Biphenyl: . Butylbenzene: . Octane: .
Step 5. Sodium. One sodium atom for each halide molecule: .
Answer: of butylbenzene, of biphenyl, of octane and of sodium.
Check. Even in the best case only half of the coupled molecules are the wanted product, so the Wurtz-Fittig reaction is a poor way to make a pure compound.
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