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In the laboratory, ethyne is made by adding water to calcium carbide…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

In the laboratory, ethyne is made by adding water to calcium carbide. The figure shows what a student then does with the gas. The letters A to E stand for compounds you must work out.

A reaction scheme in two rows. Row 1: calcium carbide, CaC2, with water written over the arrow, gives A. A, with the words red hot iron tube, 873 K over and under the arrow, gives B. B, with CH3Cl over the arrow and anhydrous AlCl3 under it, gives C. Row 2: A, with H2 and palladium on charcoal over the arrow and poisoned with quinoline under it, gives D. D, with Br2 in CCl4 over the arrow, gives E. The letters A to E stand for unknown compounds; no structures are drawn for them.

A sample of commercial calcium carbide has a mass of $20.0\ \text{g}$ and contains $80.0\%$ of $\text{CaC}_2$ by mass. The rest of the sample does not react with water. The gas A is split into two equal parts. The first part is passed through the red hot iron tube, and only $36.0\%$ of it reacts there to give B. All of this B is then treated with chloromethane and anhydrous $\text{AlCl}_3$ , and the monosubstituted product C forms with a yield of $80.0\%$ . The second part of A takes up exactly the hydrogen it needs over the poisoned palladium catalyst and gives D. All of D is passed into bromine in $\text{CCl}_4$ and gives E. Take the molar volume at STP as $22.4\ \text{L mol}^{-1}$ . Atomic masses in $\text{g mol}^{-1}$ : H 1, C 12, Ca 40, Br 80.

What volume of A at STP does the whole sample give?
What mass of B forms?
What mass of C forms?
What volume of $\text{H}_2$ at STP does the second part of A take up?
What mass of E forms?

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.

Given. $20.0\ \text{g}$ of sample with $80.0\%$ $\text{CaC}_2$ , $M(\text{CaC}_2) = 64\ \text{g mol}^{-1}$ . Molar volume $22.4\ \text{L mol}^{-1}$ at STP.

Plan. Name each letter first. A is ethyne, B is benzene, C is methylbenzene (toluene), D is ethene and E is 1,2-dibromoethane. Then follow the moles along each branch.

(a) The gas A. The pure carbide has mass $20.0\ \text{g} \times 0.800 = 16.0\ \text{g}$ . So $n(\text{CaC}_2) = \dfrac{16.0\ \text{g}}{64\ \text{g mol}^{-1}}$ , which is $0.250\ \text{mol}$ . Since $\text{CaC}_2 + 2\text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{C}_2\text{H}_2$ , A is ethyne, $0.250\ \text{mol}$ . Its volume is $0.250\ \text{mol} \times 22.4\ \text{L mol}^{-1} = 5.60\ \text{L}$ .

(b) Benzene. Each part holds $0.1250\ \text{mol}$ of ethyne. In the red hot iron tube, $3\text{C}_2\text{H}_2 \rightarrow \text{C}_6\text{H}_6$ . The ethyne that reacts is $0.1250\ \text{mol} \times 0.360 = 0.045\ \text{mol}$ . It gives $\dfrac{0.045\ \text{mol}}{3} = 0.0150\ \text{mol}$ of benzene. Its mass is $0.0150\ \text{mol} \times 78\ \text{g mol}^{-1} = 1.17\ \text{g}$ .

(c) Toluene. Friedel-Crafts alkylation puts a methyl group on the ring: $\text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \rightarrow \text{C}_6\text{H}_5\text{CH}_3 + \text{HCl}$ . With an $80.0\%$ yield, the toluene is $0.0150\ \text{mol} \times 0.800 = 0.0120\ \text{mol}$ . Its mass is $0.0120\ \text{mol} \times 92\ \text{g mol}^{-1} = 1.10\ \text{g}$ .

(d) Hydrogen. Lindlar's catalyst adds one mole of $\text{H}_2$ per mole of ethyne and stops at ethene. So $n(\text{H}_2) = 0.1250\ \text{mol}$ , and the volume is $0.1250\ \text{mol} \times 22.4\ \text{L mol}^{-1} = 2.80\ \text{L}$ .

(e) 1,2-Dibromoethane. $\text{CH}_2=\text{CH}_2 + \text{Br}_2 \rightarrow \text{BrCH}_2\text{CH}_2\text{Br}$ , with $M = 188\ \text{g mol}^{-1}$ . Mass of E $= 0.1250\ \text{mol} \times 188\ \text{g mol}^{-1} = 23.5\ \text{g}$ .

Answer: (a) $5.60\ \text{L}$ . (b) $1.17\ \text{g}$ . (c) $1.10\ \text{g}$ . (d) $2.80\ \text{L}$ . (e) $23.5\ \text{g}$ .

Check. Follow the carbon. The pure carbide holds $16.0\ \text{g} \times \tfrac{24}{64} = 6.00\ \text{g}$ of carbon. Half of it, $3.00\ \text{g}$ , goes each way. Benzene is $\tfrac{72}{78}$ carbon by mass, so $3.00\ \text{g} \times 0.360 \times \tfrac{78}{72} = 1.17\ \text{g}$ . Each $\text{C}_2$ unit, $24\ \text{g}$ , ends in $188\ \text{g}$ of E, so $3.00\ \text{g} \times \tfrac{188}{24} = 23.5\ \text{g}$ . Both agree.

Watch out. If you take D as ethane, the hydrogen doubles to $5.60\ \text{L}$ , and ethane would not take up any bromine. The poisoned catalyst is there to stop at the alkene. Using the whole $20.0\ \text{g}$ as carbide gives $7.00\ \text{L}$ in (a).

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