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A chemist is given a liquid hydrocarbon A with an open chain…

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

A chemist is given a liquid hydrocarbon A with an open chain. Its vapour density is $42$ . When $1.26\ \text{g}$ of A burns completely in oxygen, it gives $3.96\ \text{g}$ of $\text{CO}_2$ and $1.62\ \text{g}$ of $\text{H}_2\text{O}$ . A quickly decolourises a red-brown solution of bromine in $\text{CCl}_4$ . To find where its double bond sits, the chemist treats A with ozone and then with zinc dust and water. The figure shows the two products, drawn as skeletal structures.

A reaction scheme. An open-chain alkene, written only as the letter A, reacts with (i) ozone, O3, and then (ii) zinc dust and water, as written over and under the arrow. The arrow leads to two products drawn as skeletal structures and joined by a plus sign: butanone, a four carbon chain with C=O on its second carbon, and ethanal, a two carbon chain ending in an aldehyde group, CHO.

One mole of A gives one mole of each product. Atomic masses in $\text{g mol}^{-1}$ : H 1, C 12, O 16, Br 80.

How many carbon atoms does one molecule of A have?
How many hydrogen atoms does one molecule of A have?
What mass of butanone does $2.10\ \text{g}$ of A give on ozonolysis?
What mass of ethanal does the same $2.10\ \text{g}$ of A give?
What mass of bromine does $2.10\ \text{g}$ of A decolourise?

Use the molar mass from the vapour density, and the masses of carbon dioxide and water, to fix the formula before you look at the products.

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.

Given. Vapour density $42$ , so $M = 2 \times 42 = 84\ \text{g mol}^{-1}$ . Burning $1.26\ \text{g}$ gives $3.96\ \text{g}$ of $\text{CO}_2$ and $1.62\ \text{g}$ of water.

(a) and (b) The formula. The sample is $\dfrac{1.26\ \text{g}}{84\ \text{g mol}^{-1}} = 0.015\ \text{mol}$ . Carbon: $\dfrac{3.96\ \text{g}}{44\ \text{g mol}^{-1}} = 0.090\ \text{mol}$ of $\text{CO}_2$ , so $0.090\ \text{mol}$ of C. Hydrogen: $\dfrac{1.62\ \text{g}}{18\ \text{g mol}^{-1}} = 0.090\ \text{mol}$ of water, so $0.18\ \text{mol}$ of H. Per mole of A: $\dfrac{0.090}{0.015} = 6$ carbon atoms and $\dfrac{0.18}{0.015} = 12$ hydrogen atoms. A is $\text{C}_6\text{H}_{12}$ . With an open chain, this is an alkene with one double bond, which fits the bromine test.

Which alkene? Write butanone and ethanal with their oxygen atoms facing each other: $\text{CH}_3\text{CH}_2\text{C(CH}_3)=\text{O}$ and $\text{O}=\text{CHCH}_3$ . Remove the oxygen atoms and join the two carbons. A is $\text{CH}_3\text{CH}_2\text{C(CH}_3)=\text{CHCH}_3$ , 3-methylpent-2-ene.

(c) Butanone. $n(\text{A}) = \dfrac{2.10\ \text{g}}{84\ \text{g mol}^{-1}} = 0.0250\ \text{mol}$ . Butanone, $M = 72\ \text{g mol}^{-1}$ : $0.0250\ \text{mol} \times 72\ \text{g mol}^{-1} = 1.80\ \text{g}$ .

(d) Ethanal. $M = 44\ \text{g mol}^{-1}$ : $0.0250\ \text{mol} \times 44\ \text{g mol}^{-1} = 1.10\ \text{g}$ .

(e) Bromine. One $\text{Br}_2$ adds to the one double bond: $0.0250\ \text{mol} \times 160\ \text{g mol}^{-1} = 4.00\ \text{g}$ .

Answer: (a) $6$ . (b) $12$ . (c) $1.80\ \text{g}$ . (d) $1.10\ \text{g}$ . (e) $4.00\ \text{g}$ .

Check. A mass balance. Ozonolysis puts one oxygen atom on each carbon of the double bond. So $0.0250\ \text{mol}$ of A gains $0.80\ \text{g}$ of oxygen. Then $2.10\ \text{g} + 0.80\ \text{g} = 2.90\ \text{g}$ . The products: $1.80\ \text{g} + 1.10\ \text{g} = 2.90\ \text{g}$ . The carbon atoms agree too: $4 + 2 = 6$ .

Watch out. Cyclohexane is also $\text{C}_6\text{H}_{12}$ , but it does not decolourise bromine and gives no ozonide. And A has cis and trans forms: both give the same two products, so ozonolysis cannot tell them apart.

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