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Chemists save time by writing bond-line formulas…

Contributed by StemPy Lab

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

Chemists save time by writing bond-line formulas. In such a formula, a carbon atom sits at every corner and at the end of every line that has no other symbol. Hydrogen atoms on carbon are not drawn. Each carbon carries as many hydrogen atoms as it needs to make four bonds. A single line is a single bond, a double line a double bond and a triple line a triple bond. Other atoms, such as oxygen, are written with their symbols.

The figure shows the bond-line formula of a compound.

A bond-line formula of an open-chain compound joined to a ring. From the left: a carbon-carbon triple bond at the end of the chain, then a carbon that carries a methyl branch, then a carbon-carbon double bond, then a carbon with a double bond to an oxygen atom, O. This carbonyl carbon is joined to one corner of a four membered ring of carbon atoms. Hydrogen atoms are not drawn.

Put back every hydrogen atom first. Look carefully at the ring, the carbon with the methyl branch and the end of the triple bond. Atomic masses in $\text{g mol}^{-1}$ : H 1, C 12, O 16.

Write the molecular formula of the compound. What is its molar mass?
How many σ bonds does one molecule have? Count the $\text{C}-\text{H}$ bonds too.
How many π bonds does it have?
How many of its carbon atoms are $sp$ hybridised?
How many are $sp^2$ hybridised?
How many are $sp^3$ hybridised?

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. Name the carbons C-1 to C-11: C-1 and C-2 make the triple bond, C-3 carries the methyl carbon C-4, C-5 and C-6 make the double bond, C-7 is the carbonyl carbon, and C-8 to C-11 form the ring.

Step 1. Hydrogen atoms. C-1 has one H (the end of the triple bond). C-2 has none. C-3 has one, the methyl C-4 has three. C-5 and C-6 have one each. C-7 has none. The ring corner C-8 has one, and C-9, C-10 and C-11 have two each. In all $1 + 1 + 3 + 1 + 1 + 1 + 6 = 14$ hydrogen atoms. The formula is $\text{C}_{11}\text{H}_{14}\text{O}$ , so $M = 11(12) + 14(1) + 16$ $= 162\ \text{g mol}^{-1}$ .

Step 2. σ bonds. Each bond has exactly one σ bond. There are $14$ $\text{C}-\text{H}$ bonds, $11$ carbon-carbon bonds (seven in the chain and four in the ring) and one carbon-oxygen bond. So there are $14 + 11 + 1 = 26$ σ bonds.

Step 3. π bonds. The triple bond has two, the $\text{C}=\text{C}$ bond one and the $\text{C}=\text{O}$ bond one: $4$ π bonds.

Step 4. Hybridisation. C-1 and C-2 each carry two π bonds: $2$ $sp$ carbons. C-5, C-6 and C-7 each carry one: $3$ $sp^2$ carbons. C-3, C-4 and the four ring carbons carry none: $6$ $sp^3$ carbons.

Answer: (a) $\text{C}_{11}\text{H}_{14}\text{O}$ , $162\ \text{g mol}^{-1}$ . (b) $26$ . (c) $4$ . (d) $2$ . (e) $3$ . (f) $6$ .

Check. A molecule with $A$ atoms and one ring has $A - 1 + 1 = A$ σ bonds. Here $A = 11 + 14 + 1 = 26$ , which matches Step 2. Also $\tfrac{1}{2}(2 \times 11 + 2 - 14) = 5$ , which is one ring plus four π bonds.

Watch out. The carbonyl carbon C-7 is $sp^2$ even though it has no hydrogen, and the ring corner C-8 is $sp^3$ even though it sits next to the $\text{C}=\text{O}$ group. Only the π bonds a carbon itself takes part in decide its hybridisation.

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