Strategy: from a formula to the possible structures

Start with a count

Suppose that you are given the formula $\text{C}_4\text{H}_8$ and asked for its structures. Do not start by drawing. Start by counting.

An alkane with $4$ carbon atoms is $\text{C}_4\text{H}_{10}$. It has the most hydrogen that $4$ carbon atoms can hold. Our formula has $10 - 8 = 2$ hydrogens fewer. Each pair of missing hydrogens means one ring or one double bond.

Two rules make the count work for every formula.

• A halogen atom counts as a hydrogen atom. $\text{C}_2\text{H}_5\text{Cl}$ is counted as $\text{C}_2\text{H}_6$.
• Oxygen does not change the count. $\text{C}_3\text{H}_6\text{O}$ is counted as $\text{C}_3\text{H}_6$.

A count with a half in it, or below zero, means that the formula cannot be a neutral molecule of those atoms. $\text{C}_3\text{H}_9$ would give $-0.5$, so there is no such compound.

Then draw in order

With a count of $0$ the carbon atoms can only form chains and branches. Draw them in this order, so that you do not miss one or draw one twice.

1. The longest chain. All the carbon atoms in a row.
2. One carbon shorter. Put the extra carbon atom as a branch on each different carbon of the chain. The first and the last place give the same shape (turn the chain round), and so do the places that are the same distance from each end.
3. Two carbons shorter. Put a two-carbon branch, or two one-carbon branches, in the places that are different.

Example, $\text{C}_5\text{H}_{12}$.

• Five in a row: pentane.
• Four in a row, with a $\text{CH}_3$ on the second carbon: 2-methylbutane. (On the third carbon is the same shape, turned round.)
• Three in a row, with two $\text{CH}_3$ on the middle carbon: 2,2-dimethylpropane.

That makes $3$, as in the chapter.

Example, $\text{C}_4\text{H}_8$, count $1$. First the double bonds: but-1-ene, but-2-ene and 2-methylpropene, which is $3$. Then the rings: cyclobutane, and a three-carbon ring with a $\text{CH}_3$ on it. That makes $5$ structures in all.

You should find that from seven carbon atoms on, each extra carbon roughly doubles the number of structures.

A clue from a reaction

A formula with oxygen may be an alcohol, an acid, or something else. A clue narrows it. If the compound gives hydrogen when sodium is added, it has an –OH group. If it fizzes with sodium carbonate, it is an acid.

Practise with new numbers

Try it in Python

This program counts the rings and double bonds of some formulas.

Remember these

• Rings and double bonds $= \dfrac{(2\text{C} + 2) - \text{H}}{2}$. A halogen counts as a hydrogen. Oxygen does not count.
• Draw in order: the longest chain first, then a branch on each different carbon, and do not repeat a shape.
• A clue from a reaction tells the functional group.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. How many rings and double bonds does $\text{C}_9\text{H}_{12}$ have in all?

2. Count the rings and double bonds in $\text{C}_3\text{H}_5\text{Br}$.

3. Why is 3-methylbutane not a new isomer of pentane?

4. A compound $\text{C}_2\text{H}_6\text{O}$ gives hydrogen with sodium. Which structure is it?