Counting structural isomers

Counting is a method, not a guess

Questions in this topic ask "how many". How many isomers has this formula? How many of them are alcohols? How many are optically active? A student who lists structures at random misses some and counts others twice. A student with a method gets the number every time.

This lesson gives the method for structural isomers, which differ in how the atoms are joined. The next lesson counts stereoisomers. The third ranks ions and acids by stability.

Step 1: the degree of unsaturation

Before drawing anything, find how many rings and multiple bonds the formula allows. The degree of unsaturation, DoU, is

where $C$, $H$, $N$ and $X$ count the carbon, hydrogen, nitrogen and halogen atoms. Oxygen and sulfur do not change it.

Each ring adds $1$. Each double bond adds $1$. Each triple bond adds $2$.

Worked example. Find the DoU of $\text{C}_4\text{H}_8$ and of $\text{C}_6\text{H}_5\text{Cl}$.

1. First formula. $\dfrac{8 + 2 - 8}{2} = 1$. There is one ring or one double bond.
2. Second formula. $\dfrac{12 + 2 - 5 - 1}{2} = 4$. A benzene ring has one ring and three double bonds, which makes $4$.

Step 2: alkanes, by their skeleton

An alkane has $\text{DoU} = 0$, so only the carbon skeleton can change. Two drawings are the same molecule if one can be turned into the other without breaking a bond. A chain drawn bent is the same chain drawn straight.

Worked example. Count the isomers of $\text{C}_6\text{H}_{14}$.

1. Longest chain first. Six carbons in a row give hexane.
2. Shorten the chain to five. The sixth carbon is a methyl group. On carbon $1$ it only makes the chain six long again. On carbons $2$ and $3$ it gives 2-methylpentane and 3-methylpentane. Carbons $4$ and $5$ repeat these, because the chain can be read from the other end.
3. Shorten the chain to four. Two methyl groups go on carbons $2$ and $3$. Both on carbon $2$ give 2,2-dimethylbutane. One on each gives 2,3-dimethylbutane. Both on carbon $3$ repeat 2,2.
4. Stop. A shorter chain cannot hold six carbons. The total is $5$.

Two rules save time. A methyl group never goes on the end carbon, and an ethyl group never goes on carbon $2$. Each of these only lengthens the chain. For seven carbons, the first ethyl group that is allowed is on carbon $3$ of a chain of five.

The counts for $1$ to $10$ carbons are $1, 1, 1, 2, 3, 5, 9, 18, 35, 75$. You do not need to memorise past hexane, but you should know $2, 3, 5, 9$ for four, five, six and seven carbons. The count grows faster than any steady multiple.

Step 3: one group on an alkane, the fast count

Replace one hydrogen of an alkane by a group $\text{X}$, such as $-\text{Cl}$, $-\text{OH}$, $-\text{NH}_2$ or $-\text{CHO}$. The number of products is the number of different kinds of carbon that carry hydrogen, added up over all the alkane isomers.

For $\text{C}_5\text{H}_{12}$: pentane has $3$ kinds, 2-methylbutane has $4$, and 2,2-dimethylpropane has $1$. Together that is $8$ alkyl groups of formula $\text{C}_5\text{H}_{11}$.

This table counts a lot of things at once:

• A halide, alcohol or primary amine $\text{C}_n\text{H}_{2n+1}\text{X}$ has as many isomers as there are groups with $n$ carbons. So $\text{C}_4\text{H}_9\text{Cl}$ has $4$.
• An aldehyde or acid $\text{R-CHO}$ or $\text{R-COOH}$ has $\text{R}$ with one carbon fewer than the formula. So $\text{C}_5\text{H}_{10}\text{O}$ has $4$ aldehydes.
• An ether $\text{R-O-R}'$ splits the carbons in two. For $\text{C}_4\text{H}_{10}\text{O}$, a methyl and a propyl give $1 \times 2 = 2$ ethers, and two ethyl groups give $1$.

Worked example. Count the isomers of $\text{C}_4\text{H}_{10}\text{O}$ that are alcohols or ethers.

1. DoU. $\dfrac{8 + 2 - 10}{2} = 0$, so no ring and no double bond.
2. Alcohols. Four butyl groups give $4$: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol.
3. Ethers. Methoxypropane, 2-methoxypropane and ethoxyethane give $3$.
4. Total. $7$ structural isomers. Butan-2-ol also has two mirror-image forms, so $8$ if those count.

Esters, acids and carbonyl compounds

An ester $\text{R-COO-R}'$ needs the carbons of $\text{R}$, the one carbonyl carbon and the carbons of $\text{R}'$ to add up to the formula. For formate esters, $\text{R}$ is just $\text{H}$.

Worked example. Count the acids and esters of $\text{C}_4\text{H}_8\text{O}_2$.

1. DoU. $\dfrac{8 + 2 - 8}{2} = 1$. This is one $\text{C=O}$ and no ring.
2. Acids. $\text{R-COOH}$ with three carbons in $\text{R}$ and the carboxyl carbon. There are $2$ propyl groups, so $2$ acids: butanoic acid and 2-methylpropanoic acid.
3. Esters. Carbons in $\text{R}$ plus carbons in $\text{R}'$ must be $3$. With $\text{R} = \text{H}$ and a propyl group, there are $2$ esters. With methyl and ethyl, ethyl ethanoate. With ethyl and methyl, methyl propanoate. That is $4$.
4. Total. $2 + 4 = 6$.

Rings and benzene

A ring uses one DoU, so $\text{C}_4\text{H}_8$ has more members than the alkenes alone.

Worked example. Count the structural isomers of $\text{C}_4\text{H}_8$.

1. Alkenes. But-1-ene, but-2-ene and 2-methylpropene give $3$.
2. Rings. Cyclobutane and methylcyclopropane give $2$.
3. Total. $5$. But-2-ene has a cis form and a trans form, so the count is $6$ if stereoisomers are included.

On a benzene ring, count positions around the ring:

Try it in Python

A computer can count alkane skeletons without any tricks. The program builds each skeleton by adding one carbon to the one before. It then writes each as a nested pattern of brackets, so two drawings of the same molecule give the same pattern.

Common mistakes

• Counting one molecule twice. A bent chain and a straight chain are the same. Name each structure and compare the names.
• Putting a methyl on the end carbon. It only makes a longer chain, which you already counted.
• Forgetting functional group isomers. A formula with oxygen can be an alcohol, an ether, a carbonyl compound or an ester.
• Forgetting rings. Whenever the DoU is $1$ or more, ask whether a ring fits.
• Adding stereoisomers to a structural count. Read whether the question wants structural isomers only.

Remember these

• $\text{DoU} = \dfrac{2C + 2 + N - H - X}{2}$. A ring or a double bond is $1$. A triple bond is $2$.
• Alkane isomers for $4, 5, 6, 7$ carbons: $2, 3, 5, 9$.
• Alkyl groups with $1$ to $5$ carbons: $1, 1, 2, 4, 8$. They count halides, alcohols, amines, aldehydes and acids.
• Benzene with three groups: $3$, $6$ or $10$, by how many are alike.

Check yourself

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

1. What is the degree of unsaturation of $\text{C}_7\text{H}_7\text{NO}_2$?

2. How many aldehydes have the formula $\text{C}_6\text{H}_{12}\text{O}$?

3. How many structural isomers, with or without a ring, have the formula $\text{C}_3\text{H}_5\text{Cl}$?