Structural Isomerism in Alkenes

25 min

"Butene" is not enough

A chemical supplier that sells a cylinder labelled only "butene" has not said what is inside. There are three different alkenes with the formula $\mathrm{C_4H_8}$ , and a plastics factory that adds a little but-1-ene to ethene, to make a tougher kind of poly(ethene), cannot use 2-methylpropene instead. In alkanes, isomers come only from branching. Alkenes have a second way to differ: the double bond can sit in different places.

Definition

Position isomers: structural isomers with the same carbon chain but the functional group in a different position. Example: but-1-ene and but-2-ene.

Chain isomers: structural isomers with different carbon chains (branched or unbranched). Example: but-1-ene and 2-methylpropene.

Finding every isomer of $\mathrm{C_5H_{10}}$

Work in order, as for alkanes, but now place the double bond in every possible position on each chain.

Chain of 5. The double bond starts at carbon 1 or carbon 2: pent-1-ene, pent-2-ene. A bond starting at carbon 3 or 4 is the same as 2 or 1 counted from the other end.
Chain of 4 with a $\mathrm{CH_3}$ branch on carbon 2 (counting from the end where the double bond is). The double bond can be between carbons 1 and 2: 2-methylbut-1-ene; or between carbons 2 and 3: 2-methylbut-2-ene.
Chain of 4 with the branch on carbon 3: double bond between 1 and 2 gives 3-methylbut-1-ene. A double bond between 3 and 4 is 2-methylbut-1-ene again, numbered from the other end.
Chain of 3 with two branches on carbon 2: carbon 2 would need four bonds to carbon plus part of a double bond. Impossible.

So there are five alkene isomers of $\mathrm{C_5H_{10}}$ .

Worked example

Problem. State whether each pair are position isomers, chain isomers or the same compound: (a) but-1-ene and but-2-ene; (b) pent-1-ene and 3-methylbut-1-ene; (c) $\mathrm{CH_2{=}CHCH_2CH_3}$ and $\mathrm{CH_3CH_2CH{=}CH_2}$ .

Solution. (a) Same chain of four, double bond moved: position isomers. (b) Five-carbon chain against a branched four-carbon chain, double bond at carbon 1 in both: chain isomers. (c) Number each from the end nearer the double bond: both are but-1-ene, drawn from opposite ends. Same compound.

Worked example

Problem. Correct these names: (a) 3-methylbut-2-ene; (b) 2-methylbut-3-ene; (c) 2-ethylprop-1-ene.

Solution. (a) $\mathrm{CH_3CH{=}C(CH_3)CH_3}$ . Numbered from the other end the double bond still starts at carbon 2, and the branch gets carbon 2: 2-methylbut-2-ene. (b) $\mathrm{CH_3CH(CH_3)CH{=}CH_2}$ . Numbering must start at the end nearer the double bond: 3-methylbut-1-ene. (c) $\mathrm{CH_2{=}C(C_2H_5)CH_3}$ . The longest chain containing the double bond runs through the ethyl group and has four carbons: 2-methylbut-1-ene.

Going further: two forms of but-2-ene

Because a C=C bond cannot rotate, but-2-ene exists in two forms that cannot turn into one another. They have the same structural formula, $\mathrm{CH_3CH{=}CHCH_3}$ , but a different arrangement in space, so they are not structural isomers; they are called cis–trans (or geometric) isomers. Their boiling points differ slightly, about 4 °C for the cis form and 1 °C for the trans form.

Warning

Common mistakes. (1) The double bond gets the lowest number first; only then are branches numbered. (2) "Pent-3-ene" and "pent-4-ene" do not exist as names: they are pent-2-ene and pent-1-ene numbered from the wrong end. (3) When you list isomers, name every structure: two drawings with the same name are one compound counted twice.

Tip

Count isomers in two passes. First draw every carbon skeleton, as for the alkanes. Then slide the double bond along each skeleton and name each result; cross out repeated names.

Mr Ruel Tuition: Chain Isomers and Position Isomers of Alkenes, Isomers of Pentene. Watch how the isomers of pentene are found by moving the double bond and branching the chain, and compare them with Figure 2.2.

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Practice

Define position isomerism and chain isomerism, with one example of each.
Draw displayed formulas of the three alkene isomers of $\mathrm{C_4H_8}$ and name them.
Name: (a) $\mathrm{CH_3CH_2CH{=}CHCH_3}$ ; (b) $\mathrm{CH_2{=}C(CH_3)CH_2CH_3}$ ; (c) $\mathrm{(CH_3)_2C{=}CHCH_3}$ .
Classify each pair as position isomers, chain isomers or the same compound: (a) pent-1-ene and pent-2-ene; (b) but-1-ene and 2-methylpropene; (c) $\mathrm{CH_3CH{=}CHCH_2CH_3}$ and $\mathrm{CH_3CH_2CH{=}CHCH_3}$ .
Correct the names: (a) 1-methylbut-1-ene; (b) 3-methylbut-3-ene.
Explain why there is no alkene called "2,2-dimethylpropene".
Draw the cis and trans forms of but-2-ene. Explain why they cannot change into each other at room temperature.
A student says propene has a position isomer, "prop-2-ene". Is the student right? Explain.

Show answersHide answers

Position: same chain, functional group in another position (but-1-ene, but-2-ene). Chain: different carbon skeleton (but-1-ene, 2-methylpropene).
See Figure 2.1.
(a) pent-2-ene; (b) 2-methylbut-1-ene; (c) 2-methylbut-2-ene.
(a) position isomers; (b) chain isomers; (c) the same compound, pent-2-ene, written from opposite ends.
(a) $\mathrm{CH_3CH{=}CHCH_2CH_3}$ : the methyl lengthens the chain, so it is pent-2-ene. (b) $\mathrm{CH_3CH_2C(CH_3){=}CH_2}$ : numbered from the end nearer the double bond, the branch is on carbon 2: 2-methylbut-1-ene.
Carbon 2 would carry two $\mathrm{CH_3}$ groups, a bond to carbon 3 and a double bond to carbon 1: six bonds. Carbon can only form four.
See Figure 2.3. The C=C double bond cannot rotate, so the groups stay on their own sides.
No. In a chain of three, a double bond between carbons 2 and 3 is between carbons 1 and 2 counted from the other end. It is the same compound, propene.

Sources and credits

Video: Structural Isomerism, Chain Isomers and Position Isomers of Alkenes, Isomers of Pentene, Mr Ruel Tuition, embedded from YouTube.
Written by Ganzaa. The diagrams, worked examples and Practice items were drawn and written by Ganzaa; the boiling points are standard reference values, rounded.

1"Butene" is not enough

2Finding every isomer of C5H10 \mathrm{C_5H_{10}} C5​H10​

3Going further: two forms of but-2-ene

"Butene" is not enough

Finding every isomer of $\mathrm{C_5H_{10}}$

Going further: two forms of but-2-ene