Question
Draw all the structural isomers of C₄H₁₀O. Classify each isomer by functional group.
Solution — Step by Step
The formula C₄H₁₀O has a degree of unsaturation (DoU) = .
Zero DoU means no double bonds, no triple bonds, and no rings. All isomers will be saturated (single bonds only). The oxygen with zero DoU means it’s either an alcohol (–OH) or an ether (–O–).
The carbon skeleton can be a 4-carbon chain (butyl) or a branched 3-carbon chain (with a methyl branch). Place –OH on each distinct carbon position:
n-Butanol family (straight chain):
- 1-Butanol: CH₃–CH₂–CH₂–CH₂–OH
- 2-Butanol: CH₃–CH₂–CH(OH)–CH₃
Branched chain (2-methylpropan-1-ol family):
- 2-Methylpropan-1-ol (isobutanol): (CH₃)₂CH–CH₂–OH
- 2-Methylpropan-2-ol (tert-butanol): (CH₃)₃C–OH
Ethers have the oxygen between two carbon groups. The two groups on either side of O must account for all 4 carbons:
- Diethyl ether: CH₃–CH₂–O–CH₂–CH₃ (two ethyl groups)
- Methyl propyl ether: CH₃–O–CH₂–CH₂–CH₃ (methyl + n-propyl)
- Methyl isopropyl ether: CH₃–O–CH(CH₃)₂ (methyl + isopropyl)
| Isomer | IUPAC Name | Class |
|---|---|---|
| CH₃CH₂CH₂CH₂OH | Butan-1-ol | Primary alcohol |
| CH₃CH₂CH(OH)CH₃ | Butan-2-ol | Secondary alcohol |
| (CH₃)₂CHCH₂OH | 2-Methylpropan-1-ol | Primary alcohol |
| (CH₃)₃COH | 2-Methylpropan-2-ol | Tertiary alcohol |
| CH₃CH₂OCH₂CH₃ | Ethoxyethane | Ether |
| CH₃OCH₂CH₂CH₃ | 1-Methoxypropane | Ether |
| CH₃OCH(CH₃)₂ | 2-Methoxypropane | Ether |
Total: 4 alcohols + 3 ethers = 7 structural isomers of C₄H₁₀O.
Why This Works
Structural isomers share the same molecular formula but differ in the connectivity of atoms. For C₄H₁₀O, the oxygen can be bonded to one carbon (alcohol –OH) or inserted between two carbons (ether –O–). Within alcohols, we vary the carbon skeleton and the position of –OH. Within ethers, we vary how we split the 4 carbons across the two sides of oxygen.
The systematic approach is: (1) identify possible functional groups, (2) enumerate all carbon skeletons, (3) place the heteroatom at each distinct position.
JEE and CBSE Class 12 questions frequently ask you to “draw all isomers” of a given formula. The trick is to be systematic — work through functional groups one at a time, then carbon skeletons, then positional changes. If you jump around, you’ll miss isomers or duplicate them.
Alternative Method
Use the degree of unsaturation as a filter first. DoU = 0 rules out aldehydes, ketones, carboxylic acids, and any other oxygen-containing group with a C=O bond. That immediately leaves only alcohols and ethers — saving time in an exam.
Common Mistake
A very common error is to miss one of the ether isomers (usually methyl isopropyl ether) or to count n-propyl and isopropyl ethers with methyl as the same structure. Always verify by checking the carbon connectivity — CH₃–O–CH₂CH₂CH₃ and CH₃–O–CH(CH₃)₂ are different structures even though both have methyl + propyl carbons.