Question
Draw all possible structural isomers of the molecular formula C₄H₁₀. Name each isomer using IUPAC rules.
Solution — Step by Step
For C₄H₁₀, the degree of unsaturation = (2×4 + 2 − 10)/2 = 0. Zero unsaturation means no double bonds, no triple bonds, no rings — we’re looking at pure alkane chain isomers only. This tells us exactly what kind of structures to expect before we draw a single line.
Place all 4 carbons in a straight chain. This gives us n-butane:
CH₃ — CH₂ — CH₂ — CH₃IUPAC name: butane. Count the hydrogens: the two end carbons each get 3H, the two middle carbons each get 2H. Total = 3+2+2+3 = 10. Matches C₄H₁₀. ✓
Take a 3-carbon backbone (propane) and attach the remaining 1 carbon as a methyl branch. The branch can only go on C-2 (the middle carbon) — putting it on C-1 or C-3 just gives you the straight chain again.
CH₃
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CH₃—CH—CH₃IUPAC name: 2-methylpropane (common name: isobutane). Count hydrogens: one central CH with 1H, two −CH₃ groups with 3H each, and one more −CH₃ branch. Total = 1 + 3 + 3 + 3 = 10. ✓
A 2-carbon chain means 2 remaining carbons both need to branch off ethane. That forces a C with 4 carbon attachments, which is fine, but when you draw it fully: it becomes 2-methylpropane again just written differently. No new isomer here. We’re done.
Total structural isomers of C₄H₁₀: 2
- n-Butane (butane)
- 2-Methylpropane (isobutane)
Why This Works
Structural (chain) isomers have the same molecular formula but different carbon skeletons. For C₄H₁₀, we systematically reduce the main chain length and check if branching gives a genuinely new arrangement.
The key insight: two structures are the same isomer if you can rotate or flip one to match the other. n-Butane written as CH₃CH₂CH₂CH₃ and reversed CH₃CH₂CH₂CH₃ is still one compound — so we don’t count mirror-written versions separately.
This systematic approach — longest chain first, then reduce by one carbon and branch — guarantees we don’t miss any isomer and don’t double-count. It’s the same method NCERT expects in board exams and JEE tests for larger formulas like C₅H₁₂ (3 isomers) and C₆H₁₄ (5 isomers).
Alternative Method
Use the branch counting approach instead of chain-length reduction.
Ask: how many carbons are in branches off the main chain?
- 0 branch carbons → all 4 in chain → n-butane
- 1 branch carbon → main chain has 3C, 1C branches off → 2-methylpropane
Can we have 2 branch carbons? That would mean a 2-carbon chain with two 1-carbon branches — draw it out and it collapses to 2-methylpropane written differently. So again, just 2 isomers.
For C₅H₁₂ (the next alkane), use the same method: 5-carbon chain (pentane), 4-carbon chain with 1-methyl branch (2-methylbutane), and 3-carbon chain with two methyl branches (2,2-dimethylpropane). Always work top-down from the longest chain.
Common Mistake
Students often draw n-butane in a bent or angled form and count it as a different isomer from the straight-chain form. A zigzag drawing of n-butane is the same compound — the carbon backbone is still an unbranched 4-carbon chain. Structural isomers differ in connectivity (which carbons are bonded to which), not in 3D shape or how you draw the chain on paper.
Also watch out for naming: 2-methylpropane is correct, not 1-methylpropane. If the branch is on C-1 of a 3-carbon chain, that carbon becomes part of the main chain, making it a 4-carbon chain (butane) — IUPAC rules force you to choose the longest chain as the parent.