Skip to content
Problems · Example 9.1

Q.Write structures of different chain isomers of alkanes corresponding to the molecular formula C 6H14. Also write their IUPAC names.

CBSENCERTSubjective· 3mImportance★★★★★est
1% · 1/90 Questions
✓ Free question

Chain isomers differ only in the arrangement of the carbon skeleton. For CX6HX14\ce{C6H14}, we can build five distinct carbon frameworks: one straight chain, three with a single branch, and one with two branches. The isomers are n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane.

Understanding Chain Isomerism

Chain isomerism arises when compounds share the same molecular formula but differ in the way their carbon atoms are connected—the skeleton itself. Think of it as rearranging the same set of carbon atoms into different branching patterns. For alkanes, this is the only type of structural isomerism possible because we have no functional groups to move around.

The molecular formula CX6HX14\ce{C6H14} tells us we have six carbons and fourteen hydrogens, satisfying the alkane formula CXnHX2n+2\ce{C_nH_{2n+2}}. Our task is to sketch every possible way to connect six carbon atoms, then name each structure systematically.


Systematic Construction of Isomers

1. Start with the longest possible chain: six carbons in a row.

This gives us the unbranched isomer:

CHX3−CHX2−CHX2−CHX2−CHX2−CHX3\ce{CH3-CH2-CH2-CH2-CH2-CH3}

IUPAC name: n-hexane (or simply hexane)

The prefix n- stands for "normal," indicating a straight chain with no branches.


2. Reduce the main chain to five carbons and attach one methyl branch.

Now we have a pentane backbone with a CHX3\ce{CH3} group attached. Where can we place this methyl group?

  • On carbon 2:

CHX3−CH(CHX3)−CHX2−CHX2−CHX3\ce{CH3-CH(CH3)-CH2-CH2-CH3}

Number from the end that gives the substituent the lowest number. Here, numbering from the left gives the methyl group position 2.

IUPAC name: 2-methylpentane

  • On carbon 3:

CHX3−CHX2−CH(CHX3)−CHX2−CHX3\ce{CH3-CH2-CH(CH3)-CH2-CH3}

Numbering from either end gives position 3 for the methyl group.

IUPAC name: 3-methylpentane

Watch out

Placing the methyl on carbon 1 or carbon 4 doesn't create new isomers—they're identical to 2-methylpentane and 3-methylpentane respectively, just drawn backwards. Always check for symmetry.


3. Reduce the main chain to four carbons and attach two methyl branches.

We now have a butane backbone with two CHX3\ce{CH3} groups. Consider the possible arrangements:

  • Both methyls on carbon 2:

CHX3−C(CHX3)X2−CHX2−CHX3\ce{CH3-C(CH3)2-CH2-CH3}

This creates a quaternary carbon (carbon bonded to four other carbons).

IUPAC name: 2,2-dimethylbutane

  • Methyls on carbons 2 and 3:

CHX3−CH(CHX3)−CH(CHX3)−CHX3\ce{CH3-CH(CH3)-CH(CH3)-CH3}

The two methyl groups are on adjacent carbons.

IUPAC name: 2,3-dimethylbutane

Tip

Placing both methyls on carbon 3 would be identical to 2,2-dimethylbutane (just renumbered). Placing them on carbons 1 and 2, or 1 and 3, would either extend the main chain or duplicate an existing structure.


4. Check for three-carbon chains with three methyl branches.

A propane backbone with three methyls would require:

CHX3−C(CHX3)X3\ce{CH3-C(CH3)3}

But this only accounts for five carbons total—we'd need to attach another carbon somewhere, which would extend the chain back to four carbons. No new isomer emerges here.


Summary Table

StructureIUPAC Name
CHX3−CHX2−CHX2−CHX2−CHX2−CHX3\ce{CH3-CH2-CH2-CH2-CH2-CH3}n-hexane
CHX3−CH(CHX3)−CHX2−CHX2−CHX3\ce{CH3-CH(CH3)-CH2-CH2-CH3}2-methylpentane
CHX3−CHX2−CH(CHX3)−CHX2−CHX3\ce{CH3-CH2-CH(CH3)-CH2-CH3}3-methylpentane
CHX3−C(CHX3)X2−CHX2−CHX3\ce{CH3-C(CH3)2-CH2-CH3}2,2-dimethylbutane
CHX3−CH(CHX3)−CH(CHX3)−CHX3\ce{CH3-CH(CH3)-CH(CH3)-CH3}2,3-dimethylbutane

Each structure represents a unique connectivity of the six carbon atoms. No amount of rotation or flipping will convert one into another—they are true isomers.

✓Final answer

The five chain isomers of CX6HX14\ce{C6H14} are n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane, with structures and names given in the table above.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.