The Numbering Game
What does this tell you about the numbering decision?
CH3-CH2-CH(CH3)-CH2-CH3
Either direction gives the same locant, so 3-methylpentane is unambiguous.
The purpose of numbering is to obtain the lowest appropriate locants. Here both directions produce 3.
2-methylbutane
CH3-CH(CH3)-CH2-CH3
Pentane and 2-methylbutane both have molecular formula C₅H₁₂.
What single structural observation proves that they are structural isomers rather than two drawings of the same molecule?
Their atoms have different connectivity/carbon skeletons.
A student names: CH₃CH₂CH(OH)CH₃ as butan-3-ol.
Diagnose and repair.
Butan-2-ol. Number from the end that gives –OH the lower locant.
Ethane and ethanol both contain two carbon atoms, yet ethanol behaves very differently.
Why is carbon count alone a poor predictor of their properties?
Ethanol contains the –OH functional group, which introduces polarity and strong intermolecular attractions such as hydrogen bonding. Ethane does not.
CH₃–CH(CH₃)–CH₂–CH₂–CH₃
One student chooses a four-carbon parent because it is the longest horizontal section in their drawing.
Name the compound correctly and identify their first mistake.
2-methylpentane. The longest continuous chain contains five carbons. Parent-chain selection depends on connectivity, not how the structure happens to be drawn.
pent-2-ene
CH3-CH=CH-CH2-CH3
A student draws straight-chain butane horizontally and then draws the same four-carbon chain bent into a U shape.
They announce:
“I found both isomers of C₄H₁₀!”
What exactly is wrong?
Both drawings represent butane. Changing the orientation of a drawing does not change atomic connectivity.
A student proposes 2-ethylbutane for a branched alkane.
Before checking anything else, what should make you suspicious?
The supposed ethyl branch may form part of a longer continuous parent chain.
Repair: The corresponding structure is named 3-methylpentane when the longest chain is selected.
A student finds the formula C₅H₁₀ in a database.
They conclude:
“The compound is an alkene.”
Give one alternative class consistent with the formula.
A cycloalkane.
Both an alkene with one C=C and a saturated monocyclic cycloalkane can fit CₙH₂ₙ.
A six-carbon alkene has:
Numbering from the other direction would give C=C at carbon 4 and methyl at carbon 2.
Which numbering direction wins?
The first: 5-methylhex-2-ene. The multiple bond must receive the lower possible locant. You do not sacrifice hex-2-ene merely to give the methyl group a smaller number.
2-methylpropan-1-ol
HO-CH2-CH(CH3)-CH3
CH3-CH2-CH2-OH - propan-1-ol
CH3-CH(OH)-CH3 - propan-2-ol.
Question: What stays the same, and what changes?
Molecular formula, functional group and carbon skeleton remain the same; the position of –OH changes.
A student numbers an alkene so that:
The reverse direction gives:
They choose the first because: “2 is smaller than 5.”
What has gone wrong?
They compared substituent locants before giving the multiple bond its required lower locant. The second numbering direction is correct.
Two compounds have:
A third compound has:
Classify both relationships.
First pair = chain structural isomers.
Third relationship = geometric stereoisomers.
A cyclohexane ring has an ethyl group and a methyl group attached to adjacent carbons.
Explain why neither ethane nor any part of the side chains should become the parent.
The six-carbon ring is the parent, because it contains more carbons than either attached chain. The groups are substituents.
1-ethyl-3-methylcyclohexane
Drawing on board
A student says: “Every alkene can have cis and trans forms because C=C cannot freely rotate.”
Destroy this argument using propene, CH₃–CH=CH₂.
Restricted rotation is necessary but not sufficient. In propene, the terminal double-bonded carbon has two identical H substituents, so two distinct cis/trans arrangements cannot be produced.
A student says: “Cyclopropane and propene cannot be isomers because one is cyclic and the other is not.”
Evaluate the claim.
The claim is false. Both have molecular formula C₃H₆, but their atoms are connected differently. They are ring–chain structural isomers.
A chemist tells you only: “My compound has formula C₄H₈ and contains a C=C bond.”
A student immediately claims it has cis–trans isomers.
What structural information must you request before deciding?
Determine which groups are attached to each carbon of the C=C. Each double-bonded carbon must have two different substituents for the cis–trans situation studied in class.
CH3-CH=C(CH3)-CH2-CH3
Give the systematic name and justify the numbering direction.
The longest chain containing C=C has five carbons. Numbering gives the double bond locant 2, which takes precedence over obtaining a smaller substituent locant. 3-methyl pent-2-ene
A molecule must satisfy all these clues:
Six-carbon parent chain
One C=C
Double bond receives locant 2
Two methyl substituents
Substituents occur at C3 and C4
Draw and name it.
3,4-dimethylhex-2-ene.
Consider but-2-ene.
Your team must establish, in the correct logical sequence, why two geometric isomers are possible.
ALL four ideas:
Full 500 only for the complete argument.
A student writes: “C₄H₈ must be butene. Because it contains a double bond, it must also show cis–trans isomerism.”
Identify two separate logical failures.
Failure 1: C₄H₈ does not prove the presence of C=C; the formula can also represent a cycloalkane.
Failure 2: Even after establishing an alkene, the presence of C=C alone does not prove cis–trans isomerism. The substituents on each C of the double bond must be examined.
My molecular formula is C₅H₁₀.
I am not cyclic.
My longest chain contains five carbons.
My double bond is not terminal.
I can exist as cis and trans forms.
Question: Who am I?
Pent-2-ene.