Visual walkthrough — IUPAC nomenclature — alkanes, alkenes, alkynes, aromatics, alcohols, ethers, aldehydes, ketones, acids, esters, amines,

We link the parent IUPAC nomenclature and lean on Functional Groups Overview, Priority and Seniority Rules (CIP vs IUPAC), and Degree of Unsaturation along the way.

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Step 1 — Draw the skeleton and label nothing yet

WHAT. Before any naming, we draw the molecule as a chain of carbon "beads" and mark two special features: a double bond (two lines between two carbons) and an –OH group (an oxygen with a hydrogen hanging off a carbon).

WHY. A name is a description of a picture. If we cannot see the picture cleanly — every carbon, every bond, every attachment — we cannot describe it. So the very first move is to make the picture unambiguous.

PICTURE. Look at the figure. Each grey circle is a carbon atom. The double line on the left (pale-yellow) is the C=C. The blue OH is our functional group. Notice we have not written any numbers yet — numbering is a decision, and we haven't earned the right to make it.

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Step 2 — Find every functional feature and rank them

WHAT. We list the features competing for attention: a C=C double bond and an –OH (alcohol). Then we ask: which one is senior?

WHY. The parent note's seniority table exists so that exactly one feature becomes the suffix (the ending of the name) and gets first pick of the low numbers. Without a fixed ranking, two people could name the same molecule differently. Here the contest is –OH vs C=C.

$\text{seniority:}\underset{\text{alcohol, suffix }-\mathbf{ol}}{\underbrace{-OH}}>\underset{\text{unsaturation, infix }-\mathbf{en}-}{\underbrace{C=C}}$

• The under-brace on the left says: –OH wins, so it becomes the suffix -ol.
• The under-brace on the right says: the double bond is not the boss; it becomes an infix -en- that sits inside the name but does not command the numbering.

PICTURE. In the figure, a chalk crown sits on the OH (the "king") and the C=C wears a smaller badge (a "servant"). The arrow shows the –OH claiming the suffix slot. This ranking comes straight from Priority and Seniority Rules (CIP vs IUPAC).

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Step 3 — Trace the longest chain that CONTAINS the senior group

WHAT. We find the longest unbroken path of carbons — with the rule that this path must include the carbon carrying –OH.

WHY. The chain length becomes the root word (parent note's table: 4 C → but, 5 C → pent, ...). Two conditions must both hold: (a) longest possible, and (b) it passes through the senior group, because the senior group has to live on the main chain to become a suffix.

PICTURE. Trace the pink highlighted path in the figure: it runs through all six carbons, C=C at one end, OH near the other. Count the beads on the pink path: 6 carbons.

$\underset{\text{longest chain}}{\underbrace{6\text{ carbons}}}⟶\text{root}=\text{hex}$

The 6 is literally the number of grey beads on the highlighted trail; hex is its dictionary word.

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Step 4 — Number the chain so the SENIOR group gets the lowest locant

WHAT. A chain has two ends. We can number left→right or right→left. We choose the direction that gives –OH the smallest number.

WHY. This is the single most important numbering rule. The senior suffix group (–OH here) always claims the lowest possible locant — before we worry about the double bond, before substituents, before anything else.

Let us test both directions. Call the OH carbon's number $n$.

$$\text{left}\to \text{right}:OH\text{ on }C5,\text{right}\to \text{left}:OH\text{ on }C2.$$

• Left→right gives $n=5$.
• Right→left gives $n=2$.
• Rule says pick the smaller $n$: choose $n=2$, i.e. number from the OH end (right→left).

PICTURE. The figure shows the two numbering arrows side by side. The winning arrow (blue) starts at the OH-end so OH lands on C2; the losing arrow (faded) starts at the double-bond end and pushes OH out to C5. The blue arrow wins because $2<5$.

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Step 5 — Read off the double-bond locant under the chosen numbering

WHAT. With numbering fixed (OH = C2, counting from the right), we now locate the C=C. It sits between C5 and C6. We cite the lower carbon: 5.

WHY. The direction is already locked by the senior group — we do not re-number to help the double bond. We simply report where the double bond ended up. A double bond spanning carbons $k$ and $k+1$ is always cited by the lower number $k$.

$$C=C\text{ between }C5\text{ and }C6\Rightarrow \text{locant}=\min (5,6)=5\Rightarrow \text{-5-en-}$$

• $\min (5,6)$: we take the lower of the two carbons the bond connects.
• -5-en-: the 5 says where, the en says double bond.

PICTURE. In the figure, the frozen blue numbers from Step 4 stay put; only the double bond gets a spotlight. Its two carbons read 5 and 6, so we tag it with the lower one, 5.

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Step 6 — Assemble the slots into one name

WHAT. We now slot the pieces into the parent note's template: $\underset{\text{(none here)}}{\underbrace{\text{prefix}}}+\underset{\text{hex}}{\underbrace{\text{root}}}+\underset{\text{-5-en-}}{\underbrace{\text{-locant-en-}}}+\underset{\text{-2-ol}}{\underbrace{\text{-locant-ol}}}$

Stitching them together with hyphens between number-and-letter and reading straight through:

$\boxed{\text{hex-5-en-2-ol}}$

WHY. Each slot answers one question:

• hex → how many carbons? (6)
• 5-en → where is the double bond? (starts at C5)
• 2-ol → where is the senior group and what is it? (–OH at C2)

PICTURE. The figure builds the name left-to-right like train wagons clicking together: hex (the engine length) → 5-en (a coupling badge) → 2-ol (the crowned king wagon). Read the whole train and you can rebuild the molecule exactly.

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Step 7 — Edge & degenerate cases (so no scenario surprises you)

WHAT. Three "what if" variations of the same skeleton, each with its own decision.

Case (a) — symmetric molecule, both ends tie. If OH sat exactly in the middle and both numbering directions gave the same OH locant, we move to the next tiebreak: give the double bond the lower number. (The ladder from Step 4 kicks down a rung.)

Case (b) — no senior group at all (pure alkene). Remove the OH; only the C=C remains. Now the double bond is the top-ranked feature, so now we number to minimise the C=C locant. Same molecule-skeleton, opposite numbering direction — this is exactly why the OH mattered.

Case (c) — degenerate: a single carbon or no chain feature. If the "chain" is one carbon bearing –OH, it is simply methanol; there is no double bond to place and no direction to choose. The algorithm still runs; some steps just return "nothing to do."

WHY. The contract of an algorithm is that it handles every input, including boring ones. Showing the ties and the no-group case proves the rules never leave you stranded.

PICTURE. The figure stacks the three cases as three mini-boards: (a) a balanced see-saw (tie → bond decides), (b) the OH erased and the numbering arrow flipping to the bond, (c) a lone carbon labelled methanol.

Recall Why does removing the OH flip the numbering direction?

With OH present, OH is senior and claims C2 (numbering from the OH end). Remove OH and the double bond becomes the top feature, so we number from the other end to give the bond the lowest locant. Same skeleton, different boss → different direction.

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The one-picture summary

WHAT. One board that compresses all seven steps: skeleton → rank groups → longest chain → number from the senior group → read the bond → assemble → boxed name.

Recall Feynman retelling (plain words)

Picture a little train of six carbon wagons. One wagon wears a crown — that's the –OH, the king (it beat the double bond in the seniority contest). Rule: always start counting seats from the king's wagon, because the king must sit in the lowest-numbered seat he can. Counting from the king puts him in seat 2. Now we look around: there's a tight double-coupling between the last two wagons — seats 5 and 6 — so we tag it "5-en" using the lower seat. The train is six wagons long, so its family name is "hex". Click the pieces together — hex (six long) + 5-en (double bond starts at 5) + 2-ol (king –OH at 2) — and you get hex-5-en-2-ol. Anyone in the world reading that name rebuilds the exact same train. And if the king (OH) were removed, the double bond would become the new boss and we'd count from the other end instead — that's the whole secret of numbering direction.

Recall Q: Final IUPAC name of

$C{H}_2=CH-C{H}_2-C{H}_2-CH(OH)-C{H}_3$? hex-5-en-2-ol — 6 carbons (hex); –OH is senior so numbered from that end → C2 (-2-ol); double bond then falls at C5 (-5-en-).

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