Alcohols — preparation, acidity (pKa ~16), oxidation (PCC, Jones, K₂Cr₂O₇), Lucas test: Visual walkthrough

We will lean on the classification idea from the parent topic, and where oxidation gives an aldehyde or ketone we are stepping into the world of Carbonyl Compounds — Aldehydes and Ketones.

Step 1 — What a carbon looks like, and what "the carbinol carbon" means

WHAT: A carbon atom always makes four bonds — no more, no less. On the carbinol carbon, one of those four bonds goes to the oxygen. The other three go to some mix of hydrogen atoms and other carbons.

WHY it matters: The whole ladder is decided by a single question — how many of those three remaining bonds go to hydrogen? Hold that thought; it is the key to everything.

PICTURE: Look at the carbinol carbon in the middle. One bond (teal) climbs to $O - H$ . The other three bonds are drawn as empty slots — each will be filled by either an $H$ or a carbon $R$ .

Figure — Alcohols — preparation, acidity (pKa ~16), oxidation (PCC, Jones, K₂Cr₂O₇), Lucas test

$carbinol carbon C has 4 bonds: \to O - H 1 + \to H or C 3$

The term $C$ is the one carbon we care about.
The " $1 \to O - H$ " is fixed — every alcohol has it.
The " $3$ " is the budget we will spend on hydrogens vs. carbons.

Step 2 — Classification is just "how many H's are left on that carbon"

WHAT: We fill the three slots from Step 1. The parent note counted carbons ("1°, 2°, 3°"). Here we count the hydrogens instead — it is the same count read backwards ( $3 - carbons = hydrogens$ ), but the hydrogen count is what the reaction actually feels.

WHY count hydrogens: In one line — oxidation removes hydrogens from this carbon. So the number of hydrogens available on the carbinol carbon is literally the number of rungs the alcohol can climb.

PICTURE: Three carbinol carbons side by side. Count the little grey $H$ atoms on each: two, then one, then zero. The teal $O - H$ is the same on all three.

Carbon–hydrogen count on the carbinol carbon

1° has two H, 2° has one H, 3° has zero H.

Step 3 — What "oxidation" physically does: swap two H's for one double bond

WHAT: Take the H off the oxygen. Take one H off the carbon. The carbon and oxygen now each have a spare bond to share, so they form a second bond between them: a carbonyl, $C = O$ .

WHY a double bond forms: Removing an H from the oxygen leaves oxygen with a spare bond; removing an H from the carbon leaves the carbon with a spare bond. Two spare bonds facing each other pair up — that is the double bond. No new atoms are added; we only subtracted two H's and reused the electrons.

PICTURE: The two red H's are struck out; a burnt-orange double bond snaps into place between C and O. This one move is the only thing oxidation ever does — we just repeat it.

Step 4 — A primary alcohol has TWO H's, so it climbs TWICE

WHAT — first rung: Apply Step 3 once. We spend the $O - H$ hydrogen and one of the two carbon hydrogens. Result: a carbon double-bonded to O, still carrying one hydrogen — that is an aldehyde, $R - C H O$ .

WHAT — second rung: The aldehyde carbon still has one $C - H$ . In the presence of water, the $C = O$ adds water to become a gem-diol (two OH's on one carbon), which regenerates a fresh $C - H$ / $O - H$ pair — exactly the setup Step 3 loves. Oxidise again → a carbon bearing $C = O$ and $O - H$ : that is a carboxylic acid, $R - C O O H$ .

WHY water decides the top rung: Water is what lets the aldehyde become the gem-diol that has a removable $C - H$ . Dry conditions (PCC) → no gem-diol → the ladder freezes at the aldehyde. Wet + strong (Jones, $K_{2} C r_{2} O_{7} / H^{+}$ ) → gem-diol forms → climb all the way to the acid. This is the carboxylic acid endpoint.

PICTURE: A staircase. Bottom step: alcohol (2 H). Middle step: aldehyde (1 H) — a dotted "PCC stops here / dry" gate. Top step: acid (0 H) — reached only through the blue "water" archway.

Step 5 — A secondary alcohol has ONE H, so it climbs ONCE and stops

WHAT: Remove the $O - H$ hydrogen and the single $C - H$ hydrogen; form $C = O$ . The carbon that carries the $C = O$ is flanked by two carbons and no hydrogen — that is a ketone, $R_{2} C = O$ .

WHY it stops even with a strong oxidant, even in water: To climb again we would need a $C - H$ on the carbonyl carbon (Step 3's rule). A ketone's carbonyl carbon has none — its other two bonds go to carbons. Water cannot help; there is simply nothing to remove. So mild or strong, dry or wet, 2° → ketone, full stop.

PICTURE: A two-step staircase with a solid wall after the ketone — the second rung the primary alcohol used is missing here because there is no H to spend.

Step 6 — The tertiary edge case: ZERO H's, so it can't even start

WHAT: Try to oxidise $R_{3} C O H$ . The oxidant looks for a $C - H$ on the carbinol carbon to pull off. All three non-oxygen bonds go to carbons. No $C - H$ → no $C = O$ can form → no reaction.

WHY this is a degenerate case, not a slow one: It is not "hard" or "slow" — it is structurally forbidden. Forcing it (very harsh acid + heat) doesn't gently oxidise; it breaks a C–C bond and shatters the molecule, which is a different, destructive process, not the clean ladder.

PICTURE: A carbinol carbon ringed by three carbons and an $O - H$ , with a big plum "no $C - H$ here" crossed circle. Beside it a flat ground — no staircase exists.

Step 7 — The colour signal: how the oxidant announces it worked

WHAT: The two hydrogens the carbon lost are gained (as electrons) by chromium. Chromium goes from $C r^{6 +}$ (orange dichromate) to $C r^{3 +}$ (green). So orange → green is the ladder-climb, seen from chromium's side.

WHY there is no colour change for 3°: If the alcohol can't lose hydrogens (Step 6), chromium gains nothing, stays $C r^{6 +}$ , stays orange. So the colour test also distinguishes tertiary alcohols: they leave the solution orange.

PICTURE: Left flask orange ( $C r^{6 +}$ , before), right flask green ( $C r^{3 +}$ , after), a burnt-orange arrow labelled "electrons from the C–H's" linking them.

The one-picture summary

Everything collapses to one question at the carbinol carbon: how many H's can I remove? Two → aldehyde → (with water) acid. One → ketone, stop. Zero → nothing happens.

Recall Feynman retelling — say it in plain words

Picture one carbon holding an $O - H$ and three other bonds. Oxidation is a simple trick you can play only if that carbon is holding a hydrogen: yank the hydrogen off the oxygen, yank a hydrogen off the carbon, and let the two of them snap into a double bond, $C = O$ . That's it — the whole chapter is just repeating this trick. A primary alcohol is holding two hydrogens, so you can play the trick twice: first you get an aldehyde (one hydrogen still on), and if there's water around to reshuffle it, you play it a second time and land on a carboxylic acid. A secondary alcohol holds only one hydrogen, so you play the trick once, get a ketone, and now the carbon is out of hydrogens — game over, no matter how strong your reagent. A tertiary alcohol holds zero hydrogens, so you can't even start; the reagent just sits there and the solution stays orange because chromium never got any hydrogens to swallow. Count the hydrogens on that one carbon and you already know the ending.