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Alcohols & Ethers

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Alcohols & Ethers - Complete Interactive Lesson

Part 1: Alcohol Nomenclature & Properties

Alcohols & Ethers — Nomenclature and Physical Properties

Part 1 of 7 — Alcohol Nomenclature, Hydrogen Bonding & Acidity

An alcohol is any compound bearing a hydroxyl group ( $\text{-OH}$ ) bonded to an $sp^3$ (saturated) carbon. The general formula is $\text{R-OH}$ . That single functional group dominates the molecule's chemistry far out of proportion to its size: it makes the carbon electrophilic, the oxygen nucleophilic, and the $\text{O-H}$ proton both hydrogen-bonding and weakly acidic.

Alcohols are classified by the substitution at the carbinol carbon (the carbon bearing the $\text{-OH}$ ):

Class	Carbons on the carbinol C	Example
Primary ( $1°$ )	one (or zero)	ethanol, $\text{CH}_3\text{CH}_2\text{OH}$

This $1°/2°/3°$ label is not bookkeeping for its own sake — it predicts oxidation outcomes, $\text{S}_\text{N}1$ versus $\text{S}_\text{N}2$ behavior, and dehydration mechanism, all of which appear in later parts of this suite.

IUPAC Naming: Four Steps

To name an alcohol by IUPAC rules:

Find the longest chain that contains the carbon bearing $\text{-OH}$ . The hydroxyl-bearing carbon must be in the parent chain even if a longer chain exists elsewhere.
Replace the alkane "-e" with "-ol." Hexane $\rightarrow$ hexanol.
Number so the $\text{-OH}$ gets the lowest possible locant. The hydroxyl outranks alkyl groups, double bonds, and halogens for low numbers.
Cite substituents alphabetically with their locants.

Worked example. Name the compound $\text{CH}_3\text{CH(OH)CH}_2\text{CH(CH}_3\text{)CH}_3$ .

Checkpoint — Classification & Naming

Hydrogen Bonding and Physical Properties

The $\text{O-H}$ bond is strongly polarized because oxygen is far more electronegative than hydrogen. Each alcohol molecule can therefore both donate a hydrogen bond (through its $\text{O-H}$ ) and accept one (through oxygen's lone pairs). This network of intermolecular hydrogen bonds explains the two signature physical properties of alcohols:

1. Elevated boiling points. Compare molecules of similar molar mass:

Compound	Approx. MW	Boiling point	Dominant IMF
propane	44	$-42\,°\text{C}$

Acidity and $\text{p}K_\text{a}$ Trends

Alcohols are weakly acidic. Removing the $\text{O-H}$ proton gives an alkoxide ( $\text{RO}^-$ ):

Checkpoint — Properties, H-Bonding & Acidity

Exit Ticket — Part 1 Synthesis

Part 2: Alcohol Synthesis

Preparation of Alcohols

Part 2 of 7 — Synthesizing Alcohols

There are four workhorse routes to alcohols that every organic student must own. Three build the $\text{C-OH}$ bond by adding across a multiple bond or reducing a carbonyl; one, the Grignard reaction, simultaneously forms a new carbon-carbon bond while installing the hydroxyl. Choosing among them is a question of (a) what starting material you have and (b) what regiochemistry or carbon count you need.

Method	Reagents	Key feature
Acid-catalyzed hydration	alkene, $\text{H}_3\text{O}^+$

Part 3: Alcohol Reactions

Reactions of Alcohols

Part 3 of 7 — Substitution, Dehydration & Oxidation

The hydroxyl group is a chemical liability in one key respect: $\text{OH}^-$ is a terrible leaving group. Almost every alcohol reaction is, at heart, a strategy for getting around that problem — either by protonating the $\text{-OH}$ to make water (a good leaving group), by converting it to a better leaving group with a reagent like $\text{SOCl}_2$ , or by removing the whole group as part of an elimination or oxidation.

Part 4: Ether Nomenclature & Synthesis

Ethers: Structure, Naming, and Synthesis

Part 4 of 7 — Ether Nomenclature & the Williamson Synthesis

An ether has an oxygen bonded to two carbon groups: $\text{R-O-R'}$ . The C-O-C angle is roughly tetrahedral (about $110°$ in dimethyl ether), and oxygen still carries two lone pairs. Critically, an ether has no $\text{O-H}$ bond — so, unlike alcohols, ethers cannot donate hydrogen bonds to one another. They can only accept hydrogen bonds (from water or alcohols). This single structural fact explains their two defining traits:

Low boiling points, close to those of comparable alkanes. Diethyl ether boils at $35\,°\text{C}$ , near pentane () and far below 1-butanol (), even though all three have similar mass.

Part 5: Ether Reactions

Reactions of Ethers and Epoxides

Part 5 of 7 — Ether Cleavage, Epoxide Synthesis & Ring-Opening

Ordinary ethers are famously unreactive — that is exactly why they make good solvents (Part 4). But two situations break that calm:

Acidic cleavage of a normal ether by hot, concentrated $\text{HI}$ or $\text{HBr}$ .
Epoxides — three-membered cyclic ethers whose enormous ring strain (about $114\,\text{kJ/mol}$ ) makes them dramatically more reactive than any open-chain ether.

The contrast is the headline of this part: the same C-O-C linkage is essentially inert when relaxed and explosively reactive when forced into a strained three-membered ring.

Acidic Cleavage of Ordinary Ethers

Because ethers have no good leaving group and no acidic proton, only a strong acid with a good nucleophilic counterion will cleave them. The classic conditions are :

Part 6: Problem-Solving Workshop

Problem-Solving Workshop

Part 6 of 7 — Putting the Reactions Together

Real exam questions rarely ask you to recite a single reaction. They ask you to predict a product, choose between competing reagents, or design a multi-step synthesis. This part trains those skills by working four representative problems end to end. The recurring strategy is always the same three questions:

What functional group am I starting with, and what class is it? ( $1°/2°/3°$ matters constantly.)
What does the reagent do, and does the substrate class change the mechanism or outcome?
Is there a regiochemistry or selectivity decision (Markovnikov vs anti-Markovnikov, Zaitsev, acid vs base epoxide opening, aldehyde vs acid oxidation)?

Keep those three questions in view and most multi-step problems unravel cleanly.

Problem 1 — Predicting Oxidation Products

Prompt. A mixture contains 1-butanol, 2-butanol, and 2-methylpropan-2-ol. Each is treated separately with (a) PCC and (b) Jones reagent. Give all eight products.

Reasoning by class:

1-butanol (primary). PCC (mild, anhydrous) stops at the aldehyde, butanal. Jones (strong, aqueous) pushes through to the carboxylic acid, butanoic acid.
Both oxidants give the , butan-2-one. A secondary alcohol cannot be oxidized past the ketone — there is no further H on that carbon to remove.

Part 7: Synthesis & Review

Synthesis & Comprehensive Review

Part 7 of 7 — Tying the Alcohol & Ether Suite Together

Across this unit, alcohols emerged as the central hub of aliphatic functional-group chemistry. They can be made from alkenes, carbonyls, and Grignard reagents, and they can be converted into alkyl halides, alkenes, aldehydes, ketones, acids, ethers, and esters. Master a single organizing idea — the oxidation-state ladder — and most of the unit collapses into one coherent map.

This final part reviews the connections, walks a multi-step synthesis that chains several reactions, and tests integrated reasoning.

The Oxidation Ladder: One Idea That Organizes Everything

Carbon at the functional position climbs a ladder as it gains bonds to oxygen (oxidation) and descends as it gains bonds to hydrogen (reduction):

$\text{alkane} \rightarrow \text{alcohol} \rightarrow \text{aldehyde/ketone} \rightarrow \text{carboxylic acid}$

CH3CH(OH)CH2CH(CH3)CH3

Longest chain through the $\text{-OH}$ carbon: five carbons $\rightarrow$ pentane backbone, so pentanol.
Numbering from the end nearest the $\text{-OH}$ puts hydroxyl on C2 and a methyl branch on C4.
Result: 4-methylpentan-2-ol (equivalently 4-methyl-2-pentanol).

Common trap: numbering from the wrong end to give the methyl a lower number. The $\text{-OH}$ always claims the lowest locant first; substituent numbers fall out afterward. Starting from the other end would mislabel it "2-methylpentan-4-ol," which is wrong because $2+4$ is not how you decide — the principal characteristic group ( $\text{-OH}$ ) takes priority.

When $\text{-OH}$ is not the senior group (e.g., a carboxylic acid is also present), it is demoted to the "hydroxy-" prefix instead of the "-ol" suffix.

All three have nearly identical mass, yet ethanol boils about $100\,°\text{C}$ higher than propane. Dimethyl ether — an isomer of ethanol with the same formula $\text{C}_2\text{H}_6\text{O}$ — boils far lower because its hydrogens are bonded to carbon and cannot serve as H-bond donors; it can only accept.

2. Water solubility that fades with chain length. Methanol, ethanol, and propanol are miscible with water in all proportions because the $\text{-OH}$ hydrogen-bonds to water. As the nonpolar hydrocarbon tail grows, the molecule becomes increasingly "greasy" and solubility drops sharply — 1-butanol is only partially miscible, and 1-octanol is essentially insoluble.

Key idea: boiling point tracks the strength of intermolecular attraction, while water solubility tracks the balance between the polar $\text{-OH}$ head and the nonpolar tail.

\text{ROH} + \text{H}_2\text{O} \rightleftharpoons \text{RO}^- + \text{H}_3\text{O}^+

ROH + H_{2} O ⇌ RO^{-} + H_{3} O^{+}

The $\text{p}K_\text{a}$ of a simple alcohol is about $16$ – $18$ , placing it on the acidity ladder between water ( $\text{p}K_\text{a}\approx 15.7$ ) and a terminal alkyne ( $\text{p}K_\text{a}\approx 25$ ). A lower $\text{p}K_\text{a}$ means a stronger acid and a more stable conjugate base.

Three structural factors shift alcohol acidity, all by changing how well the alkoxide anion's negative charge is stabilized:

Inductive electron withdrawal. Electronegative atoms near the $\text{-OH}$ pull electron density away from the oxygen, spreading the negative charge of $\text{RO}^-$ and stabilizing it. 2,2,2-trifluoroethanol ( $\text{p}K_\text{a}\approx 12.5$ ) is far more acidic than ethanol ( $\text{p}K_\text{a}\approx 16$ ) because three fluorines drain charge through the $\sigma$ framework. The effect weakens with distance.
Alkyl substitution (in solution). tert-Butanol ( $\text{p}K_\text{a}\approx 18$ ) is a weaker acid than methanol ( $\text{p}K_\text{a}\approx 15.5$ ). Electron-donating alkyl groups slightly destabilize the alkoxide, and the bulky -butoxide is also poorly solvated, which further raises its .
Resonance. This is the dominant effect when available. A phenol ( $\text{p}K_\text{a}\approx 10$ ) is roughly a million times more acidic than cyclohexanol ( $\text{p}K_\text{a}\approx 18$ ) because the phenoxide negative charge is delocalized onto the aromatic ring. A (–) is more acidic still: its carboxylate spreads the charge equally over two oxygens.

Worked example — rank by acidity. Order these from most to least acidic: ethanol, 2-chloroethanol, tert-butanol, phenol.

Phenol wins decisively (resonance into the ring, $\text{p}K_\text{a}\approx 10$ ). 2-Chloroethanol is next (inductive withdrawal by Cl lowers it to $\approx 14$ ). Ethanol follows at $\approx 16$ . tert-Butanol is the weakest acid ( $\approx 18$ ) because of electron-donating alkyl groups and poor solvation of its bulky alkoxide.

Final order (strongest $\rightarrow$ weakest acid): phenol $>$ 2-chloroethanol $>$ ethanol $>$ tert-butanol.

Adding Water Across an Alkene: Regiochemistry Is Everything

Two complementary strategies convert an alkene into an alcohol, and they give opposite regiochemistry.

Markovnikov addition places the $\text{-OH}$ on the more substituted carbon (the one that would bear the more stable carbocation). Two reagent systems achieve this:

Acid-catalyzed hydration ( $\text{H}_2\text{O}$ , catalytic $\text{H}_2\text{SO}_4$ ) proceeds through a free carbocation. It follows Markovnikov orientation but, because a discrete cation forms, it is prone to hydride and alkyl shifts (rearrangement) when a more stable cation is one step away.
Oxymercuration-demercuration ( $\text{Hg(OAc)}_2$ , $\text{H}_2\text{O}$ ; then $\text{NaBH}_4$ ) also gives the Markovnikov alcohol but routes through a bridged that resists rearrangement. When you need the Markovnikov product , this is the method of choice.

anti-Markovnikov addition places the $\text{-OH}$ on the less substituted carbon. Hydroboration-oxidation ( $\text{BH}_3 \cdot \text{THF}$ ; then $\text{H}_2\text{O}_2$ , ) does this. Boron, being less electronegative than carbon, is the electrophile; it adds to the less hindered carbon, and after oxidation the ends up there too. The addition is also (H and OH add to the same face).

Worked example. What is the major product when 1-methylcyclohexene is treated with (a) $\text{H}_3\text{O}^+$ versus (b) $\text{BH}_3$ then ?

(a) Markovnikov: the $\text{-OH}$ goes onto the more substituted ring carbon (the one bearing the methyl), giving 1-methylcyclohexan-1-ol, a tertiary alcohol.
(b) anti-Markovnikov: the $\text{-OH}$ goes onto the adjacent, less substituted carbon, giving 2-methylcyclohexan-1-ol, a secondary alcohol.

Same alkene, opposite reagents, different constitutional isomers. The exam reflex is: see " $\text{BH}_3$ " and immediately flip your regiochemistry to anti-Markovnikov.

Checkpoint — Hydration Regiochemistry

Reducing Carbonyls to Alcohols

A second family of routes starts from a carbonyl ( $\text{C=O}$ ) and adds hydride ( $\text{H}^-$ ) to give an alcohol. No new carbon-carbon bond forms; the carbon skeleton is unchanged.

Sodium borohydride ( $\text{NaBH}_4$ ) is a mild hydride source. It reduces aldehydes (to $1°$ alcohols) and ketones (to $2°$ alcohols) but is too gentle to touch esters or carboxylic acids. It tolerates water and alcohol solvents.
Lithium aluminium hydride ( $\text{LiAlH}_4$ ) is a powerful hydride source. It reduces aldehydes, ketones, esters, carboxylic acids, amides, and nitriles. It reacts violently with water, so the reaction is run in dry ether and then quenched with aqueous acid in a separate step (written as a "2)" over the arrow).

The oxidation-state logic is symmetric with Part 3: reduction adds $\text{H}^-$ and walks the carbon down the oxidation ladder (acid $\rightarrow$ aldehyde $\rightarrow$ alcohol), exactly reversing oxidation.

Worked example. Predict the products:

Butanal $+\,\text{NaBH}_4 \rightarrow$ butan-1-ol (a $1°$ alcohol).
Cyclohexanone $+\,\text{NaBH}_4 \rightarrow$ (a alcohol).

Selectivity trap: if a molecule contains both a ketone and an ester and you want to reduce only the ketone, use $\text{NaBH}_4$ . Reaching for $\text{LiAlH}_4$ would reduce both.

The Grignard Reaction: Building the Carbon Skeleton

A Grignard reagent ( $\text{RMgX}$ ) is made by inserting magnesium into a carbon-halogen bond in dry ether. This umpolung trick inverts the carbon's normal polarity: the once-electrophilic carbon of $\text{R-X}$ becomes a powerful carbon nucleophile ( $\text{R}^{\delta-}$ , essentially a carbanion). That nucleophilic carbon attacks the electrophilic carbonyl carbon, and after an aqueous workup ( $\text{H}_3\text{O}^+$ ) you isolate an alcohol with a brand-new C-C bond.

The class of alcohol you get depends on the carbonyl partner:

Carbonyl partner	Alcohol class produced
formaldehyde ( $\text{CH}_2\text{O}$ )	primary ( $1°$ )
any other aldehyde	secondary ( $2°$ )
ketone	tertiary ()

Because the Grignard carbon is also a strong base, it is destroyed by any acidic proton — $\text{O-H}$ , $\text{N-H}$ , $\text{S-H}$ , terminal alkyne $\text{C-H}$ . You therefore cannot prepare or use a Grignard in the presence of a free hydroxyl, water, or a carboxylic acid; the reagent would simply deprotonate the substrate instead of adding to it.

Worked example — retrosynthesis. Design a Grignard synthesis of 2-phenylbutan-2-ol, $\text{Ph-C(OH)(CH}_3\text{)(CH}_2\text{CH}_3\text{)}$ .

This is a tertiary alcohol, so the carbonyl partner must be a ketone. Disconnect either C-C bond to the carbinol carbon. One clean option: combine phenylmagnesium bromide ( $\text{PhMgBr}$ ) with butan-2-one ( $\text{CH}_3\text{COCH}_2\text{CH}_3$ ). The phenyl nucleophile adds to the ketone carbonyl; aqueous workup delivers the tertiary alcohol. (Equivalently, ethyl- or methyl-Grignard plus the appropriate aryl ketone would also work.)

Notice the central planning move: a tertiary target alcohol always points back to a ketone electrophile; a secondary target points back to a non-formaldehyde aldehyde.

Checkpoint — Reduction & Grignard

Exit Ticket — Part 2 Synthesis

Three transformation families dominate:

Substitution — replace $\text{-OH}$ with a halide to make an alkyl halide.
Dehydration — eliminate $\text{H}$ and $\text{OH}$ to make an alkene.
Oxidation — convert the $\text{C-OH}$ into a $\text{C=O}$ (aldehyde, ketone, or carboxylic acid).

Substitution: Alcohols to Alkyl Halides

To swap $\text{-OH}$ for a halide, you must either protonate or activate the oxygen.

With hydrohalic acids ( $\text{HX}$ ). Reactivity follows $\text{HI} > \text{HBr} > \text{HCl}$ , and crucially the substrate class dictates the mechanism:

Tertiary (and most secondary) alcohols react by $\text{S}_\text{N}1$ : protonation gives an oxocarbenium-then-carbocation, and halide traps it. Because a carbocation forms, $3°$ substrates can rearrange, and the reaction is fast.
Primary alcohols react by $\text{S}_\text{N}2$ and need the more forcing conditions; concentrated $\text{HCl}$ with $\text{ZnCl}_2$ (the Lucas reagent) is the classic example.

The Lucas test exploits this: a $3°$ alcohol turns cloudy almost instantly, a $2°$ alcohol within minutes, and a $1°$ alcohol not at all at room temperature — a quick diagnostic for alcohol class.

With $\text{SOCl}_2$ or $\text{PBr}_3$ . These reagents convert the $\text{-OH}$ into a superb leaving group without forming a free carbocation, so they avoid rearrangement and are preferred for and alcohols:

$\text{R-OH} + \text{SOCl}_2 \rightarrow \text{R-Cl}$ (plus $\text{SO}_2$ and gas).

Trap: using $\text{HCl}$ on a neopentyl or other rearrangement-prone primary/secondary alcohol can scramble the skeleton. When skeletal integrity matters, reach for $\text{SOCl}_2$ or $\text{PBr}_3$ .

Checkpoint — Substitution

Dehydration: Alcohols to Alkenes

Heating an alcohol with a strong, non-nucleophilic acid (concentrated $\text{H}_2\text{SO}_4$ or $\text{H}_3\text{PO}_4$ ) eliminates water to form an alkene. The acid protonates the $\text{-OH}$ so it can leave as neutral water.

Mechanism depends on substrate class:

$3°$ and $2°$ alcohols dehydrate by E1: water leaves to form a carbocation, then a base removes a $\beta$ -hydrogen. Carbocations mean rearrangement is possible.
$1°$ alcohols dehydrate by E2: there is no stable primary cation, so loss of water and the $\beta$ -hydrogen is concerted. These require the harshest conditions.

Ease of dehydration tracks carbocation stability: $3° > 2° > 1°$ .

Regiochemistry follows Zaitsev's rule: the major product is the more substituted (more stable) alkene.

Worked example. Dehydration of 2-methylbutan-2-ol with hot $\text{H}_2\text{SO}_4$ .

The tertiary cation is already present after water leaves. It has $\beta$ -hydrogens on two different neighboring carbons. Removing a $\beta$ -H from the $\text{CH}_2$ gives 2-methylbut-2-ene (trisubstituted), while removing one from a terminal $\text{CH}_3$ gives (disubstituted). By Zaitsev, the .

Contrast with oxymercuration/hydration: dehydration (alcohol $\rightarrow$ alkene, lose water) and acid-catalyzed hydration (alkene $\rightarrow$ alcohol, add water) are the same equilibrium read in opposite directions. Le Chatelier decides: dilute acid and excess water drive hydration; concentrated acid and heat (distilling off the volatile alkene) drive dehydration.

Oxidation: Climbing the Carbonyl Ladder

Oxidation removes $\text{H}$ atoms (or adds $\text{O}$ ), raising the carbon's oxidation state. The outcome is governed entirely by alcohol class and choice of oxidant:

Alcohol	Mild oxidant (PCC, Swern, DMP)	Strong oxidant (Jones, $\text{CrO}_3$ , $\text{KMnO}_4$ )
primary ( $1°$ )	aldehyde ( $\text{RCHO}$ )	carboxylic acid ( $\text{RCOOH}$ )
secondary ( $2°$ )	ketone	ketone (stops there)
tertiary ( $3°$ )

Two principles do all the work:

A $3°$ alcohol cannot be oxidized under normal conditions: its carbinol carbon has no $\text{C-H}$ bond to remove, and oxidation of an alcohol requires losing the H on the carbon bearing $\text{-OH}$ .
For a $1°$ alcohol, the oxidant choice decides whether you stop at the aldehyde or push to the acid. Strong aqueous oxidants hydrate the aldehyde and oxidize it onward to the carboxylic acid. To stop at the aldehyde, you must use an anhydrous, mild oxidant such as PCC (pyridinium chlorochromate) — no water means the aldehyde hydrate cannot form, so oxidation halts.

Worked example. Predict products:

1-propanol $+\,\text{PCC} \rightarrow$ propanal (stops at the aldehyde; anhydrous).
1-propanol $+\,\text{Jones reagent } (\text{CrO}_3, \text{H}_2\text{SO}_4, \text{H}_2\text{O}) \rightarrow$ (pushed to the acid).

Signature trap: asking for the product of a primary alcohol with a strong oxidant and listing "aldehyde." Strong, aqueous oxidants overshoot the aldehyde and deliver the carboxylic acid. The aldehyde is the PCC answer.

Checkpoint — Dehydration & Oxidation

Exit Ticket — Part 3 Synthesis

Excellent solvents. Because the oxygen lone pairs accept hydrogen bonds and solvate cations, yet the molecule is largely nonpolar, ethers dissolve a wide range of organic compounds and are nearly inert. Diethyl ether and THF (tetrahydrofuran) are the standard solvents for Grignard and hydride reactions precisely because they do not react with those reagents.

Naming Ethers

Two systems are common:

Common names cite the two groups attached to oxygen alphabetically, then add "ether": $\text{CH}_3\text{-O-CH}_2\text{CH}_3$ is ethyl methyl ether; $\text{(CH}_3\text{CH}_2\text{)}_2\text{O}$ is diethyl ether.

IUPAC names treat the smaller $\text{R-O}$ piece as an alkoxy substituent on the larger parent chain. The smaller group plus oxygen becomes "(name)oxy":

$\text{CH}_3\text{-O-}$ = methoxy
$\text{CH}_3\text{CH}_2\text{-O-}$ =

Worked example. Name $\text{CH}_3\text{CH}_2\text{CH}_2\text{-O-CH}_3$ .

The larger group is propyl (three carbons); the smaller is methyl.
Common name: methyl propyl ether.
IUPAC: the propane chain is the parent and $\text{CH}_3\text{O-}$ is a methoxy substituent on C1, giving 1-methoxypropane.

Cyclic ethers have their own names: oxirane (the three-membered epoxide, covered in Part 5), oxetane (four-membered), tetrahydrofuran / THF (five-membered), and tetrahydropyran / 1,4-dioxane among the six-membered rings.

Checkpoint — Structure & Naming

The Williamson Ether Synthesis

The most reliable, general way to make an ether is the Williamson ether synthesis: an alkoxide ( $\text{RO}^-$ ) displaces a halide (or sulfonate) from an alkyl halide by $\text{S}_\text{N}2$ .

$\text{R-O}^- + \text{R'-X} \rightarrow \text{R-O-R'} + \text{X}^-$

Because the key step is $\text{S}_\text{N}2$ , every $\text{S}_\text{N}2$ rule applies, and they dictate how you must plan the synthesis:

The alkyl halide must be methyl or primary. Secondary halides give poor yields (elimination competes); tertiary halides give only elimination (E2), no ether at all, because the strongly basic alkoxide attacks a $\beta$ -hydrogen rather than the crowded carbon.
The alkoxide can be primary, secondary, or tertiary — its steric bulk sits on the nucleophile, not on the carbon being attacked, so it tolerates branching.

This asymmetry is the single most tested idea in ether synthesis. For an unsymmetrical ether, there are two conceivable alkoxide-plus-halide pairings, and you must choose the one that puts the less hindered group on the alkyl halide.

Worked example — choosing the right disconnection. Synthesize tert-butyl methyl ether (MTBE), $\text{(CH}_3\text{)}_3\text{C-O-CH}_3$ .

Two options exist on paper:

(A) tert-butoxide $+\,\text{CH}_3\text{I}$ (methyl halide). The alkoxide is tertiary, but the halide is methyl, which cannot eliminate. $\text{S}_\text{N}2$ succeeds. ✓
(B) methoxide $+$ (tertiary halide). The alkoxide is small, but the halide is tertiary, so methoxide acts as a base and gives — no ether. ✗

Correct route: option (A). Always put the alkoxide on the more hindered group and the leaving group on the methyl/primary carbon.

The alkoxide itself is typically generated by deprotonating an alcohol with a strong base — $\text{NaH}$ , $\text{Na}$ metal, or $\text{KOtBu}$ — before adding the alkyl halide.

Crown Ethers and Practical Notes

Crown ethers are large cyclic polyethers (for example 18-crown-6, an 18-membered ring containing six oxygens). The ring of inward-pointing oxygen lone pairs forms a cavity sized to encapsulate a specific metal cation — 18-crown-6 fits $\text{K}^+$ snugly, while 15-crown-5 prefers the smaller $\text{Na}^+$ .

Their power is in phase-transfer catalysis: by wrapping the cation in a lipophilic shell, a crown ether drags an otherwise insoluble ionic salt (like $\text{KF}$ or $\text{KMnO}_4$ ) into a nonpolar organic solvent. The cation is sequestered, leaving a "naked," poorly solvated anion that is far more nucleophilic and reactive than it would be in water. This is a clean illustration of how solvation governs reactivity, the same theme seen with bulky alkoxides in Part 1.

Safety footnote, important in real labs: ethers slowly react with atmospheric oxygen to form explosive peroxides on standing. Old bottles of diethyl ether or THF must be tested and never distilled to dryness.

Summary of the synthesis logic: to build an ether, use Williamson and put the leaving group on a methyl or primary carbon; to use an ether, exploit its inertness as a solvent; and to transport an ion into organic media, wrap it in a crown ether to free a hyper-reactive naked anion.

Checkpoint — Williamson & Crown Ethers

Exit Ticket — Part 4 Synthesis

excess hot $\text{HI}$ (or $\text{HBr}$ )

$\text{R-O-R'} + 2\,\text{HI} \rightarrow \text{R-I} + \text{R'-I} + \text{H}_2\text{O}$

The mechanism: the acid protonates the ether oxygen (making it a neutral, leaving-group-ready alcohol/alkyl piece), and iodide — a strong nucleophile — attacks. Which carbon iodide attacks depends on the substrate, mirroring Part 3 substitution logic:

If both groups are methyl/primary, iodide attacks by $\text{S}_\text{N}2$ at the less hindered carbon.
If one group is tertiary, benzylic, or allylic, the protonated ether ionizes to a stabilized carbocation and reacts by $\text{S}_\text{N}1$ ; iodide then traps the cation. The more substituted carbon becomes the alkyl iodide, and the other fragment leaves as the alcohol (which, with excess HI, is converted onward to a second alkyl iodide).

Worked example. tert-Butyl methyl ether $+$ excess $\text{HI}$ .

Protonation, then ionization at the tertiary carbon (a stable $3°$ cation) gives tert-butyl iodide; methanol is released and, with excess HI, becomes methyl iodide. Products: tert-butyl iodide $+$ methyl iodide. Note the tertiary group goes $\text{S}_\text{N}1$ — iodide does not attack the crowded tertiary carbon directly.

Making Epoxides

There are two standard syntheses, and they are conceptual mirror images:

1. Epoxidation of an alkene with a peroxyacid (mCPBA, meta-chloroperoxybenzoic acid). A single concerted step delivers one oxygen across the double bond with syn stereochemistry and full retention of alkene geometry. A cis-alkene gives the cis-epoxide.

2. Intramolecular Williamson synthesis from a halohydrin. Treat a $\beta$ -halo alcohol (a halohydrin) with base. The base deprotonates the $\text{-OH}$ to an alkoxide, which then performs an intramolecular $\text{S}_\text{N}2$ on the adjacent carbon bearing the halide, closing the three-membered ring. This requires the alkoxide oxygen and the leaving group to reach an anti-periplanar arrangement.

Notice that route 2 is simply the Williamson ether synthesis (Part 4) folded back on itself: an alkoxide displacing a halide by $\text{S}_\text{N}2$ , except the nucleophile and electrophile live in the same molecule.

Checkpoint — Cleavage & Epoxide Synthesis

Epoxide Ring-Opening: The Regiochemistry Showdown

Epoxide ring-opening is one of the most heavily tested ideas in the course because the regiochemistry flips depending on conditions. The strained ring is opened by a nucleophile, but which carbon the nucleophile attacks changes between acidic and basic media.

Basic / nucleophilic conditions (strong nucleophile, no acid): the nucleophile attacks by a clean $\text{S}_\text{N}2$ at the less hindered (less substituted) carbon, because sterics control $\text{S}_\text{N}2$ . Backside attack inverts that carbon.

Acidic conditions (acid catalyst protonates the epoxide oxygen first): the nucleophile attacks the more hindered (more substituted) carbon. Why the reversal? Protonation gives the C-O bonds substantial positive character; the more substituted carbon bears the larger share of that $\delta+$ (it would form the more stable carbocation), so it becomes the electrophilic target. The transition state has $\text{S}_\text{N}1$ -like character, even though both C-O bonds are still partly intact.

Condition	Site of attack	Why
Basic (e.g. $\text{NaOCH}_3$ , $\text{NaOH}$ , $\text{RMgX}$ , $\text{LiAlH}_4$ )

In both cases the nucleophile and the resulting $\text{-OH}$ end up trans (anti) across the former ring, because attack is always backside (anti) to the breaking C-O bond.

Worked example — the canonical comparison. Open 2,2-dimethyloxirane (isobutylene oxide; one carbon is quaternary-substituted with two methyls, the other is a $\text{CH}_2$ ) with methanol.

Basic methoxide ( $\text{NaOCH}_3$ ): attack at the less substituted $\text{CH}_2$ . Product: $\text{(CH}_3\text{)}_2\text{C(OH)-CH}_2\text{-OCH}_3$ — the methoxy ends up on the substituted carbon and the on the substituted (tertiary) carbon.

Same epoxide, same nucleophile, opposite regiochemistry — entirely because of acid vs base. This single example is worth memorizing as a template.

Checkpoint — Ring-Opening Regiochemistry

Exit Ticket — Part 5 Synthesis

2-butanol (secondary).

2-methylpropan-2-ol (tertiary). No reaction with either oxidant. The carbinol carbon has no C-H bond, so oxidation is impossible.

Alcohol	PCC	Jones
1-butanol ( $1°$ )	butanal	butanoic acid
2-butanol ( $2°$ )	butan-2-one	butan-2-one
tert-butanol ( $3°$ )	no reaction	no reaction

The whole problem reduces to two rules: (1) tertiary alcohols do not oxidize, and (2) for primary alcohols, mild/anhydrous (PCC) stops at the aldehyde while strong/aqueous (Jones) goes to the acid.

Checkpoint — Oxidation & Selectivity

Problem 2 — Designing a Grignard Synthesis

Prompt. Propose a synthesis of 1-phenylethanol, $\text{Ph-CH(OH)-CH}_3$ , using a Grignard reaction. Then explain why a free phenol could not be present during the reaction.

Retrosynthesis. 1-Phenylethanol is a secondary alcohol (the carbinol carbon bears one H, one phenyl, one methyl). A secondary alcohol comes from a Grignard adding to a non-formaldehyde aldehyde. Disconnect at either C-C bond to the carbinol carbon:

Route A: $\text{PhMgBr}$ (phenyl nucleophile) $+$ acetaldehyde ( $\text{CH}_3\text{CHO}$ ), then $\text{H}_3\text{O}^+$ . The phenyl adds to the aldehyde carbonyl; workup gives 1-phenylethanol. ✓
Route B: $\text{CH}_3\text{MgBr}$ (methyl nucleophile) $+$ benzaldehyde ( $\text{PhCHO}$ ), then $\text{H}_3\text{O}^+$ . Equally valid. ✓

Either route works because both partners are aldehydes (one H on the carbonyl carbon) giving the secondary product.

Why no free phenol (or any $\text{-OH}$ ). A Grignard reagent is a strong base. A phenol $\text{O-H}$ (or any hydroxyl) would protonate and destroy the carbanion in an instant acid-base reaction, before it could add to the carbonyl. If the target molecule needs a free $\text{-OH}$ elsewhere, you must either install it after the Grignard step or use a protecting group.

The planning reflex: secondary target $\Rightarrow$ aldehyde electrophile; and scan the molecule for any acidic O-H, N-H, or terminal alkyne C-H that would quench the reagent.

Problem 3 — Epoxide Regiochemistry Under Two Conditions

Prompt. 1-Methyl-1,2-epoxycyclohexane (an epoxide where one ring carbon also bears a methyl, making it the more substituted carbon) is opened with methanol. Compare the product under (a) acidic and (b) basic conditions.

Set-up. One epoxide carbon is the more substituted (quaternary-ish, bearing the methyl); the adjacent carbon is the less substituted (a $\text{CH}$ ).

(a) Acidic ( $\text{CH}_3\text{OH}, \text{H}_2\text{SO}_4$ ): protonation puts the larger $\delta+$ on the more substituted carbon, so methanol attacks there. The $\text{-OCH}_3$ lands on the carbon bearing the methyl; the $\text{-OH}$ ends up on the less substituted carbon — trans to the new $\text{OCH}_3$ .
(b) Basic ( $\text{NaOCH}_3$ ): methoxide is a strong nucleophile and attacks by $\text{S}_\text{N}2$ at the less hindered carbon. The $\text{-OCH}_3$ lands on the less substituted carbon; the tertiary ends up on the more substituted carbon — again .

Result: the constitutional placement of $\text{OCH}_3$ versus $\text{OH}$ is opposite in the two conditions, while the trans relationship is preserved in both.

Two takeaways carried from Part 5: (1) acid sends the nucleophile to the more substituted carbon, base to the less substituted; (2) the nucleophile and -OH are always anti (trans), since attack is backside in both regimes.

Checkpoint — Synthesis Design & Epoxides

Exit Ticket — Part 6 Synthesis

Moving right is oxidation (remove H / add O): use PCC, Jones, $\text{CrO}_3$ , $\text{KMnO}_4$ .
Moving left is reduction (add H): use $\text{NaBH}_4$ or $\text{LiAlH}_4$ .

The class of the alcohol fixes where on the ladder it can go:

Alcohol	One rung up (mild)	Two rungs up (strong)
primary	aldehyde	carboxylic acid
secondary	ketone (top of its track)	ketone
tertiary	cannot oxidize	cannot oxidize

Reduction simply reverses these arrows: $\text{LiAlH}_4$ takes an acid all the way down to a primary alcohol; $\text{NaBH}_4$ takes a ketone to a secondary alcohol. Seeing oxidation and reduction as two directions on one ladder, rather than as separate lists of reagents, is the single highest-leverage mental model in the unit.

The Alcohol Reaction Map (Recall Check)

Use this as a self-quiz: for each transformation, can you state the reagent from memory?

Making alcohols (Part 2):

alkene $\rightarrow$ Markovnikov alcohol: oxymercuration ( $\text{Hg(OAc)}_2$ , $\text{H}_2\text{O}$ ; $\text{NaBH}_4$ ) or acid hydration
alkene $\rightarrow$ anti-Markovnikov alcohol: hydroboration ( $\text{BH}_3$ ; $\text{H}_2\text{O}_2/\text{OH}^-$ )
carbonyl $\rightarrow$ alcohol: $\text{NaBH}_4$ or $\text{LiAlH}_4$
carbonyl $+$ new C-C bond $\rightarrow$ alcohol: $\text{RMgX}$ , then $\text{H}_3\text{O}^+$

Reactions of alcohols (Part 3):

alcohol $\rightarrow$ alkyl halide: $\text{HX}$ , or $\text{SOCl}_2$ / $\text{PBr}_3$ (no rearrangement)

Ethers & epoxides (Parts 4-5):

alcohol $\rightarrow$ ether: Williamson (alkoxide $+$ methyl/ $1°$ halide, $\text{S}_\text{N}2$ )
ether $\rightarrow$ two alkyl halides: excess hot /

Checkpoint — The Big Map

Multi-Step Synthesis: Chaining the Reactions

Prompt. Starting from 1-butene and any needed reagents, synthesize butanoic acid ( $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ ).

The target is a four-carbon carboxylic acid with the acid at a terminal (primary) position. Working backward, a carboxylic acid at a terminal carbon comes from oxidizing a primary alcohol. So we need 1-butanol first — and to put the $\text{-OH}$ on the terminal carbon of 1-butene, we need anti-Markovnikov hydration.

Forward synthesis:

1-butene $\rightarrow$ 1-butanol. $\text{BH}_3 \cdot \text{THF}$ ; then $\text{H}_2\text{O}_2$ , . Hydroboration installs the anti-Markovnikov, i.e. on the terminal carbon, giving the alcohol 1-butanol. (Acid hydration would give 2-butanol — the wrong isomer.)

Why each choice matters: the regiochemistry decision (step 1, anti-Markovnikov to get a primary alcohol) and the oxidant strength decision (step 2, strong to reach the acid) are exactly the two judgment calls this unit drills. Swap either and you get the wrong product.

A clean two-step route, and every arrow is a reaction from this suite: hydroboration (Part 2) then strong oxidation (Part 3), organized by the oxidation ladder (Part 7).

Checkpoint — Multi-Step Reasoning

Exit Ticket — Unit Synthesis

\text{p}K_\text{a}

\text{p}K_\text{a}\approx 4

H_{2} O_{2} / OH^{-}

Methyl benzoate

+\,\text{LiAlH}_4

, then

\text{H}_3\text{O}^+ \rightarrow

benzyl alcohol

+

methanol. (

\text{NaBH}_4

would leave the ester untouched.)

3\,\text{R-OH} + \text{PBr}_3 \rightarrow 3\,\text{R-Br}

2-methylbut-1-ene

trisubstituted 2-methylbut-2-ene predominates

2-propanol

+\,\text{Jones} \rightarrow

acetone (a ketone; cannot oxidize further).

2-methylpropan-2-ol

+\,\text{Jones} \rightarrow

no reaction (tertiary; no C-H on the carbinol carbon).

+ (CH_{3})_{3} C-Br

only isobutylene by E2

Acidic methanol ( $\text{CH}_3\text{OH}, \text{H}_2\text{SO}_4$ ): attack at the more substituted carbon. Product:

\text{(CH}_3\text{)}_2\text{C(OCH}_3\text{)-CH}_2\text{-OH}

— now the methoxy is on the more substituted carbon and the

\text{-OH}

on the

\text{CH}_2

alcohol

\rightarrow

alkene: conc.

\text{H}_2\text{SO}_4

, heat (Zaitsev; E1 for

2°/3°

, E2 for

1°

)