← Explore exam analytics

Alcohols Phenols and Ethers · TOPIC LESSON

Ether Synthesis and Reactions

Choose which carbon is attacked, preserve the oxygen-bearing fragment, and distinguish ordinary ethers from strained epoxides.

JEE Main · JEE Advanced · NEET UG46 min study guide10 indexed sections

BY THE END, YOU CAN

  • Read ether connectivity and distinguish aryl, alkyl, vinyl and cyclic cases.
  • Design Williamson syntheses that favour substitution over elimination.
  • Predict acidic cleavage from substrate class and identify the oxygen-retaining fragment.
  • Explain epoxide formation and condition-dependent ring-opening regioselectivity.
  • Track allyl atoms through a Claisen rearrangement.
  • Recognize crown-ether coordination and the limits of ether inertness.

BEFORE YOU START

  • SN1/SN2/E1/E2 mechanisms and stereochemical inversion.
  • Alcohol and phenol acidity, alkoxides and carbonyl atom counting.
  • Curved electron-pair arrows and basic aromaticity.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. Read the two groups attached to oxygen

An ether has an oxygen bonded to two organic groups, R–O–R′. The oxygen is ordinarily neutral with two sigma bonds and two lone pairs. An alcohol instead has an O–H bond, and a peroxide has an O–O bond. An ester contains an acyl group bonded to oxygen and follows carbonyl chemistry rather than simply behaving as a dialkyl ether.

For a simple name such as 1-ethoxypropane, the ethoxy group is –OCH₂CH₃ attached to a propyl framework. Write the full connectivity before deciding which fragment is methyl, primary, secondary or tertiary. The group on each side is classified at the carbon directly attached to oxygen, not at a remote branch.

An aryl ether has oxygen directly attached to an aromatic carbon, as in anisole C₆H₅OCH₃. A benzyl ether has an intervening CH₂, as in C₆H₅CH₂OR. Those are mechanistically different: the aryl C–O bond is not a normal SN2 site, while a benzylic carbon can be highly reactive in substitution.

An ether oxygen can accept hydrogen bonds and coordinate Lewis acids, but an ordinary ether has no O–H group to donate hydrogen bonds. Consequently, an ether and its isomeric alcohol can have different boiling behaviour. Solubility depends on the balance between oxygen’s interactions and the size of the hydrocarbon portion; one oxygen does not make every ether completely water-miscible.

Cyclic ethers keep the same oxygen connectivity but add ring geometry. Tetrahydrofuran has a five-membered ring with one oxygen, while an epoxide has a three-membered O–C–C ring. The strain of an epoxide creates reactivity absent from many larger unstrained ethers. Do not apply every rule for diethyl ether to an epoxide.

02 / UNDERSTAND & APPLY

2. Design a Williamson reaction around an accessible electrophile

Williamson ether synthesis joins an alkoxide or phenoxide nucleophile to a suitable alkyl electrophile through SN2 displacement. In its simplest form, RO⁻+R′X→ROR′+X⁻. The oxygen’s lone pair forms the new O–C bond while the C–X pair leaves with X. The nucleophilic oxygen initially has one carbon bond; the ether product has two.

The electrophilic carbon should usually be methyl or an unhindered primary centre. Secondary substrates can give substantial elimination, and tertiary substrates cannot undergo ordinary backside SN2 substitution. A bulky alkoxide can still react successfully at methyl, while a small alkoxide cannot make a tertiary electrophile SN2-compatible.

For an unsymmetrical ether, there are often two formal disconnections. Choose the one that places the less hindered carbon on the leaving-group-bearing fragment. “More hindered alkoxide plus less hindered halide” is a useful starting rule, but also inspect resonance, neighbouring branching and competing functional groups.

Aryl and vinyl halides do not undergo ordinary Williamson SN2 at their sp² carbon. To make an aryl alkyl ether, use the phenoxide as nucleophile and put the leaving group on the alkyl side. Other aryl-ether coupling or activated aromatic-substitution methods exist, but they are different mechanisms and require their own conditions.

An electrophile described as primary can still be unusually hindered. A neopentyl centre has a crowded adjacent quaternary carbon, which slows backside attack. Thus classification alone is not enough; examine the approach path. Sulfonate leaving groups can substitute for halides in suitable Williamson substrates.

ORIGINAL WORKED EXAMPLE

Choose the workable disconnection

Design a Williamson route to ethyl cyclohexyl ether and explain why the reversed pairing is less attractive.

  1. The target is cyclohexyl–O–CH₂CH₃.
  2. Choose cyclohexoxide as nucleophile and an ethyl halide or suitable ethyl sulfonate as electrophile.
  3. The new bond forms at a primary ethyl carbon.
  4. The reversed pairing would require ethoxide attacking a secondary cyclohexyl electrophile, where elimination and geometric hindrance compete.

Answer Cyclohexoxide + a suitable ethyl electrophile is the preferred SN2 disconnection.

03 / UNDERSTAND & APPLY

3. Track the carbon that changes its bond

In an SN2 ether formation at a stereogenic electrophilic carbon, backside attack inverts that carbon’s geometry. A stereocentre on the alkoxide fragment is not inverted merely because its oxygen reacts; its existing C–O bond remains intact. R/S labels must be reassigned after substituent priorities change rather than automatically swapped.

An intramolecular Williamson reaction can form a cyclic ether from a molecule containing an alkoxide and a suitably placed leaving group. Ring size is counted along the new cycle including oxygen. A chain O⁻–CH₂–CH₂–CH₂–CH₂–X can close to a five-membered ring containing O and four carbons.

Five- and six-membered rings are often accessible because they balance geometry and strain favourably, but concentration, competing intermolecular reaction and substrate structure also matter. Very small rings can form when the substrate is appropriately arranged, while very large ring closure competes with intermolecular chain growth. No single ring-size rule replaces the actual structure.

Halohydrins provide an important special case: deprotonating OH lets oxygen attack the adjacent leaving-group-bearing carbon intramolecularly to make an epoxide. This step is an SN2 ring closure and requires the appropriate backside geometry. The carbon losing the halide inverts; the carbon originally bonded to oxygen retains its bond.

An epoxide can also be prepared by oxygen transfer to an alkene with a suitable peroxyacid. That concerted addition preserves the relative stereochemical information of the alkene substituents in the resulting epoxide. Halohydrin closure and direct alkene epoxidation are different preparations that can converge on the same ring type.

ORIGINAL WORKED EXAMPLE

Count a cyclic ether correctly

What ring is formed by intramolecular SN2 closure of ⁻O–CH₂–CH₂–CH₂–CH₂–Br, assuming ring closure dominates?

  1. The attacking O is already connected to the first carbon of a four-carbon chain.
  2. It forms a bond to the terminal carbon bearing Br.
  3. The closed loop contains O plus four carbons, for five ring atoms.

Answer Tetrahydrofuran, a five-membered cyclic ether.

04 / UNDERSTAND & APPLY

4. Alcohol dehydration and alkene addition have different scope

Acid-catalysed intermolecular dehydration of suitable primary alcohols can give symmetrical ethers. One alcohol is protonated, another alcohol oxygen attacks the carbon bearing the activated water group, and deprotonation restores a neutral ether. The catalyst is regenerated and the overall balance is 2ROH→ROR+H₂O.

A mixture of two different alcohols can produce both symmetrical ethers and the mixed ether. Without selectivity information, this is not a clean general synthesis of one unsymmetrical product. The count of possible products is not the same as their experimental ratio; ratios depend on concentrations, relative rates and reversibility.

Secondary and tertiary alcohols often eliminate to alkenes under acidic dehydration conditions rather than giving clean ethers. The product choice is governed by competition between substitution and elimination. Simply lowering or raising a memorized temperature is not a universal predictor for every substrate.

Alkoxymercuration followed by reduction adds an alcohol across an alkene with common Markovnikov ether connectivity and avoids an ordinary freely rearranging carbocation pathway. It is the ether analogue of hydration with water. The alcohol supplies the OR group, while the alkene supplies the carbon framework that receives H and OR.

Acid-catalysed addition of an alcohol to a suitable alkene can also give an ether through cationic chemistry, but rearrangement, elimination and reversibility must be considered. A preparation question should name the product structure and mechanism assumptions rather than treating all alcohol-plus-alkene combinations as equally selective.

ORIGINAL WORKED EXAMPLE

Enumerate a mixed-alcohol product set

Under an idealized intermolecular dehydration comparison, methanol and propan-1-ol can combine in all pairings. List the three distinct ether connectivities, without assuming yields.

  1. Two methanol molecules give CH₃OCH₃.
  2. Two propan-1-ol molecules give CH₃CH₂CH₂OCH₂CH₂CH₃.
  3. One of each gives CH₃OCH₂CH₂CH₃.
  4. Reversing the order of the unlike pair gives the same connectivity, not a fourth ether.

Answer Dimethyl ether, di-n-propyl ether and 1-methoxypropane; no product ratio follows without kinetic assumptions.

05 / UNDERSTAND & APPLY

5. Protonation turns an ether fragment into a better leaving group

An ordinary neutral ether is resistant to attack by many nucleophiles because expelling an alkoxide is unfavourable. Protonating oxygen forms an oxonium ion and allows a neutral alcohol to leave when a carbon–oxygen bond breaks. Strong hydrogen halides such as HI and HBr are common textbook reagents for cleavage under suitable conditions.

For an ether with accessible methyl or primary alkyl groups, halide often attacks the less hindered carbon through an SN2 pathway. The attacked carbon becomes an alkyl halide, while the oxygen remains attached to the other carbon fragment and becomes an alcohol. This oxygen-tracking rule is more dependable than memorizing which product is written first.

The first cleavage does not necessarily describe the endpoint under excess reagent and conditions that also convert the resulting alcohol to a halide. For a suitable dialkyl ether, extended cleavage can give two alkyl-halide fragments plus water. A question specifying limited reagent or asking for the initial cleavage products must be distinguished from a complete reaction sequence.

When a tertiary, benzylic or allylic group can support favourable ionization, cleavage may occur on that side through SN1-like or E1 competition. Tertiary alkyl ethers can therefore break at the more substituted group even when the other side is methyl or primary. The identity of the acid and nucleophile affects whether capture or alkene formation dominates.

An aryl–alkyl ether ordinarily cleaves at the alkyl C–O bond in the standard HI/HBr substitution picture. The aryl C–O bond is not displaced by ordinary backside SN2 at an aromatic sp² carbon, and a free phenyl cation is not the normal alternative. Thus anisole gives phenol plus methyl halide in this transformation, rather than an aryl halide plus methanol.

Diaryl ethers are correspondingly resistant to simple alkyl-type ether cleavage. This does not mean no reagent or catalytic method can ever break an aryl–oxygen bond; it specifies the limit of the ordinary introductory hydrogen-halide mechanism.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

SN2 cleavage of protonated methyl butyl etherIodide lone pair attacks methyl in protonated methyl n-butyl ether. The methyl-oxygen bond electron pair moves onto oxonium oxygen, which retains its bonds to H and butyl. Products are methyl iodide and butan-1-ol. Curved arrows begin at the iodide electron pair and breaking bond.Cleavage at methyl retains O on butylShown after protonation of ether oxygenI⁻CH₃O⁺HBuI lone pair → methyl; C—O bond pair → OCH₃—I + HO—BuBu = CH₂CH₂CH₂CH₃Iodomethane + butan-1-olThe oxygen never transfers to the methyl product.Initial cleavage only; further alcohol reaction dependson reagent amount and conditions.A methyl carbon is not a stereocentre.
The protonated ether is shown after acid activation. This figure describes the initial SN2 cleavage, before any subsequent conversion of the alcohol under excess hydrogen halide. Bu denotes n-butyl.

ORIGINAL WORKED EXAMPLE

Follow oxygen in an unsymmetrical cleavage

Predict the initial main products of HI cleavage of methyl n-butyl ether under conditions favouring SN2 cleavage, before further reaction of the alcohol.

  1. Protonate the ether oxygen to activate the C–O bonds.
  2. Iodide attacks the methyl carbon, the least hindered site.
  3. The methyl–oxygen bond breaks, but the oxygen stays bonded to n-butyl.
  4. The products are CH₃I and CH₃CH₂CH₂CH₂OH.

Answer Iodomethane and butan-1-ol as the initial cleavage products.

ORIGINAL WORKED EXAMPLE

Aryl–alkyl cleavage is not aryl substitution

Predict the standard HI cleavage products of 1-propoxybenzene.

  1. The ether is Ph–O–CH₂CH₂CH₃.
  2. Ordinary SN2 attack occurs at the primary propyl carbon, not at the aromatic carbon.
  3. The oxygen remains on the phenyl fragment.

Answer Phenol and 1-iodopropane.

06 / UNDERSTAND & APPLY

6. Ring strain allows nucleophiles to open an epoxide

An epoxide’s three-membered ring has strained bond geometry. Opening relieves strain, making substitution feasible even though an ordinary unactivated ether would resist the same nucleophile. Under basic or strongly nucleophilic conditions, attack generally occurs at the less hindered epoxide carbon through backside displacement.

The nucleophile forms a bond to one ring carbon while that carbon’s C–O bond breaks and its electron pair goes to oxygen. The other C–O bond remains. After suitable protonation, the original epoxide oxygen becomes an OH group on the carbon that was not attacked. This is the central atom-tracking rule for product prediction.

At the attacked stereogenic carbon, geometry inverts. The unbroken C–O bond preserves its local configuration unless another step changes it. In a cyclic substrate, the new nucleophile and the resulting OH commonly have an anti relationship from backside opening. A flat chain formula alone does not convey that stereochemistry.

Water, alkoxides, amines, hydride donors and suitable carbon nucleophiles can produce different products through epoxide opening, depending on conditions. An alkoxide gives a beta-alkoxy alcohol after protonation; a carbon nucleophile extends the carbon skeleton. Protonation of the final alkoxide is a separate workup step when the reaction medium cannot supply the proton compatibly.

Reaction of a Grignard reagent with ethylene oxide followed by workup provides a primary alcohol with two more carbons than the reagent’s R group: R–CH₂–CH₂–OH. Compare formaldehyde addition, which extends by one carbon. These are distinct homologation patterns that should not be confused.

ORIGINAL WORKED EXAMPLE

Open an unsymmetrical epoxide with methoxide

Predict the main product of methoxide opening of 1,2-epoxybutane under basic conditions, followed by protonation.

  1. The epoxide spans terminal C1 and secondary C2 of the butyl chain.
  2. Methoxide attacks less hindered C1, breaking its bond to epoxide oxygen.
  3. The original oxygen remains on C2 and becomes OH on workup.
  4. The product connectivity is CH₃OCH₂CH(OH)CH₂CH₃.

Answer 1-Methoxybutan-2-ol; inversion applies at the attacked centre if it is stereogenic.

07 / UNDERSTAND & APPLY

7. Acidic activation changes the balance of steric and electronic effects

Acid-catalysed epoxide opening begins with protonation of oxygen. A neutral nucleophile such as water or an alcohol can then attack the activated ring, and deprotonation gives the product while regenerating acid. The transition state combines backside-attack geometry with unequal positive-charge development at the ring carbons.

The shorthand “acid attacks the more substituted carbon” is not universal for every epoxide and nucleophile. Regioselectivity depends on substitution, resonance stabilization, nucleophile and medium. A tertiary or strongly resonance-stabilized epoxide carbon can be preferentially attacked because it better supports developing positive charge; primary/secondary cases can show competing steric control or mixtures.

Do not draw an unconstrained, freely rotating carbocation as the default for every acid-opened epoxide. Backside displacement can retain a strong stereochemical preference even when the more substituted carbon is attacked. “SN1-like electronic preference” does not mean complete SN1 stereochemical scrambling.

For 2,2-dimethyloxirane, one ring carbon is CH₂ and the other carries two methyl groups. Under a strongly nucleophilic basic opening, attack at CH₂ gives NuCH₂C(OH)(CH₃)₂ after workup. Under suitable acidic conditions favouring attack at the tertiary carbon, the connectivity is HOCH₂C(Nu)(CH₃)₂. The original oxygen ends up on opposite carbons in these two products.

Acidic hydrolysis of an epoxide gives a vicinal diol, while opening by an alcohol gives an alkoxy alcohol. When the incoming and original oxygen groups are chemically identical in a diol, the constitutional product may hide which oxygen came from the ring. Isotopic labels or stereochemistry can reveal that distinction.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Condition-dependent opening of a tertiary-substituted epoxide2,2-dimethyloxirane has O bonded to CH2 and a ring carbon bearing two methyl groups. Basic nucleophilic opening attacks CH2 and leaves oxygen on the tertiary carbon, giving Nu-CH2-C(OH)(CH3)2 after protonation. Suitable acidic opening can favour the tertiary carbon, giving HO-CH2-C(Nu)(CH3)2.Epoxide opening: follow the original O2,2-DimethyloxiraneOCH₂CCH₃CH₃less hinderedtertiary carbonBasic / strong-nucleophile modelAttack at CH₂; break its C—O bond.Nu—CH₂—C(OH)(CH₃)₂Original O becomes OH on substituted carbon.Suitable acidic-opening conditionsAttack favours the tertiary carbon in this case.HO—CH₂—C(Nu)(CH₃)₂Original O remains on CH₂ and becomes OH.Proton transfer/workup gives the displayed neutral products.This acid preference is substrate- and condition-specific.
Connectivity comparison for this specified epoxide. Nu labels the incoming group; proton transfers and counterions are omitted. Acidic regioselectivity is not a universal more-substituted rule for every epoxide. Neither attacked carbon here gives a new stereocentre with two identical methyl groups.

08 / UNDERSTAND & APPLY

8. Allyl aryl ethers can rearrange through a six-atom electron cycle

The thermal Claisen rearrangement of an allyl aryl ether is a [3,3]-sigmatropic process, not hydrogen-halide cleavage. A new carbon–carbon bond forms between the terminal allyl carbon and an ortho ring carbon while the oxygen–allyl bond breaks. A nonaromatic cyclohexadienone intermediate then tautomerizes to an ortho-allylphenol when an appropriate ortho position is available.

Label the allyl chain O–C1–C2=C3. In the new aryl–allyl bond, the ring bonds to C3, and the side-chain double bond becomes C2=C1. For an unsubstituted allyl group, the endpoint formula may look like a simple migration, but a label or substituent shows that the allyl end has switched.

The core rearrangement uses a six-membered cyclic transition-state arrangement and three moving electron pairs. It forms and breaks bonds concertedly rather than releasing an allyl cation that later attacks anywhere. The following tautomerization is a separate step that restores aromaticity in the aryl example.

If the two ortho positions are nonequivalent, regioisomeric outcomes can depend on substitution and conditions. When both ortho positions are blocked, some substrates can give para products through additional rearrangement sequences; that is not the same one-step ortho atom map. State the substrate and do not apply an unqualified “always ortho” rule.

Claisen rearrangement of an allyl vinyl ether instead leads to a gamma,delta-unsaturated carbonyl compound. It shares the [3,3] electron-counting pattern but is not a phenol-forming aryl rearrangement. This reaction is also distinct from the Claisen condensation of esters, which is carbonyl/enolate chemistry despite the shared name.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Allyl-end transposition in aryl Claisen rearrangementStarting phenyl allyl ether is labeled at O-linked allyl C1 as carbon13. In the ortho Claisen product the phenolic ring forms its new carbon-carbon bond to original terminal C3, followed by C2 double-bonded to labeled C1. Original oxygen becomes phenolic OH. Ring aromaticity is represented with circles; the intermediate and cyclic transition state are not drawn.Claisen rearrangement switches allyl endsTrack O—C1—C2=C3O¹³CH₂—CH=CH₂C1C2C3The labeled O-linked carbon is C1.heat[3,3] step, then rearomatizing tautomerizationOHCH₂CH¹³CH₂C3C2C1New aryl bond attaches to C3.C1 label becomes the terminal alkene carbon.Original O remains on the ring as phenolic OH.Available ortho position assumed; arrows show net sequence.
Net atom-mapping diagram, not the complete pericyclic electron-flow mechanism. The rearrangement step is followed by tautomerization restoring aromaticity. An available ortho site is assumed. The isotope label tracks a carbon without changing its chemical role.

ORIGINAL WORKED EXAMPLE

Track a labeled allyl end

An allyl phenyl ether has connectivity Ph–O–¹³CH₂–CH=CH₂. In the usual ortho Claisen product, where is the label?

  1. Call the O-linked labeled carbon C1, the middle alkene carbon C2 and terminal carbon C3.
  2. The new ortho aryl–carbon bond forms to C3.
  3. The shifted side chain is ring–CH₂(C3)–CH(C2)=¹³CH₂(C1).
  4. The original ether oxygen remains bonded to the aryl carbon and becomes phenolic OH after tautomerization.

Answer The label is on the terminal alkene carbon of the ortho-allylphenol side chain.

09 / UNDERSTAND & APPLY

9. Ether lone pairs explain coordination, while stability has limits

Ether oxygen lone pairs can donate to Lewis acids and metal cations. This helps explain why certain ethers serve as coordinating media for organometallic chemistry: they can bind a metal without presenting an alcohol O–H proton. Such coordination changes speciation and reactivity, so “solvent” does not always mean a completely uninvolved molecule.

Crown ethers contain several ether oxygens in a ring. In the name 18-crown-6, eighteen is the total ring-atom count and six is the oxygen count. Oxygen donors can surround a cation, with binding depending on cavity size, conformation, solvent and ion identity. The common association of 18-crown-6 with K⁺ is a useful example, not a claim that it binds no other ion.

Complexing a cation can alter ion pairing and the effective reactivity or solubility of its counteranion. The anion should not be imagined as literally unsolvated under all conditions. Crown-ether effects depend on the complete solvent and salt system, and very strong binding is not automatically optimal for every transport or catalytic task.

Ethers can protect an alcohol oxygen against some later transformations. A methyl ether is comparatively resistant to many basic conditions, while a benzyl or silyl protection strategy offers different ways to remove the group. Choosing protection requires checking both installation and removal against all other functional groups; “ether protection” is not a single universal reaction.

Many ethers can form hazardous peroxides during exposure to air over time. This is oxidation chemistry involving O–O-containing products, not conversion to a normal alcohol. The educational consequence is that “ethers are inert” must mean resistant to particular reagents under stated conditions, not immune to every reaction. Suspect peroxide-forming materials require trained institutional handling; no processing procedure is implied here.

10 / UNDERSTAND & APPLY

10. Solved practice: choose the attacked carbon and retain the right oxygen bond

For each problem, circle the electrophilic carbon and draw the C–O bond that breaks or forms. In epoxide opening, only one ring C–O bond breaks. In ordinary aryl–alkyl ether cleavage, the alkyl side is the standard substitution site. In a Claisen rearrangement, neither SN2 selection nor a free carbocation gives the right atom map.

These original teaching cases specify the relevant pathway or conditions where product prediction would otherwise be ambiguous. They are conceptual reaction exercises rather than handling instructions.

ORIGINAL PRACTICE · 1

Which Williamson pairing makes anisole by ordinary SN2: phenoxide plus methyl iodide, or methoxide plus bromobenzene?

Reveal answer & reasoning

Phenoxide plus methyl iodide.

Methyl is an accessible SN2 carbon; bromobenzene’s aromatic sp² carbon is not an ordinary SN2 substrate.

ORIGINAL PRACTICE · 2

Predict the main competing pathway for ethoxide with a tertiary alkyl halide.

Reveal answer & reasoning

E2 elimination is strongly favoured over ordinary SN2 ether formation.

Backside attack at tertiary carbon is blocked, while the strong base can remove a suitable beta hydrogen.

ORIGINAL PRACTICE · 3

An alkoxide oxygen on an enantiopure secondary-carbon fragment attacks methyl iodide. Does the stereogenic carbon on that original alkoxide fragment invert?

Reveal answer & reasoning

No.

The new bond forms between O and methyl. The original stereogenic C–O bond is not broken.

ORIGINAL PRACTICE · 4

What are initial standard HBr cleavage products of phenyl ethyl ether?

Reveal answer & reasoning

Phenol and bromoethane.

Attack occurs at ethyl carbon, leaving oxygen on the phenyl fragment.

ORIGINAL PRACTICE · 5

Can a tertiary alkyl ether cleave at its more substituted carbon under suitable acid conditions?

Reveal answer & reasoning

Yes.

Ionization or elimination involving a stabilized tertiary cationic pathway can override the less-hindered SN2 rule.

ORIGINAL PRACTICE · 6

What is the primary alcohol product from n-propylmagnesium halide plus ethylene oxide and workup?

Reveal answer & reasoning

Pentan-1-ol.

The three-carbon propyl group gains the two-carbon epoxide unit, giving RCH₂CH₂OH with five carbons.

ORIGINAL PRACTICE · 7

Why can an epoxide open with a nucleophile under conditions where a normal unstrained ether resists?

Reveal answer & reasoning

Opening relieves substantial three-membered-ring strain.

The strain reduction changes the energetic balance for breaking one C–O bond while forming the nucleophile–carbon bond.

ORIGINAL PRACTICE · 8

Under basic opening of an unsymmetrical epoxide, which carbon is generally attacked?

Reveal answer & reasoning

The less hindered ring carbon.

The pathway is usually SN2-like, so backside accessibility controls the common regioselectivity.

ORIGINAL PRACTICE · 9

Does acid-catalysed epoxide opening universally involve a freely rotating carbocation and racemization?

Reveal answer & reasoning

No.

The transition state can have carbocation-like electronic character while retaining backside-attack geometry and stereochemical specificity.

ORIGINAL PRACTICE · 10

How many carbon and oxygen atoms are in the ring of 18-crown-6?

Reveal answer & reasoning

Twelve carbon atoms and six oxygen atoms.

The ring contains eighteen atoms total; subtract the six oxygens. In the ordinary structure the carbon units are methylene groups.

ORIGINAL PRACTICE · 11

Does the aryl Claisen rearrangement of PhOCH₂CH=CH₂ add or remove carbon atoms?

Reveal answer & reasoning

Neither.

The nine-carbon skeleton is rearranged; the allyl group makes a new ring C–C bond and oxygen becomes phenolic after tautomerization.

ORIGINAL PRACTICE · 12

After methoxide opens an epoxide and the alkoxide is protonated, must the product contain only one oxygen?

Reveal answer & reasoning

No; it contains two oxygen atoms.

One oxygen came from the epoxide and becomes OH, while the methoxide oxygen remains in the new ether linkage.

KEEP THE BIG IDEAS

Quick revision

  • Williamson synthesis is an SN2 design problem at the electrophilic carbon.
  • Track which fragment retains oxygen during acidic cleavage.
  • Aryl C–O bonds do not undergo ordinary alkyl-type SN2 displacement.
  • Epoxide strain enables ring opening; acid and basic conditions can give different regioselectivity.
  • Claisen rearrangement switches the allyl endpoint and is distinct from ether cleavage.
  • Ether oxygen coordinates cations, but chemical resistance is always condition-specific.
Back to the top ↑

READ WITH CONTEXT

References & scope

The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.

Explore this topic in context

Analytics use a separately reviewed, incomplete historical corpus. Opening a topic does not change its counts or denominator.

KEEP EXPLORING

Your next idea is waiting.

Back to Chemistry index →Browse all topic notes