Organic Chemistry 2 · Aromatic systems
Aromaticity and cyclic conjugation: practice problems
62 problems, each with its worked solution. Open one to read every step, or filter by tag.
Start drawing: Aromaticity and cyclic conjugation
Tap one to draw it yourself with every arrow checked, or watch the steps. No sign-up.
- C1 protonation of azulene to the azulenium ion
- Diels–Alder dimerization of cyclobutadiene (syn dimer)
- Kinetic bromination of cyclooctatetraene
- mCPBA epoxidation of cyclooctatetraene
- Protonation of cyclooctatetraene to the homotropylium ion
- Deprotonation of cyclopenta-1,3-diene by sodium hydride
- SN2 methylation of cyclopentadienide with methyl iodide
- Ionisation of 3-chlorocyclopropene with aluminium chloride
- Deprotonation of fluorene by n-butyllithium
Draw the mechanism 17
- C1 protonation of azulene to the azulenium ionDraw the protonation of azulene at C1 by trifluoroacetic acid that gives the azulenium ion with a tropylium-like seven-membered ring.
- Diels–Alder dimerization of cyclobutadiene (syn dimer)Draw the concerted [4+2] cycloaddition in which one cyclobutadiene adds across another to give syn-tricyclo[4.2.0.0^2,5]octa-3,7-diene.
- Kinetic bromination of cyclooctatetraeneDraw the low-temperature bromination of cyclooctatetraene through the bromohomotropylium ion to cis-7,8-dibromocycloocta-1,3,5-triene.
- mCPBA epoxidation of cyclooctatetraeneDraw the concerted mCPBA epoxidation of one C=C of tub-shaped cyclooctatetraene to cis-9-oxabicyclo[6.1.0]nona-2,4,6-triene.
- Protonation of cyclooctatetraene to the homotropylium ionDraw the C-protonation of cyclooctatetraene by sulfuric acid that gives the homoaromatic homotropylium ion.
- Deprotonation of cyclopenta-1,3-diene by sodium hydrideDraw the proton transfer from the CH2 of cyclopenta-1,3-diene to hydride and one resonance form of the aromatic cyclopentadienide ion.
- SN2 methylation of cyclopentadienide with methyl iodideDraw the SN2 attack of the cyclopentadienide carbon lone pair on methyl iodide that gives 5-methylcyclopenta-1,3-diene and iodide.
- Ionisation of 3-chlorocyclopropene with aluminium chlorideDraw the transfer of chloride from 3-chlorocyclopropene to AlCl3 and the resonance form of the aromatic cyclopropenyl cation.
- Deprotonation of fluorene by n-butyllithiumDraw the proton transfer from the C9–H of fluorene to n-butyllithium that forms the aromatic fluorenyl anion.
- Deprotonation of indene by sodium amideDraw the proton transfer from the CH2 of indene to the amide ion and the resonance form that moves the charge onto the other five-ring carbon.
- N-Methylation of 1-methyl-1H-imidazole by methyl iodideDraw the SN2 attack of the N3 lone pair of 1-methyl-1H-imidazole on methyl iodide that forms the aromatic 1,3-dimethylimidazolium ion.
- Protonation of pyridine by HClDraw the proton transfer from HCl to the nitrogen of pyridine and explain why pyridine, unlike pyrrole, keeps its aromaticity when protonated on nitrogen.
- Triphenylcyclopropenium ion from its methyl ether with HBF4Draw the protonation of 3-methoxy-1,2,3-triphenylcyclopropene and the loss of methanol that forms the aromatic triphenylcyclopropenium ion.
- O-Protonation of tropone to the hydroxytropylium ionDraw the protonation of the tropone oxygen by HCl and the resonance step that shows the hydroxytropylium ion as an aromatic ring cation.
- Ionisation of 7-bromocyclohepta-1,3,5-trieneDraw the heterolysis of the C–Br bond of 7-bromocyclohepta-1,3,5-triene in water and a resonance form of the tropylium ion.
- Hydride abstraction from cycloheptatriene by the trityl cationDraw the hydride transfer from the CH2 of cyclohepta-1,3,5-triene to the trityl cation and a resonance form of the tropylium ion.
- Tropylium + methoxide to 7-methoxycyclohepta-1,3,5-trieneDraw methoxide adding to the tropylium ion to give nonaromatic 7-methoxycyclohepta-1,3,5-triene, the reverse of tropylium formation.
Predict the product 3
- Choose the less disruptive lone pairPredict relative basicity from the aromatic role of a lone pair.
- Cyclooctatetraene + 2 K: the COT dianionPredict the structure of the product formed when cyclooctatetraene is reduced by potassium.
- Which imidazole nitrogen is basic?Predict the protonation site of 1H-imidazole from which nitrogen lone pair is part of the aromatic sextet.
Classify 11
- Which nitrogen pair participates?Assign the pyrrole lone pair to its aromatic circuit.
- Pick the aromatic ringIdentify the aromatic compound among six-membered rings.
- [18]AnnuleneClassify [18]annulene.
- CyclobutadieneClassify cyclobutadiene as aromatic, antiaromatic or nonaromatic.
- CycloheptatrieneClassify cycloheptatriene.
- CyclooctatetraeneClassify cyclooctatetraene and explain its shape.
- The four criteriaState the criteria a molecule must meet to be aromatic.
- Benzene's two Kekulé contributors comparedClassify the two Kekulé structures of benzene by their contribution to the hybrid.
- Where is the lone pair of pyridine?Classify the orbital that holds the nitrogen lone pair of pyridine and say whether it belongs to the aromatic sextet.
- Is thiophene aromatic?Classify thiophene as aromatic, antiaromatic or nonaromatic by counting the sulfur lone pair in its π system.
- Which ion is aromatic?Use π-electron counts to pick the one aromatic ion among four planar cyclic carbocations and carbanions.
Compare 5
- Same framework, different electron occupancyCompare the planar cyclopentadienyl anion and cation.
- Rank hydrogenation heats: cyclohexene to benzeneOrder six-membered rings by heat released on hydrogenation to cyclohexane.
- Naphthalene resonance energy per ringCompare the resonance energy per ring of naphthalene with that of benzene.
- Pyridine vs pyrrole: which nitrogen is more basic?Compare the basicity of pyridine and pyrrole using lone-pair involvement in aromaticity.
- Rank cation stability: aromatic vs antiaromaticRank three conjugated carbocations by stability using aromatic, nonaromatic and antiaromatic π-electron counts.
Calculate 7
- An empty orbital can complete the circuitCount electrons in a conjugated carbocation.
- Cyclooctatetraene dianionCount π electrons in the cyclooctatetraene dianion.
- Cyclopentadienyl anionCount π electrons in the cyclopentadienyl anion.
- Cyclopropenyl cationCount π electrons in the cyclopropenyl cation.
- How many π electrons in naphthalene?Count the π electrons of naphthalene without counting the shared ring-fusion bond twice.
- How many π electrons in pyrrole?Count the π electrons in the aromatic system of pyrrole, including the nitrogen lone pair.
- Benzene resonance energy from hydrogenation heatsCalculate benzene's resonance energy from heats of hydrogenation.
Find the error 4
- A missing orbital in a five-membered ringDistinguish nonaromaticity from antiaromaticity when conjugation is interrupted.
- A ten-electron count is not the whole proofIdentify missing evidence in an aromaticity argument.
- [10]AnnuleneDiagnose why [10]annulene is not aromatic despite 10 π electrons.
- Cyclopropene CH2 acidity and the cyclopropenyl anionDiagnose why cyclopropene is a very weak carbon acid.
Interpret the evidence 8
- Reading a flat drawing cautiouslyInterpret how nonplanarity avoids antiaromaticity.
- [18]Annulene inner protons: NMR shiftPredict the chemical shift of the inner protons of [18]annulene.
- Benzene C–C bond lengths: equal 1.39 Å bondsInterpret benzene's equal C–C bond lengths as evidence for a resonance hybrid.
- Frost circle MO diagram for benzeneInterpret the Frost circle MO diagram of benzene.
- Frost circle for the cyclopentadienyl anionUse a Frost circle to explain why the cyclopentadienyl anion is aromatic.
- Frost circle MO diagram for cyclobutadieneInterpret the Frost circle MO diagram of square cyclobutadiene.
- Furan's two oxygen lone pairsInterpret how the two lone pairs on the oxygen of furan are used, one in the π system and one in the ring plane.
- Benzene 1H NMR at 7.3 ppm: the ring currentExplain the downfield chemical shift of aromatic protons with the ring current.
Plan a synthesis 1
- Make a conjugated anionPlan a charge change that establishes aromatic conjugation.
Explain why 6
- Protonation without losing the π circuitExplain why pyridine protonation need not destroy aromaticity.
- Do not add overlapping sextetsDiagnose double counting in a fused aromatic system.
- Azulene: why the 5-7 ring isomer is blue and polarExplain the dipole moment of azulene with aromatic ion rings.
- Why cyclooctatetraene adopts a tub shapeExplain the nonplanar tub geometry of cyclooctatetraene.
- Cyclobutadiene reactivity: rapid dimerizationExplain the extreme reactivity of cyclobutadiene from its antiaromaticity.
- Why cyclopentadiene's CH2 has a pKa near 16Explain the unusual acidity of cyclopentadiene.
Stuck on your own homework?
Snap it — first scan free, no sign-up
Scanning every week? Scan Standard: for about problems