Organic Chemistry interview questions

What a strong answer covers

• Identify the acidic proton in each compound and the corresponding conjugate base stability.
• Ethanol (pKa ~16) is least acidic because the ethoxide anion is destabilized by the electron-donating alkyl group and lack of resonance.
• Acetic acid (pKa ~4.8) is more acidic due to resonance stabilization of the acetate anion.
• Phenol (pKa ~10) is less acidic than acetic acid but more than ethanol because the phenoxide anion is resonance-stabilized, though the aromatic ring is less effective than a carbonyl.
• Trifluoroacetic acid (pKa ~0) is most acidic because the three fluorine atoms strongly inductively withdraw electron density, stabilizing the conjugate base.

Where people lose the point

• ×Assuming phenol is more acidic than carboxylic acids because of aromaticity.
• ×Ignoring inductive effects of fluorine and focusing only on resonance.
• ×Forgetting that alkyl groups are electron-donating and destabilize anions.

What a strong answer covers

• SN2 is a one-step concerted mechanism with backside attack, leading to inversion of stereochemistry; rate depends on both substrate and nucleophile (second-order).
• SN1 is a two-step mechanism involving carbocation formation, leading to racemization; rate depends only on substrate (first-order).
• SN2 favored by primary substrates, strong nucleophiles, polar aprotic solvents (e.g., DMSO, acetone), and good leaving groups.
• SN1 favored by tertiary substrates, weak nucleophiles, polar protic solvents (e.g., water, alcohols), and good leaving groups.
• Steric hindrance and carbocation stability are key determinants; rearrangements possible in SN1.

Where people lose the point

• ×Stating that SN1 always gives complete racemization (often slight inversion due to ion pairing).
• ×Confusing the role of solvent: polar protic stabilizes carbocations, polar aprotic enhances nucleophilicity.
• ×Forgetting that allylic and benzylic substrates can undergo SN1 even if primary due to resonance-stabilized carbocations.

What a strong answer covers

• Toluene has a methyl group, an electron-donating activator, so nitration is faster than benzene and gives ortho and para products.
• The methyl group stabilizes the cationic intermediate through hyperconjugation and inductive donation, with ortho/para attack favored.
• Nitrobenzene has a nitro group, a strong electron-withdrawing deactivator, so nitration is slower and gives meta product.
• The nitro group destabilizes the ortho/para intermediates (positive charge adjacent to electron-withdrawing group) but less so for meta.
• Consider steric effects: ortho product may be minor for bulky groups, but methyl is small so both ortho and para form.

Where people lose the point

• ×Assuming halogens are meta directors because they are deactivators (they are ortho/para directors due to resonance).
• ×Forgetting that nitrobenzene reacts slower than benzene, not faster.
• ×Ignoring that multiple substitution can occur under harsh conditions.

What a strong answer covers

• Draw 2-butanol: CH3-CH(OH)-CH2-CH3. The chiral center is C2, bonded to OH, CH3, CH2CH3, and H.
• Prioritize groups by CIP rules: OH (highest, O atomic number 8), then CH2CH3 (C bonded to C,H,H), then CH3 (C bonded to H,H,H), then H (lowest).
• Orient the molecule so the lowest priority group (H) points away from you.
• Trace the path from 1 (OH) to 2 (CH2CH3) to 3 (CH3). If clockwise, it's R; if counterclockwise, S.
• For (R)-2-butanol, the path is clockwise when H is pointing away.

Where people lose the point

• ×Forgetting to orient the lowest priority group away before tracing.
• ×Misapplying CIP rules when comparing CH2CH3 vs CH3 (CH2CH3 has higher priority because the first carbon is bonded to another carbon).
• ×Confusing R/S with d/l or +/- notations.

What a strong answer covers

• Protonation of the alkene: H+ adds to the less substituted carbon to form the more stable secondary carbocation (Markovnikov's rule).
• Bromide ion attacks the carbocation from either face, leading to a racemic mixture of 2-bromopropane (no stereocenter, so not applicable).
• For a more substituted alkene like 2-butene, the carbocation is secondary and bromide attack gives a racemic mixture of 2-bromobutane (stereocenter at C2).
• In the presence of peroxides, radical mechanism gives anti-Markovnikov addition due to bromine radical addition to the less substituted carbon.
• Show curved arrows: π bond attacks H, forming C-H bond and carbocation; then Br- attacks carbocation.

Where people lose the point

• ×Applying Markovnikov's rule without considering carbocation rearrangement (e.g., hydride shift).
• ×Forgetting that the reaction is stereospecific only in the radical case (anti addition).
• ×Drawing the bromide attacking before the carbocation is fully formed (concerted addition is not correct for HBr).

What a strong answer covers

• Steric factors: aldehydes have only one alkyl group (or hydrogen) attached to the carbonyl carbon, while ketones have two, causing more steric hindrance in the tetrahedral intermediate.
• Electronic factors: alkyl groups are electron-donating, which reduces the electrophilicity of the carbonyl carbon in ketones.
• The tetrahedral intermediate from aldehydes is less crowded and more stable.
• Aldehydes also have a hydrogen on the carbonyl carbon, which can stabilize the transition state through hyperconjugation? Actually, the hydrogen is smaller and less electron-donating than an alkyl group.
• In general, the order of reactivity is: formaldehyde > aldehyde > ketone > ester > amide > carboxylic acid.

Where people lose the point

• ×Claiming that aldehydes are more reactive solely because of electronic effects, ignoring sterics.
• ×Forgetting that resonance in amides and esters further reduces reactivity.
• ×Assuming all ketones have the same reactivity (cyclic ketones are more reactive due to ring strain).

What a strong answer covers

• Hückel's rule: a planar, cyclic, fully conjugated molecule with 4n+2 π electrons (n = 0,1,2,...) is aromatic; with 4n π electrons is antiaromatic; if nonplanar or not fully conjugated, it is nonaromatic.
• Cyclobutadiene has 4 π electrons (4n, n=1), is planar and conjugated, so it is antiaromatic and highly unstable.
• Benzene has 6 π electrons (4n+2, n=1), is planar and conjugated, so it is aromatic and very stable.
• Cyclooctatetraene has 8 π electrons (4n, n=2), but it adopts a nonplanar tub-shaped conformation to avoid antiaromaticity, so it is nonaromatic.
• Other examples: cyclopentadienyl anion (6 π e-) is aromatic; cyclopentadienyl cation (4 π e-) is antiaromatic.

Where people lose the point

• ×Assuming all cyclic conjugated molecules are aromatic.
• ×Forgetting that antiaromatic molecules distort to become nonaromatic if possible.
• ×Misapplying the rule to ions without counting electrons correctly.

What a strong answer covers

• Draw the substrate: 2-bromo-2-methylbutane has a tertiary bromide at C2, with adjacent C1 (CH3) and C3 (CH2CH3).
• E2 requires anti-periplanar geometry; the base abstracts a β-hydrogen anti to the leaving group.
• With a bulky base like tert-butoxide, steric hindrance favors abstraction of the less hindered β-hydrogen (from C1), leading to the less substituted alkene (Hofmann product): 2-methyl-1-butene.
• With a small base like ethoxide, the more substituted alkene (Zaitsev product) predominates: 2-methyl-2-butene.
• Consider that the substrate is tertiary, so E2 competes with SN1/E1; strong bulky base favors E2.

Where people lose the point

• ×Assuming Zaitsev's rule always applies, ignoring base bulkiness.
• ×Forgetting that E2 requires anti-periplanar arrangement, which may limit accessible β-hydrogens.
• ×Confusing Hofmann elimination (from quaternary ammonium salts) with Hofmann product in E2.

What a strong answer covers

• Ethyl acetate has four distinct proton environments: CH3COO (acetyl methyl), OCH2 (methylene), CH3CH2 (terminal methyl), and no exchangeable protons.
• Acetyl methyl: singlet, δ ~2.0 ppm, integration 3H, no adjacent protons.
• OCH2: quartet, δ ~4.1 ppm, integration 2H, coupled to adjacent CH3 (3 neighbors → n+1=4).
• Terminal CH3: triplet, δ ~1.3 ppm, integration 3H, coupled to OCH2 (2 neighbors → n+1=3).
• Total integration 3+2+3=8H, consistent with molecular formula C4H8O2.

Where people lose the point

• ×Forgetting that the acetyl methyl is a singlet because it has no adjacent hydrogens.
• ×Misassigning the chemical shift of OCH2 (should be downfield due to oxygen).
• ×Assuming coupling between non-adjacent protons (e.g., acetyl methyl and OCH2).

What a strong answer covers

• Target: 2-phenyl-2-butanol (Ph-C(OH)(CH3)(CH2CH3)). Retrosynthetic disconnection: a Grignard addition to a ketone.
• Disconnect the C–C bond between the phenyl and the carbinol carbon: phenyl Grignard (PhMgBr) + 2-butanone.
• Alternatively, disconnect between the carbinol carbon and ethyl group: ethyl Grignard + acetophenone.
• Synthesize PhMgBr from bromobenzene and Mg in dry ether; 2-butanone is commercially available.
• Reaction: PhMgBr + 2-butanone → after acidic workup, 2-phenyl-2-butanol. Consider side reactions: enolization if ketone is added to Grignard too quickly; use inverse addition.

Where people lose the point

• ×Forgetting that Grignard reagents are destroyed by protic solvents or water.
• ×Choosing a disconnection that requires an unstable starting material.
• ×Ignoring that the tertiary alcohol product can undergo dehydration under acidic workup.

What a strong answer covers

• Protecting groups temporarily mask reactive functional groups to prevent unwanted reactions during subsequent steps.
• For alcohols: silyl ethers (e.g., TBS, TBDPS) are common; they are stable to many conditions and removed by fluoride (TBAF) or acid.
• For amines: Boc (tert-butoxycarbonyl) is common; stable to base and mild acid, removed by strong acid (TFA).
• Other examples: acetals/ketals for diols, benzyl ethers for alcohols (removed by hydrogenolysis).
• Criteria for a good protecting group: easy to install, stable to intended reaction conditions, selectively removable in high yield.

Where people lose the point

• ×Choosing a protecting group that is not orthogonal to other protecting groups in the molecule.
• ×Forgetting that protecting group removal must not affect other functional groups.
• ×Assuming all silyl ethers have the same stability (TMS is labile, TBS is more robust).

What a strong answer covers

• Stability order: methyl < primary < secondary < tertiary < allylic (and benzylic).
• Alkyl radicals are stabilized by hyperconjugation and inductive effects from adjacent alkyl groups; more substituted radicals are more stable.
• Allylic and benzylic radicals are further stabilized by resonance delocalization of the unpaired electron into the π system.
• Tertiary radicals are more stable than secondary, which are more stable than primary, due to increased hyperconjugation.
• This trend parallels carbocation stability but radicals are neutral and less sensitive to electronic effects.

Where people lose the point

• ×Confusing radical stability with carbocation stability (radicals are neutral, carbocations are charged).
• ×Forgetting that allylic radicals have two resonance structures, making them exceptionally stable.
• ×Assuming vinyl and aryl radicals are stable (they are very unstable due to sp2 hybridization).

What a strong answer covers

• Base deprotonates the α-carbon of one acetaldehyde to form an enolate (nucleophile).
• The enolate attacks the carbonyl carbon of a second acetaldehyde (electrophile) to form a β-hydroxy aldehyde (aldol addition product).
• The aldol addition product is often in equilibrium; heating or base can promote dehydration (condensation) via E1cb mechanism.
• Dehydration involves deprotonation of the α-carbon, elimination of hydroxide, and formation of a conjugated enal (crotonaldehyde).
• Overall: 2 CH3CHO → CH3CH(OH)CH2CHO → CH3CH=CHCHO + H2O.

Where people lose the point

• ×Forgetting that the first step is reversible and requires a base strong enough to form the enolate.
• ×Assuming the aldol addition product is always isolated; it often dehydrates under reaction conditions.
• ×Confusing aldol condensation with Claisen condensation (esters instead of aldehydes/ketones).

What a strong answer covers

• Diels-Alder is a [4+2] cycloaddition between a conjugated diene (1,3-butadiene) and a dienophile (maleic anhydride).
• The mechanism is concerted, with a cyclic transition state; no intermediates. Electron flow: diene π bonds form new σ bonds to dienophile.
• Stereochemistry: the reaction is stereospecific—cis dienophile gives cis product; endo rule favors the endo product when the dienophile has electron-withdrawing groups.
• Regiochemistry: for unsymmetrical dienes/dienophiles, 'ortho' rule applies; electron-withdrawing groups on dienophile direct to the 2-position of the diene.
• The product is a cyclohexene derivative; maleic anhydride gives a bicyclic adduct after hydrolysis.

Where people lose the point

• ×Forgetting that the diene must be in s-cis conformation for the reaction to occur.
• ×Assuming the reaction is stepwise (it is concerted).
• ×Ignoring the endo/exo selectivity, which is kinetic and favored by secondary orbital interactions.

What a strong answer covers

• Cyclohexanone has a carbonyl group (C=O) with a strong absorption around 1715 cm⁻¹ (ketone, non-conjugated).
• Cyclohexanol has an alcohol O–H group with a broad absorption around 3200–3600 cm⁻¹.
• Cyclohexanol also has C–O stretch around 1000–1200 cm⁻¹, but the O–H is most diagnostic.
• Cyclohexanone lacks O–H absorption; cyclohexanol lacks C=O absorption.
• Both have C–H stretches around 2850–3000 cm⁻¹, so those are not distinguishing.

Where people lose the point

• ×Confusing the broad O–H stretch with the sharp N–H stretch.
• ×Forgetting that the C=O stretch for a ketone is around 1715 cm⁻¹, while for an aldehyde it's ~1725 cm⁻¹ and for an ester ~1735 cm⁻¹.
• ×Assuming that the absence of O–H automatically means a ketone (could be an ether).

What a strong answer covers

• Cyclopentadienyl anion has 6 π electrons (4n+2, n=1) and is planar, cyclic, and fully conjugated, so it is aromatic.
• The anion is stabilized by resonance and is a common ligand in organometallic chemistry (e.g., ferrocene).
• Cyclopentadienyl cation has 4 π electrons (4n, n=1) and is planar, cyclic, and fully conjugated, so it is antiaromatic.
• Antiaromaticity makes the cation highly unstable and difficult to observe.
• The difference is one electron: the anion has a filled π shell, the cation has a half-filled shell leading to destabilization.

Where people lose the point

• ×Forgetting that antiaromatic compounds are destabilized relative to open-chain analogs.
• ×Assuming that all charged cyclic conjugated systems are aromatic.
• ×Misapplying Hückel's rule by not counting electrons correctly (e.g., forgetting the negative charge adds two electrons).

What a strong answer covers

• Target: PhCH2CH2COCH3. Retrosynthetic disconnection: a β-keto ester synthesis or a Grignard reaction.
• Route 1: Benzyl bromide + Mg → PhCH2MgBr (Grignard). Then react with ethylene oxide to give PhCH2CH2CH2OH, oxidize to aldehyde, then add methyl Grignard and oxidize? That's longer.
• Better route: Use acetoacetic ester synthesis. Alkylate ethyl acetoacetate with benzyl bromide, then hydrolyze and decarboxylate to give PhCH2CH2COCH3.
• Steps: (1) NaOEt, ethyl acetoacetate; (2) benzyl bromide → alkylated β-keto ester; (3) NaOH, H2O, heat → hydrolysis and decarboxylation → 4-phenyl-2-butanone.
• Alternative: PhCH2MgBr + acetaldehyde → secondary alcohol, then oxidize to ketone? That gives PhCH2CH(OH)CH3, oxidation gives PhCH2COCH3 (1-phenyl-2-propanone), not the target. So acetoacetic ester is better.

Where people lose the point

• ×Using a Grignard on an ester to make a ketone (over-addition to tertiary alcohol).
• ×Forgetting that the acetoacetic ester synthesis requires a strong base and careful control of pH during decarboxylation.
• ×Choosing a route that gives a regioisomer (e.g., 1-phenyl-2-propanone instead of 4-phenyl-2-butanone).