How to use these organic chemistry practice questions
These organic chemistry practice questions are designed to test recognition, prediction and explanation rather than recall alone. For each one, choose an option and then check the explanation. Pay attention to the distractor that is most tempting: in an exam, that is often the error your reasoning needs to correct.
The set covers substitution, elimination, carbonyl chemistry, aromatic substitution, spectroscopy, acidity and synthesis. The questions are independent, so you can work through them in order or return to the topics where your accuracy is lowest.
1. Predicting an SN2 product
Start with a reaction where the substrate, nucleophile and solvent all point towards one mechanism. The key features are a primary carbon, a strong nucleophile and a polar aprotic solvent.
A question in MySummaries is marked like this:
1-bromopropane is heated with excess sodium iodide in dry acetone. What is the major organic product?
1-iodopropane is formed by an SN2 substitution: iodide attacks the primary carbon bearing bromine and bromide leaves. Prop-1-ene is an elimination product, but elimination is disfavoured for a primary substrate under these conditions. 2-iodopropane would require rearrangement or substitution at the wrong carbon, and water or hydroxide would be needed for propan-1-ol.
The practical check is to identify the carbon attached to the leaving group before thinking about the reagent. Primary substrates usually make backside substitution easier than elimination.
2. Reducing a carbonyl compound
Reagents must be matched to the functional group they can actually reduce. Sodium borohydride is a milder hydride reagent commonly used for aldehydes and ketones.
Here is the second question:
What is the major product when propanal is treated with NaBH4 in methanol, followed by aqueous work-up?
NaBH4 reduces the aldehyde carbonyl of propanal to a primary alcohol, giving propan-1-ol. Propan-2-ol would require reduction of a ketone with the carbonyl at carbon 2. Propanoic acid is an oxidation product, while propane would require removal of oxygen as well as reduction of the carbonyl.
A useful exam habit is to preserve the carbon skeleton first. NaBH4 changes C=O into C–OH; it does not move the carbonyl or remove a carbon.
3. Electrophilic aromatic substitution
In aromatic nitration, the reacting electrophile is generated from concentrated nitric and sulfuric acids. Knowing the attacking species is more reliable than memorising the overall reaction alone.
Which species is the electrophile that attacks benzene during nitration with concentrated HNO3 and H2SO4?
The nitronium ion, NO2+, is the electrophile in nitration. It is generated when sulfuric acid protonates nitric acid and water is lost. Nitrate is a nucleophilic counter-ion rather than the species that forms the new bond to the ring. H2NO3+ is an intermediate in electrophile generation, not the principal attacking species.
When answering mechanism questions, name the role as well as the formula: NO2+ accepts electron density from the aromatic ring, producing the sigma complex before aromaticity is restored.
4. Choosing the major elimination product
Elimination questions require both the alkene position and, where relevant, its stereochemistry. With a secondary haloalkane, a small strong base and heat, the more substituted alkene is usually favoured.
2-bromobutane is heated with sodium ethoxide in ethanol. Which is the major alkene product?
E2 elimination gives mainly but-2-ene because it is the more substituted alkene, and the E isomer is more stable than the Z isomer because the larger groups are on opposite sides. But-1-ene is less substituted. 2-methylpropene cannot be formed without changing the carbon skeleton.
Draw the carbon bearing the leaving group, then inspect each adjacent carbon for a removable hydrogen. This prevents selecting an alkene that cannot form from the starting structure.
5. Following an aldol reaction
Aldol questions often test whether you can distinguish the initial addition product from the dehydrated product. The reaction conditions matter: heating commonly drives dehydration to an α,β-unsaturated carbonyl compound.
Acetaldehyde undergoes self-aldol condensation followed by heating. What is the major dehydrated product?
Two acetaldehyde molecules first form 3-hydroxybutanal. Heating removes water from this β-hydroxy aldehyde to give but-2-enal, also called crotonaldehyde. 3-Hydroxybutanal is the addition product before dehydration, while butan-2-one has the carbonyl in the wrong position.
The fastest route is to join two acetaldehyde carbon skeletons, retain the aldehyde carbonyl, and place the new C=C bond between the α and β carbons after water is removed.
6. Reading an IR spectrum
Spectroscopy questions reward approximate recognition of a functional-group signal, not false precision. A broad, very high-wavenumber absorption is particularly useful for identifying an O–H group.
An IR spectrum shows a very broad absorption from 2500–3300 cm−1 and a strong absorption near 1710 cm−1. Which functional group is most consistent with these signals?
A carboxylic acid has a broad O–H absorption extending roughly from 2500 to 3300 cm−1 and a strong C=O absorption near 1710 cm−1. An aldehyde has a carbonyl signal but not this broad acid O–H band. Amines have narrower N–H absorptions, and alkenes do not account for the strong carbonyl peak.
Use the combination of peaks, not one peak in isolation. The broad acid O–H band plus C=O is much stronger evidence than either signal alone.
7. Extending a carbon chain with a Grignard reagent
Grignard reactions are a common synthesis method because the carbon attached to magnesium behaves as a nucleophile. Carbon dioxide is an electrophile that adds one carbon before acidic work-up.
Methylmagnesium bromide reacts with CO2 and is then treated with dilute acid. What product is formed?
The methyl carbon attacks CO2, forming a carboxylate with two carbons. Acidic work-up gives ethanoic acid. Methanol would result from a different electrophile such as formaldehyde. Propanoic acid has one carbon too many, and methanal is not the product of carboxylation.
Count carbons before and after the reaction. CO2 contributes one carbon, so a one-carbon Grignard reagent gives a two-carbon carboxylic acid.
8. Directing effects in aromatic substitution
Substituents affect both the rate and position of a second electrophilic aromatic substitution. An alkoxy group donates electron density by resonance and directs incoming electrophiles to the ortho and para positions.
Nitration of anisole gives mainly which pair of constitutional isomers?
The methoxy group is activating and ortho/para-directing through resonance donation. Both ortho and para products form, but the para product is commonly favoured because it has less steric crowding between the nitro and methoxy groups. Meta substitution is not the electronically favoured pathway, and the ortho position remains available.
Separate electronic direction from steric preference. Resonance tells you ortho or para; steric effects often help decide which of those two is the larger product.
9. Comparing acidity
Acidity comparisons become clearer when you compare the stability of the conjugate bases. A more stable conjugate base corresponds to a stronger acid and a lower pKa.
Which compound is the stronger acid under ordinary aqueous comparison?
Ethanol is the strongest acid in this list, with a pKa of about 16. Its conjugate base is stabilised by the electronegative oxygen atom. Ethyne is more acidic than ethene or ethane because its sp-hybridised carbon holds negative charge more effectively, but it is still much less acidic than an alcohol.
For a quick ranking, look for charge stabilisation by electronegativity, resonance, induction and hybridisation. Do not rank hydrocarbons by molecular size alone.
10. Interpreting proton NMR splitting
A final question combines chemical environment and integration. Ethyl acetate contains three distinct proton environments, and neighbouring proton counts determine the splitting patterns.
Which set of 1H NMR signals is most consistent with ethyl ethanoate, CH3COOCH2CH3?
The acetyl CH3 is a singlet integrating to 3H because it has no neighbouring hydrogens across the carbonyl. The OCH2 group is a quartet from three neighbouring CH3 hydrogens and integrates to 2H. The terminal CH3 is a triplet from two neighbouring CH2 hydrogens and integrates to 3H. The other patterns do not match the three proton environments and their neighbours.
For splitting, count equivalent hydrogens on adjacent atoms using the n+1 rule as a first approximation. Then check that the integrations add to the molecular formula.
What to do with your score
A result below 70% usually means the issue is not simply a missing fact. Look at the type of error: mechanism selection, structure drawing, reagent recognition, spectroscopy or quantitative comparison. Rework the relevant reaction or spectrum immediately, then attempt a new question on the same skill later.
This review view ranks the areas where marks are being lost:
Do not reread all of organic chemistry after one error. Write a short rule that would have prevented it, apply that rule to a fresh structure, and then return the missed question after a delay.
A repeated error can be reduced to one targeted prompt:
You lost this mark twice: when comparing ethanol, ethyne, ethene and ethane, which conjugate-base feature explains ethanol's lower pKa?
How MySummaries helps
You can turn your own organic chemistry notes into a revision board, generate question sets from that material, and use the weak-area results to decide which reactions or spectra to practise next. Start at MySummaries.