Choosing the right reagent for an organic chemistry conversion depends on three things: what functional group you’re starting with, what functional group you need to end up with, and whether you need to control selectivity along the way. This question appears on nearly every organic chemistry exam, and the key to answering it is recognizing the type of transformation (oxidation, reduction, substitution, elimination, or carbon-carbon bond formation) and then matching it to the reagent that does exactly that job.
Below is a practical walkthrough of the most common conversions and the specific reagents that accomplish each one.
Oxidation: Alcohols to Aldehydes, Ketones, or Acids
The single most important distinction in oxidation chemistry is whether you need a mild or strong oxidant. Mild oxidants like PCC (pyridinium chlorochromate), Dess-Martin periodinane, and the Swern oxidation will convert a primary alcohol to an aldehyde and stop there. Strong oxidants like chromic acid and potassium permanganate will push a primary alcohol all the way to a carboxylic acid. If the question shows a primary alcohol becoming an aldehyde, you need PCC, DMP, or Swern conditions. If it becomes a carboxylic acid, reach for chromic acid or potassium permanganate.
Secondary alcohols become ketones with either type of oxidant, since ketones can’t be oxidized further under normal conditions. Tertiary alcohols don’t oxidize at all because there’s no hydrogen on the carbon bearing the hydroxyl group.
Reduction: Matching the Reagent to the Target
Reducing agents exist on a spectrum from aggressive to highly selective, and picking the wrong one will over-reduce your product or destroy other functional groups in the molecule.
- Lithium aluminum hydride (LiAlH₄) is the least selective. It reduces carboxylic acids, esters, amides, nitriles, ketones, aldehydes, epoxides, and alkyl halides. If you need to reduce almost anything to an alcohol or amine, this is the reagent.
- Sodium borohydride (NaBH₄) is much milder. It reduces aldehydes and ketones to alcohols but leaves esters, carboxylic acids, nitriles, and epoxides untouched. When combined with cerium chloride (Luche conditions), it selectively reduces the carbonyl of an alpha,beta-unsaturated ketone without touching the carbon-carbon double bond.
- DIBAL-H occupies a useful middle ground. At low temperature, it reduces esters and amides to aldehydes rather than all the way to alcohols. It also converts nitriles to aldehydes through an intermediate imine. If a conversion shows an ester becoming an aldehyde, DIBAL-H at low temperature is almost certainly the answer.
- Lithium borohydride (LiBH₄) selectively reduces esters in the presence of carboxylic acids, amides, and nitriles, filling a niche between NaBH₄ and LiAlH₄.
- Borane (BH₃) reduces carboxylic acids to alcohols while leaving esters, amides, and halides intact, which is the opposite selectivity of NaBH₄.
The exam trick is usually about selectivity. If the starting material has two reducible groups and only one changes, the answer is whichever reagent is selective enough to leave the other group alone.
Alcohol to Alkyl Halide Conversions
Converting an alcohol to a halide requires replacing the hydroxyl group, and the reagent depends on which halide you need. For chlorides, thionyl chloride (SOCl₂) is the standard choice. For bromides, phosphorus tribromide (PBr₃) with pyridine is most common. These reagents activate the hydroxyl group as a leaving group and deliver the halide in one step.
Alcohols can also be converted to sulfonate esters first (using mesyl chloride, tosyl chloride, or triflyl chloride), which turns the poor hydroxyl leaving group into an excellent one. A subsequent reaction with a nucleophilic halide then completes the conversion. This two-step approach gives you more control, especially when you need to invert stereochemistry at the carbon center.
Substitution vs. Elimination: Reagent and Substrate Together Decide
When a conversion shows an alkyl halide reacting, the product might result from substitution (replacing the halide with a new group) or elimination (forming a double bond). The reagent, the substrate, and the solvent together determine which pathway wins.
Strong nucleophiles that are weak bases, such as iodide, cyanide, azide, thiolates, and acetate, favor SN2 substitution, especially in polar aprotic solvents. Strong bases, particularly bulky ones like potassium tert-butoxide, favor E2 elimination. Weak nucleophiles that are also weak bases, like water or methanol, tend to promote SN1 and E1 pathways in protic solvents.
Substrate structure matters just as much. Primary alkyl halides almost always undergo SN2 with a good nucleophile. Tertiary alkyl halides cannot undergo SN2 due to steric hindrance, so they proceed through SN1 or E1 with weak nucleophiles, or E2 with strong bases. Secondary substrates can go either way, making the choice of reagent and solvent the deciding factor. If you see a secondary halide with a strong, unhindered nucleophile in a polar aprotic solvent, SN2 is the answer. If you see a bulky base like tert-butoxide, E2 wins.
Addition to Alkenes: Controlling Regiochemistry
When a conversion adds atoms across a double bond, the question is usually testing whether you know Markovnikov vs. anti-Markovnikov selectivity.
Adding HBr, HCl, or HI to an alkene follows Markovnikov’s rule: the halogen ends up on the more substituted carbon. This happens because the reaction proceeds through the more stable carbocation intermediate. If the conversion instead shows the halogen on the less substituted carbon, the answer is HBr with peroxides, which switches the mechanism to a radical pathway and gives anti-Markovnikov addition. This peroxide-driven reversal only works reliably with HBr, not HCl or HI.
Hydroboration-oxidation (BH₃ followed by hydrogen peroxide and base) adds water across a double bond with anti-Markovnikov regiochemistry and syn stereochemistry. Acid-catalyzed hydration (dilute acid, water) gives Markovnikov addition of water. So if a conversion shows an alcohol forming on the less substituted carbon of a former alkene, hydroboration-oxidation is the reagent set. If the alcohol is on the more substituted carbon, acid-catalyzed hydration is the answer.
Halogenation with bromine or chlorine alone adds across the double bond with anti stereochemistry, producing the trans product. This is worth remembering when the conversion specifies stereochemistry in the product.
Carbon-Carbon Bond Formation
If the product has more carbon atoms than the starting material, a carbon-carbon bond-forming reaction occurred. The most commonly tested is the Grignard reaction.
A Grignard reagent forms when an alkyl or aryl halide reacts with magnesium metal in anhydrous diethyl ether. The resulting organometallic compound then reacts with electrophilic carbons. Adding a Grignard reagent to formaldehyde gives a primary alcohol. Adding it to any other aldehyde gives a secondary alcohol. Adding it to a ketone gives a tertiary alcohol. Adding it to carbon dioxide gives a carboxylic acid (after acidic workup).
The critical detail is that everything must be completely dry. Grignard reagents react instantly with water, destroying the reagent before it can do anything useful. If an exam answer includes a Grignard reaction, the implied conditions always include anhydrous solvent and an inert atmosphere.
Carbonyl Interconversions
Several named reagents handle conversions between different oxidation states of carbonyl compounds. Converting an aldehyde or ketone to an oxime uses hydroxylamine hydrochloride. Dehydrating that oxime to a nitrile requires phosphorus pentoxide. Going the other direction, reducing a nitrile back to an aldehyde uses DIBAL-H at low temperature.
Esterification, converting a carboxylic acid to an ester, is accomplished by Fischer esterification: heating the acid with an alcohol in the presence of an acid catalyst like sulfuric acid. The reaction is reversible, so excess alcohol or removal of water drives it to completion.
Amide bond formation from a carboxylic acid and an amine can be achieved with coupling agents or Lewis acid catalysts. Titanium and zirconium-based catalysts promote this condensation effectively, with zirconium tetrachloride giving yields of 62 to 99 percent at moderate temperatures when molecular sieves are present to absorb the water byproduct.
How to Identify the Conversion Type on an Exam
Start by comparing the starting material and product. Identify which functional group changed and whether the carbon count stayed the same. Then classify the transformation: is it an oxidation (loss of hydrogen or gain of oxygen), a reduction (gain of hydrogen or loss of oxygen), a substitution (one group replaced by another), an elimination (loss of atoms to form a double bond), or an addition (atoms added across a double bond)?
Once you’ve classified it, the reagent choices narrow dramatically. An oxidation of a secondary alcohol to a ketone could use PCC, DMP, Jones reagent, or chromic acid, and any of those would be correct. A reduction of a ketone to an alcohol in the presence of an ester requires NaBH₄ specifically, because LiAlH₄ would reduce both groups. The conversion shown in the question always contains enough clues to eliminate all but one or two reagent options, and recognizing the transformation type is the fastest way to find them.