Difficulty: Intermediate to Advanced | Prerequisites: Nucleophilic addition to aldehydes and ketones, carbonyl electrophilicity, basics of carboxylic acid derivatives (see companion notes on SNAc).
Once you know how to build and interconvert carboxylic acid derivatives, the next question is how to move between oxidation states: aldehydes, alcohols, ketones, and carboxylic acids. Reduction (adding H) and oxidation (removing H or adding O) are the tools for that, and Grignard or organolithium reagents let you form new C-C bonds at the same time. This set of reactions is the backbone of the multi-step synthesis problems that dominate Exam 3. Hemiacetal/acetal chemistry and reductive amination round out the picture by showing how carbonyls react with alcohols and amines under controlled conditions.
LiAlH₄ is the sledgehammer that reduces almost any carbonyl all the way to an alcohol (or an amine, from an amide). DIBAL-H is gentler and can stop at the aldehyde stage. PCC oxidises a primary alcohol to an aldehyde without going further. Grignard and organolithium reagents add carbon to carbonyls. Hemiacetals and acetals form when alcohols attack aldehydes or ketones. Reductive amination turns a carbonyl into an amine in one pot.
LiAlH₄ (lithium aluminium hydride, LAH)
A powerful, non-selective reducing agent. Delivers H⁻ to the carbonyl carbon. Reduces carboxylic acids, esters, amides, aldehydes, and ketones. Requires anhydrous conditions, followed by aqueous workup (H₂O). Think of it as: the brute-force reducer that takes everything down to the alcohol (or amine) level.
DIBAL-H (diisobutylaluminium hydride)
A milder reducing agent that can be controlled by temperature. At low temperature (−78 °C), it reduces an ester to an aldehyde (stops at one reduction). At room temperature or with excess, it goes all the way to the alcohol. In simple terms, DIBAL-H is the precision version of LAH.
NaBH₃CN (sodium cyanoborohydride)
A mild, selective reducing agent used specifically in reductive amination. It reduces imines (C=N) but does not reduce carbonyls (C=O) under the same conditions, which is why the reaction works in one pot.
PCC (pyridinium chlorochromate)
An oxidising agent that converts a primary alcohol to an aldehyde and stops there (does not over-oxidise to the carboxylic acid). Also oxidises secondary alcohols to ketones. Think of it as: the gentle oxidiser.
CrO₃/H⁺ (Jones reagent) and H₂CrO₄ (chromic acid)
Stronger oxidising agents that take primary alcohols all the way to carboxylic acids. Secondary alcohols go to ketones.
Grignard reagent (RMgBr)
An organometallic nucleophile that adds a carbon group to a carbonyl. PhMgBr adds a phenyl group; CH₃MgBr adds a methyl group. After aqueous workup (H₃O⁺), you get an alcohol. The new C-C bond is the key.
Organolithium reagent (RLi, e.g., H₃C-Li)
Similar to Grignard but more reactive. Adds an alkyl group to a carbonyl to give an alcohol after workup.
Gilman reagent (lithium dialkylcuprate, R₂CuLi, e.g., Me₂CuLi)
A milder organometallic that performs 1,2-addition to acyl chlorides to give ketones without over-adding. In simple terms, it adds one R group and stops, which Grignard reagents sometimes do not.
Hemiacetal
The product of one equivalent of alcohol adding to an aldehyde or ketone. Contains both an OH and an OR on the same carbon. Usually in equilibrium with the open-chain form.
Acetal
The product formed when a hemiacetal reacts with a second equivalent of alcohol under acid catalysis, replacing the OH with a second OR group. Acetals are stable under basic conditions and are used as protecting groups for carbonyls.
Reductive amination
A one-pot, two-step process: an amine condenses with an aldehyde or ketone to form an imine (C=N), which is then reduced (typically by NaBH₃CN) to give an amine product.
Imine (Schiff base)
A compound containing a C=N double bond, formed by condensation of a primary amine with an aldehyde or ketone, with loss of water.
Suzuki coupling
A palladium-catalysed cross-coupling reaction between a vinyl or aryl halide and an organoboron compound (boronic acid or boronate ester), using Pd(PPh₃)₄ as catalyst and a base (e.g., OH⁻). The reaction retains the stereochemistry of both coupling partners.
LiAlH₄ (LAH) then H₂O workup:
Reduces an ester → primary alcohol
Reduces a carboxylic acid → primary alcohol
Reduces an aldehyde → primary alcohol
Reduces a ketone → secondary alcohol
Reduces an amide → amine (the C=O becomes CH₂, and the nitrogen stays attached)
Reduces a nitrile (R-CN) → primary amine (RCH₂NH₂)
DIBAL-H then H₂O workup:
Reduces an ester → aldehyde (at −78 °C, 1 equivalent)
Can also reduce all the way to an alcohol with excess or warmer conditions
Key exam application (Q2): an ester treated with 1. DIBAL-H, 2. H₂O gives an aldehyde
NaBH₃CN:
Used exclusively in reductive amination to reduce the imine intermediate
Does not reduce simple aldehydes or ketones under these conditions
PCC:
Primary alcohol → aldehyde (stops here)
Secondary alcohol → ketone
Does not affect tertiary alcohols or isolated double bonds
CrO₃/H⁺ or H₂CrO₄ (Jones oxidation):
Primary alcohol → carboxylic acid (goes all the way)
Secondary alcohol → ketone
Exam Q4 tested both: PCC to go from cyclohexanol to cyclohexanone, then Jones-type oxidation where needed
Grignard (RMgBr) + carbonyl, then H₃O⁺:
Formaldehyde + RMgBr → primary alcohol
Aldehyde + RMgBr → secondary alcohol
Ketone + RMgBr → tertiary alcohol
Ester + 2 equiv RMgBr → tertiary alcohol (two R groups added)
Organolithium (RLi) + carbonyl, then H₃O⁺:
Same products as Grignard, but RLi is more reactive
Exam Q2: aldehyde + H₃C-Li then H₃O⁺ gives a secondary alcohol
Gilman reagent (Me₂CuLi) + acyl chloride, then H₂O:
Gives a ketone (adds one methyl group, replaces Cl)
Key point: the cuprate is mild enough to stop at the ketone and not add a second time
NaCN is a nucleophile that displaces a halide via SN2 to install a nitrile (R-Br + NaCN → R-CN)
Nitrile + H₃O⁺ (acid hydrolysis) → carboxylic acid
Nitrile + LAH, then H₂O → primary amine
The nitrile is a one-carbon extension of the chain, which makes it a useful synthetic tool
An aldehyde or ketone + one equivalent of alcohol → hemiacetal (equilibrium)
Hemiacetal + second equivalent of alcohol, acid catalyst → acetal (thermodynamic product)
Exam Q6: a lactol (cyclic hemiacetal) treated with CH₃OH and H⁺ gives a mixed acetal where the OH is replaced by OCH₃
Exam Q10: intramolecular hemiacetal formation from a hydroxy aldehyde (HO-chain-CHO) under H₃O⁺ gives a cyclic hemiacetal
Step-by-step mechanism:
Nucleophilic addition: the amine nitrogen attacks the aldehyde carbonyl carbon, forming a tetrahedral intermediate (carbinolamine)
Proton transfer: the nitrogen picks up a proton, and the oxygen becomes OH
Dehydration: loss of water gives the imine (C=N)
Reduction: NaBH₃CN delivers H⁻ to the imine carbon, giving an alkoxide-like intermediate on nitrogen
Protonation (from solvent, e.g., CH₃OH): gives the final amine product
In the benzaldehyde + methylamine example, the product is N-methylbenzylamine (PhCH₂NHCH₃)
An imine treated with H⁺/H₂O hydrolyses back to the carbonyl (ketone or aldehyde) plus the free amine
This is the reverse of imine formation
A vinyl bromide + arylboronic acid (or boronate ester), Pd(PPh₃)₄ catalyst, base (OH⁻)
Forms a new C-C bond between the vinyl and aryl groups
Stereochemistry is retained: a cis vinyl bromide gives a cis alkene product with the new aryl group
The exam review's synthesis problems chain together many of the reactions above. The general approach:
Identify the target's functional groups and the carbon skeleton
Work backwards (retrosynthetic analysis) to identify which bond was formed last
Match each disconnection to a known reaction
Common sequences from the exam:
PhBr → PhCN (NaCN, SN2) → PhCOOH (H₃O⁺) → PhCOCl (SOCl₂) → PhCOCH₃ (Me₂CuLi) → PhCH(OH)CH₃ (LAH) or PhCHO (DIBAL-H)
Cyclohexanol → cyclohexanone (PCC) → 1-methylcyclohexanol (CH₃MgBr then H₃O⁺) → oxidation/Grignard further along
Students often confuse LiAlH₄ and DIBAL-H. LAH reduces esters all the way to primary alcohols. DIBAL-H at low temperature stops at the aldehyde. Getting this wrong costs you the entire synthesis question.
A common error is thinking PCC will oxidise a primary alcohol to a carboxylic acid. It will not. PCC stops at the aldehyde. Use Jones reagent (CrO₃/H⁺) if you need the carboxylic acid.
Students forget that Grignard reagents add twice to esters. One equivalent of PhMgBr does not give you a ketone from an ester; it gives a tertiary alcohol (two phenyl groups). If you want a ketone from an acyl derivative, use a Gilman reagent (R₂CuLi) with an acyl chloride.
In reductive amination, students sometimes draw NaBH₃CN reducing the aldehyde directly. Under reductive amination conditions, NaBH₃CN is selective for the imine. The aldehyde reacts with the amine first.
⚠️ The LAH vs DIBAL-H distinction appears in nearly every multi-step synthesis problem. Know which one to pick and why.
⚠️ PCC vs Jones oxidation is a classic exam question. If the answer key shows an aldehyde, the reagent must be PCC, not CrO₃.
⚠️ The reductive amination mechanism is a frequent "write the complete mechanism" question. Practise drawing every arrow.
⚠️ Suzuki coupling: know the conditions (Pd(PPh₃)₄, OH⁻) and that stereochemistry is retained. A cis starting material gives a cis product.
⚠️ Multi-step synthesis problems are worth a large share of the exam marks. Practise working backwards from the product.
Fill in the blank: To reduce an ester to an aldehyde, use ________ at low temperature.
Answer: DIBAL-H (at −78 °C).
True or False: PCC can oxidise a primary alcohol to a carboxylic acid.
Answer: False. PCC stops at the aldehyde.
Fill in the blank: The Gilman reagent Me₂CuLi reacts with an acyl chloride to give a ________.
Answer: Ketone (methyl ketone).
True or False: In reductive amination, NaBH₃CN reduces the aldehyde directly to an alcohol.
Answer: False. It selectively reduces the imine intermediate.
Fill in the blank: Suzuki coupling uses a ________ catalyst and retains the stereochemistry of both coupling partners.
Answer: Palladium (Pd(PPh₃)₄).
Q: PhCH₂Br is treated with NaCN, then the product is treated with H₃O⁺. What is the final product?
A: Phenylacetic acid (PhCH₂COOH). NaCN gives the nitrile (PhCH₂CN), and acid hydrolysis converts it to the carboxylic acid.
Q: An ester (PhCOOCH₃) is treated with 1. DIBAL-H, 2. H₂O. What is the product?
A: Benzaldehyde (PhCHO). DIBAL-H reduces the ester to the aldehyde and stops.
Q: Cyclohexanol is treated with PCC, then the product is treated with CH₃MgBr followed by H₃O⁺. What is the final product?
A: 1-Methylcyclohexanol. PCC gives cyclohexanone, and the Grignard adds a methyl group.
Q: An acyl chloride (PhCOCl) is treated with 1. Me₂CuLi, 2. H₂O. What is the product?
A: Acetophenone (PhCOCH₃). The Gilman reagent adds one methyl group to give the ketone.
Q: Write the product of the Suzuki coupling between a cis-vinyl bromide and phenylboronic acid with Pd(PPh₃)₄/OH⁻.
A: A cis-substituted styrene derivative. The cis geometry of the vinyl bromide is retained, and the phenyl group replaces the bromine.
Q: Benzaldehyde is treated with methylamine, then NaBH₃CN. What is the product?
A: N-Methylbenzylamine (PhCH₂NHCH₃). The imine forms first, then NaBH₃CN reduces it.
Reduction and oxidation reactions connect to everything in Org Chem I. Grignard additions tie back to SN2 and nucleophilicity from the first half of the course. Hemiacetal and acetal chemistry leads directly into carbohydrate chemistry (the next set of notes), since sugars exist as cyclic hemiacetals. Reductive amination is the biochemical route to amino acids and neurotransmitters, so this mechanism appears again in biochemistry courses.
LiAlH₄, lithium aluminium hydride, LAH, DIBAL-H, diisobutylaluminium hydride, NaBH₃CN, sodium cyanoborohydride, PCC, pyridinium chlorochromate, Jones oxidation, CrO₃, chromic acid, Grignard reagent, RMgBr, PhMgBr, CH₃MgBr, organolithium, RLi, Gilman reagent, Me₂CuLi, lithium dimethylcuprate, hemiacetal, acetal, lactol, protecting group, reductive amination, imine, Schiff base, carbinolamine, Suzuki coupling, Pd(PPh₃)₄, palladium-catalysed cross-coupling, boronic acid, vinyl halide, stereoretention, retrosynthetic analysis, multi-step synthesis, Org Chem I Exam 3