You want to oxidize a primary alcohol to an aldehyde and stop there — you do not want the carboxylic acid. Which reagent should you use?
AKMnO₄ in aqueous acid — a strong oxidant that ensures complete oxidation
BJones reagent (CrO₃/H₂SO₄) — reliable for primary alcohols
CPCC in anhydrous CH₂Cl₂ — stops at the aldehyde because no water is present to allow further oxidation
DSOCl₂ — converts the alcohol to an acid chloride in one step
PCC (pyridinium chlorochromate) in anhydrous dichloromethane is the standard reagent for stopping oxidation of a primary alcohol at the aldehyde stage. The mechanism requires water to hydrate the aldehyde to a gem-diol, which is what gets further oxidized to the carboxylic acid. Anhydrous conditions prevent this hydration, so the reaction stops at the aldehyde. Jones reagent and KMnO₄ (options A and B) are strong oxidants in aqueous conditions that carry the reaction all the way to the carboxylic acid. SOCl₂ (option D) does not oxidize — it converts an alcohol to an alkyl chloride by replacing OH with Cl.
Question 2 Multiple Choice
Why do tertiary alcohols resist oxidation under conditions that oxidize primary and secondary alcohols?
ATertiary alcohols are more stable than primary or secondary alcohols, making them thermodynamically resistant to oxidation
BTertiary alcohols have a higher boiling point, so common oxidants do not reach them at standard temperatures
COxidation of an alcohol requires removing a hydrogen from the carbon bearing the OH group; tertiary alcohols have no such hydrogen, so the oxidation cannot proceed
DTertiary alcohols form stable carbocations that resist further reaction
The oxidation of an alcohol to a carbonyl compound requires removing an α-hydrogen — the hydrogen attached to the carbon bearing the –OH. In a primary alcohol (R-CH₂-OH), there are two such hydrogens; in a secondary alcohol (R₂CH-OH), there is one. In a tertiary alcohol (R₃C-OH), the carbon bearing the hydroxyl has no hydrogens — it is bonded to three carbon substituents instead. There is nothing to remove, so the oxidation mechanism has no available pathway. This is a structural argument, not a thermodynamic or kinetic one: it's not that oxidation is slow or unfavorable for tertiary alcohols — it simply cannot proceed by the required mechanism.
Question 3 True / False
When SOCl₂ converts an alcohol to an alkyl chloride, the product has inverted configuration at the carbon that bore the hydroxyl group.
TTrue
FFalse
Answer: True
True. The mechanism proceeds through a chlorosulfite intermediate (R-O-SOCl), which activates the oxygen as a leaving group. Chloride ion then attacks the carbon in an SN2 backside attack, inverting the configuration at that carbon — the same Walden inversion seen in all SN2 reactions. This stereochemical outcome is one of the main reasons SOCl₂ is preferred over acid-catalyzed methods (using HCl): acid catalysis often proceeds through a carbocation intermediate (SN1-like), which gives a mixture of configurations. SOCl₂ provides a predictable, clean inversion, making it valuable in synthesis when stereochemical control is needed.
Question 4 True / False
Acid-catalyzed dehydration of a tertiary alcohol proceeds by an E2 mechanism because the tertiary carbon is more hindered.
TTrue
FFalse
Answer: False
False. Tertiary alcohol dehydration proceeds by E1, not E2. Acid protonates the –OH to give –OH₂⁺ (water as a leaving group). Water then departs to form a stable tertiary carbocation — the most stable type of carbocation — which subsequently loses a proton from an adjacent carbon to give the alkene. The stability of the tertiary carbocation drives the E1 pathway; a strong non-nucleophilic base is not required. E2 requires a strong base to remove the β-hydrogen simultaneously with leaving group departure, avoiding carbocation formation. The common misconception is that bulkier (more substituted) substrates 'favor' E2 because of steric arguments, but in acid-catalyzed reactions, the determining factor is mechanism, not bulk.
Question 5 Short Answer
Every major reaction of alcohols is a strategy for solving the same fundamental problem. What is that problem, and how does acid-catalyzed dehydration solve it?
Think about your answer, then reveal below.
Model answer: The fundamental problem is that –OH is a poor leaving group: hydroxide (HO⁻) is a strong base and will not spontaneously depart from a carbon in any of the standard substitution or elimination mechanisms. Acid-catalyzed dehydration solves this by protonating the –OH group (using H₂SO₄ or H₃PO₄) to convert it to –OH₂⁺ (water). Water is an excellent leaving group — it is a weak base and departs readily. For a tertiary alcohol, this allows a carbocation to form (E1), which then loses a proton to give the alkene. The same logic underlies all alcohol reactions: every reagent — SOCl₂, PBr₃, TsCl, or protonation — is a method for converting the poor –OH leaving group into something that will actually leave.
Understanding the leaving-group problem as the unifying concept makes alcohol chemistry coherent rather than a list of memorized reactions. Once a student sees that every reaction is answering 'how do we activate the OH?' the choice of reagent becomes logical rather than arbitrary: proton acid for elimination/SN1, SOCl₂ or PBr₃ for halide substitution with stereochemical control, TsCl for flexible activation before SN2 with any nucleophile. The reagents are different answers to the same question.