Alcohol Oxidation Control and Synthesis Strategy

Strong aqueous oxidants like acidic KMnO₄ or Na₂Cr₂O7/H₂SO₄ oxidise a primary alcohol all the way to a carboxylic acid, because the aldehyde intermediate is further hydrated in water and oxidised again. Pyridinium chlorochromate (PCC) is different: it is used in anhydrous, non-aqueous conditions (typically dichloromethane), so the aldehyde intermediate has no water available to hydrate into a geminal diol, and with no diol to oxidise further, the reaction stops cleanly at the aldehyde. A secondary alcohol oxidised with PCC stops at the ketone, exactly as it would with a stronger oxidant, since a ketone has no further oxidation state to reach without breaking a carbon-carbon bond. Whenever a question specifies PCC (or names it as pyridinium chlorochromate) acting on a primary alcohol, expect the aldehyde as the final product, not the carboxylic acid a stronger aqueous oxidant would give.

An aryl halide substituent on a benzene ring is not just a functional group in its own right — it can also be installed purely to control WHERE a later electrophilic substitution (like Friedel-Crafts alkylation) happens, by its directing effect, and then removed afterward once it has served that purpose. The removal trick uses the same halide: convert the aryl halide to its Grignard reagent (Mg in dry ether), which replaces the C-X bond with a C-MgX bond, and then quench that Grignard reagent with water or aqueous alcohol, which protonates the carbanion and replaces C-MgX with a simple C-H bond — net effect, the halogen is gone and that ring position is now just hydrogen. Because AlCl₃ (the Lewis acid catalyst needed for Friedel-Crafts alkylation) is incompatible with a Grignard reagent, the Friedel-Crafts step must always be carried out on the neutral aryl halide BEFORE the halide is converted to the reactive, moisture- and Lewis-acid-sensitive Grignard reagent — do the electrophilic substitution first, form and quench the Grignard last.

Diborane in THF followed by H₂O2/KOH adds H and OH across a C=C in an anti-Markovnikov, syn fashion: boron attaches to the less hindered, less substituted alkene carbon, oxidation replaces that boron with OH at the same carbon, and the hydrogen ends up on the more substituted carbon. For a trisubstituted ring alkene with a phenyl group on one alkene carbon and an H on the other, the OH therefore lands on the CH carbon, NOT on the carbon bearing the phenyl group. The resulting secondary alcohol is then oxidised by acidic dichromate to a ketone, which is the final product, so the C=O sits one carbon away from the CH(Ph) carbon and that carbon stays an sp₃ CH. A tertiary alcohol at the phenyl-bearing carbon would be the Markovnikov (acid-catalysed hydration) product, and dichromate cannot oxidise it.

A phenolic OH activates the ortho and para positions (C₂, C₄ and C₆). In 3-tert-butylphenol the bulky tert-butyl group at C₃ crowds the two positions beside it: C₂ (squeezed between OH and tert-butyl) and C₄ (next to tert-butyl), while C₆ (ortho to OH and para to tert-butyl) is open. The size of the halogen then decides how far substitution goes. Small, reactive chlorine can reach all three activated positions to give the 2,4,6-trichloro compound. Bromine, which is larger, takes C₄ and C₆ but skips the most crowded C₂ (4,6-dibromo). Iodine, the largest and least reactive, substitutes only the open C₆ (a single iodo group). So with a bulky group present, expect fewer positions to be substituted as the halogen gets bigger. Read answer options by position, not by name: check where each halogen sits relative to the OH and the tert-butyl group.