Everything from the 2 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
Aldehydes and ketones are carbonyl compounds: both contain C=O. In an aldehyde (RCHO, names ending in -al) the C=O is at the end of the chain with an H attached; in a ketone (RCOR′, names ending in -one) it is bonded to two carbons. A primary alcohol is oxidised by acidified K2Cr2O7 to an aldehyde, which must be distilled off as it forms or it is oxidised further to a carboxylic acid (as happens under reflux). A secondary alcohol gives a ketone, which resists further oxidation. NaBH4 (or LiAlH4) reduces aldehydes back to primary alcohols and ketones back to secondary alcohols.
An aldehyde has the C=O at the end of the chain, a ketone in the middle. Distillation gives the aldehyde, reflux the acid. Ketones are not oxidised by acidified dichromate, so a ketone never goes on to an acid. In equations, oxidation uses [O] and reduction uses 2[H]; NaBH4 reduces C=O but does not reduce C=C.
The C=O bond is polar, so the carbon is δ+ and is attacked by nucleophiles. With HCN and a KCN catalyst, the CN− ion attacks the carbon, the π electrons move onto oxygen, and the O− then takes H+ from HCN: a 2-hydroxynitrile forms. Because the C=O group is planar, attack from either face is equally likely, so a chiral product is a racemic mixture. 2,4-DNPH gives an orange-yellow precipitate with any aldehyde or ketone. Only aldehydes reduce Fehling's (blue to brick-red Cu2O) and Tollens' (silver mirror) and turn acidified dichromate green. Alkaline iodine gives yellow CHI3 with the CH3CO– group.
HCN adds by nucleophilic addition, not electrophilic: the attacking species is CN−, and it bonds through carbon. The product is racemic because the C=O is planar, not because the reaction is SN1. 2,4-DNPH does not distinguish aldehydes from ketones; Fehling's and Tollens' do. Fehling's turns brick-red, Tollens' gives silver; do not swap them. Propanal gives no CHI3: it has no CH3CO– group.