Aldehydes and ketones

In butanal the C=O carbon carries an H and one alkyl group, which is the CHO group of an aldehyde; the name ends in -al. Butanone has its C=O bonded to two carbon atoms, so it is a ketone (-one). Propanoic acid contains C=O but as part of a COOH group, which is a carboxylic acid, not a carbonyl compound in this sense. Propan-2-ol has no C=O at all; it is the secondary alcohol that oxidises to propanone.

All four have similar numbers of electrons, so similar instantaneous dipole-induced dipole forces. The C=O bond is polar (O is more electronegative), adding permanent dipole-dipole attractions, so the carbonyls boil well above butane. Hydrogen bonding needs H bonded to N, O or F; aldehydes and ketones have no O-H, so their molecules cannot hydrogen bond to each other, while propan-1-ol can. The M_r values are almost identical (58 to 60), so instantaneous dipole forces cannot explain the gap. Boiling separates molecules and breaks no covalent bonds.

Ethanal boils far below ethanol (78 °C), so in distillation apparatus it leaves the flask as soon as it forms and escapes further oxidation: CH₃CH₂OH + [O] → CH₃CHO + H₂O. Under reflux the ethanal condenses back into the flask and excess dichromate oxidises it to ethanoic acid. Adding NaBH₄ would reduce any carbonyl back to the alcohol, undoing the oxidation. A sealed flask with excess oxidant keeps the ethanal in contact with the dichromate, so it is oxidised further.

CH₃CH₂CH₂CHO + 2[H] → CH₃CH₂CH₂CH₂OH. The C=O was on the end carbon, so the OH ends up there too: aldehydes reduce to primary alcohols. Butan-2-ol would come from butanone, where the C=O is on C2. Butanoic acid is an oxidation product, not a reduction product. NaBH₄ reduces C=O to CH-OH; it does not remove the oxygen completely to give an alkane.

Oxidation turns the CH(OH) carbon of a secondary alcohol into C=O without changing its position, so pentan-2-ol gives pentan-2-one. Pentan-1-ol is primary and gives pentanal (distilled) or pentanoic acid (reflux). Pentan-3-ol is secondary but gives pentan-3-one, with the C=O in the middle. 2-methylbutan-2-ol is tertiary, has no H on the C-OH carbon, and is not oxidised.

n(butanone) = 2.88 / 72.0 = 0.0400 mol. Butanone + 2[H] gives butan-2-ol, 1 : 1, so mass = 0.0400 × 74.0 = 2.96 g. The product is 2 mass units heavier per molecule because two H atoms are added, so the mass rises slightly. 2.88 g assumes the mass is unchanged. 2.80 g subtracts the hydrogen instead of adding it, as if the reaction were an oxidation. 5.92 g doubles the answer by wrongly treating 2[H] as doubling the moles of product.

NaBH₄ acts as a source of hydride ions, H⁻, which attack the δ+ carbon of the polar C=O; the C=C bond is non-polar and electron-rich, so it repels nucleophiles and is untouched. The ketone becomes a secondary alcohol and the C=C stays: CH₂=CHCH(OH)CH₃. Butan-2-ol would need both double bonds reduced (H₂/Ni would do that). Butanone would need the C=C reduced and the C=O kept, the opposite of what NaBH₄ does. The OH stays on C2, where the C=O was, so a primary alcohol cannot form.

The aldehydes are butanal, CH₃CH₂CH₂CHO, and 2-methylpropanal, (CH₃)₂CHCHO. The only ketone is butanone, CH₃COCH₂CH₃; putting C=O on C3 gives the same molecule, and a branched C₄ ketone is impossible. That makes 3. Answering 2 misses the branched aldehyde. Answering 4 counts 'butan-3-one' as different from butanone. Answering 5 counts alcohols with C=C, such as but-3-en-1-ol, which share the formula but are not carbonyl compounds.