C reacts with phosphorus pentachloride to give steamy fumes, so is an alcohol – it does not have enough oxygen to be a carboxylic acid. The steamy fumes are HCl.

2C₃H₇OH + PCl₅ à 2 C₃H₇Cl + 2HCl + POCl₃

Since C can be oxidised with acidified potassium dichromate solution in two stages it is a primary alcohol; D is therefore an aldehyde (confirmed by reaction with both 2,4-dinitrophenylhydrazine and ammoniacal silver nitrate) and E a carboxylic acid (confirmed by reaction and with sodium hydrogen carbonate). A is therefore a primary halogenoalkane. The compounds must be A, 1-chloropropane, D propanal, E propanoic acid.

C à D: CH₃CH₂CH₂OH + [O] à CH₃CH₂CHO + H₂O

D à E: CH₃CH₂CHO + [O] à CH₃CH₂COOH

A à C: CH₃CH₂CH₂Cl + NaOH à CH₃CH₂CH₂OH + NaCl

CH₃CH₂COOH + NaHCO₃ à CH₃CH₂COONa + H₂O + CO₂

Since 1-chloropropane will react with benzene in the presence of anhydrous aluminium chloride to give B, C₉H₁₂, but the structure of this compound is not what might have been expected, it is clearly not propylbenzene C₆H₅CH₂CH₂CH₃. At this stage it is best to identify F, G and H.

Since F is an isomer of 1-chloropropane, it must be 2-chloropropane since this is the only other structure possible. On reaction with sodium hydroxide solution this gives the secondary alcohol G, which can be oxidised to a ketone H; this must therefore be propanone. The ketone structure is confirmed by the positive 2,4-DNP reaction, the absence of the ammoniacal silver nitrate reaction, and the inability of acidified potassium dichromate solution to oxidise it further.

F à G: CH₃CH(Cl)CH₃ + NaOH à CH₃CH(OH)CH₃ + NaCl

G à H: CH₃CH(OH)CH₃ + [O] à CH₃COCH₃ + H₂O

The ketone must contain the CH₃CO group since it gives the iodoform reaction:

CH₃COCH₃ + 3I₂ + 4NaOH à CH₃COONa + CHI₃ + 3NaI + 3H₂O

Since B can be made from benzene and F, 2-chloropropane, as the expected product, B is 2-phenylpropane, C₆H₅CH(CH₃)₂. (This compound is also known as isopropylbenzene and as cumene; it is used industrially since on oxidation with oxygen it produces propanone and phenol, both very important organic chemicals.)

The fact that 2-phenylpropane is also obtained (as the major, not the only, product) from 1-chloropropane and benzene, there must have been some sort of rearrangement during the reaction. The electrophile is, in the case of 1-chloropropane, CH₃CH₂CH₂⁺. This is a primary carbocation, and rearranges to the more stable CH₃CH⁺CH₃ before the benzene ring is attacked.

Molecular rearrangements are common; they are found not only during reactions, but in the mass spectrometer as well. You will not find them in A level questions – but it’s worth knowing that they exist.

On heating with alkaline potassium manganate(VII) solution 2-phenylpropane loses most of its side-chain and gives sodium benzoate C₆H₅COONa (I); this is soluble in water. On acidification, the benzoate ion, a strong base (benzoic acid being a weak acid), accepts a proton and gives insoluble benzoic acid C₆H₅COOH, which is J.