Answer 6

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 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 D: CH3CH2CH2OH + [O] CH3CH2CHO + H2O

D E: CH3CH2CHO + [O] CH3CH2COOH

A C: CH3CH2CH2Cl + NaOH CH3CH2CH2OH + NaCl

CH3CH2COOH + NaHCO3 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 G: CH3CH(Cl)CH3 + NaOH CH3CH(OH)CH3 + NaCl

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

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

CH3COCH3 + 3I2 + 4NaOH 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.


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