Answer 1

Compound W, C₃H₈O, reacts with hot concentrated phosphoric acid to give X, C₃H₆

This is a dehydration reaction - water has been lost - suggesting that W is an alcohol. It could be either propan-1-ol or propan-2-ol:

CH₃CH(OH)CH₃ à CH₃CH=CH₂ + H₂O

..... C₃H₆ which rapidly decolourises a solution of bromine in tetrachloromethane.

C₃H₆ is an alkene; X is therefore propene:

CH₃CH=CH₂ + Br₂ à CH₃CH(Br)CH₂Br

On treatment with ethanoyl chloride, W gives Y, C₅H₁₀O₂.

Alcohols treated with ethanoyl chloride give esters and hydrogen chloride;

CH₃CH(OH)CH₃ + CH₃COCl à CH₃COOCH(CH₃)₂ + HCl

Oxidation of W with potassium dichromate in dilute sulphuric acid gives Z, C₃H₆O,

Oxidation of alcohols gives either aldehydes or ketones, depending on whether the alcohol is primary or secondary.

CH₃CH(OH)CH₃ + [O] à CH₃COCH₃ + H₂O

..... Z, C₃H₆O, which gives an orange precipitate with 2,4-dinitrophenylhydrazine but does not react with an ammoniacal solution of silver nitrate.

The 2,4-DNP test shows the presence of a carbonyl group, but the lack of reactivity with ammoniacal silver nitrate shows that Z is a ketone and not an aldehyde; the original alcohol W must have been a secondary alcohol, so is propan-2-ol. This has been used in the eqautions above, but note that this is on the basis of the last piece of information.

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