Since compound A, C₄H₈O, gives the iodoform reaction it must contain the group CH₃CO or CH₃CH(OH). A gives a precipitate with 2,4-dinitrophenylhydrazine but no reaction with ammoniacal silver nitrate, so it must be a ketone. It therefore contains the CH₃CO group, which means it is CH₃CH₂COCH₃ or butanone. B is sodium propanoate, CH₃CH₂COONa:

CH₃CH₂COCH₃ + 4NaOH + 3I₂ à CHI₃ + CH₃CH₂COONa + 3NaI + 3H₂O

A, being a ketone, is reduced with lithium aluminium hydride in dry ether to give a secondary alcohol C, CH₃CH₂CH(OH)CH₃, butan-2-ol.

CH₃CH₂COCH₃ + 2[H] à CH₃CH₂CH(OH)CH₃

This is chiral since any given molecule is non-superimposable on its mirror image. Note that this is a necessary and sufficient condition for chirality, there being no exceptions to this requirement. The product of the reaction would be a racemic mixture, i.e. have equimolar amounts of the (+) and (-) forms of the molecule.

With sodium chloride in 50% sulphuric acid butan-2-ol gives 2-chlorobutane, CH₃CH₂CH(Cl)CH₃.

CH₃CH₂CH(OH)CH₃ + HCl à CH₃CH₂CH(Cl)CH₃ + H₂O

With magnesium in dry ether 2-chlorobutane gives the Grignard reaction to produce CH₃CH₂CH(MgCl)CH₃. This, with solid carbon dioxide followed by dilute hydrochloric acid gives a carboxylic acid H, which must be CH₃CH₂CH(COOH)CH₃.

If 2-chlorobutane, D, is heated with potassium hydroxide dissolved in ethanol an elimination reaction occurs to give the cis- and trans- forms of but-2-ene CH₃CH=CHCH₃, E, which is the major product. F is the minor product and is but-1-ene, CH₂=CHCH₂CH₃.