Compound A, C₄H₈O₂, gives an orange precipitate with 2,4-dinitrophenylhydrazine reagent, and a silver mirror with ammoniacal silver nitrate. A is therefore a carbonyl compound and is an aldehyde since it is a reducing agent. Because it gives the iodoform reaction it has the group CH₃CO- or CH₃CH(OH)- which accounts for one oxygen atom; the other is part of the -CHO group. A has four carbon atoms, so is probably CH₃CH(OH)CH₂CHO. A compound with the CH₃CO- group would not have enough hydrogen atoms.

A is converted to B, C₄H₆O, on treatment with acid. Comparison of this formula with that of A shows that water has been lost, so the acid is either concentrated sulphuric or phosphoric acid. Since B may exist as a cis- or trans- isomer it has a C=C double bond, with the groups on each carbon not being the same. The structure of A suggested above fits:

A  à   B:                     CH₃CH(OH)CH₂CHO à    CH₃CH=CHCHO + H₂O

B reacts with hydrogen chloride to give C, C₄H₇OCl, and with chlorine to give D, C₄H₆OC1₂ , both of these clearly being addition reactions:

B    à     C:                CH₃CH=CHCHO + HCl à CH₃CH(Cl)CH₂CHO or CH₃CH₂CH(Cl)CHO

B    à     D:                 CH₃CH=CHCHO + Cl₂ à CH₃CH(Cl)CH(Cl)CHO

The reduction reaction with C to give E, C₄H₉OCl, converts the aldehyde group to an alcohol, which is therefore CH₃CH(Cl)CH₂CH₂OH or CH₃CH₂CH(Cl)CH₂OH. Reaction of this with sodium hydroxide converts the -Cl to -OH in F. F gives the iodoform reaction, so must contain CH₃CH(OH)-, the reaction being

E    à     F:                CH₃CH(Cl)CH₂CH₂OH + OH ^- à CH₃CH(OH)CH₂CH₂OH + Cl ^-

This shows that C is the first structure given in the equation above.

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