Answer 3

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:

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

B        D:                 CH3CH=CHCHO + Cl2 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        F:                CH3CH(Cl)CH2CH2OH + OH - CH3CH(OH)CH2CH2OH + Cl -

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

Problem 3       Organic problems contents     Home page