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If somebody could explain these please?
Hydrogen bromide: (Organic halogen compound formed) HBr (g) will react and the Br will substitute the OH group.
Alkaline iodine: (No organic halogen compound formed). This is the iodoform test to test for the presence (appearance of yellow ppt) of certain structures of alcohol and carbonyls. There should not be a reaction.
Ethanoyl chloride : (No organic halogen compound formed)Acyl chloride would react with the alcohol groups and amine groups to form ester and amide bonds respectively. HCl (g) would be released.
For D, refluxing with NaOH (aq) would hydrolyze the ester bond -OCOCH3What's wrong in E and D?View attachment 49817
For E, the phenol group is not reactive enough to react with CH3COOH (and conc H2SO4). CH3COCl is required to form ester bonds with phenol groups.
The clue to refer to is that the compound would form a weakly alkaline solution.Why is it C and not B,anyone??View attachment 49816
The methoxide CH3CO- will form a strong alkaline solution.
CH3CO- is a much stronger base than H2O, it will readily accept hydrogen atoms from water molecules.
CH3CO- + H2O --> CH3OH + OH-