Question

How do you explain the following ?
(a) Presence of an aldehyde group in glucose
(b) Presence of a primary alcoholic group in glucose

Hint:

Aldehyde → oxime/cyanohydrin/Tollens'; primary −OH → HNO₃ gives saccharic acid.

Answer 1

(i)
Start by checking the charge on platinum. The sulphate outside is $\mathrm{SO_4^{2-}}$, so the complex ion must be $+2$. Inside, each ethane-1,2-diamine (en) is neutral and the two chlorides give $-2$, so platinum sits at $+4$. Now name the parts in alphabetical order: chlorido comes before the ethane-1,2-diamine, and 'bis' is used for the two en groups because en itself contains a number. Putting it together gives dichloridobis(ethane-1,2-diamine)platinum(IV) sulphate.

(ii)
Here the complex is the negative ion, $\mathrm{[CoF_4]^{2-}}$, balanced by two ammonium ions. Whenever the complex part is an anion, the metal name takes the '-ate' ending, so cobalt becomes cobaltate. Four fluorides give $-4$, and since the whole ion is $-2$, cobalt is $+2$. The positive ammonium is named first, then the anion. The full name is ammonium tetrafluoridocobaltate(II).