Question

State Kohlrausch's law of independent migration of ions. With the help of a curve, explain why it is not easy to determine $\mathrm{\Lambda^\circ_m}$ for weak electrolytes by extrapolating the concentration – molar conductivity curve, as it is for strong electrolytes.

Hint:

Limiting molar conductivity = sum of ion contributions; weak electrolytes give a steep, non-extrapolatable curve.

Answer 1

(a)
To show glucose carries a $\mathrm{-CHO}$ group, we lean on reactions that only an aldehyde group performs. Glucose joins with hydroxylamine to make an oxime and grabs one molecule of HCN to make a cyanohydrin, both being typical aldehyde additions across the C to O double bond. On top of that, glucose acts as a reducing sugar: it turns Tollens' reagent into a silver mirror and reduces Fehling's solution to a red precipitate. Only a free aldehyde behaves this way, so glucose must contain a $\mathrm{-CHO}$ group.

(b)
To pin down a primary alcohol group, the key clue is how the chain ends react. When glucose is treated with acetic anhydride it makes a penta-acetate, telling us it has five $\mathrm{-OH}$ groups in all. More tellingly, when glucose is oxidised with strong nitric acid, both end carbons get converted into $\mathrm{-COOH}$, giving saccharic acid, a diacid. One end was already an aldehyde, but the other end could only become a $\mathrm{-COOH}$ if it had started as a primary $\mathrm{-CH_2OH}$ group. So glucose contains a primary alcoholic group.