Question

An antifreeze solution is prepared by dissolving 31 g of ethylene glycol (Molar mass = 62 g $\mathrm{mol^{-1}}$) in 600 g of water. Calculate the freezing point of the solution. ($\mathrm{K_f}$ for water = 1·86 K kg $\mathrm{mol^{-1}}$)

Hint:

moles = 31/62 = 0.5; m = 0.5/0.6 = 0.833; ΔT_f = 1.86 × 0.833.

Answer 1

The idea here is simple: we cannot see the groups inside glucose directly, so we test for them by reactions. Each reaction below only works if a certain group is present, so a positive result proves that group is there.

(a) answer: To show glucose has an aldehyde group, $\mathrm{-CHO}$, we use reactions that only an aldehyde gives. Glucose reacts with hydroxylamine, $\mathrm{NH_2OH}$, to form an oxime, and it adds exactly one molecule of HCN to form a cyanohydrin. Both of these are typical of a carbonyl carbon that has an attached hydrogen, that is an aldehyde. On top of that, glucose reduces Tollens' reagent (giving a silver mirror) and Fehling's solution (giving a red precipitate), and only an aldehyde group does this so easily. All these results together prove glucose contains a $\mathrm{-CHO}$ group.

(b) answer: To show glucose has a primary alcohol group, $\mathrm{-CH_2OH}$, we look at two things. First, glucose reacts with acetic anhydride to form a penta-acetate, meaning it has five $\mathrm{-OH}$ groups in all (some secondary and one primary). Second, and more telling, when glucose is oxidised with strong nitric acid, both end carbons turn into $\mathrm{-COOH}$ groups, giving a dicarboxylic acid called saccharic acid (glucaric acid). One end was already the aldehyde, but for the other end to become a $\mathrm{-COOH}$ it must have started as a primary alcohol $\mathrm{-CH_2OH}$, because only a primary alcohol can be oxidised all the way to a carboxylic acid. This proves the presence of a primary alcoholic group.