Question

Sodium bisulfite is used during the extraction to procedure to react with unreacted aldehyde according to the following equation: a) Why is this done? Answer this question by explaining what would happen (and why) without bisulfite extraction versus with bisulfite extraction, assuming that not all of the aldehyde used during the reaction reacts. Focus on the isolated product from the reaction (2-methyl-3-heptanol) b) Saturated sodium bisulfite in water contains 42 g of sodium bisulfite per 100 mL of solution. What is the molar concentration of saturated sodium bisulfite?

Answer 1

To get rid of excess aldehyde, it is reacted with NaHSo_3. So, that aldehyde reacts with NaHSO_3 and separates from the reaction mixture as while crystals. rightarrow polar compound, not soluble in organic medium So, separates out from medium as while crystals. \r\n So, without extraction there will remain the excess aldehyde in the organic medium as impurities. 42 g of NaHSo_3 (molar mass of NaHSo_3 rightarrow 104.061 g/mol) therefore 42 g = 42 g/104.061 g/mol = 0.404 mol in 100 ml therefore In 1000 ml 4.04 mol of NaHSo_3 therefore molar conc 4.04 moles/Litre = 4.04(m)