We did an experiment with the synthesis of n-butyl bromide, using n-butyl alcohol, NaBr, and H2SO4. The question is: Aqueous sodium bicarbonate was used to wash the crude n-butyl bromide. A) What was the purpose of this wash? Give Equations. I know it was used to remove any remaining acid or n-butyl alcohol, but I don't know what it means by give equations? Please help. Also, B) Why would it be undesirable to wash the crude halide with aqueous sodium hydroxide? I think this has something to do with them possibly being able to react? Please Help.
Thanks, but what happened to the S in the equation? There needs to be one in the products to balance.
It is supposed to say, "Na2SO4" not "Na2CO3"
A) The purpose of washing the crude n-butyl bromide with aqueous sodium bicarbonate is to remove any remaining acid or n-butyl alcohol that might be present in the product. Sodium bicarbonate (NaHCO3) is a weak base that can react with acidic impurities to form water-soluble salts, making it easier to separate them from the organic product.
The equation for the reaction between sodium bicarbonate and sulfuric acid (H2SO4) is:
NaHCO3 + H2SO4 → NaHSO4 + H2O + CO2
In this reaction, sodium bicarbonate reacts with sulfuric acid to form sodium hydrogen sulfate, water, and carbon dioxide.
Similarly, n-butyl alcohol can react with sodium bicarbonate to form an alcoholate salt and water:
C4H9OH + NaHCO3 → C4H9ONa + H2O + CO2
By washing the crude n-butyl bromide with sodium bicarbonate solution, any remaining acid and alcohol will form water-soluble salts that can be separated from the organic product.
B) It would be undesirable to wash the crude halide with aqueous sodium hydroxide (NaOH) because sodium hydroxide is a strong base that can react with the n-butyl bromide. This reaction is called an elimination reaction, and it would result in the formation of an alkene and sodium bromide:
C4H9Br + NaOH → C4H8 + NaBr + H2O
The undesirable consequence here is the formation of an alkene (C4H8) instead of the desired n-butyl bromide (C4H9Br). This reaction is known as a β-elimination reaction, where the halide is eliminated from the alkyl group to form a double bond. Therefore, washing the crude n-butyl bromide with aqueous sodium hydroxide could lead to product loss and the formation of undesired by-products.
The bicarbonate is used to wash the product because you want to remove the H2SO4 from the reaction.
H2SO4 + NaHCO3 ==> H2O + CO2 + Na2CO3 but you need to balance it.
Washing the alkyl halide with a strong base such as NaOH risks dehydrohalogenation (removal of HX---in this case HBr) to produce the alkene.