Question

If you were performing a reaction under acidic conditions (where 1.1 equivalents of the acid are used) and you had 6.5 g of your carboxylic acid, how much alcohol would you need to bring the reaction to completion? (Show your calculations, this is for a Fischer esterification reaction producing octyl butyrate).

Answer 1

Solution: In a Fischer esterification reaction, 1 mole of the carboxylic acid is reacted with 1 mole of alcohol in acidic medium, to obtain the ester. The balanced reaction (in acid as catalyst) can be written as:

R-COOH + R'OH $\rightarrow$ RCOOR'

Now, in the problem given, The ester to be made is octyl butyrate. Therefore, the carboxylic acid needed is Butyric acid (C₄H₈O₂, M = 88 g/mol) and alcohol needed is octanol (C₈H₁₈O, M = 130 g/mol).

Note that the acid (taken as catalyst) only increases the rate of the reaction, and its amount (1.1 equivalents) does not determine the yield of the reaction. The alcohol or carboxylic acid will be the limiting agents.

Here 6.5 g of carboxylix acid i.e. butyric acid is taken. Therefore moles of butyric acid taken = (6.5 g)/ (88 g/mol) = 0.074

For reaction to go upto completion, 0.074 mole of alcohol i.e. octanol should be present.

$\Rightarrow$amount of octanol required = (0.074 mol) x (130 g/mol)    = 9.6 g            .... Answer

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