
Interpretation: To draw the structures of the products that are expected for each reaction given.
Concept Introduction: The

Answer to Problem 56A
CH3COOCH3+H2OHCl→CH3COOH+CH3OH
CH3CH2COOCH2CH2CH3+H2OH+→CH3COOH+CH3CH2CH2OH
HCOOCH2CH3+KOH→HCOOH+CH3CH2OH
Explanation of Solution
Methyl ethanoate on hydrolysis will produce Ethanoic acid and methanol. Similarly, CH3CH2COOCH2CH2CH3 which is an ester will also form acid and alcohol as a product. The acid is ethanoic acid and the alcohol is propanol. HCOOCH2CH3 on reaction with a strong base like KOH will produce formic acid and ethanol. This clearly shows that ester on reduction will produce primary alcohols as the alcohols produced in all three reactions are primary.
This reduction of ester will take place in two steps. In the first step, the ester will be converted
Chapter 23 Solutions
Chemistry 2012 Student Edition (hard Cover) Grade 11
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