Predict and balance the following organic reactions.
1. Ethanol(ethyl alcohol) and methanoic acid(formic acid) are mixed and warmed.
2. Ethanoic acid is combined with propanol.
An organic acid is RCOOH (where R can be anything except a functional group).
An alcohol is R'OH (where R' is anything except a functional group and the prime is used to show that it MAY be different than the first R).
RCOOH + R'OH ==> RCOOR' + H2O
Now all you need to do is fill in the R and R' groups. This is a general formula.
To predict and balance organic reactions, we need to consider the functional groups present in each reactant and the possible reactions they can undergo. Let's analyze each reaction and determine the predicted products:
1. Ethanol and methanoic acid are mixed and warmed:
Ethanol (CH3CH2OH) is an alcohol, and methanoic acid (HCOOH) is a carboxylic acid. When an alcohol reacts with a carboxylic acid, it typically undergoes an esterification reaction. In this case, ethanol will react with methanoic acid to form an ester called ethyl methanoate (CH3COOCH2CH3) and water (H2O).
The balanced equation for this reaction is:
CH3CH2OH + HCOOH β CH3COOCH2CH3 + H2O
2. Ethanoic acid is combined with propanol:
Ethanoic acid (CH3COOH) is also a carboxylic acid, whereas propanol (CH3CH2CH2OH) is an alcohol. Similar to the previous reaction, the combination of a carboxylic acid and alcohol will likely result in an esterification reaction. Ethanoic acid will react with propanol to form a new ester called propyl ethanoate (CH3COOCH2CH2CH3) and water (H2O).
The balanced equation for this reaction is:
CH3COOH + CH3CH2CH2OH β CH3COOCH2CH2CH3 + H2O
By analyzing the functional groups and the reactivity of the reactants, we can predict the products and then balance the equations accordingly.