Formation of compounds (Organic Chemistry)

[ScrubsFan]

Write balanced equations to show the formation of each of the following compounds. Name the reactants in each case, and show clearly the removal of the water molecule.

a) butyl propanoate

b) propyl methanoate


Im not exactly sure how to answer this question... below are pics of the structural formulas.

a) butyl propanoate - CH3CH2COOCH2CH2CH2CH3

b) propyl methanoate - HCOOCH2CH2CH3

I know the reaction will be a deydration (Elimination) seeing as the water molecule is removed. I am lost on where to go from here though; and even if what I have done so far is right.

 

[Cesium]

Both of those compounds are esters. They are derived from an alcohol and a carboxyllic acid, and water is, as you noticed, removed.

The general reaction for forming esters is: R-COOH + R'OH --> R-COOR' + H2O

a) butanol + propionic acid
b) propanol + methanoic acid

You also generally need an acid catalyst like sulfuric acid to drive this reaction to completion.

 

[Blueshift5]

To form butyl propanoate, the reactants are butanol (C4H9OH) and propanoic acid (C3H6O2). The reaction is a condensation reaction, where the hydroxyl group (-OH) from butanol combines with the carboxyl group (-COOH) from propanoic acid, releasing a water molecule (H2O). The balanced equation is:

CH3CH2CH2CH2OH + CH3CH2COOH → CH3CH2COOCH2CH2CH2CH3 + H2O

To form propyl methanoate, the reactants are propanol (C3H7OH) and methanoic acid (HCOOH). The reaction is also a condensation reaction, where the hydroxyl group (-OH) from propanol combines with the carboxyl group (-COOH) from methanoic acid, releasing a water molecule (H2O). The balanced equation is:

CH3CH2CH2OH + HCOOH → HCOOCH2CH2CH3 + H2O

In both cases, the water molecule is removed from the reactants to form the ester compounds. This reaction is known as esterification and is a common process in organic chemistry to form esters.