Esters — practice questions

(a) First, identify the structures of the reactants. Butan-1-ol is $CH_3CH_2CH_2CH_2OH$ and propanoic acid is $CH_3CH_2COOH$. The reaction removes the –OH from the carboxylic acid and the –H from the alcohol's hydroxyl group to form water. The remaining parts join together. \n The ester link is –COO–. The part from the acid is $CH_3CH_2CO-$ and the part from the alcohol is $-OCH_2CH_2CH_2CH_3$. \n Displayed formula: \n H H O H H H H \n | | || | | | | \n H-C-C-C-O-C-C-C-C-H \n | | | | | | \n H H H H H H \n (b) To name the ester, we use the 'alkyl alkanoate' format. \n The alkyl part comes from the alcohol, butan-1-ol, so it is 'butyl'. \n The alkanoate part comes from the carboxylic acid, propanoic acid, so it is 'propanoate'. \n Therefore, the name of the ester is butyl propanoate.

(a) The general formula of the ester is RCOOR'. We know the alcohol is methanol ($CH_3OH$), so the R' group is methyl ($CH_3$). \n The formula is $RCOOCH_3$. The total formula is $C_4H_8O_2$. \n Let's subtract the atoms from the $-COOCH_3$ part: $C_2H_3O_2$. \n $C_4H_8O_2 - C_2H_3O_2 = C_2H_5$. \n So, the R group is an ethyl group, $C_2H_5$. \n The carboxylic acid is therefore $C_2H_5COOH$, which is propanoic acid. \n The ester is methyl propanoate. \n (b) Alkaline hydrolysis of methyl propanoate ($CH_3CH_2COOCH_3$) with sodium hydroxide ($NaOH$) will produce the sodium salt of the carboxylic acid and the alcohol. \n Products: Sodium propanoate and methanol. \n Equation: \n $CH_3CH_2COOCH_3(aq) + NaOH(aq) \rightarrow CH_3CH_2COONa(aq) + CH_3OH(aq)$