Problem 70 - Chemistry An Atoms First Approach

Question

Complete the following reactions. a. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}+\mathrm{CH}_{3} \mathrm{OH} \rightarrow\) b. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{HCOOH} \rightarrow\)

Step-by-Step Solution

Answer

a. \(\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H} + \mathrm{CH}_{3}\mathrm{OH} \rightarrow \mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{CH}_{3} + \mathrm{H}_{2}\mathrm{O}\) b. \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH} + \mathrm{HCOOH} \rightarrow \mathrm{HCO}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3} + \mathrm{H}_{2}\mathrm{O}\)

Step 1: Reaction 1

The first reaction involves acetic acid (\(\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H}\)) and methanol (\(\mathrm{CH}_{3}\mathrm{OH}\)). When these two molecules react, they undergo an esterification reaction. In this reaction, the -OH group from the acetic acid and the -H from methanol will combine to form water, and the resulting ester will be formed.

Step 2: Product of Reaction 1

Since water is formed from the -OH group of acetic acid and the -H of methanol, the ester formed will be between the remaining groups: \(\mathrm{CH}_{3}\mathrm{CO}_{2}\) (from acetic acid) and \(\mathrm{CH}_{3}\) (from methanol). Combining these, we get the ester \(\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{CH}_{3}\). So, the complete reaction can be written as: \[\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H} + \mathrm{CH}_{3}\mathrm{OH} \rightarrow \mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{CH}_{3} + \mathrm{H}_{2}\mathrm{O}\]

Step 3: Reaction 2

The second reaction involves propanol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH}\)) and formic acid (\(\mathrm{HCOOH}\)). These two molecules will also undergo an esterification reaction, similar to the previous reaction. The -OH group from formic acid and the -H from propanol will combine to form water, and the resulting ester will be formed.

Step 4: Product of Reaction 2

Since water is formed from the -OH group of formic acid and the -H of propanol, the ester formed will be between the remaining groups: \(\mathrm{HCO}_{2}\) (from formic acid) and \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\) (from propanol). Combining these, we get the ester \(\mathrm{HCO}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3}\). So, the complete reaction can be written as: \[\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH} + \mathrm{HCOOH} \rightarrow \mathrm{HCO}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3} + \mathrm{H}_{2}\mathrm{O}\]