Q. 14.79 - An Introduction to General, Organic, and Biological Chemistry

Question

Propyl acetate is the ester that gives the odor and smell of pears. 

a. Draw the condensed structural formula for propyl acetate.

b. Write the balanced chemical equation for the formation of propyl acetate.

c. Write the balanced chemical equation for the acid hydrolysis of propyl acetate.

d. Write the balanced chemical equation for the base hydrolysis of propyl acetate with $\mathrm{NaOH}$.

e. How many milliliters of a $0.208\mathrm{MNaOH}$ solution is needed to completely hydrolyze (saponify)

 $1.58 \mathrm{g}$of propyl   acetate?

Step-by-Step Solution

Answer

Part (a)    

                   

Part (b)

   

           

Part (c)

               

Part (d)

                   

Part (e)     $74.4 \mathrm{mL} \text{ of }0.208 \mathrm{M} \mathrm{NaOH}$

(Part a) step-1 given information

The propyl acetate structural formula in its simplest form.

Part (a) step-2 Explanation

The propylacetate structure can be expressed as follows:

              

Part (b) step-3 given information

Formula for the synthesis of propyl acetate in a balanced chemical equation.

Part (b) step-4 Explanation

•   Propyl acetate can be produced by reacting 1-propanol with ethanoic acid. 

          

  

                                

Part (c) step-5 given information

 The balanced chemical equation for propyl acetate acid hydrolysis.

step-6 Explanation (part-c)

            

Part (d) step-7 given information

1-propanol and carboxylic acid salt are the results of propyl acetate hydrolysis with NaOH.

Part (d) step-8 Explanation

                  

Part (e) step-9 given information

  $$\begin{gathered} \text{ Given ;} \\ 1.58 \mathrm{g}\text{ propyl acetate }\left({\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2\right)\text{ and }0.208\mathrm{MNaOH} \end{gathered}$$

                Need MI NaOH 

               ${\mathrm{gC}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2$$\text{ Molar moles }{\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2\text{ Mole-mole moles NaoH }$

                factor of mass

                Molarity ML NaOH Factor L NaOH Metric

Part (e) step-10 Explanation

•    $1\text{ mole }{\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2 = 102{ \mathrm{g}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2$

•  $\frac{102{\mathrm{gC}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2}{1{\mathrm{moleCC}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2} \text{ and } \frac{1{\mathrm{moleC}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2}{102 {\mathrm{gC}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2}$

• $\text{ I mole }{\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2 = 1\text{ mole }\mathrm{NaOH}$

• $\frac{1 {\mathrm{mole}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2}{\text{ Imole }\mathrm{NaOH}} \text{ and } \frac{1\text{ mole }\mathrm{NaOH}}{1\text{ mole }{\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2}$

• $1 \mathrm{LNaOH} = 0.208\text{ moles }\mathrm{NaOH}$

• $\frac{0.208\text{ moles }\mathrm{NaOH}}{\mathrm{ILNaOH}}\text{ and }\frac{1 \mathrm{LNaOH}}{0.208\text{ moles }\mathrm{NaOH}}$

• $1 \mathrm{LNaOH} = 1000\mathrm{mlNaOH}$

• $\frac{1000\mathrm{mlNaOH}}{1 \mathrm{LNaOH}} \text{ and } \frac{1 \mathrm{LNaOH}}{1\mathrm{mlNaOH}}$
  

Therefore,

                    =$1.58{\mathrm{gC}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2\times \frac{1 {\mathrm{mole}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2}{102{\mathrm{gC}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2}\times \frac{1\text{ mole }\mathrm{NaOH}}{1\text{ mole }{\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2}\times \frac{1 \mathrm{LNaOH}}{0.208 \mathrm{moleNaOH}}\times \frac{1000\mathrm{mlNaOH}}{1 \mathrm{LNaOH}}$

                    .$=74.4\mathrm{ml} \text{ of }0.208\mathrm{M} \mathrm{NaOH}$