Chapter 22, Problem 22.45P

Draw the product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with each reagent. With some no reaction occurs.

a. ${\text{NaHCO}}_{\text{3}}$

b. $\text{NaOH}$

c. ${\text{SOCL}}_{\text{2}}$

d. $\text{NaCl}$

e. ${\text{NH}}_{\text{3}}\text{(1 equiv)}$

f. ${\text{NH}}_{\text{3}}\text{,}△$

g. ${\text{CH}}_{\text{3}}{\text{OH, H}}_{\text{2}}{\text{SO}}_{\text{4}}$

h. ${\text{CH}}_{\text{3}}{\text{OH, OH}}^{-}$

i. ${\text{[1]NaOH; [2]CH}}_{\text{3}}\text{COCl}$

j. ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{, DCC}$

k. ${\text{[1]SOCL}}_{\text{2}}{\text{; [2]CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\text{(excess)}$

l. ${\text{[1]SOCL}}_{\text{2}}\text{; [2]}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$

Interpretation Introduction

(a)

Interpretation: The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{NaHCO}}_{\text{3}}$ is to be drawn.

Concept introduction: Carboxylic acids react with ${\text{NaHCO}}_{\text{3}}$ to form sodium salts of carboxylic acid.

Answer to Problem 22.45P

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{NaHCO}}_{\text{3}}$ is

Explanation of Solution

Carboxylic acids react with ${\text{NaHCO}}_{\text{3}}$ to form sodium salts of carboxylic acid. Phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ reacts with ${\text{NaHCO}}_{\text{3}}$ to form sodium-$2$-phenylacetate as depicted below.

Figure 1

Conclusion

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{NaHCO}}_{\text{3}}$ is drawn in Figure 1.

Interpretation Introduction

(b)

Interpretation: The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with $\text{NaOH}$ is to be drawn.

Concept introduction: Carboxylic acids react with $\text{NaOH}$ to form sodium salts of carboxylic acid.

Answer to Problem 22.45P

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with $\text{NaOH}$ is

Explanation of Solution

Carboxylic acids react with $\text{NaOH}$ to form sodium salts of carboxylic acid. Phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ reacts with $\text{NaOH}$ to form sodium-$2$-phenylacetate as depicted below.

Figure 2

Conclusion

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with $\text{NaOH}$ is drawn in Figure 2.

Interpretation Introduction

(c)

Interpretation: The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{SOCL}}_{\text{2}}$ is to be drawn.

Concept introduction: Carboxylic acids react with ${\text{SOCL}}_{\text{2}}$ to form substituted products in which hydroxyl group is replaced by chlorine atom.

Answer to Problem 22.45P

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{SOCL}}_{\text{2}}$ is

Explanation of Solution

Carboxylic acids react with ${\text{SOCL}}_{\text{2}}$ to form substituted products in which hydroxyl group is replaced by chlorine. Phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ reacts with ${\text{SOCL}}_{\text{2}}$ to form $2$-phenylacetyl chloride as depicted below.

Figure 3

Conclusion

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{SOCL}}_{\text{2}}$ is drawn in Figure 3.

Interpretation Introduction

(d)

Interpretation: The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with $\text{NaCl}$ is to be drawn.

Concept introduction: Carboxylic acids does not react with $\text{NaCl}$.

Answer to Problem 22.45P

No product is formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with $\text{NaCl}$.

Explanation of Solution

Carboxylic acids does not react with $\text{NaCl}$. No product is formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with $\text{NaCl}$ as depicted below.

Figure 4

Conclusion

No product is formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with $\text{NaCl}$ as shown in Figure 4.

Interpretation Introduction

Interpretation: The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{NH}}_{\text{3}}\text{(1 equiv)}$ is to be drawn.

(e)

Concept introduction: Carboxylic acids react with ${\text{NH}}_{\text{3}}\text{(1 equiv)}$ to form ammonium salts of carboxylic acid.

Answer to Problem 22.45P

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{NH}}_{\text{3}}\text{(1 equiv)}$ is

Explanation of Solution

Carboxylic acids react with ${\text{NH}}_{\text{3}}\text{(1 equiv)}$ to form ammonium salts of carboxylic acid. Phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ reacts with ${\text{NH}}_{\text{3}}\text{(1 equiv)}$ to form ammonium-$2$-phenylacetate as depicted below.

Figure 5

Conclusion

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{NH}}_{\text{3}}\text{(1 equiv)}$ is drawn in Figure 5.

Interpretation Introduction

Interpretation: The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{NH}}_{\text{3}}\text{,}\Delta$ is to be drawn.

(f)

Concept introduction: Carboxylic acids react with ${\text{NH}}_{\text{3}}\text{,}\Delta$ to form amides.

Answer to Problem 22.45P

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{NH}}_{\text{3}}$ is

Explanation of Solution

Carboxylic acids react with ${\text{NH}}_{\text{3}}\text{,}\Delta$ to form amides. Phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ reacts with ${\text{NH}}_{\text{3}}\text{,}\Delta$ to form $2$-phenylacetamide as depicted below.

Figure 6

Conclusion

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{NH}}_{\text{3}}\text{,}\Delta$ is drawn in Figure 6.

Interpretation Introduction

Interpretation: The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{CH}}_{\text{3}}{\text{OH, H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is to be drawn.

(g)

Concept introduction: Carboxylic acids react with alcohols in acidic medium to form esters.

Answer to Problem 22.45P

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{CH}}_{\text{3}}{\text{OH, H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is

Explanation of Solution

Carboxylic acids react with ${\text{CH}}_{\text{3}}{\text{OH, H}}_{\text{2}}{\text{SO}}_{\text{4}}$ to form esters. Phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ reacts with ${\text{CH}}_{\text{3}}{\text{OH, H}}_{\text{2}}{\text{SO}}_{\text{4}}$ to form methyl-$2$-phenylacetate as depicted below.

Figure 7

Conclusion

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{CH}}_{\text{3}}{\text{OH, H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is drawn in Figure 7.

Interpretation Introduction

Interpretation: The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{CH}}_{\text{3}}{\text{OH, OH}}^{-}$ is to be drawn.

(h)

Concept introduction: Carboxylic acids react with alcohols in basic medium to form carboxylate ions.

Answer to Problem 22.45P

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{CH}}_{\text{3}}{\text{OH, OH}}^{-}$ is

Explanation of Solution

Carboxylic acids react with ${\text{CH}}_{\text{3}}{\text{OH, OH}}^{-}$ to form carboxylate ions. Phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ reacts with ${\text{CH}}_{\text{3}}{\text{OH, OH}}^{-}$ to form $2$-phenylacetate as depicted below.

Figure 8

Conclusion

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{CH}}_{\text{3}}{\text{OH, OH}}^{-}$ is drawn in Figure 8.

Interpretation Introduction

Interpretation: The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{[1]NaOH; [2]CH}}_{\text{3}}\text{COCl}$ is to be drawn.

(i)

Concept introduction: Carboxylic acids react with acid chlorides in presence of strong base to form anhydrides.

Answer to Problem 22.45P

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{[1]NaOH; [2]CH}}_{\text{3}}\text{COCl}$ is

Explanation of Solution

Carboxylic acids react with ${\text{[1]NaOH; [2]CH}}_{\text{3}}\text{COCl}$ to form anhydrides. Phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ reacts with ${\text{[1]NaOH; [2]CH}}_{\text{3}}\text{COCl}$ to form acetic-$2$-phenylacetic anhydride as depicted below.

Figure 9

Conclusion

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{[1]NaOH; [2]CH}}_{\text{3}}\text{COCl}$ is drawn in Figure 9.

Interpretation Introduction

Interpretation: The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{, DCC}$ is to be drawn.

(j)

Concept introduction: Carboxylic acids react with ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{, DCC}$ to form amides.

Answer to Problem 22.45P

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{, DCC}$ is

Explanation of Solution

Carboxylic acids react with ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{, DCC}$ to form amides. Phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ reacts with ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{, DCC}$ to form N-methyl-$2$-phenylacetamide as depicted below.

Figure 10

Conclusion

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{, DCC}$ is drawn in Figure 10.

Interpretation Introduction

Interpretation: The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{[1]SOCL}}_{\text{2}}{\text{; [2]CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\text{(excess)}$ is to be drawn.

(k)

Concept introduction: Carboxylic acids react with ${\text{[1]SOCL}}_{\text{2}}{\text{; [2]CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\text{(excess)}$ to form amides.

Answer to Problem 22.45P

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{[1]SOCL}}_{\text{2}}{\text{; [2]CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\text{(excess)}$ is

Explanation of Solution

Carboxylic acids react with ${\text{[1]SOCL}}_{\text{2}}{\text{; [2]CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\text{(excess)}$ to form amides. Phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ reacts with ${\text{[1]SOCL}}_{\text{2}}{\text{; [2]CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\text{(excess)}$ to form $2$-phenyl-N-propylacetamide as depicted below.

Figure 11

Conclusion

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{[1]SOCL}}_{\text{2}}{\text{; [2]CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\text{(excess)}$ is drawn in Figure 11.

Interpretation Introduction

(l)

Interpretation: The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{[1]SOCL}}_{\text{2}}\text{; [2]}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ is to be drawn.

Concept introduction: Carboxylic acids react with ${\text{[1]SOCL}}_{\text{2}}\text{; [2]}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ to form esters.

Answer to Problem 22.45P

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{[1]SOCL}}_{\text{2}}\text{; [2]}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ is

Explanation of Solution

Carboxylic acids react with ${\text{[1]SOCL}}_{\text{2}}\text{; [2]}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ to form esters. Phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ reacts with ${\text{[1]SOCL}}_{\text{2}}\text{; [2]}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ to form isopropyl-$2$-phenylacetate as depicted below.

Figure 12

Conclusion

The product formed when phenylacetic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\right)$ is treated with ${\text{[1]SOCL}}_{\text{2}}\text{; [2]}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ is drawn in Figure 12.