Chapter 24, Problem 24.8P

Draw the products formed in the crossed aldol reaction of phenylacetaldehyde $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CHO}\right)$ with each compound: (a) ${\text{CH}}_{\text{2}}{\left(\text{COOEt}\right)}_{\text{2}}$; (b) ${\text{CH}}_{\text{2}}{\left({\text{COCH}}_3\right)}_{\text{2}}$; (c) ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}\text{CN}$.

Interpretation Introduction

(a)

Interpretation: The product that is formed in the given crossed aldol reaction of phenylacetaldehyde $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CHO}\right)$ with ${\text{CH}}_{\text{2}}{\left(\text{COOEt}\right)}_{\text{2}}$ is to be drawn.

Concept introduction: The crossed aldol reaction takes place between the two different carbonyl compounds that is between two different aldehydes or two different ketones or one aldehyde and one ketone. The crossed aldol reaction takes place only if the carbonyl compound contains an acidic $\text{α}-\text{hydrogen}$ atom.

Answer to Problem 24.8P

The product that is formed in the given crossed aldol reaction of phenylacetaldehyde $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CHO}\right)$ with ${\text{CH}}_{\text{2}}{\left(\text{COOEt}\right)}_{\text{2}}$ is,

Explanation of Solution

The product that is formed in the given crossed aldol reaction is shown below.

Figure 1

In this crossed aldol reaction, phenylacetaldehyde reacts with ${\text{CH}}_{\text{2}}{\left(\text{COOEt}\right)}_{\text{2}}$ in the presence of strong base, sodium ethoxide to form the unsaturated aldehydic compound. In this reaction, the base abstracts a proton from ${\text{CH}}_{\text{2}}{\left(\text{COOEt}\right)}_{\text{2}}$ that result in the formation of carbanion. Then, that carbanion reacts with phenylacetaldehyde to form a $\text{β}-\text{hydroxy}$ compound. The $\text{β}-\text{hydroxy}$ compound losses water molecule that leads to the formation of the desired unsaturated product.

Conclusion

The product that is formed in the given crossed aldol reaction of phenylacetaldehyde $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CHO}\right)$ with ${\text{CH}}_{\text{2}}{\left(\text{COOEt}\right)}_{\text{2}}$ is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation: The product that is formed in the given crossed aldol reaction of phenylacetaldehyde $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CHO}\right)$ with ${\text{CH}}_{\text{2}}{\left({\text{COCH}}_3\right)}_{\text{2}}$ is to be drawn.

Concept introduction: The crossed aldol reaction takes place between the two different carbonyl compounds that is between two different aldehydes or two different ketones or one aldehyde and one ketone. The crossed aldol reaction takes place only if the carbonyl compound contains an acidic $\text{α}-\text{hydrogen}$ atom.

Answer to Problem 24.8P

The product that is formed in the given crossed aldol reaction of phenylacetaldehyde $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CHO}\right)$ with ${\text{CH}}_{\text{2}}{\left({\text{COCH}}_3\right)}_{\text{2}}$ is,

Explanation of Solution

The product that is formed in the given crossed aldol reaction is shown below.

Figure 2

In this crossed aldol reaction, phenylacetaldehyde reacts with ${\text{CH}}_{\text{2}}{\left({\text{COCH}}_3\right)}_{\text{2}}$ in the presence of strong base, sodium ethoxide to form the unsaturated aldehydic compound. In this reaction, the base abstracts a proton from ${\text{CH}}_{\text{2}}{\left({\text{COCH}}_3\right)}_{\text{2}}$ that result in the formation of carbanion. Then, that carbanion reacts with phenylacetaldehyde to form a $\text{β}-\text{hydroxy}$ compound. The $\text{β}-\text{hydroxy}$ compound losses water molecule that leads to the formation of the desired unsaturated product.

Conclusion

The product that is formed in the given crossed aldol reaction of phenylacetaldehyde $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CHO}\right)$ with ${\text{CH}}_{\text{2}}{\left({\text{COCH}}_3\right)}_{\text{2}}$ is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation: The product that is formed in the given crossed aldol reaction of phenylacetaldehyde $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CHO}\right)$ with ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}\text{CN}$ is to be drawn.

Concept introduction: The crossed aldol reaction takes place between the two different carbonyl compounds that is between two different aldehydes or two different ketones or one aldehyde and one ketone. The crossed aldol reaction takes place only if the carbonyl compound contains an acidic $\text{α}-\text{hydrogen}$ atom.

Answer to Problem 24.8P

The product that is formed in the given crossed aldol reaction of phenylacetaldehyde $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CHO}\right)$ with ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}\text{CN}$ is,

Explanation of Solution

The product that is formed in the given crossed aldol reaction is shown below.

Figure 3

In this crossed aldol reaction, phenylacetaldehyde reacts with ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}\text{CN}$ in the presence of strong base, sodium ethoxide to form the unsaturated aldehydic compound. In this reaction, the base abstracts a proton from ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}\text{CN}$ that result in the formation of carbanion. Then, that carbanion reacts with phenylacetaldehyde to form a $\text{β}-\text{hydroxy}$ compound. The $\text{β}-\text{hydroxy}$ compound losses water molecule that leads to the formation of the desired unsaturated product.

Conclusion

The product that is formed in the given crossed aldol reaction of phenylacetaldehyde $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CHO}\right)$ with ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}\text{CN}$ is shown in Figure 3.