Chapter 17, Problem 17.56P

Interpretation Introduction

(a)

Interpretation:

The synthesis of the given alkene from an alkyl halide and a ketone or aldehyde is to be shown.

Concept introduction:

Wittig reactions generate the carbon double($\text{C = C}$) bond by joining the two carbon-containing groups - one from the Wittig reagent and the second from the aldehyde or ketone. In a Wittig reaction, the carbonyl ($\text{C=O}$) bond of aldehyde or ketone is converted into a ($\text{C = C}$) bond. In step one, the negatively charged C atom from the Wittig reagent attacks on the aldehyde and produces a betaine, that is species in which negative charge present on O atom and a positive charge on P atom. In step two, a bond is formed between a positively charge P atom and negatively charged O atom. This results in an oxaphosphetane that contains a four-membered ring. Due to the ring strain, oxaphosphetane converts into the alkene and triphenylphosphine oxide in the final step.

Answer to Problem 17.56P

The synthesis of the given alkene from an alkyl halide and a ketone or aldehyde is shown below:

Explanation of Solution

The given structure of alkene is

In the above structure, the $\text{C=C}$ is formed via a Wittig reaction. The transformation of aldehyde or ketone into a $\text{C=C}$ bond in Wittig reaction, first disconnects the $\text{C=C}$ by undoing a retrosynthetic analysis. The $\text{C=C}$ bond is symmetrical and produces a part of $\text{C=O}$ aldehyde and nucleophile, negatively charged carbon, which is the Wittig reagent.

By undoing a Wittig reaction in retrosynthetic analysis, $\text{C=C}$ was part of the benzaldehyde and Wittig reagent. Lastly, undo Wittig reagent formation to determine the alkyl halide. In this reaction, an alkyl bromide is used, but alkyl chloride or iodide would be a suitable choice as well.

In the forward direction, the synthesis of the given alkene from an alkyl halide and benzaldehydewould appears as follows:

Conclusion

The synthesis of given alkene from an alkyl halide and a ketone or aldehyde is shown.

Interpretation Introduction

(b)

Interpretation:

The synthesis of the given alkene from an alkyl halide and a ketone or aldehyde is to be shown.

Concept introduction:

Wittig reactions generate the carbon double($\text{C = C}$) bond by joining the two carbon-containing groups. The one from the Wittig reagent and the second from the aldehyde or ketone. In a Wittig reaction, the carbonyl ($\text{C=O}$) bond of aldehyde or ketone is converted into a ($\text{C = C}$) bond. In step one, the negatively charged C atom from the Wittig reagent attacks the aldehyde and produces a betaine, that is, the species in which negative charge is present on O atom and a positive charge on P atom. In step two, a bond is formed between a positively charge P atom and negatively charged O atom. This results in an oxaphosphetane that contains a four-membered ring. Due to the ring strain, oxaphosphetane converts into the alkene and triphenylphosphine oxide in the final step.

Answer to Problem 17.56P

The synthesis of the givenunsymmetricalkene from an alkyl halide and a ketone or aldehyde by two different ways as shown below.

Method 1:

Method 2:

Explanation of Solution

The given structure of alkene is

In the above structure, the $\text{C=C}$ is formed via a Wittig reaction. The transformation of aldehyde or ketone into a $\text{C=C}$ bond in Wittig reaction, first disconnect the $\text{C=C}$ by undoing a retrosynthetic analysis. The $\text{C=C}$ bond is unsymmetrical compound and produces a one-half part of $\text{C=O}$ ketone or an aldehydeandanother half would have come from nucleophile, negatively charged carbon, which is the Wittig reagent. It is an unsymmetrical compound, so we can undo a Wittig reaction in two different ways.

Method 1:

By undoing a Wittig reaction in retrosynthetic analysis, $\text{C=C}$ was part of thecyclohexanone and Wittig reagent. Next, undoing Wittig reagent formation to determine the alkyl halide. In this reaction, an alkyl bromide is used, but alkyl chloride or iodide would be a suitable choice as well.

In the forward direction, the synthesis of the given alkene from an alkyl halide and a ketone would appear as follows:

Method 2:

By undoing a Wittig reaction in retrosynthetic analysis, $\text{C=C}$ was part of an aldehydeand Wittig reagent. Next, undoing Wittig reagent formation to determine the alkyl halide. In this reaction, an alkyl bromide is used, but alkyl chloride or iodide would be a suitable choice as well.

In the forward direction, the synthesis of the given alkene from an alkyl halide and an aldehyde would appear as follows:

Conclusion

The synthesis of the givenunsymmetricalkene from an alkyl halide and a ketone or an aldehyde by two different ways is shown.

Interpretation Introduction

(c)

Interpretation:

The synthesis of the given alkene from an alkyl halide and a ketone or aldehyde is to be shown.

Concept introduction:

Wittig reactions generate the carbon double($\text{C = C}$) bond by joining the two carbon-containing groups. The one from the Wittig reagent and the second from the aldehyde or ketone. In a Wittig reaction, the carbonyl ($\text{C=O}$) bond of aldehyde or ketone is converted into a ($\text{C = C}$) bond. In step one, the negatively charged C atom from the Wittig reagent attacks on the aldehyde and produces a betaine, that is species in which negative charge present on O atom and a positive charge on P atom. In step two, a bond is formed between a positively charge P atom and negatively charged O atom. This results in an oxaphosphetane that contains a four-membered ring. Due to the ring strain which converts into the alkene and triphenylphosphine oxide in the final step.

Answer to Problem 17.56P

The synthesis of the givenunsymmetricalkene from an alkyl halide and a ketone or aldehyde by two different ways as shown below.

Method 1:

Method 2:

Explanation of Solution

The given structure of alkene is

In the above structure, the $\text{C=C}$ is formed via a Wittig reaction. The transformation of aldehyde or ketone into a $\text{C=C}$ bond in Wittig reaction, first disconnect the $\text{C=C}$ by undoing a retrosynthetic analysis. The $\text{C=C}$ bond is unsymmetrical compound and produces a one-half part of $\text{C=O}$ ketone or an aldehyde and another half would have come from nucleophile, negatively charged carbon, which is the Wittig reagent. It is an unsymmetrical compound, so we can undo a Wittig reaction in two different ways.

Method 1:

By undoing a Wittig reaction in retrosynthetic analysis, $\text{C=C}$ was part of thebenzaldehyde and Wittig reagent. Next step is the undoing of Wittig reagent formation to determine the alkyl halide. In this reaction, an alkyl bromide is used, but alkyl chloride or iodide would be a suitable choice as well.

In the forward direction, the synthesis of the given alkene from an alkyl halide and benzaldehyde would appear as follows:

Method 2:

By undoing a Wittig reaction in retrosynthetic analysis, $\text{C=C}$ was part of the ketone and Wittig reagent. Next step is the undoing of the Wittig reagent formation to determine the alkyl halide. In this reaction, an alkyl bromide is used, but alkyl chloride or iodide would be a suitable choice as well.

In the forward direction, the synthesis of the given alkene from an alkyl halide and a ketone would appear as follows:

Conclusion

The synthesis of the given unsymmetric alkene from an alkyl halide and a ketone or an aldehyde by two different ways is shown.