Chapter 18, Problem 18.98P

Interpretation Introduction

(a)

Interpretation:

By using ${\text{IR, }}^{\text{1}}{\text{H NMR and }}^{\text{13}}\text{C NMR}$ spectra, structure of ${\text{C}}_{\text{5}}{\text{H}}_{\text{8}}{\text{O and C}}_{\text{10}}{\text{H}}_{\text{14}}\text{O}$ is to be drawn.

Concept introduction:

For the analysis of the molecular formula of organic molecules, the sites of unsaturation, also known as IHD is used for the determination of the total number of ring and $\text{π}$ bonds in the structure.

In condensation reaction, the loss of water takes place.

Absence of $\text{OH}$ band in IR indicates that the product is beta-hydroxy carbonyl compound which is also a product of Aldol condensation reaction.

Absence of aldehyde proton at $\text{8-10 ppm}$ indicates the presence of ketone in the product as well as the reactant.

Absence of alkene proton $\text{5-6 ppm}$ indicates that there is an absence of protons on alkene carbons.

Answer to Problem 18.98P

Structure of the ${\text{C}}_{\text{5}}{\text{H}}_{\text{8}}{\text{O and C}}_{\text{10}}{\text{H}}_{\text{14}}\text{O}$ is

Explanation of Solution

IHD of ${\text{C}}_{\text{5}}{\text{H}}_{\text{8}}\text{O}$ is $2$ and for ${\text{C}}_{\text{10}}{\text{H}}_{\text{14}}\text{O}$ it is $\text{4}$.

Comparing the formulas ${\text{C}}_{\text{5}}{\text{H}}_{\text{8}}{\text{O and C}}_{\text{10}}{\text{H}}_{\text{14}}\text{O}$ we can observe that the product results from a combination of two reactant molecules, then loss of water occurs. So, this indicates an Aldol condensation reaction.

${\text{C}}_{\text{5}}{\text{H}}_{\text{8}}{\text{O + C}}_{\text{5}}{\text{H}}_{\text{8}}\text{O }\to {\text{ C}}_{\text{10}}{\text{H}}_{\text{16}}{\text{O}}_{\text{2}}\to {\text{ C}}_{\text{10}}{\text{H}}_{\text{14}}{\text{O + H}}_{\text{2}}\text{O}$

The compound ${\text{C}}_{\text{10}}{\text{H}}_{\text{14}}\text{O}$ is the product of an Aldol condensation reaction; it is $\text{α,β-unsaturated carbonyl compound}$. It is not $\text{β-hydroxy carbonyl compound}$ because there is absence of $\text{OH}$ band in IR. Absence of proton at $\text{8-10 ppm}$ indicates that there is no aldehyde proton. So, the reactant is ketone.

From the proton NMR, there is no alkene proton signal at $\text{5-6 ppm}$ so, there are no protons on alkene carbons. Notice that no signal in the proton NMR integrates to three protons –all of them integrate to either two protons or four protons. So, the product contains no methyl groups, and the same for ketone reactants.

The NMR signal for the product is clear; there are five chemically distinct groups of protons. All the NMR signals in the product are $\text{1-3 ppm}$. This indicates the alkene ${\text{CH}}_{\text{2}}$ protons near the double bond of the same type.

So, the structure of product is

So, the reactant ketone with formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{8}}\text{O}$ with $\text{2}$ IHD and no methyl group is cyclopentanone

So, the structure of the ${\text{C}}_{\text{5}}{\text{H}}_{\text{8}}\text{O}$ (reactants) and ${\text{C}}_{\text{10}}{\text{H}}_{\text{14}}\text{O}$ product are

Conclusion

By using the given ${\text{IR, }}^{\text{1}}{\text{H NMR and }}^{\text{13}}\text{C NMR}$ data structure of the reactant and product is determined.

Interpretation Introduction

(b)

Interpretation:

Mechanism of the given Aldol condensation reaction is to be determined.

Concept introduction:

In Aldol reaction, $\text{C-C}$ bond is formed. In this reaction, the electrophilic substitution at the alpha carbon in enolate takes place.

In Aldol reaction, dimerization of an aldehyde or ketone to $\text{β-hydroxy aldehyde or ketone}$ is carried out in the presence of a strong base.

For this reaction, the presence of $\text{α}$ hydrogen is essential on at least one reactant.

In the heterolysis step, water is removed and alkene ($\text{α,β-unsaturated carbonyl compound}$) is formed.

Answer to Problem 18.98P

Mechanism of the given reaction is

Explanation of Solution

In the first step, the most acidic proton at alpha position in cyclopentanone is abstracted by the base. Hydroxide ion ${\text{(OH}}^{-})$ abstracts the $\text{α}$ hydrogen which results in the formation of an enolate anion.

In the second step, nucleophilic addition takes place; the enolate anion acts as a nucleophile, and it will attack the second molecule of cyclopentanone. In this step, a new $\text{C-C}$ bond is formed between two species.

The third step is proton transfer. ${\text{O}}^{\text{-}}$ is protonated by water resulting in the formation of $\text{β-hydroxy ketone}$. Hydroxide ion ${\text{(OH}}^{-})$ is regenerated.

In fourth step $\text{α}$ proton is abstracted by the hydroxide ion ${\text{(OH}}^{-})$ resulting in the formation of carbanion at $\text{α}$ position.

Last step of the reaction is heterolysis in which water is removed and $\text{α,β-unsaturated carbonyl compound}$ is formed.

Complete and detailed mechanism for the given reaction is

Conclusion

By using Aldol condensation reaction, the mechanism of the given reaction is determined.