Question

Following is a retrosynthetic analysis for the synthesis of the herbicide ( \(S\) )-Metolachlor from 2-ethyl-6-methylaniline, chloroacetic acid, acetone, and methanol. Show reagents and experimental conditions for the synthesis of Metolachlor from these four organic starting materials. Your synthesis will most likely give a racemic mixture. The chiral catalyst used by Novartis for reduction in Step 2 gives \(80 \%\) enantiomeric excess of the \(S\) enantiomer.

Short answer

In order to synthesize (S)-Metolachlor from the given starting materials, the following steps are involved: 1. Formation of amide from 2-ethyl-6-methylaniline and chloroacetic acid using Schotten-Baumann conditions. 2. Reduction of the amide to amine using a chiral catalyst like Shibasaki's catalyst, providing an 80% enantiomeric excess of the (S)-enantiomer. 3. Aldol condensation between the (S)-amine and acetone using a base, forming a new carbon-carbon bond and a ketone. 4. Reduction of the ketone to alcohol using methanol and a suitable catalyst such as sodium borohydride. Following these steps will enable the synthesis of (S)-Metolachlor from the provided starting materials.

Step-by-step solution

Formation of amide from 2-ethyl-6-methylaniline and chloroacetic acid

To form the amide from 2-ethyl-6-methylaniline and chloroacetic acid, we can treat the amine with chloroacetic acid under Schotten-Baumann conditions. In this process, the amine will react with chloroacetic acid to form an amide bond. Consequently, the first step will be: 2-ethyl-6-methylaniline + chloroacetic acid \(\rightarrow\) 2-ethyl-6-methylaniline-chloroacetamide

Reduction of amide to amine using a chiral catalyst

Next, we'll need to reduce the amide formed in step 1 to an amine. This can be accomplished by performing a reduction of the amide using a chiral catalyst such as Shibasaki's catalyst. The chiral catalyst will provide enantioselectivity, resulting in an 80% enantiomeric excess of the (S) enantiomer. The reduction step will be: 2-ethyl-6-methylaniline-chloroacetamide + chiral catalyst \(\rightarrow\) 2-ethyl-6-methylaniline-(S)-chloroacetamine (80% ee)

Aldol condensation with acetone

In the next step, we will perform an aldol condensation between the 2-ethyl-6-methylaniline-(S)-chloroacetamine obtained in Step 2 and acetone. The aldol condensation can be promoted by using a base, usually an alkoxide. This reaction will form a new carbon-carbon bond and a ketone in the final product. The aldol condensation step will be: 2-ethyl-6-methylaniline-(S)-chloroacetamine + acetone + base \(\rightarrow\) (S)-Metolachlor-ketone

Reduction of ketone to alcohol using methanol

Finally, we will reduce the ketone formed in step 3 to an alcohol using methanol as the reducing agent. This can be achieved by performing a reduction in the presence of a suitable catalyst, such as an aluminum alkoxide like sodium borohydride, or with a hydride donor. The ketone reduction step will be: (S)-Metolachlor-ketone + methanol + catalyst \(\rightarrow\) (S)-Metolachlor By following these steps and using the appropriate experimental conditions, we can synthesize (S)-Metolachlor from the four specified organic starting materials.

Key concepts

Retrosynthetic Analysis

Retrosynthetic analysis is a pivotal strategy in organic synthesis where the desired compound is systematically deconstructed into simpler precursor molecules. This approach helps to identify the starting materials and the types of reactions required to build the complex structure. In the context of (S)-Metolachlor synthesis, retrosynthetic analysis allows chemists to map out a multi-step pathway from 2-ethyl-6-methylaniline, chloroacetic acid, acetone, and methanol to the target molecule.

Key steps might include the formation of an amide bond, chiral catalyst reduction, aldol condensation, and ketone reduction. Each of these transformations is chosen for its ability to introduce the necessary functional groups, construct the carbon skeleton, and establish the correct stereochemistry. This ordered backward-thinking provides a clear blueprint that guides the forward synthetic journey.

Amide Bond Formation

Amide bond formation is an essential reaction in organic chemistry, often forming the backbone of peptides and proteins. When synthesizing (S)-Metolachlor, the amide bond is created between 2-ethyl-6-methylaniline and chloroacetic acid. This can be achieved using reagents and conditions known as Schotten-Baumann conditions which often include the use of an acid chloride and a base such as sodium hydroxide.

The amide linkage is a result of a nucleophilic substitution reaction where the amine group attacks an activated carboxylic acid derivative, leading to the expulsion of a leaving group and the formation of an amide. This bond is robust and withstands many subsequent chemical transformations, providing a solid framework for the rest of the synthetic route.

Chiral Catalyst Reduction

Chiral catalysts play a critical role in enantioselective synthesis, which aims to produce a single enantiomer of a chiral molecule. The challenge in synthesizing (S)-Metolachlor is the creation of a compound with a specific stereochemistry. To achieve this, the amide reduction step utilizes a chiral catalyst such as Shibasaki's catalyst. It promotes the formation of the (S) enantiomer with a significant enantiomeric excess (ee), in this case, 80%.

The choice of a chiral catalyst ensures that the reaction proceeds with a preference for one enantiomer, allowing the synthesis to yield a product with the desired chirality. This enantiomer is often more effective or safer due to the way it interacts with biological systems, indicating the importance of this method in pharmaceutical manufacturing.

Aldol Condensation

Aldol condensation is a fundamental reaction for creating carbon-carbon bonds and is crucial for building complex molecular architectures. In the synthesis of (S)-Metolachlor, aldol condensation involves the reaction of 2-ethyl-6-methylaniline-(S)-chloroacetamine with acetone.

This reaction typically requires a strong base, such as an alkoxide, to deprotonate the acetone, making it a highly reactive enolate ion. The enolate then attacks an aldehyde or a ketone, in this case, the carbonyl group of the amine compound, to form a β-hydroxy ketone. The product of this reaction further undergoes dehydration to yield an enone, which is a key intermediate leading towards the final structure of (S)-Metolachlor.

Ketone Reduction

The last step in the creation of (S)-Metolachlor is the reduction of a ketone to an alcohol. Ketone reduction is a common and vital step in the synthesis of many organic molecules. This reaction involves the addition of hydrogen to the carbon-oxygen double bond of the ketone, transforming it into a secondary alcohol.

In practice, this reduction can be conducted with various reducing agents like sodium borohydride or with methanol under catalytic conditions. The choice of reagents and conditions can influence the outcome of the reaction, and it is critical to ensure that the stereochemistry of the product is preserved. This step solidifies the molecular framework of the final product, thereby completing the synthesis of (S)-Metolachlor.