Question

Show the product expected when the following unsaturated \(\delta\)-ketoester is treated with each reagent. CCOC(=O)CC=CC(C)=O (a) \(\underset{\mathrm{Hd}, \mathrm{EtOH}}{\mathrm{H}(1 \mathrm{~mol})}\) (b) \(\frac{\mathrm{NaBH}_{4}}{\mathrm{CH}_{3} \mathrm{OH}}\) (c) \(\frac{1 . \mathrm{LiAlH}_{4}, \mathrm{THF}}{2 \cdot \mathrm{H}_{2} \mathrm{O}}\) (d) \(\underset{2 . \mathrm{H}_{2} \mathrm{O}}{ } \underset{\text { DIBALH, }-78^{\circ}}{\longrightarrow}\)

Short answer

Answer: (a) CH3COC(=O)CH2CH2CH(O)=O (b) CH3COC(=O)CH2CH=CHCH(OH)2 (c) CH3CH(OH)CH2CH=CHCH(OH)2 (d) CH3CH(O)CH2CH=CHC(O)=O

Step-by-step solution

Finding the structure of the given compound

The given compound is represented by the SMILES notation "CCOC(=O)CC=CC(C)=O". Converting this into a chemical structure, the compound is: CH3COC(=O)CH2CH=CHC(O)=O Now, we will find the product formed for each reagent.

(a) Reaction with 1 mol H2 and Pd/C catalyst in EtOH

This reagent contains hydrogen gas (H2) and a palladium catalyst on carbon (Pd/C) in an ethanol (EtOH) solvent. This combination of reagents indicates a hydrogenation reaction. However, hydrogenation will only reduce the carbon-carbon double bond and will not affect the carbonyl groups. The product of this reaction would be CH3COC(=O)CH2CH2CH(O)=O.

(b) Reaction with NaBH4 in MeOH

Sodium borohydride (NaBH4) is a mild reducing agent that can reduce carbonyl groups to their corresponding alcohols. In this case, NaBH4 preferentially reduces the more reactive ketone group while leaving the ester group unaffected. The product of this reaction would be CH3COC(=O)CH2CH=CHCH(OH)2.

(c) Reaction with LiAlH4 in THF followed by water quenching

Lithium aluminum hydride (LiAlH4) is a strong reducing agent capable of reducing both ketones and esters to alcohols. First, the LiAlH4 reacts with the ketone and ester groups in tetrahydrofuran (THF), followed by the addition of H2O to quench the reaction. The product of this reaction would be CH3CH(OH)CH2CH=CHCH(OH)2.

(d) Reaction with DIBALH at -78°C followed by water quenching

Diisobutylaluminum hydride (DIBALH) selectively reduces esters to aldehydes without affecting ketones when used in low temperatures (like -78°C). After the reduction, water is added to quench the reaction. The product of this reaction would be CH3CH(O)CH2CH=CHC(O)=O.

Key concepts

Ketone and Ester Reduction

Reduction reactions are key processes in organic chemistry, particularly when dealing with ketones and esters. These compounds are prevalent in natural and synthetic organic compounds.

Ketones have the general structure R2C=O and are reduced to secondary alcohols (R2CHOH) using common reducing agents like sodium borohydride (NaBH4) or lithium aluminum hydride (LiAlH4). Ester reduction, however, is a bit more complex. Esters (R1COOR2) can be reduced to alcohols (R1CH2OH) and one molecule of R2OH. The choice of reducing agent is crucial for determining the outcome.

Selectivity in Ketone and Ester Reduction

Sodium borohydride is a milder reagent suitable for reducing more reactive carbonyl compounds like ketones, whereas lithium aluminum hydride is more potent and can reduce both ketones and esters. Selective reduction of an ester over a ketone can be achieved with reagents like diisobutylaluminum hydride (DIBALH), which, when used at low temperatures, preferentially reduces the ester to an aldehyde.

Understanding these nuances helps in predicting and controlling the products of reduction reactions in organic synthesis.

SMILES Notation

SMILES (Simplified Molecular Input Line Entry System) notation is an efficient way to represent chemical structures using simple ASCII strings. This method enables chemists and various computational chemistry software to quickly and clearly communicate complex molecules.

Encoding structures in SMILES involves representing atoms, bonds, ring closure, and branching without resorting to drawing the chemical structure. For example, the SMILES string 'CCOC(=O)CC=CC(C)=O' illustrates a specific molecule where 'CC' indicates a carbon chain, while '=O' implies a carbonyl group. Transforming these strings into chemical structures allows chemists to visualize the molecule and helps to predict the behavior of the compound in various reactions.

Application of SMILES in Organic Chemistry

SMILES notation is especially useful in computational chemistry for database searches, chemical information storage, and facilitating digital communication among chemists worldwide.

Hydrogenation Reaction

A hydrogenation reaction involves the addition of hydrogen atoms (H2) to unsaturated compounds, usually at a carbon-carbon double bond or a carbonyl group. Catalysts such as palladium on carbon (Pd/C) or platinum (Pt) are typically employed to speed up the reaction.

For example, the hydrogenation of a double bond is achieved by treating the unsaturated compound with hydrogen gas in the presence of a catalyst. This process converts alkenes into alkanes, a transformation signifying the saturation of the compound.

Understanding Catalyst Roles in Hydrogenation

Within the context of organic synthesis, the catalyst not only speeds up the reaction but can also influence the selectivity of the reaction, depending on its properties. When dealing with multifunctional molecules, chemists must carefully choose reaction conditions to achieve selectivity in hydrogenation reactions.

Selective Reduction

Selective reduction in organic chemistry refers to the targeted reduction of specific functional groups within a molecule while leaving other groups intact. Achieving selectivity is a matter of choosing the right reducing agent and reaction conditions.

For instance, DIBALH at low temperatures selectively reduces esters to aldehydes, a handy reaction when the presence of sensitive ketones or alkenes requires preservation. The careful management of reagent, temperature, and solvent plays a critical role in steering the reaction toward the desired product.

Strategies for Selective Reduction

Organic chemists often use protecting groups to shield certain functional groups from reduction or employ steric hindrance to prevent certain areas of a molecule from reacting. As a result, selectivity is controllable, making it a powerful tool in the synthesis of complex molecules.