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The total number of enolates that can be formed, when 2 - Butanone is treated with a base is

Short Answer

Expert verified
Two enolates can be formed from 2-butanone.

Step by step solution

01

Understanding Enolates

Enolates are formed from ketones, aldehydes, or esters in the presence of a strong base. The base abstracts an hydrogen from the -carbon, generating an enolate ion.
02

Identify -carbons in 2-Butanone

In 2-butanone, the -carbons are the carbons adjacent to the carbonyl group. For 2-butanone, these -carbons are located at the 1st and 3rd carbon positions relative to the carbonyl carbon.
03

Determine Possible Enolate Formation Sites

Since 2-butanone has 2 -carbons (1st and 3rd positions), abstraction of a proton (hydrogen) from either of these -carbons can lead to enolate formation.
04

Calculate Total Enolates

Each deprotonation at these different -carbons generates a distinct enolate. Therefore, 2 distinct enolates can be formed: one from the 1st -carbon and another from the 3rd -carbon.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enolates
Enolates are a fascinating component of organic chemistry and play a crucial role in many reactions. They are ions formed when a base deprotonates an alpha-carbon in compounds containing carbonyl groups, such as ketones, aldehydes, and esters. This deprotonation results in a resonance-stabilized anion with negative charge localized mainly on the oxygen, while maintaining correspondence with the alpha-carbon.

Enolates are highly reactive species. They are excellent nucleophiles, meaning they have a strong tendency to donate an electron pair to an electrophile. This unique property makes enolates a valuable intermediate in many synthetic transformations, including aldol condensation and Michael addition.

Understanding the formation and reactivity of enolates is fundamental for anyone delving into organic synthesis. The ability to control which alpha-carbon becomes deprotonated is essential, leading to precision in creating complex molecules.
2-butanone
2-butanone, also known as methyl ethyl ketone (MEK), is an example of a simple ketone. It is valuable as an industrial solvent and also serves as a great example molecule when studying enolate chemistry.

Structurally, 2-butanone consists of a four-carbon chain with a carbonyl group (7C=O8) at the second carbon. Its molecular formula is C t 4 H 8 O, and it is a clear, colorless liquid at room temperature. The presence of the carbonyl group is what primarily defines its reactivity.

In 2-butanone, the carbon adjacent to the carbonyl carbon is the 1alpha carbon, which makes it a site for deprotonation. This susceptibility to reaction allows 2-butanone to serve as a good study subject for examining the behavior and properties of enolates.
Carbonyl Group
The carbonyl group, with the structure C=O, is one of the most significant functional groups in organic chemistry. Found in aldehydes, ketones, carboxylic acids, and derivatives, the carbonyl carbon is sp2 hybridized and planar, making it a reactive center.

The carbonyl group owes its reactivity to its polar nature. Electron density is pulled towards the electronegative oxygen, resulting in a partial positive charge on the carbon. This makes the carbonyl carbon a good electrophile, ready to react with nucleophiles like enolates.

Additionally, the 1alpha hydrogens, or hydrogens attached to the carbon adjacent to the carbonyl, become slightly acidic due to resonance stabilization of the resulting enolate. Upon deprotonation, this resonance significantly stabilizes the resulting species, simplifying the process of forming enolates and supporting a variety of organic reactions.
Proton Abstraction
Proton abstraction is a crucial step in the formation of enolates and involves removing a hydrogen ion (proton) from the 1alpha-carbon of a carbonyl compound by a base. This process is pivotal because it activates the species for enolate formation and subsequent reactions.

A strong base is needed for this purpose, as the 1alpha-hydrogens are only slightly acidic. Common bases used include sodium hydroxide (NaOH) or even stronger ones like lithium diisopropylamide (LDA). The base effectively pulls the proton away, generating a carbanion, which resonates with the carbonyl group to form a stable enolate ion.

Understanding how proton abstraction works is key to mastering the further reactivity of carbonyl compounds. With careful choice of base and conditions, chemists can selectively form enolates from specific 1alpha-carbons, controlling the pathway and forms of the reaction products.

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Most popular questions from this chapter

In the following sequence of reactions: Tolouene \(\stackrel{\mathrm{KMnO}_{4}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{SOCl}_{2}}{\longrightarrow} \mathrm{B} \frac{\mathrm{H}_{2} / \mathrm{Pd}}{\mathrm{BaSO}_{4}} \mathrm{C}\), the product \(\mathrm{C}\) is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\)

Which is not true about acetophenone? (1) Reacts with \(\mathrm{I}_{2} / \mathrm{NaOH}\) to form iodoform (2) Reacts with Tollen's reagent to form silver mirror (3) On oxidation with alkaline \(\mathrm{KMnO}_{4}\) followed by hydrolysis gives benzoic acid (4) Reacts to form 2,4 -dinitrophenyl-hydrazone (a) 2 and 4 (b) 2 only (c) 1 and 4 (d) 4 only

Match the following List II (i) Anhydrous \(\mathrm{ZnCl}_{2}\) (ii) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{\theta}\) (iii) \(\mathrm{P}\) (iv) \(\mathrm{LiAlH}_{4}\) The correct matching is: The correct matching is: 1 2 3 4 (a) (ii) (iii) (iv) (i) (b) (ii) (iii) (i) (iv) (c) (iii) (ii) (iv) (i) (d) (i) (iii) (iv) (ii)

In the Cannizzaro reaction given below, \(2 \mathrm{Ph}-\mathrm{CHO} \stackrel{\mathrm{OH}^{-}}{\longrightarrow} \mathrm{Ph}-\mathrm{CH}_{2} \mathrm{OH}+\mathrm{PhCO}_{2}^{-}\) the slowest step is (a) the attack of \(-\mathrm{OH}\) at the carbonyl group (b) the transfer of hydride ion to the carbonyl group (c) the abstraction of proton from the carboxylic acid (d) the deprotonation of \(\mathrm{Ph}-\mathrm{CH}_{2} \mathrm{OH}\).

Match the following: List I (Reaction) 1\. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}+\mathrm{NaOH} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) 2\. \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{CH}_{3} \mathrm{Cl} \stackrel{\text { Anhyd. } \mathrm{AICl}_{3}}{+\mathrm{Cd} / \mathrm{BaSO}} \stackrel{\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}\). 3\. \(\mathrm{RCOCl}+\mathrm{H}_{2} \stackrel{\mathrm{Pd} / \mathrm{BaSO}_{4}}{\longrightarrow} \mathrm{RCHO}\) 4\. \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{OH}+\mathrm{NaOH}+\mathrm{CO}_{2} \stackrel{\text { 1. heat, 2. } \mathrm{H}^{\prime}}{\longrightarrow}\) List II (Name of reactions) (i) Rosenmund reduction (ii) Kolbe's reaction (iii) Claisen reaction (iv) Cannizzaro reaction (v) Friedal-Craft's reaction The correct matching is: The correct matching is: 1 2 3 4 (a) (iv) (v) (i) (ii) (b) (v) (i) (ii) (iii) (c) (v) (iii) (ii) (i) (d) (iv) (ii) (iii) (i)

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