Textbook Question
Given the value of Keq for the following acid–base reactions, identify the weakest acid and the weakest base.
(a)
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3. Acids and Bases
Acid Base Equilibrium
5:02 minutesProblem 22
Textbook Question
Explain why an amide ion cannot be used to form a carbanion from an alkane in a reaction that favors products.
Verified step by step guidance1
Step 1: Begin by understanding the nature of the amide ion (NH₂⁻). The amide ion is a strong base, meaning it has a high tendency to accept protons (H⁺) from other molecules. This property is crucial in determining its reactivity in deprotonation reactions.
Step 2: Consider the acidity of alkanes. Alkanes are hydrocarbons with single bonds between carbon atoms, and their hydrogen atoms are bonded to sp³-hybridized carbons. The C-H bond in alkanes is nonpolar and has a very high bond dissociation energy, making the hydrogen atoms extremely weakly acidic.
Step 3: Compare the basicity of the amide ion to the acidity of the alkane. For a reaction to favor the formation of products, the base must be strong enough to deprotonate the acid. However, the amide ion is not strong enough to overcome the very low acidity of the alkane's hydrogen atoms. This means the equilibrium of the reaction would strongly favor the reactants rather than the products.
Step 4: Analyze the stability of the carbanion that would form if deprotonation occurred. Carbanions formed from alkanes are highly unstable because the negative charge is localized on an sp³-hybridized carbon atom, which is not electronegative enough to stabilize the charge. This instability further discourages the formation of a carbanion from an alkane.
Step 5: Conclude that the combination of the weak acidity of alkanes and the instability of the resulting carbanion makes it thermodynamically unfavorable for an amide ion to deprotonate an alkane in a reaction that favors products.
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Video duration:
5mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Amide Ion
An amide ion (NH2-) is a strong base and a nucleophile, typically used in organic reactions to deprotonate compounds or to react with electrophiles. However, its strong basicity can lead to side reactions, making it less effective in forming stable carbanions from alkanes, which are generally unreactive due to their saturated nature.
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Amide Nomenclature
Carbanion Stability
Carbanions are negatively charged carbon species that are highly reactive and unstable, especially when derived from alkanes. The stability of a carbanion is influenced by the surrounding groups; for alkanes, the lack of electron-withdrawing groups makes it difficult to stabilize the negative charge, thus making the formation of carbanions unfavorable in reactions that favor products.
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Reaction Favorability
In organic chemistry, the favorability of a reaction is determined by the Gibbs free energy change, which considers both enthalpy and entropy. Reactions that lead to more stable products or lower energy states are favored. Since the formation of a carbanion from an alkane using an amide ion does not lead to a more stable product, the reaction is not favored under standard conditions.
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