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The correct order of increasing basicity of the given conjugate bases \(\left(\mathrm{R}=\mathrm{CH}_{3}\right)\) is (a) \(\mathrm{RCO} \overline{\mathrm{O}}<\mathrm{HC} \equiv \overline{\mathrm{C}}<\overline{\mathrm{R}}<\overline{\mathrm{N}} \mathrm{H}_{2}\) (b) \(\overline{\mathrm{R}}<\mathrm{HC} \equiv \overline{\mathrm{C}}<\mathrm{RCO} \overline{\mathrm{O}}<\overline{\mathrm{N}} \mathrm{H}_{2}\) (c) \(\mathrm{RCO} \overline{\mathrm{O}}<\overline{\mathrm{N}} \mathrm{H}_{2}<\mathrm{HC} \equiv \overline{\mathrm{C}}<\overline{\mathrm{R}}\) (d) \(\mathrm{RCO} \overline{\mathrm{O}}<\mathrm{HC} \equiv \overline{\mathrm{C}}<\overline{\mathrm{N}} \mathrm{H}_{2}<\overline{\mathrm{R}}\)

Short Answer

Expert verified
The correct order is option (a): \\(\mathrm{RCO} \overline{\mathrm{O}}<\mathrm{HC} \equiv \overline{\mathrm{C}}<\overline{\mathrm{R}}<\overline{\mathrm{N}} \mathrm{H}_{2}\\).

Step by step solution

01

Identify the Basics of Basicity

Basicity refers to the ability of a compound to donate a pair of electrons. It is also related to the stability of the conjugate acid formed when the base accepts a proton. The more stable the conjugate base, the weaker it is as a base.
02

Understand the Structures and Their Conjugate Bases

Consider the given conjugate bases:- \(\mathrm{RCO} \overline{\mathrm{O}}\), which is the conjugate base of a carboxylic acid, and is generally less basic due to the resonance stabilization.- \(\mathrm{HC} \equiv \overline{\mathrm{C}}\), which is an acetylide ion, typically less basic.- \(\overline{\mathrm{R}}\), which is a methyl anion with high electron density, making it very basic.- \(\overline{\mathrm{N}} \mathrm{H}_{2}\), which is an amide ion, known to be a strong base.
03

Determine the Order of Basicity

The general trend for basicity based on ion structures is:1. Least basic: \(\mathrm{RCO} \overline{\mathrm{O}}\) due to its resonance stabilization and electron-withdrawing group.2. Next is \(\mathrm{HC} \equiv \overline{\mathrm{C}}\), as the acetylide ion is less basic due to its electronegative triple bond.3. Strong bases are \(\overline{\mathrm{N}} \mathrm{H}_{2}\) and \(\overline{\mathrm{R}}\). Since \(\overline{\mathrm{R}}\) is likely \(\mathrm{CH}_3^-\), making it a stronger base due to its less electronegative nature compared to an amide ion.
04

Match the Order to the Given Options

Upon analyzing, the correct order of increasing basicity is option (a): \(\mathrm{RCO} \overline{\mathrm{O}}<\mathrm{HC} \equiv \overline{\mathrm{C}}<\overline{\mathrm{R}}<\overline{\mathrm{N}} \mathrm{H}_{2}\). This arrangement understands resonance, hydrogen bonding, and electronegativity effects contributing to basicity.

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Key Concepts

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

Conjugate Bases
In chemistry, a conjugate base is what's left after an acid donates a proton. This concept is critical for understanding basicity because the stability of the conjugate base directly affects how basic a compound is. When an acid loses a proton, it forms a conjugate base, and the strength of the base is inversely related to the stability of this conjugate base.
  • If the conjugate base is highly stable, it means the original acid was strong and the base is weak.
  • Conversely, if the conjugate base is less stable, the original acid was weak, indicating a stronger base.
In the given exercise, compounds like \( \mathrm{RCO} \overline{\mathrm{O}}\) and \(\mathrm{HC} \equiv \overline{\mathrm{C}}\) have more stable conjugate bases and thus lower basicity. The resonance effect in \(\mathrm{RCO} \overline{\mathrm{O}}\) makes its conjugate base quite stable, resulting in minimal basicity properties. On the other end, \(\overline{\mathrm{R}}\) and \(\overline{\mathrm{N}} \mathrm{H}_{2}\) are less stable, thus acting as stronger bases.
Electron Density
Electron density refers to the likelihood of finding an electron in a specific area of a molecule. In the context of basicity, the electron density around an atom indicates how likely it is to donate an electron pair. The higher the electron density, the stronger the base, since more electrons are readily available for bonding with protons. For example, the methyl anion \(\overline{\mathrm{R}}\), which could be \(\mathrm{CH}_3^-\), has high electron density. This abundance of electrons makes it highly basic as it can easily donate electrons.
  • More electrons mean a greater ability to bond with protons.
  • Higher electron density is generally associated with elements of low electronegativity.
The balance between electron density and the nature of the other atoms in the molecule can dictate the basicity level. For example, even though \(\overline{\mathrm{N}} \mathrm{H}_{2}\) is strong due to its additional electrons, its capacity as a base is somewhat moderated by the electronegativity of nitrogen.
Resonance Stabilization
Resonance stabilization plays a significant role in determining the basicity of conjugate bases. It refers to the delocalization of electrons across multiple atoms, leading to increased stability of the molecule. The more stable a conjugate base becomes due to resonance, the less basic it typically is. A classic example in this context is \(\mathrm{RCO} \overline{\mathrm{O}}\), the conjugate base of a carboxylic acid. This base is considerably stabilized by resonance as the negative charge can be shared between two oxygen atoms and distributed across the carbon-oxygen bond structure.
  • Resonance stabilization disperses electron density and reduces reactivity.
  • It is less likely to donate an electron pair due to its stability, therefore being a weaker base.
In contrast, structures like \(\mathrm{HC} \equiv \overline{\mathrm{C}}\) do not benefit from resonance stabilization but are affected by the nature of the triple bond, slightly reducing basicity compared to more localized electron-rich environments. Understanding the dual impact of resonance and electron density is critical for predicting the strength of a base.

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Most popular questions from this chapter

The stability of the following carbocations decreases in the order. (1) [C]1CCCCC1 (2) [C+]1C=C1 (3) [C+]1C=CC=C1 (4) [C+]1C=CC=CC=C1 (a) \(3>2>4>1\) (b) \(4>2>3>1\) (c) \(4>3>2>1\) (d) \(4>2>1>3\)

The most stable free radical among the following is (1) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}\) (2) \(\mathrm{CH}_{3}-\dot{\mathrm{CH}}-\mathrm{CH}_{3}\) (3) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHCH}_{3}\) (4) \(\mathrm{CH}_{3} \mathrm{CH}_{2}\) (a) \(3>1>4>2\) (b) \(1>3>2>4\) (c) \(3>1>2>4\) (d) \(3>2>1>4\)

Which of the following order/statement (s) is/are correct? (a) In \(S_{N} 1\) reactions, \(\mathrm{AgNO}_{3}\) increases the rates of solvolysis (b) Soft bases have enhanced nucleophilicities towards \(\mathrm{S}_{\mathrm{N}} 2\) reactions. (c) \(\mathrm{PhSO}_{3}^{-}>\mathrm{Cl}_{3} \mathrm{CCOO}^{-}>\mathrm{CH}_{3} \mathrm{COO}^{-}>\mathrm{PhO}^{-}\)(leav- ing group ability) (d) HCOOH may lead a tertiary halide towards \(\mathrm{S}_{\mathrm{N}}^{2} .\)

Which of the following phrases is/are not correctly associated with \(\mathrm{S}_{\mathrm{N}} 1\) reaction? (a) Rate of the reaction is affected by polarity of solvent. (b) The strength of nucleophile is important in determining rate of the reaction. (c) The reaction proceeds with complete inversion of configuration. (d) Neopentyl halide gives no rearranged product by \(\mathrm{S}_{\mathrm{N}} 1\) mechanism.

Consider the following carbanions: (1) \(\mathrm{CH}_{3}-\stackrel{\Theta}{\mathrm{C}} \mathrm{H}_{2}\) (2) \(\mathrm{CH}_{3}=\stackrel{\ominus}{\mathrm{C}} \mathrm{H}\) (3) \(\mathrm{CH} \equiv \stackrel{\ominus}{\mathrm{C}}\) Correct order of stability of these carbanions in decreasing order is (a) \(1>2>3\) (b) \(2>1>3\) (c) \(3>2>1\) (d) \(3>1>2\)

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