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The correct order of increasing basicity of the given conjugate bases \(\left(\mathrm{R}=\mathrm{CH}_{3}\right)\) is |2010] (a) \(\mathrm{RCO} \overline{\mathrm{O}}<\mathrm{HC} \equiv \overline{\mathrm{C}}<\overline{\mathrm{R}}<\overline{\mathrm{N}} \mathrm{H}_{2}\) (b) \(\overline{\mathrm{R}}<\mathrm{HC}=\overline{\mathrm{C}}<\mathrm{RCO} \overline{\mathrm{O}}<\overline{\mathrm{N}} \mathrm{H}_{2}\) (c) \(\mathrm{RCO} \overline{\mathrm{O}}<\overline{\mathrm{N} H}_{2}<\mathrm{HC} \equiv \overline{\mathrm{C}}<\overline{\mathrm{R}}\) (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 (d): \(\text{RCO} \overline{\text{O}}<\text{HC} \equiv \overline{\text{C}}<\overline{\text{N}} \text{H}_{2}<\overline{\text{R}}\).

Step by step solution

01

Analyze Conjugate Bases

The bases given are the conjugate bases of different acids. These are carboxylate ion \(\text{RCO} \overline{\text{O}}\), acetylide ion \(\text{HC} \equiv \overline{\text{C}}\), methyl anion \(\overline{\text{R}}\), and amide ion \(\overline{\text{N}}\text{H}_{2}\). Identify their original acids and the stability of the conjugate bases.
02

Consider Base Stability

More stable conjugate bases are weaker and thus have less basicity. The carboxylate ion \(\text{RCO} \overline{\text{O}}\) is quite stable due to resonance within the carboxylate group, making it a weak base.
03

Analyze Electron Withdrawing Effects

The acetylide ion \(\text{HC} \equiv \overline{\text{C}}\) is less stable due to a high concentration of negative charge on carbon without stabilization mechanisms like resonance, making it stronger than \(\text{RCO} \overline{\text{O}}\).
04

Evaluate Nitrogen vs Carbon

The amide ion \(\overline{\text{N}}\text{H}_{2}\) is a stronger base than acetylide ion because nitrogen is less electronegative and holds negative charge less tightly than carbon, and is more stable than an alkyl anion like \(\overline{\text{R}}\).
05

Determine Alkyl Anion Basicity

The methyl anion \(\overline{\text{R}} = \text{CH}_3\) is highly basic due to carbon's low electronegativity and lack of stabilization by delocalization, making it the strongest base here.
06

Order of Increasing Basicity

Combine insights from previous analyses to arrange the conjugate bases in the order of increasing basicity: \[\text{RCO} \overline{\text{O}} < \text{HC} \equiv \overline{\text{C}} < \overline{\text{N}} \text{H}_{2} < \overline{\text{R}}\] which matches with option (d).

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Key Concepts

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

Stability of Conjugate Bases
Understanding the stability of conjugate bases is crucial for determining their basicity.
This stability is fundamentally related to the ability of a base to handle or delocalize the negative charge it carries.
In general, the more stable a conjugate base is, the weaker its basicity will be.

A prime example is the carboxylate ion, \(\text{RCO} \overline{\text{O}}\).
This ion owes its stability to the resonance that occurs within the carboxylate group.
Resonance allows the negative charge to be delocalized across the ion, reducing the energy of the ion and making it less likely to donate electrons compared to other bases.
Consequently, it is a weaker base compared to ones that do not stabilize the charge as effectively.

Another example is the acetylide ion, \(\text{HC} \equiv \overline{\text{C}}\).
Unlike the carboxylate ion, this ion has no resonance structures to stabilize the negative charge concentrated on carbon.
This lack of stabilization makes it less stable and thus a stronger base compared to the carboxylate ion.
Resonance and Basicity
Resonance is a mechanism by which molecules or ions can stabilize themselves by sharing electrons across multiple atoms.
This electron delocalization is significant in determining the basicity of an ion.
Resonance can effectively disperse the negative charge, leading to increased stability and decreased basicity.

An illustrative example is the carboxylate ion, \(\text{RCO} \overline{\text{O}}\), where resonance is a key feature.
The ability to spread the negative charge over multiple oxygen atoms makes this ion less reactive and, consequently, a weaker base.
In contrast, ions that cannot engage in resonance to stabilize negative charge, such as the acetylide ion \(\text{HC} \equiv \overline{\text{C}}\), display higher basicity due to higher strain from concentrated negative charge.

The relationship between resonance and basicity underscores why some ions are fundamentally weaker bases.
The presence or absence of resonance impacts how willing a base might be to participate in chemical reactions.
Electronegativity and Basicity
Electronegativity refers to an atom's tendency to attract electrons towards itself.
The basicity of a conjugate base is heavily influenced by the electronegativity of the atom holding the negative charge.

For instance, the amide ion \(\overline{\text{N}}\text{H}_{2}\) is a stronger base than the acetylide ion despite both having a negative charge.
Nitrogen, being less electronegative than carbon, holds the negative charge less tightly.
This tendency results in the amide ion being more eager to donate electrons compared to ions with a higher electronegativity, making it a stronger base.

In the case of the methyl anion \(\overline{\text{R}} = \text{CH}_3\), carbon's relatively low electronegativity and absence of stabilizing structures like resonance make it a very strong base.
  • This is due to carbon's propensity to release electrons, creating a highly reactive base.
  • Such traits highlight the interplay between electronegativity and basicity, showing that lower electronegativity often correlates with increased basic character.

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Most popular questions from this chapter

In the presence of peroxide, hydrogen chloride and hydrogen iodide do not undergo antiMarkownikoff's addition to alkenes because (a) both are highly ionic (b) one is oxidizing and the other is reducing (c) one of the steps is endothermic in both the cases (d) all the steps are exothermic in both the cases

The major product obtained on treatment of \(\mathrm{CH}_{3}\) \(\mathrm{CH}_{2} \mathrm{CH}(\mathrm{F}) \mathrm{CH}_{3}\) with \(\mathrm{CH}_{3} \mathrm{O} / \mathrm{CH}_{3} \mathrm{OH}\) is (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{OCH}_{3}\right) \mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OCH}_{3}\)

The decreasing order of nucleophilicity of HS', \(\mathrm{RCOO}, \mathrm{RCOOH}\) and \(\mathrm{ROH}\) is (a) \(\mathrm{RCOO}>\mathrm{HS}-\mathrm{RCOOH}>\mathrm{ROH}\) (b) \(\mathrm{HS}^{-}>\mathrm{RCOO}^{-}>\mathrm{RCOOH}>\mathrm{ROH}\) (c) \(\mathrm{HS}>\mathrm{RCOO}^{-}>\mathrm{ROH}>\mathrm{RCOOH}\) (d) \(\mathrm{RCOO}>\mathrm{HS}->\mathrm{ROH}>\mathrm{RCOOH}\)

Arrange \(\mathrm{CCl}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{NH}_{2}\) (a), \(\mathrm{CCl}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{NH}_{2}\) (b) and \(\mathrm{CCl}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (c)in order of decreasing basic nature. (a) \(C>B>A\) (b) \(C>A>B\) (c) \(\mathrm{A}>\mathrm{C}>\mathrm{B}\) (d) \(\mathrm{B}>\mathrm{A}>\mathrm{C}\)

Among the following, the strongest nucleophiles is (a) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH}\) (b) \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) (c) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (d) \(\mathrm{NCCH}_{2}\)

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