The correct order of increasing basicity of the given conjugate bases (R = CH3) is
(a) RCOO-

Q: Why is an $sp$ hybridized carbon more electronegative than nitrogen?

Electronegativity depends on how strongly an atom's nucleus pulls on electrons. An $sp$ orbital has 50% s-character, meaning it is spherical and penetrates closer to the nucleus compared to p-orbitals. This proximity to the positively charged nucleus allows the $sp$ carbon to stabilize a negative charge better than an $sp^3$ nitrogen, which only has 25% s-character.

Q: Why does resonance make the carboxylate ion so stable?

Resonance allows the negative charge to be delocalized, or spread out, over multiple atoms (in this case, two oxygen atoms). Spreading the charge reduces electron-electron repulsion and lowers the overall potential energy of the molecule, making it highly stable.

Q: I thought carbon is always less electronegative than nitrogen. Is that wrong?

As a general rule across the periodic table, yes, nitrogen is more electronegative than carbon. However, hybridization changes the effective electronegativity. The high s-character of an $sp$ hybridized carbon makes it an exception to this general rule when compared to an $sp^3$ hybridized nitrogen.

Q: Why does a stable conjugate base mean it's a weak base?

A base is a substance that wants to donate a pair of electrons to grab a proton ($H^+$). If a conjugate base is very stable, it is 'happy' as it is. It has no strong desire to react and grab a proton. Therefore, it acts as a weak base.

Q: What would happen if we had an alkene ($sp^2$ carbon) instead of an alkyne?

An $sp^2$ hybridized carbon has 33.3% s-character. It is more electronegative than an $sp^3$ carbon but less electronegative than an $sp^3$ nitrogen. So, a vinylic carbanion ($CH_2=CH^-$) would be less stable than $\bar{N}H_2$ but more stable than an alkyl carbanion ($R^-$).

Question

The correct order of increasing basicity of the given conjugate bases ($R = CH_3$) is

Options:

(A) $RCOO^- < HC \equiv C^- < R^- < \bar{N}H_2$

(B) $R^- < HC \equiv C^- < RCOO^- < \bar{N}H_2$

(C) $RCOO^- < \bar{N}H_2 < HC \equiv C^- < R^-$

(D) $RCOO^- < HC \equiv C^- < \bar{N}H_2 < R^-$

Sigma Percentile · Accepted Answer

### The Golden Rule of Acid-Base Chemistry When faced with a problem asking to compare the basicity of different ions, the most reliable strategy is to look at their corresponding conjugate acids, or more directly, to evaluate the **stability of the ions themselves**. The fundamental principle is simple: **A stronger base is highly reactive, which means it is less stable.** Conversely, a weak base is very stable and unreactive. Therefore, if we can determine the order of stability for these four conjugate bases, we simply reverse that order to find their basicity. ### Analyzing the Contenders Let's break down the stability of each ion by looking at where the negative charge resides and what factors help stabilize it. **1. The Carboxylate Ion ($RCOO^-$)** In the carboxylate ion, the negative charge is initially on an oxygen atom. Oxygen is highly electronegative, meaning it comfortably accommodates extra electron density. But the real magic here is **resonance**. The negative charge is delocalized over two equivalent oxygen atoms. This spreading out of charge drastically lowers the energy of the system, making the carboxylate ion exceptionally stable. **2. The Alkynyl Ion ($HC \equiv C^-$)** Here, the negative charge is on a carbon atom. Normally, carbon is not very electronegative. However, this specific carbon is part of a triple bond, meaning it is **$sp$ hybridized**. An $sp$ hybridized orbital has 50% s-character. Because s-orbitals are closer to the nucleus, a higher s-character translates to a higher effective electronegativity. In fact, an $sp$ hybridized carbon is more electronegative than an $sp^3$ hybridized nitrogen! This makes the alkynyl ion quite stable, though not as stable as the resonance-stabilized carboxylate. **3. The Amide Ion ($\bar{N}H_2$)** In the amide ion, the negative charge sits on a nitrogen atom. Nitrogen is inherently more electronegative than a standard $sp^3$ carbon, so it stabilizes the negative charge reasonably well. However, as we just established, it is less electronegative than the $sp$ hybridized carbon of the alkynyl ion. Thus, it ranks third in our stability list. **4. The Alkyl Carbanion ($R^-$)** Finally, we have the alkyl carbanion, where the negative charge is localized on an **$sp^3$ hybridized carbon**. With only 25% s-character, this carbon is the least electronegative atom among our choices. It strongly repels the extra electron density, making the alkyl carbanion highly unstable and extremely reactive. ### The Final Verdict Putting it all together, the order of **stability** is: $$RCOO^- > HC \equiv C^- > \bar{N}H_2 > R^-$$ Since basicity is inversely proportional to stability, we reverse this order to get the **increasing order of basicity**: $$RCOO^- < HC \equiv C^- < \bar{N}H_2 < R^-$$ This perfectly matches option (d). Understanding how hybridization and resonance affect electronegativity and stability is the key to mastering these types of organic chemistry problems!

Animated Solution for Chemistry - Organic Chemistry: The correct order of increasing basicity of the given conjugate bases ( $R = C H_{3}$ ) is

Select Answer:

Visualized Solution

$V i s u a l A n c h or$

$T h e G o l d e n R u l eo f B a s i c i t y$

$S t abi l i t y o f R C O O^{-}$

$S t abi l i t y o f H C \equiv C^{-}$

$S t abi l i t y o f \overset{ˉ}{N} H_{2}$

$S t abi l i t y o f R^{-}$

$F ina l C o n c l u s i o n$

The Sigma Insight: Bond Fission, Electronic Displacement and Hyperconjugation

The Golden Rule of Acid-Base Chemistry

Analyzing the Contenders

The Final Verdict

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Animated Solution for Chemistry - Organic Chemistry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The correct order of increasing basicity of the given conjugate bases (R=CH3​) is