Based strictly on the percentage of s-character in the hybridized carbon orbital, which of the follo — GOC and Organic Chemistry Basics Chemistry Question
Question
Based strictly on the percentage of s-character in the hybridized carbon orbital, which of the following is the correct decreasing order of stability for the given carbanions?
Answer: A
💡 Solution & Explanation
The stability of a carbanion increases with the electronegativity of the carbon holding the negative charge. Electronegativity increases with s-character. $RC\equiv C^-$ ($sp$, 50% s) > $R_2C=CH^-$ ($sp^2$, 33.3% s) > $R_3C-CH_2^-$ ($sp^3$, 25% s).
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