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Among the following, the least and the b...

Among the following, the least and the best `H^(Θ)` ion donor, respectively, are:

A

B

C

D

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The correct Answer is:
To determine the least and the best H⁻ ion donor among the given options, we need to analyze the effects of substituents on the aromatic compounds. Here's a step-by-step solution: ### Step 1: Identify the Compounds First, we need to identify the compounds that are being compared. In this case, we have compounds with different substituents, particularly focusing on the presence of the nitro group (NO₂). ### Step 2: Understand the Effects of Substituents - **Electron-Withdrawing Groups (EWGs)**: The nitro group (NO₂) is an electron-withdrawing group that exerts a -I (inductive) effect and a -R (resonance) effect. This means it pulls electron density away from the aromatic ring. - **Electron-Donating Groups (EDGs)**: If there are no substituents or if there are electron-donating groups, the electron density on the ring will be higher, making it easier for H⁻ ions to be released. ### Step 3: Analyze the Position of the Nitro Group The position of the nitro group relative to the hydrogen atom affects the electron density: - **Ortho NO₂**: This position has both -I and -R effects, significantly reducing electron density. - **Para NO₂**: This position also has -I and -R effects, but the distance reduces the overall impact compared to ortho. - **Meta NO₂**: This position only has the -I effect, which is less impactful than the -R effect present in ortho and para. ### Step 4: Determine the Least H⁻ Ion Donor Among the compounds with nitro groups: - The **ortho NO₂** compound experiences the strongest electron-withdrawing effects, leading to the least electron density and thus making it the least likely to donate an H⁻ ion. - Therefore, **ortho NO₂** is the least H⁻ ion donor. ### Step 5: Determine the Best H⁻ Ion Donor - The compound without any electron-withdrawing or donating groups (option D) will have a higher electron density and thus a greater tendency to donate H⁻ ions. - Therefore, **option D** is the best H⁻ ion donor. ### Conclusion The least and the best H⁻ ion donor among the options are: - **Least H⁻ ion donor**: Ortho NO₂ (Option A) - **Best H⁻ ion donor**: No substituent compound (Option D)

To determine the least and the best H⁻ ion donor among the given options, we need to analyze the effects of substituents on the aromatic compounds. Here's a step-by-step solution: ### Step 1: Identify the Compounds First, we need to identify the compounds that are being compared. In this case, we have compounds with different substituents, particularly focusing on the presence of the nitro group (NO₂). ### Step 2: Understand the Effects of Substituents - **Electron-Withdrawing Groups (EWGs)**: The nitro group (NO₂) is an electron-withdrawing group that exerts a -I (inductive) effect and a -R (resonance) effect. This means it pulls electron density away from the aromatic ring. - **Electron-Donating Groups (EDGs)**: If there are no substituents or if there are electron-donating groups, the electron density on the ring will be higher, making it easier for H⁻ ions to be released. ...
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