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The order of carbonyl stretching frequency in the IR spectra of ketone, amide and anhydride is:
- a)Anhydride > amide > ketone
- b)Ketone > amide > anhydride
- c)Amide > anhydride > ketone
- d)Anhydride > ketone > amide
Correct answer is option 'D'. Can you explain this answer?
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The order of carbonyl stretching frequency in the IR spectra of ketone...
Carbonyl stretching frequency in IR spectra
The carbonyl stretching frequency in the infrared (IR) spectra of ketone, amide, and anhydride can be used to differentiate between these functional groups. The order of carbonyl stretching frequency is as follows:
Anhydride ketone amide
Explanation
Anhydrides have the highest carbonyl stretching frequency because they have two carbonyl groups that are both highly polarized due to the electron-withdrawing effect of the adjacent oxygen atoms. The presence of two carbonyl groups also results in a more efficient coupling of the C=O bond vibrations, leading to a higher frequency.
Ketones have a slightly lower carbonyl stretching frequency than anhydrides because they have only one carbonyl group. However, the C=O bond in ketones is still highly polarized due to the electron-withdrawing effect of the adjacent carbon atoms.
Amides have the lowest carbonyl stretching frequency because the C=O bond is less polarized due to the electron-donating effect of the adjacent nitrogen atom. This results in a weaker C=O bond that vibrates at a lower frequency.
Conclusion
Therefore, the order of carbonyl stretching frequency in the IR spectra of ketone, amide, and anhydride is anhydride, ketone, amide. This information can be useful in identifying and characterizing these functional groups in organic molecules.
The carbonyl stretching frequency in the infrared (IR) spectra of ketone, amide, and anhydride can be used to differentiate between these functional groups. The order of carbonyl stretching frequency is as follows:
Anhydride ketone amide
Explanation
Anhydrides have the highest carbonyl stretching frequency because they have two carbonyl groups that are both highly polarized due to the electron-withdrawing effect of the adjacent oxygen atoms. The presence of two carbonyl groups also results in a more efficient coupling of the C=O bond vibrations, leading to a higher frequency.
Ketones have a slightly lower carbonyl stretching frequency than anhydrides because they have only one carbonyl group. However, the C=O bond in ketones is still highly polarized due to the electron-withdrawing effect of the adjacent carbon atoms.
Amides have the lowest carbonyl stretching frequency because the C=O bond is less polarized due to the electron-donating effect of the adjacent nitrogen atom. This results in a weaker C=O bond that vibrates at a lower frequency.
Conclusion
Therefore, the order of carbonyl stretching frequency in the IR spectra of ketone, amide, and anhydride is anhydride, ketone, amide. This information can be useful in identifying and characterizing these functional groups in organic molecules.
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The order of carbonyl stretching frequency in the IR spectra of ketone, amide and anhydride is:a)Anhydride > amide > ketoneb)Ketone > amide > anhydridec)Amide > anhydride > ketoned)Anhydride > ketone > amideCorrect answer is option 'D'. Can you explain this answer?
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