Answer the following questions. N N-H Compound 1 Compound 2 Identify which compound is expected to have a lower pkg. Compound 1 O Compound 2 Choose the correct explanation for the above observation. O Compound 1 is not aromatic, but when compound 1 is deprotonated, the resulting conjugate base is aromatic. Therefore, deprotonation of compound 1 results in the formation of aromaticity, which renders compound 1 much more acidic than compound 2 (compound 2 is already aromatic even without being deprotonated). O Compound 1 has more hydrogen atoms and hence it is more acidic than compound 2. O Compound 1 is a five membered ring and hence it is more acidic than compound 2. O Compound 1 has two N-H bonds, so it is more acidic than compound 2, which has only one N-H bond.

Organic Chemistry: A Guided Inquiry
2nd Edition
ISBN:9780618974122
Author:Andrei Straumanis
Publisher:Andrei Straumanis
Chapter19: Eas: Electrophilic Aromatic Substitution
Section: Chapter Questions
Problem 21E
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Answer the following questions.
N-H,
H.
Compound 1
Compound 2
Identify which compound is expected to have a lower pK.
O Compound 1
O Compound 2
Choose the correct explanation for the above observation.
O Compound 1 is not aromatic, but when compound 1 is deprotonated, the resulting conjugate base is aromatic.
Therefore, deprotonation of compound 1 results in the formation of aromaticity, which renders compound 1 much
more acidic than compound 2 (compound 2 is already aromatic even without being deprotonated).
O Compound 1 has more hydrogen atoms and hence it is more acidic than compound 2.
O Compound 1 is a five membered ring and hence it is more acidic than compound 2.
O Compound 1 has two N-H bonds, so it is more acidic than compound 2, which has only one N-H bond.
Transcribed Image Text:Answer the following questions. N-H, H. Compound 1 Compound 2 Identify which compound is expected to have a lower pK. O Compound 1 O Compound 2 Choose the correct explanation for the above observation. O Compound 1 is not aromatic, but when compound 1 is deprotonated, the resulting conjugate base is aromatic. Therefore, deprotonation of compound 1 results in the formation of aromaticity, which renders compound 1 much more acidic than compound 2 (compound 2 is already aromatic even without being deprotonated). O Compound 1 has more hydrogen atoms and hence it is more acidic than compound 2. O Compound 1 is a five membered ring and hence it is more acidic than compound 2. O Compound 1 has two N-H bonds, so it is more acidic than compound 2, which has only one N-H bond.
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