Solution Found!
Using the pKa values of the conjugate acids of the leaving
Chapter 11, Problem 3P(choose chapter or problem)
Using the \(\mathrm{pK}_\mathrm{a}\) values of the conjugate acids of the leaving groups (the \(\mathrm{pK}_\mathrm{a}\) of \(\mathrm{HBr}\) is \(⎻9\), the \(\mathrm{pK}_\mathrm{a}\) of \(\mathrm{H}_2\mathrm{O}\) is \(15.7\), and the \(\mathrm{pK}_\mathrm{a}\) of \(\mathrm{H}_3\mathrm{O}^+\) is \(⎼1.7\)), explain the difference in reactivity in substitution reactions between
Equation Transcription:
pKa
HBr
⎻9
H2O
15.7
H3O+
⎼1.7
Text Transcription:
pK_a
HBr
-9
H_2 0
15.7
H_3 O^+
-1.7
Questions & Answers
QUESTION:
Using the \(\mathrm{pK}_\mathrm{a}\) values of the conjugate acids of the leaving groups (the \(\mathrm{pK}_\mathrm{a}\) of \(\mathrm{HBr}\) is \(⎻9\), the \(\mathrm{pK}_\mathrm{a}\) of \(\mathrm{H}_2\mathrm{O}\) is \(15.7\), and the \(\mathrm{pK}_\mathrm{a}\) of \(\mathrm{H}_3\mathrm{O}^+\) is \(⎼1.7\)), explain the difference in reactivity in substitution reactions between
Equation Transcription:
pKa
HBr
⎻9
H2O
15.7
H3O+
⎼1.7
Text Transcription:
pK_a
HBr
-9
H_2 0
15.7
H_3 O^+
-1.7
ANSWER:Solution 3P
Step 1 of 2
