H p=1.39 (pH = 4.7) CH3 p=-1.29 (pH = 10.4) a morpholine enamine of a para-substituted propiophenone -2 +2 +4

Introduction to General, Organic and Biochemistry
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ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Chapter10: Organic Chemistry
Section: Chapter Questions
Problem 10.20P: 10-20 Why are the following molecular formulas impossible? (a) CH5 C2H7
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To solve this problem, you need to read the description of the Hammett s, r treatment given in Chapter 15, Problem 92. When the rate constants for the hydrolysis of several morpholine enamines of para-substituted propiophenones are determined at pH 4.7, the r value is positive; however, when the rates of hydrolysis are determined at pH 10.4, the r value is negative.
a. What is the rate-determining step of the hydrolysis reaction when it is carried out in a basic solution?
b. What is the rate-determining step of the hydrolysis reaction when it is carried out in an acidic solution?

H
p=1.39 (pH = 4.7)
CH3
p=-1.29 (pH = 10.4)
a morpholine enamine of
a para-substituted propiophenone
-2
+2
+4
Transcribed Image Text:H p=1.39 (pH = 4.7) CH3 p=-1.29 (pH = 10.4) a morpholine enamine of a para-substituted propiophenone -2 +2 +4
Expert Solution
Step 1

σ or  substituent constant :  electron-withdrawing substituents have positive substituent constants and electron- donating substituents have negative substituent constants.

When log(K0 ) vs σ is plotted the slope is called rho value (ρ )

ρ or reaction constant : negative ρ  value indicates that molecules with electron donation groups reacts faster than compounds with electron withdrowing groups.

Again  positive ρ  value indicates the eletron-withdrowing groups are reacted faster than compounds with electron donating groups.

Step 2

The  ρ  value is negative  in basic solution indicates that electron donating group increases the rate. It tells that the slowest step will be carbocation formation step ,because more negative charge on the carbon means more electron donation of the substituent and easy to protonate.

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