You have been taught that carbocations are "reactive intermediates"; true substances that are formed during a reaction that have finite but fleeting existence (due to their high reactivity). In this case, you can observe the cation by virtue of its color and it will persist for quite some time! First, why is this carbocation colored? And why does it persist for such an extended time?

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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SYNTHETIC PROCEDURE
PERFORM THIS REACTION IN THE HOOD
Scale this reaction as necessary based on the amount of triphenylmethanol you obtained
from Experiment V. Before starting the reaction, set aside a few milligrams of
triphenylmethanol for running a comparison TLC after you have isolated the product
(starting material (SM)-vs-product):
In a 25 mL Erlrnmeyer flask, add 1.0 g triphenylmethanol and grind the crystals to a fine
powder using a glass rod. Carefully add 10 mL of con. H2SO4 and stir the mixture with
the glass rod to effect complete dissolution. You should observe a red-brown to light
yellow solution. The color is due to the presence of the carbocation.
You have been taught that carbocations are "reactive intermediates"; true substances
that are formed during a reaction that have finite but fleeting existence (due to their
high reactivity). In this case, you can observe the cation by virtue of its color and it will
persist for quite some time! First, why is this carbocation colored? And why does it
persist for such an extended time?
Transcribed Image Text:SYNTHETIC PROCEDURE PERFORM THIS REACTION IN THE HOOD Scale this reaction as necessary based on the amount of triphenylmethanol you obtained from Experiment V. Before starting the reaction, set aside a few milligrams of triphenylmethanol for running a comparison TLC after you have isolated the product (starting material (SM)-vs-product): In a 25 mL Erlrnmeyer flask, add 1.0 g triphenylmethanol and grind the crystals to a fine powder using a glass rod. Carefully add 10 mL of con. H2SO4 and stir the mixture with the glass rod to effect complete dissolution. You should observe a red-brown to light yellow solution. The color is due to the presence of the carbocation. You have been taught that carbocations are "reactive intermediates"; true substances that are formed during a reaction that have finite but fleeting existence (due to their high reactivity). In this case, you can observe the cation by virtue of its color and it will persist for quite some time! First, why is this carbocation colored? And why does it persist for such an extended time?
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