:0 : | : он он, Add charges, bonds, and nonbonding electrons, where applicable, to complete the structure of the prodi :0: || H Cr O : :0 : :0 :

Chemistry
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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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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I need help with the step 5 and 6 please

 

Draw the major organic product. Do not draw non-bonding electrons.
K2Cr207
H2SO4
HOʻ
distill
What occurs in step 1 of the mechanism?
The hydroxy group of the alcohol attacks the chromium to form a pentavalent chromium.
The hydroxy group of the alcohol is protonated.
O The hydroxy group of the alcohol is deprotonated.
What occurs in step 2 of the mechanism?
O The protonated alcohol intermediate is deprotonated.
O A hydroxy group on chromium is deprotonated.
A proton a to the protonated oxygen is deprotonated.
Steps 3 and 4 involve a series of proton transfer steps. In step 3, a hydroxy group on the chromium is protonated to form a good
leaving group. In step 4, a hydroxy proton is deprotonated, a new t bond is formed and the leaving group is removed to give the
chromate ester.
Step 3
Step 4
H
H-
H.
H.
HỒ: :0,
H
H.
Cr-OH
HO: :OH
H
H.
Next, let's explore the mechanism for converting the chromate ester to the product.
Transcribed Image Text:Draw the major organic product. Do not draw non-bonding electrons. K2Cr207 H2SO4 HOʻ distill What occurs in step 1 of the mechanism? The hydroxy group of the alcohol attacks the chromium to form a pentavalent chromium. The hydroxy group of the alcohol is protonated. O The hydroxy group of the alcohol is deprotonated. What occurs in step 2 of the mechanism? O The protonated alcohol intermediate is deprotonated. O A hydroxy group on chromium is deprotonated. A proton a to the protonated oxygen is deprotonated. Steps 3 and 4 involve a series of proton transfer steps. In step 3, a hydroxy group on the chromium is protonated to form a good leaving group. In step 4, a hydroxy proton is deprotonated, a new t bond is formed and the leaving group is removed to give the chromate ester. Step 3 Step 4 H H- H. H. HỒ: :0, H H. Cr-OH HO: :OH H H. Next, let's explore the mechanism for converting the chromate ester to the product.
Step 5: The chromate ester is shown. Do not alter the structures. Draw curved arrows to show conversion of the chromic
ester to the products.
:0 :
: он
он,
Incorrect
Step 6: Add charges, bonds, and nonbonding electrons, where applicable, to complete the structure of the products.
:0 :
|
H
Cr
O :
: он
: он,
Incorrect
:
Transcribed Image Text:Step 5: The chromate ester is shown. Do not alter the structures. Draw curved arrows to show conversion of the chromic ester to the products. :0 : : он он, Incorrect Step 6: Add charges, bonds, and nonbonding electrons, where applicable, to complete the structure of the products. :0 : | H Cr O : : он : он, Incorrect :
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