A= Consider the following compounds D-HOCH CH B. E = C = F= a) Which one(s) would give a positive Tollens (silver mirror) test? A. I b) Which one(s) could be oxidized with Cro? D Which one(s) could be reduced with NaBH? E A d) Which one(s) would dissolve in water? D e) Which one(s) would have a peak in the 'H NMR spectrum which would B disappear when DiO is added?
A= Consider the following compounds D-HOCH CH B. E = C = F= a) Which one(s) would give a positive Tollens (silver mirror) test? A. I b) Which one(s) could be oxidized with Cro? D Which one(s) could be reduced with NaBH? E A d) Which one(s) would dissolve in water? D e) Which one(s) would have a peak in the 'H NMR spectrum which would B disappear when DiO is added?
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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Transcribed Image Text:8. S
A =
Consider the following compounds:
D-HOCH/CH₂
B=
E=
مله
F=
CH₂
Mey
H₂C
a) Which one(s) would give a positive Tollens (silver mirror) test?
ALE
b) Which one(s) could be axidized with Cros? D
c) Which one(s) could be reduced with NaBH? E A
d) Which one(s) would dissolve in water? D
e) Which one(s) would have a peak in the 'H NMR spectrum which would
disappear when D/O is added?
B
Expert Solution
Step 1
a) Tollen's test is given by compounds which have aldehyde groups. Out here, the compound A has a aldehydic CHO group.
b) Oxidation with CrO3 converts alcohol to aldehyde/ketone. C and D are alcohol. But C cannot be oxidised as it has no Hydrogen. So, only D can be oxidised.
c) C=O can be reduced with NaBH4, only aldehyde and ketone, not ester group. So, B cannot be reduced but A and E can be.
d) Alcohols readily dissolve in water as they can form hydrogen bonding. So, C and D
e) On adding D2O, tautomerism can occur and the Hydrogen at the alpha position can be exchanged with deuterium. So, a 1HNMR peak will disappear. Option E is correct.
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