На Br The proton NMR would have total unique proton(s) in the spectrum. The carbon NMR would have total unique carbon(s) in the spectrum Ha would have the following multiplicity: Ha would have different J values. roton nmr choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 or more arbon nmr choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 or more a multiplicity choices: Singlet, doublet, triplet, quartet, quintet, doublet of doublets, doublet of triplets, triplet of triplets, publet of quartet
На Br The proton NMR would have total unique proton(s) in the spectrum. The carbon NMR would have total unique carbon(s) in the spectrum Ha would have the following multiplicity: Ha would have different J values. roton nmr choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 or more arbon nmr choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 or more a multiplicity choices: Singlet, doublet, triplet, quartet, quintet, doublet of doublets, doublet of triplets, triplet of triplets, publet of quartet
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:For the compound below, answer the following questions:
На
Br
The proton NMR would have
• total unique proton(s) in the spectrum.
The carbon NMR would have
total unique carbon(s) in the spectrum.
Ha would have the following multiplicity:
Ha would have
different J values.
proton nmr choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 or more
carbon nmr choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 or more
ha multiplicity choices: Singlet, doublet, triplet, quartet, quintet, doublet of doublets, doublet of triplets, triplet of triplets,
doublet of quartet
ha j value choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 or more
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