a) Singlet will appear
Explanation: There are no coupling partners for the protons of methyl group (3H).
b) Doublet will appear
Explanation: The 6 protons of two methyl groups couple with 19F, multiplicity of the signal = 2nI + 1 = 2*1*1/2 + 1 = 2 (doublet)
c) Doublet will appear
Explanation: The 6 protons of two methyl groups couple with 77Se, multiplicity of the signal = 2nI + 1 = 2*1*1/2 + 1 = 2 (doublet)
d) Triplet of doublet will appear
Explanation: The 3 protons of methyl group first couple with 19F, multiplicity of the signal = 2nI + 1 = 2*2*1/2 + 1 = 3 (triplet), followed by coupling with 31P, multiplicity of the signal = 2nI + 1 = 2*1*1/2 + 1 = 2 (doublet). Hence, each line of the triplet will further split into doublet, i.e. triplet of doublet.
e) A triplet and a quartet (two different signals) will appear.
Explanation: triplet appears for the protons of 2 methyl groups (6H), quartet appears for the protons of two methylene groups (4H).
sketch and explanation please Relevant i = 12 nuclei and relative natural abundances are as follows (assume all othe...
Relevant i = 12 nuclei and relative natural abundances are as follows (assume all other nuclei are spin inactive): H (1 = 12, 100%) 295i (1 = 42, 4.7%) 19F (1 = 2, 100%) 31P (1 = 72, 100%) 77Se (1 = 72, 7.6%) Also, assume 1 > 2/> 34, there is no resolvable 4 coupling, and there is no observable coupling to 13C nuclei. 1. Using stick diagrams, sketch the TH NMR spectra for the following compounds. Assume structures...
Relevant I 2 nuclei and relative natural abundances are as follows (assu me all other nuclei are spin inactive): H(I 100%) 19F (I, 100%) 77Se (I2, 7.6%) 29Si (I 4.7%) 31P (I 2, 100%) Also, assume 'J 2J> 3J, there is no resolvable 4J coupling, and there is no observable coupling to 13C nuclei. Using stick diagrams, sketch the 'H NMR spectra for the following compounds. Assume structures are rigid in solution (give all isomers). Include all relative peak intensities...
number 2 sketch please
Relevant i = 12 nuclei and relative natural abundances are as follows (assume all other nuclei are spin inactive): 19F (1 = /2, 100%) TH (1 = 12, 100%) 295i (1 = 42, 4.7%) 31P (1 = /2, 100%) 77 Se (1 = 12, 7.6%) Also, assume 1 > 2/> 34, there is no resolvable 4 coupling, and there is no observable coupling to 13C nuclei. 1. Using stick diagrams, sketch the TH NMR spectra for...
number 2
Relevant i = 12 nuclei and relative natural abundances are as follows (assume all other nuclei are spin inactive): 19F (1 = /2, 100%) TH (1 = 12, 100%) 295i (1 = 42, 4.7%) 31P (1 = /2, 100%) 77 Se (1 = 12, 7.6%) Also, assume 1 > 2/> 34, there is no resolvable 4 coupling, and there is no observable coupling to 13C nuclei. 1. Using stick diagrams, sketch the TH NMR spectra for the following...
number 2 sktech and explanation please
Relevant i = 12 nuclei and relative natural abundances are as follows (assume all other nuclei are spin inactive): 19F (1 = /2, 100%) TH (1 = 12, 100%) 295i (1 = 42, 4.7%) 31P (1 = /2, 100%) 77 Se (1 = 12, 7.6%) Also, assume 1 > 2/> 34, there is no resolvable 4 coupling, and there is no observable coupling to 13C nuclei. 1. Using stick diagrams, sketch the TH NMR...
Using the tree approach for predicting first-order NMR spectra
for molecules and draw the requested spectra for the following
molecules #1-#3 in the space provided (do not use backside of
pages). All spectra are collected with no {1H} (proton decoupling).
The NMR active nuclei are all I = 1⁄2, i.e., dipolar, so Pascal’s
triangle applies. Consider only 1-bond coupling constants 1JAB, and
ignore ≥2- bond coupling constants (EXCEPT where noted in #2);
label the chosen coupling constant on each branch...