ent/takeCovalentActivity.do?locator-assignment-take&itakeAssignmentSessionlocato
ID: 553952 • Letter: E
Question
ent/takeCovalentActivity.do?locator-assignment-take&itakeAssignmentSessionlocators; assignment-take Splitting of a signal in a proton NMR spectrum / signal. Predict the number of lines exhibited by hydrogens at the labeled positions constants are equal.) tells us the number of chemically non-equivalent hydrogens in the impsediate vicinity of i the hydrogen giving the in a first-order NMR spectrum. (Make the approximation that all coupling / 1) The number of lines exhibited by hydrogens) a is The number of lines exhibited by hydrogen(s) b is The number of lines exhibited by hydrogen(s) e is 2) a c The number of lines exhibited by hydrogens) The mumber of lines exhibited by bydrogets)b s The number of lines exhibited by hydrogen(s) e is Submit Answer Retry Entire Group 6 more group attempts remaining Previous Next Email Instructor Save andExplanation / Answer
To get this answer you must count the number of neighbor protons and apply number of lines = n + 1
1.
a : the neighbor of this carbon has only 1 hydrogen so n = 1, number of lines = 2
b: neighbor from the left has 3 hydrogens, neighbor from the right 2 hydrogens, number of lines 3 + 2 + 1 = 6
c: neighbor from the left has 1 hydrogen, neighbor from the right 3 hydrogens, number of lines 1 + 3 + 1 = 5
2.
a: neighbor has only 1 hydrogen, number of lines = 1 + 1 = 2 lines
b: neighbor from the top has 3, from the left has 2, from the right has 3: 3 + 2 + 3 + 1 = 9 lines
c: neighbor from the left has no hydrogens, neighbor from the right has 1 hydrogen, 1 + 1 = 2 lines
3.
a: neighbor has 2 hydrogens, number of lines 2 + 1 = 3 lines
b: neighbor from the right has 3, from the left has 2, lines = 3 + 2 + 1 = 6 lines
c: neighbor from the left has 2 , from the right has 2 , number of lines = 2 + 2 + 1 = 5 lines
4.
a: the CN group has no hydroges, the neighbor from the left has 2 hydrogens: lines = 2 + 1 = 3
b: from the left we have 3 hydrogens, from the right we have 1, then lines = 3 + 1 + 1 = 5
c: the only neighbor is the carbon from the right which has 2 hydrogens, lines = 2 + 1 = 3
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