Question

I'm really struggling with this! Every time I only get about 1/2 of them right! Please help!
Splitting of a signal in a proton NMR spectrum tells us the number of chemically non-equivalent hydrogens in the immediate vicinity of the hydrogen giving the signal. Predict the number of lines exhibited by hydrogens at the labeled positions in a first-order NMR spectrum. (Make the approximation that all coupling constants are equal.) OCH The number of lines exhibited by hydrogen(s) a is The number of lines exhibited by hydrogen(s) b is The number of lines exhibited by hydrogen(s) c is 2) CN The number of lines exhibited by hydrogen(s) a is The number of lines exhibited by hydrogen(s) b is The number of lines exhibited by hydrogen(s) c is

Answer 1

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1. There is a rule to predict the number of lines in 1H NMR, i.e.

Multiplicity = (n+1) rule, where n is the number of Hydrogen on adjacent carbons,

a. The number of lines exhibited by proton a is 2.

As the number of hydrogen on neighboring carbon is 1, so multiplicity is (1+1) = 2

b. The number of lines exhibited by proton b is 2.

As the number of hydrogen on neighboring carbon is 1, so multiplicity is (1+1) = 2

c. The number of lines exhibited by proton c is 1.

As the number of hydrogen on neighboring carbon is 0, so multiplicity is (0+1) = 1

2. Multiplicity = (n+1) rule, where n is the number of Hydrogen on adjacent carbons,

a. The number of lines exhibited by proton a is 5.

As the number of hydrogen on neighboring carbon is 3+1 = 4, so multiplicity is (4+1) = 5

b. The number of lines exhibited by proton b is 2.

As the number of hydrogen on neighboring carbon is 1, so multiplicity is (1+1) = 2

c. The number of lines exhibited by proton c is 8.

As the number of hydrogen on neighboring carbon is 3+3+1=7, so multiplicity is (7+1) = 8

3. Multiplicity = (n+1) rule, where n is the number of Hydrogen on adjacent carbons,

a. The number of lines exhibited by proton a is 1.

As the number of hydrogen on neighboring carbon is 0, so multiplicity is (0+1) = 5

b. The number of lines exhibited by proton b is 3.

As the number of hydrogen on neighboring carbon is 2, so multiplicity is (2+1) = 3

c. The number of lines exhibited by proton c is 5.

As the number of hydrogen on neighboring carbon is 2+2=4, so multiplicity is (4+1) = 5

4. Multiplicity = (n+1) rule, where n is the number of Hydrogen on adjacent carbons,

a. The number of lines exhibited by proton a is 1.

As the number of hydrogen on neighboring carbon is 0, so multiplicity is (0+1) = 1

b. The number of lines exhibited by proton b is 3.

As the number of hydrogen on neighboring carbon is 2, so multiplicity is (2+1) = 3

c. The number of lines exhibited by proton c is 3.

As the number of hydrogen on neighboring carbon is 1+1=2, so multiplicity is (2+1) = 3