Draw the structure of the compound C 5 H 13 N from its proton ( 1 H) NMR spectru

Question

Draw the structure of the compound C5H13N from its proton (1H) NMR spectrum below.

First-order spin-spin splitting rules and equal coupling constants can be assumed. (Detailed analysis of any non-first order portions of the spectrum will not be required.)

Integral ratios to the nearest whole number are (left to right, ignoring the impurity peak) 4:3:6.

Can you please explain how ??

Used with permission from Aldrich Chemical Co., Inc. Solvent CDCls (impurity at 3.40 ppm) 9.0 7.5 6.0 4.5 Chemical shift, (ppm) 3.0 1.5 0.0

Explanation / Answer

First calculate the double bond equivalent of the compound C5H13N. It tells the number of multiple bond or rings in the given compound.

D.B.E = No. of C + 1 - (No. of H - No. of N / 2) = 5 +1 - (13-1/2) = 6 - 6 =0 (It means no double and triple bond or ring). It means N is present as amine group.

Now look at NMR spectrum

There are three signals, it means three types of H are present

Peak between 0 - 1.5 ppm is due to H of alkyl group

Peaks between 1.5 - 3 ppm are due CH2NR2

Peak ratio 4:3:6 corresponds that there are three types of hydrogen with 4, 3 and 6 number of hydrogen respectively.

The structure of C5H13N comes out to be

CH3-CH2-N(CH3)-CH2-CH3

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