Draw the structure of the compound C9H12 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) 5:2:2:3.

Question

Tristen Mcconnell

Chemistry

12 April, 14:28

Draw the structure of the compound C9H12 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) 5:2:2:3.

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Carley Stokes · Answer 1

Answer:Draw the structure of the compound C9H12 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) 5:2:2:3.

from A Solvent CDCls 9.0 7.5 6.0 1.5 4.5 Chemical shift, δ (ppm) 3.0 0.0

Explanation: As applicable above

Gisselle Hart · Answer 2

Isopropylbenzene

Explanation:

If you draw the structure, you can see that there are two methyl groups and in between there.

Adjacent to CH3, there are four neighbouring hydrogens, therefore, n=4+1 = 5. The same is for methyl on other side. For carbon present in benzene ring, there is 2, since there is one hydrogen on benzene per carbon.