Question

i need help with these nmr questions

.. Exercise N.5: Spectral Identification Using 'H and 1SC NMR Predict the structure of the organic compounds that give the ill and labeled 5-A through 5-C NMR spectra Problem 5-A δ1.0(3H) δ1,5 (2H) 81.7 (2H) 84.4 (2H) 87.4-8.1 (SH) TMS 'H NMR 85 В.0 75.70 6.5 ea ss so 45 40 36,30,25 20 ia e' as on CDCl TMS 20 10 SoURCE: Reprinted with permission of Aldrich Chemical. 25 20 5 10 CDCH 5. Questions Are the efic emical shifts ofattached electronegative atoms additive fene hete acec acid and methanol using 'H NMR 6. Explain 8. 3Esa pens o acs o singles inthe CW H NMR 4. How could H NMR spectroscopy be used to differentiate between methyl propanate Technique N Nleur Magnetic Resonance Spectroscopyy Problem 5-C CH O 82.2 (3H) 65,9-6.3 (3H) TMS 0.5 HNMR 5 6 045 403 3.5 3.0 25 2.0 15 1.0 05 00 CDCl TMS 1SC NMR 20 10 180 10 10 140 130 120 10 1o0s0 80 0 50 So o20 10 6 SoURCE: Reprinted with permission of Aldrich Chemical. 5. The 'H NMR spectrum of cyclohexane shows one singlet. If the sample is cooled to-80°C, the spectrum changes. Predict the spectrum of the cooled cyclohexane and explain why the spectrum changes as the solution is cooled. What is the purpose of TMS? Predict the chemical shift of the protons in methyllithium (CH;Li). 6. 7. 8. A specific proton in an organic compound has a chemical shift of 63.4 in a 60 MHz NMR spectrum. What will be the chemical shift if the spectrum is recorded using a 90 MHz NMR instrument? The coupling between two protons in an alkene is 10.8 Hz at 60 MHz. What will be the value of the coupling constant between these same two protons at 200 MHz? 9.

Answer 1

Solution :-

Molecular formula = C11H14O2

Hydrocarbon with 11 carbon atoms will have (11*2)+2 = 24 H atoms

Number of deficient hydrogen = 24-14 = 10

Number of unsaturation = deficient hydrogen / 2

                                             = 10/2

                                             = 5

In the NMR spectrum we have 5H which shows the peak at 7.4 -8.1 ppm which is nothing but the aromatic benzene ring protons

Ring is monosubstituted.

1.0 ppm peak is for the CH3 group

1.5 ppm is the 2H peak gives sextet

1.7 ppm is the 2H peak gives triplet

4.4 ppm is the 2H peak gives triplet (downfield because of CH2 attached to oxygen)

In the C NMR spectrum we have peak at 167 ppm which is for the carbonyl group

Peak in the range of 15-35 are the two CH2 group and one CH3 group

65 ppm peak is the peak for the carbon (CH2-O)

Peaks around the 130 ppm are the C=C aromatic ring carbons.

Following image shows the structure of the compound.

4.4 ppm 1.5 ppm 5 L4 1.7 ppm 1.0 ppm