Question
spectral problem its all one problem
1) identify the compounds for each group of spectra (mass, proton, carbon and IR)
2) assign the proton and carbon and IR spectra. identify at least two possible fragmentation on the mass spectrum
#6 SH (2 x CH2 and 1 x CH) 36 200 180 160 140 120 100 80 60 40
MI =102 (C = 70.6; SH - 13.7; 10 - 15.7) 100 NS-W-714 80- 10 20 30 40 50 60 70 80 90 100
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Answer #1

Hi there, unknown determined as 2-ethyl-1-butanol

Но

Elemental analysis:

C - 70.6%, H - 13.7%, O - 15.7% Total = 70.6 + 13.7+15.7 = 100%, hence no other element present.

Now divide these percentages with respective element atomic mass

C:H:O = 70.6/ 12.0107 : 15.7/ 1.00794 : 15.7/ 15.9994 = 6:14:1

so molecular formula is C6H14O and molar mass of this formula gives 102.1748 g/mol

Mass:

Molecular Ion (MI OR M+) is at m/z 102

m/z 84 is due to loss of water molecule (H2O) from M+ ion

m/z 56 (peak is due to loss of gamma hydrogen through McLafferty rearrangement)

+7 Нs — сн. — Н, Н, c— си, CH – CH2-OH нс — си, H₂C_CH/ MI m/2 102 +8 /m TOT CH2 H J (CH2 — + он C CH C Сн, н»-2 — Н2 = 5н /

IR:

3300 cm-1, strong, broad peak is due to alcohol O-H stretch

2850-3000 cm-1, strong, sharp due to sp3 C-H stretch

1380 & 1460 cm-1, medium peaks are due to sp3 C-H bending

1050 cm-1, strong due to alcohol C-O stretch

HNMR: protons are in BOLD for respective splitting

3.5 ppm, doublet, 2H due to CH-CH2-OH fragmnet

2.1 ppm, singlet, 1H due to -OH

1.2-1.5 ppm, multiplet,5H due to -CH2-CH-CH2- fragment

0.85 ppm, triplet, 6H due to -CH2-CH3 fragment twice

NMR assignments:

4- 3 2— 1— 0 CHCH CHE–SH–THE CHACHE Assign. Shift (ppm) Assign. ppm 3.5 2.1 1.2 - 1.5 0.85 WN HNMR CNMR

Hope this helped you!

Thank You So Much!

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