Question

sprectra for an unknown organic compount please help us figure out what it is

SL LED %Transmittance 630.07 992.79 1466.77 3079 21 2960.08 2874.61 1641.92 909 24 424.95 401,68 -20 | 4000 3500 3000 1500 1000 2500 2000 Wavenumbers (cm-1) Tue Mar 10 15 27 35 2020 (GMT-04:00) FIND PEAKS: Spectrum crifG7 Region 4000.00 400.00 Absolute threshold: 43.143 Sensitivity Peak Ist Position: 401.68 Position 424.95 Position 630 07 Position: 909 24 Position 992 79 Position 1379 15 Position 1466 77 Position: 1641.92 Intensity Intensity: Intensity Intensity Intensity intensity Intensity: intensity: 3.600 -1.783 37 150 2.752 11.449 38.828 18.454 10.687

Answer 1

MS data:

M+ (parent peak) = 84

According to rule 13, dividing mass with 13 we get (84/13 = 6) which indicates six CH and remaining 6 hydrogens and hence molecular formula is C6H12

Degree of Unsaturation (DOU) is also known as Double Bond Equivalent. If the molecular formula is given, DoU can be calculated as follows

DoU   =   2C+2+N−X−H

               2

C is the number of carbons

N is the number of nitrogens

X is the number of halogens (F, Cl, Br, I)

H is the number of hydrogens

Molecular formula = C6H12

Sites of unsaturation = 1

1H NMR data:

NMR peak value (ppm)	Protons	Indication
5.6	Multiplet, 1H	Alkene proton
4.6	Doublet, 2H	Alkene CH2 couling with CH
1.9	Quartet, 2H	CH2 coupling with CH3
1.5	Quintet, 2H	CH2 attached to two CH2
1.3	Sextet, 2H	CH2 attached to CH2 and CH3
0.9	Triplet, 3H	CH3 attached to CH2

13C NMR data:

NMR peak value (ppm)	Indication
140	Alkene carbon
118	Alkene carbon
32	Aliphatic carbon attached to alkene
31	Aliphatic carbon
21	Aliphatic carbon
15	Aliphatic carbon

IR data:

IR peak (cm-1)	Indication
2960	Aliphatic C-H
1641 and 1466	Alkene C=C stretch
909	Monosubstituted C-H alkene stretch

All above data confirms following structure