URGENT PLEASE :) Define the molecule & make comments and calculations on these H1 NMR and IR Spectrums just like they di
Posted: Wed May 25, 2022 7:04 am
URGENT PLEASE 
Define the molecule & make
comments and calculations on these H1 NMR and
IR Spectrums just like they did in the example
below.
THANK YOU!!!
Unknown 2 6.0 5.5 t TT T 5.0 4.5 21.69 4.0 3.5 3.0 f1 (ppm) 2.5 91¹EE {F 2.0 45'EE 1.5 265.7 -58'6+ 1.0 0.5 0.(
4000 3500 3000 2500 Wavenumbers (cm-1) 2000 1500 1000 500 2955,68 2916,43 1739,46 1204,08 1172,32 2849,63 1727,37 1464,16 1261,68 1235,79 974,04 945,95 1327,73 1027,69 % T 1104,22 1412,86 1391,42 4060,94 960,66 875,52 921,78 719,58 3850,81 836,19 632,40
An example for FTIR 3007 Transmittance 2919 2852 1743 1655 1462 1416 1376 1347- 1236 1500 1008 1168- 918 996 722 4000 3500 3000 2500 2000 1000 Wavenumber/cm¹ The bands at 3007 cm-1 and at 965 cm-1 show the relative degree of unsaturation and the trans fat component, respectively. Also, note that a free fatty acid carbonyl band at 1704 cm-1 is absent as is the strong alcohol CO band at 1043 cm-1that would indicate free glycerol. The broad OH band of free glycerol at ~3300 cm-1 is also noticeably missing. Frequency/cm-1 Assignment Intensity 3007 =C-H (cis) m 3025 =C-H (trans) VW 2919 -C-H (CH₂) vst 2852 -C-H (CH₂) vst 1743 -C=O (ester) vst 1462 C-H (CH2, CH3) m 1168 -C-O-C of ester (asym) st 1098 -C-O-C of ester (sym) m 965 -CH=CH-(trans) W 914 -CH=CH- (cis) VW 722 (CH₂)-(rock) m NOTE: vw = very weak, w = weak, m = medium, st = strong, vst = very strong. Make your assumptions according to the table.
-CH=CH- ►CHO- 4489 807 - CH,O– =CHCH₂CH= 3229 4838 414 —CH2(CO)— -CH₂CH=CHCH₂- 8360 =CH,CH2(CO) 5095 49572 —(CH2)n— -CH3 7401 5.0 4.0 3.0 2.0 1.0 Chemical Shift (ppm) Peaks that are not present should also be noted. There is no evidence for free glycerol (at 3.5-3.8 ppm) in 1H NMR spectrum (also check it with IR). The presence of free fatty acids should also be checked. This absence is more evident in the IR (1704 cm-1) for the free acid carbonyl. The integrals are divided by the number of like substituents per molecule (i.e., one glycerol backbone or three fatty chains), then divided by the number of protons in the substituent (i.e., three protons per methyl group), and finally divided by a constant (value of one proton) to normalize the integrals. Table summarizes the process and shows the final normalized integrals from which average chain length and average degree of unsaturation are derived. Proton NMR Integrals and Normalization for Determining Chain Length Position Norm Assignment Adjusted Integral Integral (ppm) Integral -CH3 -(CH₂)- |–CH,CH2(CO)- -CH₂CH=CHCH₂- 2.0 0.9 7401 +3+3=822 1.00 1.3 49572 +3+2-8262 10.06 1.6 5095 +3+2-849 1.03 8360 +3+4-697 0.85 CH,(CO)- 2.3 4838 +3+2 - 806 0.98 -CHCH₂CH- 2.8 414 +3+2 = 69 0.084 4.3 3229 +1+4-807 0.98 |-CH,O -CHO- 5.3 807 +1+1-807 0.98 -CH-CH- 5.4 4489 +3+2-748 0.91
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comments and calculations on these H1 NMR and
IR Spectrums just like they did in the example
below.
- Urgent Please Define The Molecule Make Comments And Calculations On These H1 Nmr And Ir Spectrums Just Like They Di 3 (142.86 KiB) Viewed 20 times
- Urgent Please Define The Molecule Make Comments And Calculations On These H1 Nmr And Ir Spectrums Just Like They Di 4 (221.93 KiB) Viewed 20 times
Unknown 2 6.0 5.5 t TT T 5.0 4.5 21.69 4.0 3.5 3.0 f1 (ppm) 2.5 91¹EE {F 2.0 45'EE 1.5 265.7 -58'6+ 1.0 0.5 0.(
4000 3500 3000 2500 Wavenumbers (cm-1) 2000 1500 1000 500 2955,68 2916,43 1739,46 1204,08 1172,32 2849,63 1727,37 1464,16 1261,68 1235,79 974,04 945,95 1327,73 1027,69 % T 1104,22 1412,86 1391,42 4060,94 960,66 875,52 921,78 719,58 3850,81 836,19 632,40
An example for FTIR 3007 Transmittance 2919 2852 1743 1655 1462 1416 1376 1347- 1236 1500 1008 1168- 918 996 722 4000 3500 3000 2500 2000 1000 Wavenumber/cm¹ The bands at 3007 cm-1 and at 965 cm-1 show the relative degree of unsaturation and the trans fat component, respectively. Also, note that a free fatty acid carbonyl band at 1704 cm-1 is absent as is the strong alcohol CO band at 1043 cm-1that would indicate free glycerol. The broad OH band of free glycerol at ~3300 cm-1 is also noticeably missing. Frequency/cm-1 Assignment Intensity 3007 =C-H (cis) m 3025 =C-H (trans) VW 2919 -C-H (CH₂) vst 2852 -C-H (CH₂) vst 1743 -C=O (ester) vst 1462 C-H (CH2, CH3) m 1168 -C-O-C of ester (asym) st 1098 -C-O-C of ester (sym) m 965 -CH=CH-(trans) W 914 -CH=CH- (cis) VW 722 (CH₂)-(rock) m NOTE: vw = very weak, w = weak, m = medium, st = strong, vst = very strong. Make your assumptions according to the table.
-CH=CH- ►CHO- 4489 807 - CH,O– =CHCH₂CH= 3229 4838 414 —CH2(CO)— -CH₂CH=CHCH₂- 8360 =CH,CH2(CO) 5095 49572 —(CH2)n— -CH3 7401 5.0 4.0 3.0 2.0 1.0 Chemical Shift (ppm) Peaks that are not present should also be noted. There is no evidence for free glycerol (at 3.5-3.8 ppm) in 1H NMR spectrum (also check it with IR). The presence of free fatty acids should also be checked. This absence is more evident in the IR (1704 cm-1) for the free acid carbonyl. The integrals are divided by the number of like substituents per molecule (i.e., one glycerol backbone or three fatty chains), then divided by the number of protons in the substituent (i.e., three protons per methyl group), and finally divided by a constant (value of one proton) to normalize the integrals. Table summarizes the process and shows the final normalized integrals from which average chain length and average degree of unsaturation are derived. Proton NMR Integrals and Normalization for Determining Chain Length Position Norm Assignment Adjusted Integral Integral (ppm) Integral -CH3 -(CH₂)- |–CH,CH2(CO)- -CH₂CH=CHCH₂- 2.0 0.9 7401 +3+3=822 1.00 1.3 49572 +3+2-8262 10.06 1.6 5095 +3+2-849 1.03 8360 +3+4-697 0.85 CH,(CO)- 2.3 4838 +3+2 - 806 0.98 -CHCH₂CH- 2.8 414 +3+2 = 69 0.084 4.3 3229 +1+4-807 0.98 |-CH,O -CHO- 5.3 807 +1+1-807 0.98 -CH-CH- 5.4 4489 +3+2-748 0.91