Can you please help me about the question below : NOTES YOU WILL NEED : (I want to answer my question just like in this
Posted: Wed May 18, 2022 10:04 am
Can you please help me about the question below :
NOTES YOU WILL NEED : (I want to answer my question
just like in this example, you see below)
QUESTION : (detailed answer please) 
- Make the calculations for the unknown compound, just
like in the example question above.
- Try to find out the unknown compound by looking at the
spectrums below. (you can also use average chain length &
average degree of unsaturation to define the unknown)
Thank you!
An example for FTIR hthony HECH Transmittance (CH) -CH -CH2(CO) -CH2CH=CHCH2 - =CHCH2CH= -CH2CH2(CO) 4000 3500 1000 3000 2500 2000 1500 Wavenumber / cm -1 8360 S60S 7401 49572 3229 414 4838 The bands 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-lis absent as is the strong alcohol co band at 1043 cm-1 that would indicate free glycerol. broad OH band of free glycerol at 3300 cm-1 is also noticeably missing, 50 40 30 20 10 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 H 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. m vst (ppm) Integral Frequency/cm- Assignment Intensity 3007 -C-H (cis) 3025 =C-H (trans) ww 2919 -C-H (CH) vst 2852 -C-H (CH2) 1743 -C-O (ester) vst 1462 -C-H (CH2, CH) 1168 -C-O-C of ester (asym) st 1098 -C-0-C of ester (sym) 965 -CH=CH-(trans) 914 -CH-CH-(cis) 722 (CH).- (rock) Note: vw - very weak, w - weak, m - medium, st - strong, vst - very strong. m m Proton NMR Integrals and Normalization for Determining Chain Length Position Adjusted Norm Assignment Integral Integral -CH 0.9 7401 +3 + 3 - 822 1.00 -(CH3)- 1.3 49572 +3+2-8262 10.06 -CHCHICO- 1.6 5095 +3+2-849 1.03 -CH,CH-CHCH - 2.0 8360 3-4-697 0.85 -CHICO- 2.3 4838 +3 - 2 - 806 0.98 -CHCH,CH- 2.8 414 +3+2-69 0.084 |-CH₂O 4.3 3229 +1 -4 -807 0.98 -CHO 5.3 807 -1.1 - 807 0.98 -CH-CH- 5.4 4489 +32-748 0.91 w ww m Make your assumptions according to the table.
The adjusted integrals for the chain-end methyl (0.9 ppm, 822), the methylene beta to the carbonyl (1.6 ppm, 849), the methylene alpha to the carbonyl (2.3 ppm, 806), and the methylene in the glycerol backbone (4.3 ppm, 807) should all agree since they all represent one proton integral in the triglyceride. The average proton integral for this spectrum of the oil was [(822+806 + 849 + 807)/4= 821]. This value was used to derive the normalized integral in the last column in Table. Calculation of total degree of unsaturation: The total degree of unsaturation is given by the normalized integralat 5.4 ppm or the mole fraction of 0.91 or 91%. This is confirmed by the proton integrals alpha to the unsaturation (mono or poly) or 0.85 (2.0 ppm)+0.08 (2.8 ppm) = 0.93 or 93%. This gives an average total degree of unsaturation of 92% (or 0.95). You can just mention the total degree of unsaturation for your report, like 91%. Calculation of the average chain length: The average chain length can be calculated by using a linear sum of the carbons in the chain: 1.00 +10.1+1.03 +2*(0.85)+0.98 +0.08 + 2*(0.91) = 16.7 16.7 +1.00 (1.00 is coming from carbonyl C) = 17.7. A factor of two is used if two car bons are represented by the proton integral. Your unknown is one of them on the table. After your calculations, you can easily find which one is yours. Fotorol Olive Coconut Canola Animal Vegetable for Lovric Acid Myristic Acid Palmitic Acid Stearic Acid Oleic Acid Linoleic Acid Cia Care Ci C C C C C 20 CM 7 2 85 5 60 18 11 2 10 3 50 34 27 14 50 6 3 56 26 Av Chain Length 177 13.3 17.4 17.2 16.6 Av Degree of Unsolation 0.95 0.08 1.18 0.56 1.08
OS Weing NMD and I2 Spectroscopy for the Characterization of Edtle Fess and Orts. Fort / Oil ester product lepic th-c-o-6-2 Floranta count hono) CH3 5.4 43 1pm O FTIL solid 3007 -CH(CT) 20 1.9 Int Ads. Int Noom Int 2025 - CH (Hans) 서 7401 3 13 = 822 143 - Colegler) 1.30 1048 cac letter) (CH₂) 1.3 49572 -3 +22322 10.06 45 - CHECH(trans) 46-od- (HCH₂ (CO) 16 5095 +3 +2=849 1.03 -08-2" 822+ 806+ 849+ ROZ 801 9. CH CH=CHCH-2.0 8360 +3+4=693 0.35 4 821) Sve o dere soreld triglyceride -CH) (Co)- 2.3 L838 3:2= 800 0.90 CH CH, CH=28 44 -3-2-69 0.84 54ppy total degree of S Myristic acid) Ct3-4C4), Coop - CALO 43 3223 -1- 4= 807 0.98 Palithace Cth (12) COOH 0.91= 99% 5.3 807 - 1 - 1 - 807 0.98 Joscadd> CHf_C452 - CH2Ct CH_]_cofl=CH-54 4489 +8=-2740 0.91 overage chain length chain tergh, degree/possition of was 100140.1+ 1.0311*085+298+ 0.58+ 2"0.9 16.7 wa Sottretien sphysies! 16.7 +1.30-17.7, meno, cr-gyceride content Clienical features Dofurasia CHO Sope Ryplet also compared by 0.156207) 3082 0.93 or 337. one
WA 69'10 9158 F-SEE 265.75 F 58'6b 6.0 5.5 5.0 4.5 4.0 3.5 2.5 2.0 1.5 1.0 0.5 3.0 fi (ppm) 0.C
%T 20 30 35 8 09 65 o បា 3850,81 3500 3000 2955,68 2916,43 2849,63 2500 Wavenumbers (cm-1) 1739,46 1727,37 1500 1464, 16 1412,86 1391,42 1327,73 1261,68 TZ35,79 1204,08 1172,32 1000 1104,22 1027,69 974,04 945,95 1060,94 960, 66 921,78 875,52 836,19 719,58 632,40 500
NOTES YOU WILL NEED : (I want to answer my question
just like in this example, you see below)
- Can You Please Help Me About The Question Below Notes You Will Need I Want To Answer My Question Just Like In This 1 (282.16 KiB) Viewed 44 times
- Can You Please Help Me About The Question Below Notes You Will Need I Want To Answer My Question Just Like In This 2 (81.56 KiB) Viewed 44 times
- Can You Please Help Me About The Question Below Notes You Will Need I Want To Answer My Question Just Like In This 3 (98.49 KiB) Viewed 44 times
- Make the calculations for the unknown compound, just
like in the example question above.
- Try to find out the unknown compound by looking at the
spectrums below. (you can also use average chain length &
average degree of unsaturation to define the unknown)
- Can You Please Help Me About The Question Below Notes You Will Need I Want To Answer My Question Just Like In This 4 (35.38 KiB) Viewed 44 times
An example for FTIR hthony HECH Transmittance (CH) -CH -CH2(CO) -CH2CH=CHCH2 - =CHCH2CH= -CH2CH2(CO) 4000 3500 1000 3000 2500 2000 1500 Wavenumber / cm -1 8360 S60S 7401 49572 3229 414 4838 The bands 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-lis absent as is the strong alcohol co band at 1043 cm-1 that would indicate free glycerol. broad OH band of free glycerol at 3300 cm-1 is also noticeably missing, 50 40 30 20 10 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 H 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. m vst (ppm) Integral Frequency/cm- Assignment Intensity 3007 -C-H (cis) 3025 =C-H (trans) ww 2919 -C-H (CH) vst 2852 -C-H (CH2) 1743 -C-O (ester) vst 1462 -C-H (CH2, CH) 1168 -C-O-C of ester (asym) st 1098 -C-0-C of ester (sym) 965 -CH=CH-(trans) 914 -CH-CH-(cis) 722 (CH).- (rock) Note: vw - very weak, w - weak, m - medium, st - strong, vst - very strong. m m Proton NMR Integrals and Normalization for Determining Chain Length Position Adjusted Norm Assignment Integral Integral -CH 0.9 7401 +3 + 3 - 822 1.00 -(CH3)- 1.3 49572 +3+2-8262 10.06 -CHCHICO- 1.6 5095 +3+2-849 1.03 -CH,CH-CHCH - 2.0 8360 3-4-697 0.85 -CHICO- 2.3 4838 +3 - 2 - 806 0.98 -CHCH,CH- 2.8 414 +3+2-69 0.084 |-CH₂O 4.3 3229 +1 -4 -807 0.98 -CHO 5.3 807 -1.1 - 807 0.98 -CH-CH- 5.4 4489 +32-748 0.91 w ww m Make your assumptions according to the table.
The adjusted integrals for the chain-end methyl (0.9 ppm, 822), the methylene beta to the carbonyl (1.6 ppm, 849), the methylene alpha to the carbonyl (2.3 ppm, 806), and the methylene in the glycerol backbone (4.3 ppm, 807) should all agree since they all represent one proton integral in the triglyceride. The average proton integral for this spectrum of the oil was [(822+806 + 849 + 807)/4= 821]. This value was used to derive the normalized integral in the last column in Table. Calculation of total degree of unsaturation: The total degree of unsaturation is given by the normalized integralat 5.4 ppm or the mole fraction of 0.91 or 91%. This is confirmed by the proton integrals alpha to the unsaturation (mono or poly) or 0.85 (2.0 ppm)+0.08 (2.8 ppm) = 0.93 or 93%. This gives an average total degree of unsaturation of 92% (or 0.95). You can just mention the total degree of unsaturation for your report, like 91%. Calculation of the average chain length: The average chain length can be calculated by using a linear sum of the carbons in the chain: 1.00 +10.1+1.03 +2*(0.85)+0.98 +0.08 + 2*(0.91) = 16.7 16.7 +1.00 (1.00 is coming from carbonyl C) = 17.7. A factor of two is used if two car bons are represented by the proton integral. Your unknown is one of them on the table. After your calculations, you can easily find which one is yours. Fotorol Olive Coconut Canola Animal Vegetable for Lovric Acid Myristic Acid Palmitic Acid Stearic Acid Oleic Acid Linoleic Acid Cia Care Ci C C C C C 20 CM 7 2 85 5 60 18 11 2 10 3 50 34 27 14 50 6 3 56 26 Av Chain Length 177 13.3 17.4 17.2 16.6 Av Degree of Unsolation 0.95 0.08 1.18 0.56 1.08
OS Weing NMD and I2 Spectroscopy for the Characterization of Edtle Fess and Orts. Fort / Oil ester product lepic th-c-o-6-2 Floranta count hono) CH3 5.4 43 1pm O FTIL solid 3007 -CH(CT) 20 1.9 Int Ads. Int Noom Int 2025 - CH (Hans) 서 7401 3 13 = 822 143 - Colegler) 1.30 1048 cac letter) (CH₂) 1.3 49572 -3 +22322 10.06 45 - CHECH(trans) 46-od- (HCH₂ (CO) 16 5095 +3 +2=849 1.03 -08-2" 822+ 806+ 849+ ROZ 801 9. CH CH=CHCH-2.0 8360 +3+4=693 0.35 4 821) Sve o dere soreld triglyceride -CH) (Co)- 2.3 L838 3:2= 800 0.90 CH CH, CH=28 44 -3-2-69 0.84 54ppy total degree of S Myristic acid) Ct3-4C4), Coop - CALO 43 3223 -1- 4= 807 0.98 Palithace Cth (12) COOH 0.91= 99% 5.3 807 - 1 - 1 - 807 0.98 Joscadd> CHf_C452 - CH2Ct CH_]_cofl=CH-54 4489 +8=-2740 0.91 overage chain length chain tergh, degree/possition of was 100140.1+ 1.0311*085+298+ 0.58+ 2"0.9 16.7 wa Sottretien sphysies! 16.7 +1.30-17.7, meno, cr-gyceride content Clienical features Dofurasia CHO Sope Ryplet also compared by 0.156207) 3082 0.93 or 337. one
WA 69'10 9158 F-SEE 265.75 F 58'6b 6.0 5.5 5.0 4.5 4.0 3.5 2.5 2.0 1.5 1.0 0.5 3.0 fi (ppm) 0.C
%T 20 30 35 8 09 65 o បា 3850,81 3500 3000 2955,68 2916,43 2849,63 2500 Wavenumbers (cm-1) 1739,46 1727,37 1500 1464, 16 1412,86 1391,42 1327,73 1261,68 TZ35,79 1204,08 1172,32 1000 1104,22 1027,69 974,04 945,95 1060,94 960, 66 921,78 875,52 836,19 719,58 632,40 500