3rd Year Chemistry & 4th Year Biology Questions

3rd Year Past Exam 2009-2010

1-Write a brief account on characteristic features to determine the molecular structure of an unknown organic compound using 1HNMR & Mass spectra Tools.

2-An organic lab student carried out the reaction of methyl magnesium iodide with acetone following hydrolysis to give two isomeric compounds (a & b) of formula C4H10O. During the distillation to isolate the products he forgets to mark the isolated isomers, he turned in the two products that gave 1HNMR & Mass spectra of base peak. Propose the two structures if you have A-Chemical shifts 1.28ppm (S, step high 8.8) and 3.9(S, step high 9.9),m/z=57,b- Chemical shifts 1.30ppm (t, step high 5.9 and 2.9ppm (q, step height 3.4),m/z=45

3-Deduce the two unknown compounds structures (A,C6H10 & B,C6H10O) which afforded precipitated if reacted with ammonium silver nitrate and if you are provided 1HNMR spectral data a- Chemical shifts 2.5ppm (1H), 1.5ppm (9H) for A and 1.4ppm (s,6H0,3.3ppm (s,3H).2.4ppm (s,1H)for B

4a-Propose structure of C11H16 if you are provided 1HNMR signals at 1.28ppm (s, step high=8.8), 2.28ppm(s, step high=1.9), 7.1ppm (m, step high with ratios 3:2) and m/z=91 &65

b-Suggest the two isomeric structures A,B of formulas C10H14 if you provided 1HNMR and mass spectral data as follows for: a- Chemical shifts at1.30ppm(S, 9H), 7.28ppm(s,5H),m/z=51,b-Chemical shifts at 0.88ppm (d,6H), 1.86ppm (m,1H) ,2.45ppm(d,2H),7.12ppm(s,5H) and m/z=91,65

3rd Year Chemistry  Exam. (2011)

4-Write a brief account on: a-Fundamental features of NMR and their signal splitting Pascal's Triangle of hydrogen attached to isopropyl bromide. b-Suggest the two isomeric structures (A, B) of molecular formula C8H8O if you are provided their absorption information as A, δ 7.2(s, 5H), δ9.7 (s, 1H), δ3.7(s, 2H),m/z=91,65 and B- δ 7.2 (m, 5H ), δ 2.04 (s, 3H ), m/z =51.

c-The mass spectrum of n-octane shows a prominent molecule ion peak at m/e=114.There is also a large peak at m/e= 57, but it is not the base peak. The mass spectrum of 3, 4-dimethyl hexane shows a smaller molecular ion and the peak at mass 57 is the base peak. Explain these trends in abundance of the molecular ions and the ions at mass 57 and predict the intensities of the peaks at mass 57,114 in the spectrum of 2, 2, 3, 3-tetramethyl butane.

 

A-Select & Answer two from the following: 5

1-An organic lab student carried out the reaction of CH3MgI with acetone following hydrolysis to give two isomeric compounds (A & B) of formula C4H10O. During the distillation to isolate the products he forgets to mark the vials he used to collect the fractions, he turned in the two products that gave 1HNMR & Mass spectra of base peak. Propose the two structures if you are provided that A-Chemical shifts δ 1.28ppm (S, step high 8.8) and δ 3.9(S, step high 0.9),m/z=57 and b- Chemical shifts δ 1.30ppm (t, step high 5.9) and δ 2.9 ppm (q, step high 4.1),m/z=45

2-A compound C10H12O2 resulted from the reaction mixture of 2-phenyl ethanol and acetic acid and reveals1HNMR spectral signals and mass base peak as δ 7.3(s, 5H), δ4.3(t,2H), δ2.9(t,2H), δ2.0(s,3H),m/z=91,65 A-Propose a structure and Make peak assignment  for the compound showing which protons give rise to which absorptions in the spectrum.

3-Two  isomeric compounds C6H12O2 (A,B) resulted in 1HNMR spectral signals at δ 1.3(s,9H), δ 2.5(s,3H) for A and δ 1.3(s,2H), δ 2.5(s,9H) δ 9(s,1H) for B,  A-Propose the two structures A,& B and Make peak assignment  for the compounds showing which protons give rise to which absorptions in the spectrums.

ا3rd year Chemistry Exam.2010-2011 Retrogrades Exam

Question No.4

1-Write a brief account on characteristic features to determine the molecular structure of an unknown organic compound using 1HNMR & Mass spectra Tools.

2-Deduce the two unknown compounds structures(A,C6H10 & B,C6H10O) which afforded precipitated if reacted with ammonium silver nitrate and if you are provided 1HNMR spectral data, a-Chemical shifts 2.5ppm (1H), 1.5ppm (9H) for A and 1.4ppm (s,6H0,3.3ppm (s,3H).2.4ppm (s,1H)for B

3-a-Propose structure of C11H16 if you are provided 1HNMR signals at 1.28 ppm (s, step high=8.8), 2.28ppm(s, step high=1.9),7.1ppm (m, step high with ratios 3:2) and m/z=91 &65

b-suggest the two isomeric structures A,B of formulas C10H14 if you provided 1HNMR and mass spectral data as follows a- Chemical shifts at1.30ppm(S, 9H), 7.28ppm(s,5H),m/z=51, b- Chemical shifts at 0.88ppm (d,6H), 1.86ppm (m,1H) ,2.45 ppm (d,2H),7.12ppm(s,5H) and m/z=91,65

 Half Question No.5

 

a-An organic lab student carried out the reaction of methyl magnesium iodide with acetone following hydrolysis to give two isomeric compounds (a & b) of formula C4H10O. During the distillation to isolate the products he forgets to mark the isolated isomers, he turned in the two products that gave 1HNMR & Mass spectra of base peak. Propose the two structures if you are provided: A-Chemical shifts 1.28ppm (S, step high 8.8) and 3.9(S, step high 9.9),m/z=57, b- Chemical shifts 1.30ppm (t, step high 5.9 and 2.9 ppm (q, step high 3.4),m/z=45

Suggested Exam -1

1a, Write an account on the following: -Metastable ion peak, uses of mass spectra, mass spectra fragmentation dependency, similarity of mass spectra of butane-1 &-2, graphical analyses of signal splitting in 1HNMR of sp2 carbon bearing proton attached to isopropyl bromide, cracking pattern and stability of molecular ion peak. various aspects of 1HNMR spectra determination. The priority of stability of molecular ion peak in mass spectra.

1b-Suggest general problem solving algorithm to elucidate structure for unknown compound.

2a-Write an account on the following:

a- Characteristic feather taken in consideration in 1HNMR spectra, origin of signals in1HNMR spectra, signals generation in 1HNMR spectra, magnetic environments dependency in 1HNMR spectra. Physical meaning for cps & TMS shielding and deshielding, chemical shifts in1HNMR spectra. b-The appearance of a base peak at m/e=65(phenol), 44(dimethyl amine, 57(cyclohexanol).the mass spectra of o-substituted acids and esters.

2b-The mass spectra of the isomeric structures of each C2H6O, C3H6O.

2c-Suggest the structure of C5H12O, (MW = 88.15), if you are provided the following sets of m/e data 70, 55, 42, 31, 29 and m x =43.3, Illustrate the possible fragmentation processes if you know that the Cleavage of the C-C bond next to the oxygen usually occurs.  A loss of H2O may occur as in the below spectra,

 

 

 

 

 

 

3a-How you might go about identifying an unknown organic compound depending upon the absorption information spectra.

3b-Explain fragmentation process and build your choice which spectrum represents (A, B) of molecular formula C7H16O.

Suggested Exam -2

4-The 1HNMR spectra for compound of C9H10 shows the following peaks pentet at δ 2.04,triplet at δ 2.91 and multiplet at δ 7.17.The area ratios of these peaks are 1:2:2 respectively. Propose the structure.

5-The molecular ion of alkyl halide undergo fragmentation depending on the nature of halogen atom and on the nature of alkyl group. Write the require fragmentation pathways.

6-Suggest signals1HNMR and their signals graphical splitting analyses for N-phenyl piperidine and isopropyl bromide.

Suggested Exam -3

7-The 1HNMR signals for compound of C11H16 shows the following peaks multiplet at τ 2.9, singlet at τ 7.72 and singlet at τ 8.72.The area ratios of these peaks are 3.8:2.9:8.8 respectively. Propose the structure.

8-Explain a-the fragmentation mechanism of benzyl alc..Tri ethyl amine and pentanol. Alkyl and aryl halides -signals suggestion in N-phenyl piperidine and ethyl Propyl ether. c-The appearance of a base peak at the following m/e values 31(ethyl methyl ether), 51(benzyl alc.), 91(ethyl benzene)

9-Some organic molecules proceed via α– and β- cleavage fragmentation in their mass spectra, Discuss giving examples.

Suggested Exam -4

10-The spectral data of C9H10O2 in 1HNMR and mass spectra reveals δ

3.25(multiplet), δ 9.75(quartet), δ 9.1(triplet) and m/e-51,Deduce the structure.

11-Give reasonable explanation for the appearance of the base peak for the following: m/e =51 for the mass spectra of benzyl alcohol., m/e =30 for the mass spectra of primary amine, m/e=44 for the mass spectra of dimethyl amine, m/e=57 for the mass spectra of cyclohexanol, m/e =31 for the mass spectra of ethyl methyl ether., m/e=65 for the mass spectra of phenol and butyl benzene,

M/e =79 for the mass spectra of o-toludine, m/e=91 for the mass spectra of Ethyl benzene.

12- Explain using Pascal's Triangle for the appearance of sextet and heptet 1HNMR signals for primary and secondary chloro propane.

Suggested Exam -5

13-Differentiate between the following taken in your consideration1HNMR and mass spectra for ethanol &diethyl ether and acetone & proponaldehyde

14-Alkyl halide leads to alkane and alkene fragmentation, Discuss giving the factor affecting on the mass spectra.

15-Suggest signals systems in the following compounds:

3-bromo-2-(tertiary butyloxy)-thiophene, b-2, 6-dimethyl pyridine, c-1, 1,di –fluoro  ethylene, d-methyl acrylate. 2-bromo propane, butanol, N-phenyl piperidine, o-toluidine, ethyl propyl ether.-a

Suggested Exam -6

16a-Write a brief account on characteristic features taken in our consideration in H NMR & Mass Spectra to determine organic molecule structures?

16b-Deduce the two unknown isomeric structures (A, & B ) of formula C9H10O2 if you are provided that they give the same mass spectral base peak at m/z=91 &65 and 1HNMR& spectral data as follows:

A- Chemical shifts δ 2.7ppm (t, 2H), δ 3.0 ppm (t, 2H), δ 7.2 ppm (m, 5H) and δ 9.8 ppm(s, 1H). B- Chemical shifts δ 2.9ppm (s, 2H), δ 3.3ppm (s, 3H) and δ 7.2 ppm (m, 5H).

16c-Propose structure of C9H10 if you are provided 1HNMR signals pentet at δ 2.04 ppm (2H), triplet at δ 2.91 ppm (4H) and multiplet at δ  7.17ppm (4H),

17a-An organic lab student carried out the hydrolysis of two unknown isomeric compounds (A & B) of formula C9H10O2 using aq.KOH. During the distillation to isolate the precipitated products he forgets to mark the isolated isomers, he turned in the two products that gave 1HNMR & Mass spectra of base peak.

Propose the two structures (A & B), if you have, A- Chemical shifts δ 1.30ppm (t, 3H) and δ 2.9 ppm (q, 2H), m/z=51, b- Chemical shifts δ 1.28ppm (S, 2H) and δ 3.9(S, 3H) and δ 7.2 ppm (m, 5H), M/z =91, 65

Suggested Exam -7

17b-The molecular ion in the      spectrum below is 58. What alkyl group was lost to  form the base peak at m/z 43?       

 

a) Methyl (CH3),b) Ethyl(CH3CH2),c) Propyl (CH3CH2CH2).

 18-Deduce the two unknown isomeric structures (A, & B) of formula C4H8O ,with the following sets1HNMR data and mass m/e data, A- Chemical shifts δ 2.0 ppm (s, 3H), δ 4.1 ppm (q, 2H), δ 1.3 ppm (t, 3H) and m/e=73. B- Chemical shifts δ 3.6 ppm (s, 3H), δ 2.3ppm (q, 2H) and δ 1.2 ppm (t, 3H) and m/e=59

19-Write a brief account on characteristic features to determine the molecular structure of an unknown organic compound using 1HNMR & Mass spectra Tools.

Suggested Exam -8

20-An organic lab student carried out the reaction of methyl magnesium iodide with acetone following hydrolysis to give two isomeric compounds (A & B) of formula C4H10O. During the distillation to isolate the products he forgets to mark the vials he used to collect the fractions, he turned in the two products that gave 1HNMR & Mass spectra of base peak. Propose the two structures if you are provided the following:

A-Chemical shifts δ 1.28ppm (S, step high 8.8) and δ 3.9(S, step height 9.9),m/z=57

b- Chemical shifts δ 1.30ppm (t, step high 5.9) and δ 2.9ppm (q, step high 3.4),m/z=45

21-Deduce the two unknown compounds structure (A, C6H10 & B,C6H10O) which afforded precipitated if reacted with ammonium silver nitrate and if you are provided 1HNMR spectral data

a- Chemical shifts δ 2.5ppm (1H), δ 1.5ppm (9H) for A and 1.4ppm (s,6H), δ 3.3ppm (s,3H). δ 2.4ppm (s,1H)for B

22-a-Propose structure of C11H16 if you are provided 1HNMR signals at δ 1.28ppm (s, step high=8.8), δ 2.28ppm(s, step high=1.9), δ 7.1ppm (m, step high with ratios 3:2) and m/z=91 &65

Suggested Exam -9

b-suggest the two isomeric structures (A, B) of formulas C10H14 if you provided 1HNMR and mass spectral data as follows for:

a- Chemical shifts at δ 1.30ppm(S, 9H), δ 7.28ppm(s, 5H), m/z=51

b- Chemical shifts at δ0.88 ppm(d,6H), δ 1.86ppm(m,1H) , δ 2.45ppm(d,2H) , δ 7.12 ppm (s,5H) and m/e=91,65

23-Molecular formula C4H10O reveals 1HNMR signals in step ratios 1:2:1:6 at δ4.0(singlet), δ3.4,0.9ppm(duplet) and at δ 1.8 (multiplet),suggest the structure if you provided  a base peak at m/z=31

24-The 1HNMR spectra for compound of C14H18 shows the following sets signals, singlet at δ 2.54,triplet at δ 2.91 and multiplet at δ 2.04.The area ratios of these peaks are 5.85:3.96:7.89 respectively. Suggest the structure of the compound provided.

Suggested Exam -10

25-The 1HNMR spectra for compound of C12H14 shows the following sets signals and step high ratios as follows, pentet at δ 2.04,triplet at δ 2.91 and multiplet at δ 7.17.The area ratios of these peaks are 2:4:8 respectively. Suggest the structure of the compound provided.

26- Give reasonable explanation for the similarity appearance of the base peak for butane-1 &-2 and Ethanol &diethyl ether.

27-When 2-chloro-2-methylbutane is treated with a variety of strong base, the products always seem to contain two isomers(A,B) of formula C5H10.When sodium hydroxide is used as the base, isomer A predominates. When potassium t-butoxide is used as the base, isomer B predominates. The1HNMR spectra of (A & B) reveals δ 5.1(q,1H), δ1.6(s,6H), δ1.5(d,3H) and δ4.6(s,2H), δ2.0(q,2H), δ1.7(s,3H) δ1.0(t,3H) respectively. Deduce the two isomers A, B

Suggested Exam -11

28-Account for the peaks at m/e = 87,111.126 in the mass spectra of 2,6-dimethyl-4-heptane.

29-Catalytic hydrogenation of unknown compound A gives 2,6-dimethyloctane as the only product. The mass spectrum of A shows a molecular ion at m/e= 140 and prominent peaks at m/e=57,83, suggest a structure of compound A and justify your answer.

30-The mass spectrum of n-octane dhows a prominent molecule ion peak at m/e=114.There is also  a large peak at m/e= 57, but it is not the base peak.The mass spectrum of 3,4-dimethyl hexane shows a smaller molecular ion and the peak at mass 57 is the base peak. Explain these trends in abundance of the molecular ions and the ions at mass 57 and predict the intensities of the peaks at mass 57,114 in the spectrum of 2,2,3,3-tetramethyl butane.

Suggested Exam -12

31-Draw the expected 1HNMR spectrum of methyl propionate C4H8O2 ,MW = 88.11 and point out how is differ from the spectrum of ethyl acetate. Fragments appear due to bond cleavage next to C=O (alkoxy group loss, -OR) and hydrogen rearrangements. Ethyl acetate,

 

 

 

 

 

 

 

32-Suggest the two isomeric structures (A,B) of molecular formula C8H8O if you are provided the following absorption information as follows:

A, δ 7.2(s,5H), δ9.7(s,1H), δ3.7(s,2H),m/z=91,65. B- δ 7.2 (m,5H ), δ 2.04 (s,3H),m/z=51.

33-Deduce the unknown organic structure A has molecular formula C4H8O and reveals1HNMR spectral data at δ4.7-6.5 (vinylic-H), δ4.1(q,1H). δ2.4(s, 1H).

Suggested Exam -13

34-A compound C10H12O2 resulted from the reaction mixture of 2-phenyl ethanol and acetic acid and reveals1HNMR spectral signals and mass base peak as δ 7.3(s,5H), δ4.3(t,2H), δ2.9(t,2H), δ2.0(s,3H),m/z=91,65

A-Propose a structure for compound. B-Make peak assignment showing which protons give rise to which absorptions in the spectrum.

35-For the same spectrum shown in the previous two questions , choose the compound that the spectrum represents.

 

2-methyl-2-propanol , 1-butanol,2-butanol,1-pentanol,2-methyl-1-prpanol.

Structure Determination

Using 1H NMR Spectroscopy

Example 1

An organic compound (C3H6O2) was known to be one of the following. Use its low resolution NMR spectrum to decide which it is. CH3-CH2-COOH, CH3-COO-CH3,

H-COO-CH2CH3

Notice that there are three peaks showing three different environments for the hydrogens. That eliminates methyl ethanoate as a possibility because that would only give two peaks - due to the two differently situated CH3 group hydrogens.

Does the ratio of the areas under the peaks help? Not in this case - both the other compounds would have three peaks in the ratio of 1:2:3.

Now you need to look at the chemical shifts:

 

Checking the positions of the various hydrogens in the two possible compounds against the chemical shift table gives you this pattern of shifts:

 

Comparing these with the actual spectrum means that the substance was propanoic acid, CH3CH2COOH.

Example 2

How would you use low resolution NMR to distinguish between the isomers propanone and propanal?

 

The propanone would only give one peak in its NMR spectrum because both CH3 groups are in an identical environment - both are attached to -COCH3.

The propanal would give three peaks with the areas underneath in the ratio 3:2:1.

You could refer to the chemical shift table above to decide where the peaks are likely to be found, but it isn't really necessary.

Example 3

How many peaks would there be in the low resolution NMR spectrum of the following compound, and what would be the ratio of the areas under the peaks?

 

All the CH3 groups are exactly equivalent so would only produce 1 peak. There would also be peaks for the hydrogens in the CH2group and the COOH group. There would be three peaks in total with areas in the ratio 9:2:1.

Example 4

 

Deduce the structure of a compound with the formula C3H7Br that has the following 1H NMR spectrum: 3.1 ppm (triplet; integral = 1), 1.7 ppm (sextet; integral = 1), and 1.0 ppm (triplet; integral = 1.5).

 

To Start: calculate the number of double bond equivalents.DBE = C - (H/2) + (N/2) + 1 = 3 - (7 + 1)/2 + 0/2 + 1 = 0. We conclude that the molecule has no rings or pi bonds. 1H-NMR Analysis

 

Step 1: summarize the 1H-NMR data in a table.

Chemical Shift Splitting Integral  of H Implications

3.1 ppm triplet 1,1.7 ppm sextet 1,1.0 ppm triplet 1.5,Totals 3.5

 

Step 2: The total of the integrals is 1 + 1 + 1.5 = 3.5. Because there are seven hydrogens, we conclude that each unit of the integral corresponds to two hydrogens.

Chemical Shift Splitting Integral of H Implications 3.1 ppm triplet 1 2 H,1.7 ppm sextet 1 2 H, 1.0 ppm triplet 1.5 3 H, Totals 3.5 7 H

 

Step 3: For each 1H-NMR signal, use the number of hydrogens to decide if each signal corresponds to a CH3, CH2, CH, OH, NH, etc. For example, a three hydrogen signal might arise from a CH3, 3 x CH, or 3 x OH. OH is possible only when there is enough oxygen in the formula, if the IR does not rule it out, and if the 1H-NMR signal is not split. Chemical Shift Splitting Integral # of H Implications 3.1 ppm triplet 1 2 H CH2 1.7 ppm sextet 1 2 H CH2, 1.0 ppm triplet 1.5 3 H CH3, Totals 3.5 7 H2

 

Step 4: Use the n+1 rule and the observed splitting patterns to determine all the possibilities that would give the observed 1H-NMR signals. Example: The 2H triplet at 3.1 ppm could be due to a CH2, or 2 CH groups. The CH2 or 2 x CH groups must have two neighbors, as the observed splitting pattern is a triplet. Keeping things simple, it would appear that the CH2 group has another CH2 group next to it; this accounts for the triplet. The sextet at 1.7 ppm must arise from a total of five neighboring hydrogens. Since a single carbon can’t have five hydrogens, there

must be two different sets of hydrogens that sum to five; let’s use a CH3 and a CH2 group. We make record of this in the table. (In this table the underlined hydrogens correspond to the given chemical shift.) The triplet at 1.0 ppm is determined accordingly. Chemical Shift Splitting Integral # of H Implication 3.1 ppm triplet 1 2 H CH2 in CH2CH2, 1.7 ppm sextet 1 2 H CH2 in CH2CH2CH3, 1.0 ppm triplet 1.5 3 H CH3 in CH3CH2, Totals 3.5 7H

 

Step 5: We now sum up the atoms that are underlined. We do this as a check to make sure we have used all the hydrogens and to look for atoms that do not show up in the 1H-NMR spectrum, such a COO portion of an ester Chemical Shift Splitting Integral # of H Implication 3.1 ppm triplet 1 2 H CH2 in CH2CH2,1.7 ppm sextet 1 2 H CH2 in CH2CH2CH3,1.0 ppm triplet 1.5 3 H CH3 in CH3CH2, Totals 3.5 7H CH2 + CH2 + CH3 = C3H7

 

Step 6: Atom Check. Look for unused atoms. Given formula - atoms from 1H-NMR or other sources such as IR = unused atoms. For this example: C3H7Br – C3H7 = Br. Therefore 1H-NMR data accounts for all of the atoms in the formula except the bromine. This bromine atom must be included when assembling the pieces.

 

Step 7: DBE Check. Look for unused DBEs. DBEs calculated from formula – DBE accounted for (by 1H-NMR, etc.) = unused DBEs. In this example the formula has no DBEs and the 1HNMR data does not require any DBEs, so there are no left over DBEs to consider when assembling the pieces.

 

Step 8: List and assemble the pieces. At this point you’ve done 90% of the work. Now we assemble the pieces to come up with the final structure. The pieces suggested by the 1H-NMR data are: CH2 in CH2CH2, CH2 in CH2 CH2CH3, CH3 in CH3CH2Br3

 

Also include pieces suggested by other means such as IR data. There is often many ways to go about assembling the pieces in solving NMR spectroscopy problems. Use the clues provided by the spectra, such as coupling patterns. Sometimes trial and error is all that works. In this case, the middle fragment uses all three of the available carbons and fits with the other fragments as shown below. CH2 in CH2CH2

CH2 in CH2 CH2CH3, CH3 in CH2CH3, The fragments have been reduced to CH3CH2CH2 and Br, which can only be assembled one way: CH3CH2CH2Br

 

Step 9: Check to make sure the proposed structure is consistent with the original data, i.e., chemical shift (use table), splitting, and integration. In this case, examine CH2 next Br: is the chemical shift consistent with what we expect? This process might seem long and unnecessarily complicated for such a simple molecule, but can

be very useful in organizing your thoughts for more complex examples. The process becomes much faster with practice and a little bit of experience.

Example 5:

Deduce the structure of a compound with the formula C9H12 that has the following 1H NMR spectrum: 7.40 – 7.02 ppm (multiplet, integral = 2.5); 2.57 ppm (triplet, integral = 1); 1.64 ppm (sextet, integral = 1) and 0.94 ppm (triplet, integral = 1.5).

DBE = C - (H/2) + (N/2) + 1 = 9 - (12)/2 + 0/2 + 1 = 4.

We conclude that the molecule has a benzene ring. 1H-NMR Analysis

Step 1: 1H-NMR data in a table. Chemical Shift 7.40 to 7.02 ppm

2.57 ppm, 1.64 ppm, 0.94 ppm, Splitting multiplet, triplet, sextet, triplet

Totals Integral ,2.5,1,1,1.5=6, of H Implications 4

Step 2: The total of the integrals is 1 + 1 + 1.5 + 2.5 = 6. Because there are 12 hydrogens, we conclude that each unit of the integral corresponds to two hydrogens.

Chemical Shift 7.40 to 7.02 ppm,2.57 ppm,1.64 ppm, .94 ppm, Splitting multiplet,t riplet, sextet,triplet,Totals Integral 2.5,1,1.5,6, of H,5,2,2,3,12 Implications

Step 3: Again, for each 1H-NMR signal, use the number of hydrogens to decide if each signal corresponds to a CH3, CH2, CH, OH, NH, etc. Chemical Shift: 7.40 to 7.02 ppm,2.57 ppm,1.64 ppm, 0.94 ppm, Splitting multiplet, triplet, sextet, triplet

Totals Integral 2.5, 1, 1, 1.5,6 # of H 5,2,2,3,12 Implications CH2,CH2,CH3

Step 4: As before, use the n+1 rule and the observed splitting patterns to determine all the possibilities that would give the observed 1H-NMR signals. Example: The 2H triplet at 2.57 ppm is most likely due to a CH2 bonded to another CH2 group, as the observed splitting pattern is a triplet. Determine the rest of table accordingly.

Chemical Shift 7.40 to 7.02 ppm, 2.57 ppm,1.64 ppm, 0.94 ppm, Splitting multiplet

Triplet, sextet, triplet Totals Integral: 2.5,1,1,1.5,6, # of H,5,2,2,3,12 Implications CH2 in CH2CH2, CH2 in CH2CH2CH3, CH3 in CH3CH2,C6H5, C6H5,5

 

Step 5: We now sum up the atoms that are underlined. Check to make sure we have used all the hydrogens and to look for atoms that do not show up in the 1H-NMR spectrum. Chemical Shift 7.40 to 7.02 ppm, 2.57 ppm, 1.64 ppm, 0.94 ppm Splitting multiplet, triplet, sextet, triplet, Totals Integral: 2.5,1,1,1.5,6, # of H,5,2,2,3,12, Implications CH2 in CH2CH2, CH2 in CH2CH2CH3, CH3 in CH3CH2 C6H5 + CH2 + CH2 + CH3 = C9H12

Step 6: Atom Check. Given formula - atoms from 1H-NMR or other sources such as IR = unused atoms. For this example: C9H12 – C9H12 = nothing. Therefore 1H-NMR data accounts for all of the atoms in the formula.

Step 7: DBE Check. For this example the formula has 4 DBEs and the 1H-NMR data accounts for 4 DBE. No additional DBE.

Step 8: List and assemble the pieces. The pieces suggested by the 1H-NMR data are: CH2 in CH2CH2, CH2 in CH2 CH2CH3, CH3 in CH3CH2 The fragments have been reduced to CH3CH2CH2 and mono-substituted benzene ring with a total of two open valences (places to make a bond). There is only one way to assemble pieces. Final structure is propyl benzene.

Step 9: Check to make sure the proposed structure is consistent with the original data, i.e., chemical shift (use table), splitting, and integration, and IR data (if given).

CH2 CH2 CH3, C6H5, 6

Example 6:

Deduce the structure of a compound with the formula C4H8O2 has 1H NMR data: 3.67 ppm (singlet, integral = 1.5); 2.32 ppm (quartet, integral = 1); 1.15 ppm (triplet, integral = 1.5).

DBE = C - (H/2) + (N/2) + 1 = 4 - (8)/2 + 0/2 + 1 = 1.

The molecule has either one double bond or one ring.

1H-NMR Analysis Chemical Shift: 3.67 ppm, 2.32 ppm 1.15 ppm, Splitting singlet, quartet, triplet,

Totals Integral 1.5,1,1.5=4, of H 3, 2, 3 =8

Implications CH3, CH2 in CH2CH3, CH3 in CH3CH2, CH3 + CH2 + CH3 = C3H8

Atom Check: C4H8O2 (given formula) – C3H8 (from 1H-NMR) = CO2.

DBE Check: 1H-NMR data does not account for the one DBE. Now suspect an ester functional group, since have CO2 remaining and one DBE. Being able to make inferences like this is very important and requires practice 7 List and Assemble the Pieces: C, O, O, CH3, CH2 in CH2CH3, CH3 in CH3CH2 ester have an ethyl group and a methyl group, each with an empty valence. Since there are two empty valences on the ester functional group, the ethyl and methyl groups can go on either end of the ester; thus, there are two possible ways to assemble the fragments: H3C,C,O,O,CH2CH3 A, CH3CH2,C,O,O,CH3, B Use chemical shift and splitting patterns to distinguish between these. Compound A would have a singlet integrating for three at δ ~2.2 ppm for the CH3 group bonded to the carbonyl carbon. Compound A would also have a quartet at δ ~3.8 ppm for the CH2 group bonded to the ether oxygen. Our NMR data is not consistent with these predicted chemical shifts and splitting patterns. We conclude, therefore, that compound B is the correct answer.

 

3-pentanone

 

 

 

                                                          

                                                                        

, Mass Spectrum of 4,4-dimethyl Cyclohexene

7- Deduce of structure  of  unknown compound of the molecular formula (C8H14) which undergo a retro-Diels-Alder rearrangement to give diene and alkene fragments if you are provided  its mass spectra gives base peak at (m/z=95 amu), strong both ions are relatively (m/z=54 & 56) and homologues smaller ions at m/z=81 & 67 less).

 

8-Predict 1H NMR & M.s. analyses of 1HNMR shifts [ppm ]:1.9, 4.0, 4.6 & 6.4 & mass peaks (intensities): 84(54%), 83(26%), 6(31%),55(100%), 54(23%), 41(25%), 39 (28%)  , 29(38%), 28(42%) & 27(44%) for the following

g.(3).