Solving Spectroscopy Problems Part 1 Lecture Supplement page 159 + +

Solving Spectroscopy Problems Part 1Lecture Supplement page 159

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CH3

H3C

O

Solving Spectroscopy Problems

•Logical, orderly procedure•Compare information between spectra when necessary•Conservative analysis - do not discard possibilities until 100% sure

Procedure•MS gives formula•Formula gives DBE•Use formula plus DBE to guide IR analysis of functional groups•NMR gives skeleton•Assemble the pieces•Check your work!

How to deduce structure from spectra?

Sample Problem #1Steps 1 and 2: Formula and DBE

Mass spectrumm/z = 120 (M; 100%), m/z = 121 (9.8%), and m/z = 122 (0.42%)

Formula and DBE

•Usual procedure reveals only one viable formula: C9H12

•DBE = 4 Four rings and/or pi bonds

Possible benzene ring

Sample Problem #1Step 3: Functional Groups from IR

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04000 3000 2000 1500 1000 Stretching frequency (cm-1)

Alcohol O-H:Amide/amine N-H:

Terminal alkyne C-H:

Zone 1 (3700-3200 cm-1) C9H12 DBE = 4

All absent – no peaks


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Zone 2 (3200-2700 cm-1) C9H12 DBE = 4Aryl/vinyl sp2 C-H:

Alkyl sp3 C-H:Aldehyde C-H:

Carboxylic acid O-H:

Present - peaks > 3000 cm-1

Present - peaks < 3000 cm-1

Absent - no peak ~2700 cm-1; no C=O in zone 4Absent - not broad enough; no C=O in zone 4


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Zone 3 (2300-2000 cm-1) C9H12 DBE = 4

Alkyne CC:

Nitrile CN:Both absent – no peaks


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Zone 4 (1850-1650 cm-1) C9H12 DBE = 4C=O: Absent - no strong peak; no oxygen in formula


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Zone 5 (1680-1450 cm-1) C9H12 DBE = 4Benzene ring C=C:

Alkene C=C:Present - peaks at ~1610 cm-1 and ~1500 cm-1

Absent - not enough DBE for alkene plus benzene

Sample Problem #1Step 4: C-H Skeleton from 1H-NMR

Shift Splitting Integral # of H Implications

Step 4a: Copy NMR data to table

7.40-7.02 ppm multiplet 5

2.57 ppm triplet 2

1.64 ppm sextet 2

0.94 ppm triplet 3

1H-NMR: 7.40-7.02 ppm (multiplet; integral = 5), 2.57 ppm (triplet; integral = 2), 1.64 ppm (sextet; integral = 2), and 0.94 ppm (triplet; integral = 3).



Step 4b: Divide hydrogens according to integrals


2.57 ppm triplet 2

1.64 ppm sextet 2

0.94 ppm triplet 3

Totals 12 12 H

5 H

2 H

2 H

3 H



Step 4c: Combine splitting with number of hydrogens to get implications


2.57 ppm triplet 2

1.64 ppm sextet 2

0.94 ppm triplet 3

Totals 12 12 H

5 H

2 H

2 H

3 H

C6H5 (Ph; monosubstituted benzene ring)

CH2 in CH2CH2

CH2 in CHCH2CH2 x CH in CHCH2

2 x CH in CHCHCHtwo neighbors CH2 or 2 x CH

five neighbors CH2 or 2 x CH

two neighbors CH3 or 3 x CH

CH2 in CH2CH5 2 x CH in CH3CHCH2

2 x CH in (CH)2CHCH3

2 x CH in (CH2)2CHCH

CH3 in CH3CH2 3 x CH in CHCH2

3 x CH in CHCHCH

CH2 in CH3CH2CH2



Step 4d: Select most likely implications, then total the atoms Most likely implication = least number of atoms


2.57 ppm triplet 2

1.64 ppm sextet 2

0.94 ppm triplet 3

Totals 12 12 H

5 H

2 H

2 H

3 H

C6H5 (Ph; monosubstituted benzene ring)

CH2 in CH2CH2


2 x CH in CHCHCHtwo neighbors CH2 or 2 x CH

five neighbors CH2 or 2 x CH

two neighbors CH3 or 3 x CH

CH2 in CH2CH5

CH2 in CH3CH2CH2

2 x CH in CH3CHCH2

2 x CH in (CH)2CHCH3

2 x CH in (CH2)2CHCH

CH3 in CH3CH2 3 x CH in CHCH2

3 x CH in CHCHCH

C6H5 + CH2 + CH2 + CH3 = C9H12

Sample Problem #1Step 5: Checks

Atom Check

Formula - atoms used = atoms left over

C9H12 - C9H12 (from 1H-NMR) = all atoms used

DBE Check

DBE from formula - DBE used = DBE left over

4 - 4 (benzene ring) = all DBE used

Step 5: Check to see that all atoms and DBE are used

Ph

CH2 of CH2CH2

CH2 of CH2CH2CH3

CH3 of CH3CH2

Sample Problem #1Step 6: Assembly

Pieces:

How to assemble?•Pay attention to splitting patterns•All pieces form one molecule•Trial and error•Pentavalent carbonsXXXXXXXXXXXXXXX

CH2CH2CH3

CH2CH2CH3

Step 7: Check your work!•Formula•Functional groups•Number of signals•Splitting

•Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice

Step 6: Now to put it all together...

Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice

Solving Spectroscopy Problems Part 2Lecture Supplement page 166

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CH3

H3C

O

Implications, Blind Men, and Elephants

-CH2CH2-

C

H

H

C

H

H

CH3CH2CH2-

C C

H

H

C

H

H

H

H

H

-CH3Ph-

CH2CH2CH3

+ + +

An NMR Fable

+ + +

C

H

H

H

http://en.wikipedia.org/wiki/Blind_men_and_an_elephant


Mass spectrumm/z = 118 (M; 100%) Molecular mass (lowest mass isotopes) = 118

Even number of nitrogen atoms

5.7% / 1.1% = 5.2m/z = 119 (5.7%)

m/z = 120 (0.63%)

Formula118 - (5 x 12) = 58 amu for oxygen, nitrogen, and hydrogen

Five carbon atoms

No sulfur, chlorine, or bromine

Usual procedure gives three possible formulas:•C5H10O3 DBE = 1 One ring or one pi bond•C5H14N2O Rejected: 14H does not fit NMR integration (sum of integrals = 5.0)•C5H2N4 Rejected: More than two signals in NMR; no oxygen for C=O (IR zone 4)


Absent - no strong peakAbsent - no peak; no nitrogen in formulaAbsent - no peak; no CC peak in zone 3

Zone 1 (3700-3200 cm-1) C5H10O3 DBE = 1

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Zone 2 (3200-2700 cm-1) C5H10O3 DBE = 1

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Aryl/vinyl sp2 C-H:Alkyl sp3 C-H:

Aldehyde C-H:Carboxylic acid O-H:

Absent - no peaks > 3000 cm-1


Absent - no peak ~2700 cm-1

Absent - not broad enough


Zone 3 (2300-2000 cm-1) C5H10O3 DBE = 1

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Alkyne CC:Nitrile CN:

Absent - no peak; not enough DBEAbsent - no peak; no nitrogen in formula


Zone 4 (1850-1650 cm-1) C5H10O3 DBE = 1

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C=O: Present 1741 cm-1 = ester (1750-1735 cm-1)ketone (1750-1705 cm-1)aldehyde (1740-1720 cm-1)

1741 cm-1


Zone 5 (1680-1450 cm-1) C5H10O3 DBE = 1

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Benzene ring C=C:Alkene C=C:

Absent - no peak; not enough DBEAbsent - no peak; not enough DBE for C=O plus alkene


1H-NMR: 4.22 ppm (triplet; integral = 1.0), 3.59 ppm (triplet; integral = 1.0), 3.39 ppm (singlet; integral = 1.5), 2.09 ppm (singlet; integral = 1.5)



4.22 ppm triplet 1.0


3.39 ppm singlet 1.5









Totals 5 10 H

2 H

2 H

3 H

3 H








Totals 5 10 H

2 H

2 H

3 H

3 H

two neighborsCH2 in CH2CH2

CH2 in CHCH2CHCH2 or 2 x CH2 x CH in CHCH2

2 x CH in CHCHCH

two neighbors CH2 or 2 x CHCH2 in CH2CH2


2 x CH in CHCHCH

no neighbors

no neighbors

CH3 or 3 x CH

CH3 or 3 x CH

CH3 or 3 x CH

CH3 or 3 x CH








Totals 5 10 H

2 H

2 H

3 H

3 H

two neighborsCH2 in CH2CH2

CH2 in CHCH2CHCH2 or 2 x CH2 x CH in CHCH2

2 x CH in CHCHCH

two neighbors CH2 or 2 x CHCH2 in CH2CH2


2 x CH in CHCHCH

no neighbors

no neighbors

CH3 or 3 x CH

CH3 or 3 x CH

CH3 or 3 x CH

CH3 or 3 x CH

CH2 + CH2 + CH3 + CH3 = C4H10


Atom Check


C5H10O3 - C4H10 (from 1H-NMR) - C=O (ester or ketone from IR) = 2 O

DBE Check


1 - 1 (C=O) = all DBE used



CH2 in CH2CH2

CH2 in CH2CH2

CH3

CH3

C=O (ester or ketone)OOAssembly•Coupling suggests 2 x CH2 join to form CH2CH2

•CH3 (singlet in NMR) cannot be in CH3CH2CH2

•Leaves CH3O and CH3C=O

2.09 ppm observed shift3.39 ppm observed shift

Typically 3.8 ppm

Typically 2.0-2.3 ppm

CH3C=O

CH3O

CH2CH2

Pieces


Pieces

CH3CH3O

O

Does not use all pieces

X CH3OO CH3

O

OCH3CH3O

O

•Observed CH2 chemical shifts = 4.22 and 3.59 ppm•Typical OCH2 chemical shift = 3.6 - 4.6 ppm•Typical O=CCH2 chemical shift = 2.2 - 3.0 ppm•More consistent with middle structure

Step 7: Check your work Left as a student exercise

Three ways to assemble these pieces:

CH2CH2

CH3OCH3C=O (ester or ketone)O


Mass spectrumm/z = 87 (M; 100%)

m/z = 88 (4.90%)

m/z = 89 (0.22%)

Molecular mass (lowest mass isotopes) = 87Odd number of nitrogen atoms

4.9% / 1.1% = 4.5

Formula

Usual procedure gives two formula candidates:

•C4H9NO DBE = 1 One ring or one pi bond

•C5H13N Rejected: Does not fit 1H-NMR integration (4:4:1 Integral sum = 9)

Four or five carbon atoms

No sulfur, chlorine, or bromine


Present? - weaker than usualPresent?Absent - not enough DBE; no CC in zone 3

Zone 1 (3700-3200 cm-1) C4H9NO DBE = 1

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Zone 2 (3200-2700 cm-1) C4H9NO DBE = 1

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Aryl/vinyl sp2 C-H:Alkyl sp3 C-H:

Aldehyde C-H:Carboxylic acid O-H:

Absent - no peaks > 3000 cm-1


Absent - ~2700 cm-1 present but no C=O in zone 4Absent - no C=O in zone 4; not broad enough


Zone 3 (2300-2000 cm-1) C4H9NO DBE = 1

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CC:

CN:Absent - no peaks; not enough DBE


Zone 4 (1850-1650 cm-1) C4H9NO DBE = 1

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C=O: Absent - no peaks


Zone 5 (1680-1450 cm-1) C4H9NO DBE = 1

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Benzene ring:Alkene C=C:

Absent - no peak ~1600 cm-1; not enough DBEAbsent - no peak ~1600 cm-1


1H-NMR: 3.67 ppm (triplet; integral = 4.0), 2.86 ppm (triplet; integral = 4.0), and 2.59 ppm (singlet; integral = 1.0).













Total = 9.0 9 H

4 H

4 H

1 H








Total = 9.0 9 H

4 H

4 H

1 H

two neighbors

two neighbors 2 x CH2 or 4 x CH

no neighbors

2 x CH2 in CH2CH2

2 x CH2 in CHCH2CH4 x CH in CHCH2

4 x CH in CHCHCH

2 x CH2 in CH2CH2


4 x CH in CHCHCH

CH or NH or OH

2 x CH2 or 4 x CH








Total = 9.0 9 H

4 H

4 H

1 H

two neighbors

two neighbors 2 x CH2 or 4 x CH

no neighbors

2 x CH2 in CH2CH2


4 x CH in CHCHCH

2 x CH2 in CH2CH2


4 x CH in CHCHCH

CH or NH or OH IR more consistent with NH than OH

2 x CH2 or 4 x CH

(2 x CH2) + (2 x CH2) + NH = C4H9N

X


Atom Check


C4H9NO - C4H9N (from 1H-NMR) = one oxygen Not part of a functional group that appears in IR or 1H-NMR:

An ether

DBE Check


1 - 0 = one DBE C=O, C=C, C=N absent in IR Therefore DBE = ring



Pieces

Assembly•Splitting pattern requires two sets of 2 x CH2 to become two equivalent CH2CH2

•Remaining pieces can only be assembled in one way: O N H

2 x CH2CH22 x CH2 in CH2CH2

2 x CH2 in CH2CH2

NHO (ether)one ring

Documents

Solving Spectroscopy Problems Part 1 Lecture Supplement page 159 + +