Spectra Problems

8/10/2019 Spectra Problems

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SPECTRA PROBLEMS


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HINTS for PROBLEM #1


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Find the molecular weight from molecular ion in the mass spectrum What functional groups are present from the IR ?

Count the number of types of C in the 13 C-nmr Look at the integration in the H-nmr

Use the above to get the molecular formula (make sure it matches the MW)

SPECTRA PROBLEM #1 SOLUTION

In the MS, the molecular ion occurs at m/z = 72 indicating the MW = 72 g /

mol. The IR shows a carbonyl C=O (1720 cm-1), no -OH or -NH (above 3200 cm-

1), no C=C (1600 cm-1) 13 C-nmr shows 4 types of C, including a C=O (210 ppm), and 3

hydrocarbon C (37, 29 and 8 ppm)

H-nmr

/ppm multiplicity integration assignment

2.4 quartet 2 CH2coupled to 3H

2.1 singlet 3 isolated CH3

1.1 triplet 3 CH3coupled to 2H

The following pieces can be deduced from the H-NMR spectra : an isolated

-CH3, and a -CH2CH3.

Summary....

The MS indicated MW = 72 g/mol. The IR showed the presence of a C=O bond. 13C peak at 210 ppm suggests that the C=O is an aldehyde or ketone. H nmr gives an uncoupled -CH3 and an -CH2-CH3

Note we can rule out it being an aldehyde (no peak at 9-10 ppm in the H-

NMR) Use what we have to check the molecular formula : C4H8O = 4 x 12 + 8 x 1

+ 1 x 16 = 72 g/mol

This has an IHD = 1, and is consistent with the bond of the C=O.

Altogether....

With the pieces we have : C=O, -CH2-CH3, and -CH3.

IR and 13C is consistent with a ketone.

Given these pieces, there is only one way they can fit

together: 2-butanone

or


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ethyl methyl ketone

PROBLEM #2


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Count the number of types of C in the 13 C-nmr Look at the integration in the H-nmr

Use the above to get the molecular formula (make sure it matches the MW) Use the coupling patterns in the H-nmr to identify the hydrocarbon "pieces" What is the 1H, singlet in the H-nmr ? (what did the IR show ?)


In the MS, the molecular ion occurs at m/z = 60 indicating the MW = 60 g /mol.

The IR shows an -OH (3350 cm-1) and C-O (1075 cm-1) but no C=O

(around 1700 cm-1), or C=C (around 1600 cm-1)

13 C-nmr shows 3 types of C, including a deshielded C (69 ppm), and 2other hydrocarbon C (27 and 10 ppm)

H-nmr


3.6 triplet 2 CH2coupled to 2H, deshielded

2.3 singlet 1 isolated H (-OH ?)

1.6 multiplet 2 CH2coupled to many H


The H-NMR gives us the following pieces -OH (helped by the IR) -

CH3, and 2 different -CH2-.

We can start to fit them together based on the coupling patternsas: CH3CH2CH2-

Summary....

The MS indicated MW = 60 g/mol.

The IR showed the presence of an O-H and C-O bonds. H nmr gives a CH3CH2CH2- group From the data, we can get a molecular formula of C3H8O = 3 x 12 + 8 x 1 +

1 x 16 = 60 g/mol

This has an IHD = 0, so no bonds or rings.

Altogether....

With the pieces we have : CH3CH2CH2- and -OH

IR an consistent with an alcohol.


together.


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Checking the H-nmr, the middle -CH2- is actually six lines

since it has a total of 5 similar neighbours. The -OH deshields

the other -CH2- but does not couple to it.

1-propanol


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Find the molecular weight from molecular ion in the mass spectrum What functional groups are present from the IR ? Count the number of types of C in the 13 C-nmr

Look at the integration in the H-nmr Use the above to get the molecular formula (make sure it matches the MW) Use the coupling patterns in the H-nmr to identify the hydrocarbon "pieces"


In the MS, the molecular ion occurs at m/z = 86 indicating the MW = 86 g /mol.

The IR shows an C=O (1715 cm-1) and but no -OH or -NH (around 3500

cm-1) or C-O (1250-1000 cm-1) 13 C-nmr shows 5 types of C, including a C=O (210 ppm), and 4 other

hydrocarbon C (55, 34, 22 and 18 ppm) H-nmr


2.4 triplet 2 CH2coupled to 2H, slightly deshielded


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2.1 singlet 3 isolated CH3

1.7 hextet 2 CH2coupled to 5 H


The following pieces can be deduced from the H-NMR spectra 2 different -

CH3, and 2 different -CH2-. Given these pieces, we can start to fit them together based on the coupling

patterns: an uncoupled CH3- and a CH3-CH2-CH2- group.

Summary....

The MS indicated MW = 86 g/mol. The IR showed the presence of a C=O bond.

13C peak at 210 ppm suggests that the C=O is an aldehyde or ketone. H nmr gives an uncoupled -CH3 and a -CH2-CH2-CH3 group. Note we can rule out it being an aldehyde (no peak at 9-10 ppm in the H-

NMR). Use what we have to check the molecular formula : C5H10O = 5 x 12 + 10 x

1 + 1 x 16 = 86 g/mol

This has an IHD = 1, and is consistent with the bond of the C=O.

Altogether....

With the pieces we have : C=O, -CH2-CH2-CH3, and -

CH3.

IR and 13C is consistent with a ketone.


together:

With the H-NMR chemical shifts we can see the CH3-

C=O

Checking the H-NMR, the middle -CH2- is actually six

lines since it has a total of 5 similar neighbours. The C=O

slightly deshields the other adjacent groups.

2-pentanone

or

propyl methyl ketone


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Find the molecular weight from molecular ion in the mass spectrum


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What functional groups are present from the IR ? Count the number of types of C in the 13 C-nmr

Look at the integration in the H-nmr Use the above to get the molecular formula (make sure it matches the MW)

Use the coupling patterns and chemical shifts in the H-nmr to identify the"pieces"


In the MS, the molecular ion occurs at m/z = 88 indicating the MW = 88g/mol (even, no isotope pattern for Cl or Br)

The IR shows a carbonyl C=O (1740 cm-1), and possible C-O (between

1250-1000 cm-1). No -OH or -NH (above 3200 cm-1). 13 C-nmr shows 4 peaks indicating 6 types of C, a C=O (171 ppm, probably

an acid derivative), a C-O at 60 ppm and 2 hydrocarbon C (21 and 14 ppm) H-nmr


4.1 quartet 2 CH2coupled to 3H, deshielded by O ?

2.0 singlet 3 CH3with no adjacent H, slightly deshielded


The most significant structural information in the H nmr data is that we have

an ethyl group : -CH2-CH3 most likely as an ethoxy group: -O-CH2-CH3andan isolated methyl group : -CH3

Summary....

The MS indicated MW = 88 g/mol. The IR showed the presence of C=O and C-O bonds, possibly an ester

13C peak at 171ppm suggests that the C=O is a carboxylic acid derivative,

an ester ? H nmr also gives a -O-CH2-CH3and an uncoupled -CH3.

Use this to check the molecular formula : C4H8O2= 4 x 12 + 8 x 1 + 2 x 16= 88 g/mol

So with all this information we have the following pieces: C=O, -O-CH2-

CH3, and -CH3.

Altogether....

With the pieces we have : C=O, -O-CH2-CH3, and -CH3.

IR and 13C suggest an ester so we have : CO2CH2-CH3

This only leaves the -CH3 group.

Therefore, it must be CH3-CO2CH2-CH3


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ethyl ethanoate

or

ethyl acetate


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Find the molecular weight from molecular ion in the mass spectrum What is the significance of the two peaks at 122 and 124 m/z in the MS ?

What functional groups are present from the IR ? How many types of C in the C nmr ?

How many H in the H nmr (look at the integration) ? Use the above to get the molecular formula (make sure it matches the MW) What hydrocarbon chain does the H-nmr indicate ?


In the MS, the pair of peaks of about 1:1 ratio at 122 and 124 suggest Br dueto the isotopes79Br and81Br

The molecular ion occurs at m/z = 122 indicating the MW = 122 g / mol(based on 79Br)

The IR shows has no absorbtions for the main functional groups, looks likeC-H and C-C only.

13 C-nmr shows 3 types of C, including a slightly deshielded C (35 ppm)

plus 2 other hydrocarbon C (26 and 13 ppm) H-nmr


3.4 triplet 2 CH2deshielded, coupled to 2H


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1.8 sextet 2 CH2coupled to 5H

1.0 triplet 3 CH3 coupled to 2H

Summary....

The MS indicated MW = 122 g/mol. The H nmr gives a total of 7 H, C nmr 3 C Use this to check the molecular formula : C3H7Br = 3 x 12 + 7 x 1 + 1 x 79

= 122 g/mol This confirms that we have the molecular formula

Altogether....

The coupling patterns in the H nmr are the critical

information

The patterns give a CH3next to a CH2and a CH2next toanother CH2

Since we only have 3C, this means we have a CH3CH2CH2-

With the pieces we have : CH3CH2CH2- and -Br

This can only be put together to give CH3CH2CH2-Br1-bromopropane


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What functional groups are present from the IR ? What do the peaks at 7-8 ppm in the H-nmr and at 125-145 ppm in the 13C-

nmr reveal ? Look at the integration in the H-nmr...

Use the above to get the molecular formula (make sure it matches the MW) What does "exchanges with D2O" tell you in the H-nmr ?


In the MS, the molecular ion occurs at m/z = 108 indicating the MW = 108g/mol (even, no isotope pattern for Cl or Br)

The IR shows an O-H (broad 3330 cm-1) and a C-O at 1020 cm-1. There is

no carbonyl C=O (around 1715 cm-1) 13 C-nmr shows 5 types of C, including 4 ArC between 125-145 ppm and a

deshielded C (65 ppm) H-nmr


7.2-7.4 "singlet" 5 5 ArH = monosubstituted

4.6 singlet 2 deshielded CH2no coupling

2.3 singlet 1 H no coupling (-OH ?)

The H-NMR gives us a monosubstituted benzene ring, C6H5-, a -CH2- and a

peak that could match the -OH seen in the IR.

Summary....

The MS indicated MW = 108 g/mol. The IR gave us an -OH but no C=O, so looks like some type of alcohol. The 13C and H NMR show a benzene system C6H5-.

The H NMR gives another -CH2- Use this to check the molecular formula : C7H8O = 7 x 12 + 8 x 1 + 1 x 16 =

108 g/mol

This confirms that we have the molecular formula

Altogether....

With the pieces we have : C6H5-, -CH2- and -OH

These can actually only be put together in one way....

benzyl alcohol

or

phenylmethanol


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nmr reveal ? Look at the integration in the H-nmr

Use the above to get the molecular formula (make sure it matches the MW) Count the number of types of C in the 13 C-nmr. What does this tell you ?

(particularly those between 120-180ppm)

What hydrocarbon pieces the H-nmr indicate ? Using the chemical shifts how are these pieces connected together ? Symmetry ?


In the MS, the molecular ion occurs at 164 indicating the MW = 164 g/mol(even, no isotope pattern for Cl or Br)

In the IR there are significant absorptions at 1720cm -1due to a C=O and at

1600cm-1 due to aC=C, C-O possible due to bands at 1275 and 1100cm -1.There is no OH (about 3500cm-1).

13 C-nmr shows 8 types of C. By analysis of the chemical shifts, we have a

C=O (probably an acid derivative) at 167 ppm, 4 types in the region 125-

145 ppm is typical of aromatic C, 61 ppm is typical of a C-O and those at 22

and 14 ppm most likely from hydrocarbon. H-nmr


7.8 "doublet" 2 Ar-H, must be disubstituted, most likely

para7.3 "doublet" 2

4.3 quartet 2 CH2coupled to 3H, deshielded by O ?

2.4 singlet 3CH3with no adjacent H, slightly

deshielded



a disubstituted phenyl group, an ethyl group : -CH2-CH3 most likely as anethoxy group: -O-CH2-CH3and an isolated methyl group : -CH3

Summary....

The MS indicated MW = 164 g/mol.


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The IR gave us the C=O and the C-Owhich the C nmr suggests is an acid

derviative, such as an ester rather than a ketone (typically > 200ppm).

13C peak at 167 ppm suggests that the C=O is a carboxylic acid derivative,an ester ?

Together the IR and 13C NMR suggests it is an acid derviative, such as anester rather than a ketone (typically > 200ppm). The H nmr gives us -O-CH2-CH3, -CH3and C6H4-

Use all the pieces to check the molecular formula = C10H12O2= 10 x 12 + 12x 1 + 1 x 16 = 164 g/mol

The molecular formula of C10H12O2implies an IHD = 5.

Altogether....

So with this information we have the following pieces: C=O, -O-

CH2-CH3, -CH3and C6H4-.IR and 13C suggest an ester so we have : CO2CH2-CH3The -CH2- in the ethyl group is deshielded by O and coupled to the -

CH3, while the other -CH3is isolated and only slightly deshielded

by the phenyl ring. The number of types of ArC (4) and the coupling

in the Ar region of the H nmr (7-8ppm) suggests the para

substitution pattern.

ethyl p-toluate

or

ethyl 4-methylbenzoate


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What functional groups are present from the IR ? What type of C=O does the peak at 171 ppm in the 13C-nmr suggest ?

What do the peaks at 4.5-7 ppm in the H-nmr, at 95-150 ppm in the 13C-nmrand at 1650cm-1in the IR reveal ?

Look at the integration and coupling patterns in the H-nmr to make sure

Use the above to get the molecular formula (make sure it matches the MW) What hydrocarbon pieces the H-nmr indicate ? Use the H nmr chemical shifts to decide which part is next to the C=O


In the MS, the molecular ion occurs at m/z = 114 indicating the MW = 114 g

/ mol (even, no isotope pattern for Cl or Br) The IR shows a carbonyl C=O (1770 cm-1), C=C (1650cm-1) and possible

C-O (between 1200-1000 cm-1). No -OH or -NH (above 3200 cm-1). 13 C-nmr shows 6 peaks indicating 6 types of C, a C=O (171 ppm, probably

an acid derivative), 2 types in the region of for C=C, one at141 ppm the

other at 97 ppm plus 3 other hydrocarbon C (61, 18 and 14 ppm) H-nmr



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7.4doublet of

doublets1 =CH coupled to 2 different H

4.9 doublet 1 =CH coupled to H

4.6 doublet 1 =CH coupled to H

2.4 triplet 2CH2coupled to 2H, slightlydeshielded

1.7 sextet 2 CH2coupled to 5H



an monosubstituted alkene HC=CH2(not a benzene system), a propyl group: -CH2-CH2-CH3

Summary....

The MS indicated MW = 114 g/mol. The IR showed the presence of C=O, C=C and C-O bonds.

13C peak at 171 ppm suggests that the C=O is a carboxylic acid derivative,

probably an ester : RCO2R' H nmr gives a monosubstituted alkene HC=CH2, and -CH2-CH2-CH3. Use these pieces to check the molecular formula : C6H10O2= 6 x 12 + 10 x 1

+ 2 x 16 = 114 g/mol

So with all this information we have the following pieces: -CO2-, -CH2-

CH2-CH3, and HC=CH2.

Altogether....

We have : -CO2-, -CH2-CH2-CH3, and HC=CH2.

IR and 13C suggest an ester so we have : RCO2R'

The key is to sort out which group is connected to the

carbonyl C=O and which is connected to the -O- of the

ester....

The -CH2- at 2.4 ppm is not deshielded enough to be -

OCH2- (nearer to 4 ppm), so we must have CH3-CH2-

CH2-C=O and O-HC=CH2these connect to give the

answer.

vinyl butanoate

orethenyl butanoate


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nmr reveal ? Look at the integration in the H-nmr

Use the above to get the molecular formula (make sure it matches the MW) Count the number of types of C in the 13 C-nmr. What does this tell you ?

(particularly those between 120-180ppm)

What hydrocarbon pieces the H-nmr indicate ? Using the chemical shifts how are these pieces connected together ? Symmetry ?


In the MS, the molecular ion occurs at m/z = 150 indicating the MW = 150 g/ mol.

The IR shows an C=O (1680 cm-1) and possible C-O (1250-1000 cm-1) but

no -OH or -NH (around 3500 cm-1) 13 C-nmr shows 7 types of C, including a C=O (196 ppm), and 3 ArC (163,

131, 130 and 114 ppm) plus 2 other hydrocarbon C (55 and 26 ppm) H-nmr


8.0 doublet 2 2 x ArH

7.0 doublet 2 2 x ArH

3.9 singlet 3 isolated CH3 deshielded, possibly by O

2.6 singlet 3 isolated CH3slightly deshielded

The lack of a peak at about 9 ppm rules out an aldehyde and thereforesuggests a ketone.

The peaks at 7-8 ppm in the H-nmr and the 4 peaks in the aromatic region ofthe 13C-nmr suggests a para-disubstituted benzene This gives the following pieces C6H4, C=O, and 2 different -CH3

Summary....

The MS indicated MW = 150 g/mol. The IR showed the presence of C=O and maybe C-O bonds. The13C C=O at 196 ppm is a little high for an ester, and is more indicative

of an aldehyde or a ketone.

H nmr gave the following pieces C6H4, C=O, and 2 different -CH3


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Leads to a suggested molecular formula of C9H10O = 9 x 12 + 10 x 1 + 1 x

16 = 134 g/mol.

This is 16 short, suggesting an extra O, which if we look at the H-NMR isprobably as part of -OCH3(peak at 3.9 ppm).

So with all this information we have the following pieces: C6H4, C=O, -CH3 and -OCH3

Altogether....

With the pieces we have : C6H4, C=O, -CH3 and -OCH3

IR and 13C suggest a ketone rather than an ester.

H-nmr and 13C-nmr suggest a para-disubstituted benzene

Checking the H-nmr, we can see the two uncoupled methyl groups,

one deshielded by O, the other slightly deshielded by the C=O and the

para-disubstituted benzene.

p-methoxyacetophenone


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What type of C=O does the peak at 167 ppm in the 13C-nmr suggest ? What do the peaks at 4.5-7 ppm in the H-nmr, at 120-140 ppm in the 13C-

nmrand at 1650cm

-1

in the IR reveal ? Look at the integration in the H-nmr to make sure Use the above to get the molecular formula (make sure it matches the MW)

What hydrocarbon pieces the H-nmr indicate ? Use the H nmr chemical shifts to decide which part is next to the C=O


In the MS, the molecular ion occurs at m/z = 114 indicating the MW = 114

g/mol (even, no isotope pattern for Cl or Br)

The IR shows a carbonyl C=O (1720 cm-1

), C=C (1640 cm-1

) and possibleC-O (between 1200-1000 cm-1). No -OH or -NH (above 3200 cm-1).

13 C-nmr shows 6 peaks indicating 6 types of C, a C=O (167 ppm, probably

an acid derivative), 2 types in the region of 125-140ppm for C=C, a C-O at61ppm and 2 hydrocarbon C (18 and 14 ppm)

H-nmr


6.1 "multiplet" 1 C=C-H

5.5 "multiplet" 1 C=C-H4.2 quartet 2 CH2coupled to 3H, deshielded by O ?

1.9 singlet 3 CH3with no adjacent H, slightly deshielded


The most significant structural information in the H nmr data is that we havean disubstituted alkene (not a benzene system), an ethyl group : -CH2-

CH3 most likely as an ethoxy group: -O-CH2-CH3and an isolated methyl

group : -CH3

Summary....

The MS indicated MW = 114 g/mol. The IR showed the presence of C=O, C=C and C-O bonds. 13C peak at 167 ppm suggests that the C=O is a carboxylic acid derivative H nmr also gives a disubstituted alkene, a -O-CH2-CH3and an uncoupled -

CH3. Use this to check the molecular formula : C6H10O2= 6 x 12 + 10 x 1 + 2 x

16 = 114 g/mol


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So with all this information we have the following pieces: C=O, -O-CH2-

CH3, -CH3and H-C=C-H or H2C=C.

Altogether....

With the pieces we have : C=O, -O-CH2-CH3, -CH3and H-C=C-H orH2C=C.

IR and 13C suggest an ester so we have : CO2CH2-CH3

This means the two substituents on the alkene C=C are the ester group

and the -CH3

There are three possible arrangements: 1,1- , cis-1,2- or trans-1,2-

If it were 1,2- then the methyl group would have to have an H

neighbour and therefore appear as a doublet (i.e.in a CH3-CH=

system).

Therefore, it must be the 1,1-isomer where is has no adjacent H.....

ethyl methacrylate

or

ethyl 2-methylpropenoate


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http://orgchem.colorado.edu/Spectroscopy/Problems/Problems.html

Show Unsaturation answer

C5H10O

Rule 2, omit O, gives C5H10

5 - 10/2 + 1 = 1degree of unsaturation.

Look for 1 pi bond or aliphatic ring.

Show IR answer

The band at 1716 indicates a carbonyl, probably a ketone. The bands at

3000-2850 indicate C-H alkane stretches.
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Show Structure answer

This is the structure. See if you can assign the peaks on your own.

Show NMR answer

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C7H14ORule 2, omit O, gives C7H14



Show IR answer


3000-2850 indicate C-H alkane stretches.



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Show NMR answer

Note that D and E are chemically similar enough that their peaks overlap.


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Problem 3

Formula: C4H10O

Spectroscopy Reference
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C4H10O


4 - 10/2 + 1 = 0degrees of unsaturation.

No pi bonds or rings.

Show IR answer

The broad band at 3339 indicates an O-H stretch, probably an alcohol.

The bands at 3000-2850 indicate C-H alkane stretches. The band at 1041

is C-O stretch, consistent with an alcohol.



Show NMR answer

A and B do not couple because of A's ability to H-bond.


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Problem 4

Formula: C6H14O



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C6H14O




Show IR answer

The broad band at 3350 indicates O-H stretch, probably an alcohol. The

bands at 3000-2850 indicate C-H alkane stretches. The bands from 1320-

1000 indicate C-O stretch, consistent with an alcohol.



Show NMR answer

As noted, C and D are not equivalent even though they're on the same

carbon.


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C4H8O2Rule 2, omit O, gives C4H8


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Show IR answer

The band at 1743 indicates a carbonyl, probably a saturated aliphatic

ester. The bands at 3000-2850 indicate C-H alkane stretches. The bands

in the region 1320-1000 could be due to C-O stretch, consistent with an

ester.



Show NMR answer

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Problem 6

Formula: C5H10O2



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C5H10O2Rule 2, omit O, gives C5H10


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Show IR answer

The band at 1740 indicates a carbonyl, probably a saturated aliphatic

ester. The bands at 3000-2850 indicate C-H alkane stretches. The bands

in the region 1320-1000 could be due to C-O stretch, consistent with an

ester.



Show NMR answer

C has a higher chemical shift than D because it's closer to a moreelectron-withdrawing functional group.


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Problem 7

Formula: C8H14O3



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C8H14O3



Look for 2 pi bonds or aliphatic rings, or 1 of each.

Show IR answer

The bands at 1745 and 1716 indicate that there are two carbonyls,

probably an aliphatic ester and an aliphatic ketone. The bands at 3000-2850 indicate C-H alkane stretches.




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Show NMR answer

Even though A, B, C and D are all 2H peaks, they can be distinguished by

chemical shift and splitting. B is outside the normal range for protons next

to carbonyls, because it's adjacent to both carbonyls and the combined

deshielding is higher than normal.


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Problem 8

Formula: C5H10O



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C5H10O




Show IR answer

The band at 1727 indicates a carbonyl, probably an aldehyde; analdehyde is also suggested by the band at 2719 which is likely the C-H

stretch of the H-C=O group. The bands at 3000-2850 indicate C-H alkane

stretches.



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Show NMR answer



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Problem 9

Formula: C11H14O2



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C11H14O2



Look for an aromatic ring, plus 1 other pi bond or aliphatic ring.

Show IR answer

The band at 1603 or 1894 might indicate a carbonyl outside the normal

range; the small peak around 1800 indicates it may be an aldehyde. The

band at 1510 is an aromatic ring. The bands at 3000-2850 indicate C-H

alkane stretches.



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Show NMR answer

Based on what is covered, you shouldn't need to distinguish between B

and C, but you should be able to tell that the ring is para-substituted.


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Problem 10

Formula: C3H6O2



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C3H6O2



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Show IR answer

The band at 1716 indicates a carbonyl. The wide band from 3300-2500 is

characteristic of the O-H stretch of carboxylic acids. The bands at 3000-

2850 indicate C-H alkane stretches. The bands in the region 1320-1000

indicate the C-O stretch of carboxylic acids.



Show NMR answer



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Problem 11

Formula: C8H8O2



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C8H8O2



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Look for an aromatic ring, plus another pi bond or aliphatic ring.

Show IR answer

The band at 1689 indicates a carbonyl, probably an acid. The broad dip at

2924 indicates the OH group of the acid. The band at 1589 indicates an

aromatic ring.



Show NMR answer

You are not expected to tell the aromatic Hs apart from each other, butyou should be able to tell that the ring is not para-substituted.


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Problem 12

Formula: C7H9N



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C7H9N

Rule 3, omit the N and one H, gives C7H8


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Show IR answer

The two bands at 3433 and 3354 indicate a primary amine (-NH2). The

bands at 3000-2850 indicate C-H alkane stretches. The band at 3034

indicates aromatic C-H stretch; aromatics also show bands in the regions

1600-1585 and 1500-1400 (C-C in-ring stretch), and 900-675 (C-H out-

of-plane). C-N stretch of aromatic amines would show up at 1335-1250

(there is a band in that region).



Show NMR answer

You are not expected to tell the aromatic Hs apart from each other, but

you are expected to know that the ring is not para-substituted.



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Problem 13

Formula: C5H13N



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C5H13N




Show IR answer

The two bands at 3388 and 3292 indicate a primary amine (-NH2). The

bands at 3000-2850 indicate C-H alkane stretches.




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Show NMR answer

Note that C is showing as 2 2H peaks, instead of a 4H peak. This may be

caused by a similar effect as in Problem 4 - on each carbon C, the H

closer to the N will be in a slightly different environment. If the rotation of

this carbon is hindered, these Hs may have a different chemical shift.


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Problem 14

Formula: C8H14O



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C8H14O



Look for any 2 of pi bonds or aliphatic rings.

Show IR answer


3000-2850 indicate C-H alkane stretches. Since the compound is an

alkene, one would expect to see C=C stretch at 1680-1640; these weak

bands are not seen in this IR (according to Silverstein, "the C=C

stretching mode of unconjugated alkenes usually shows moderate to

weka absorption at 1667-1640"). Since the compound is an alkene, C-H


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stretch should appear above 3000 (not seen: the absorption for this

single hydrogen must be too weak).



Show NMR answer


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Here, the two E methyl groups have different chemical shifts because

they are permanently locked into different positions with respect to the

alkene.


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Problem 15

Formula: C4H9Br



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No pi bonds and no rings.

Show IR answer

The bands at 3000-2850 indicate C-H alkane stretches. The bands in the

region 1300-1150 could indicate C-H wag (-CH2Br) of an alkyl halide.



Show NMR answer



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Problem 16

Formula: C10H14O



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Problem 17

Formula: C13H10O3



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Problem 18

Formula: C8H14

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C8H14


Look for any 2 of pi bonds or aliphatic rings.

Show IR answer

The bands at 3000-2850 indicate C-H alkane stretches. There really aren't

many other bands in the spectrum to indicate functional groups. The

compound is an alkyne; we would expect to see a carbon-carbon triple

bond stretch at 2260-2100, however, this is a weak band at best and

often does not show up on IR.



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Show NMR answer

The integral values for A, B, and C need to be multiplied by 2 to get atotal of 14 protons in the molecule. By convention, integral values on an

NMR spectrum are reported as the lowest common multiple.


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Problem 19

Formula: C9H13N



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C9H13N



Look for an aromatic ring.

Show IR answer

The bands at 3000-2850 indicate C-H alkane stretches. The band at 3028indicates C-H aromatic stretch; aromatics also show bands in the regions


of-plane). The bands in the region 1250-1020 could be due to C-N

stretch. The weak, broad banc at about 3500 could be amine N-H stretch

or it could be a slight contamination of an impurity (water) in the sample.



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Show NMR answer



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Problem 20

Formula: C9H10



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C9H10




Show IR answer

The bands at 3000-2850 indicate C-H alkane stretches. The band at 3060

indicates C-H aromatic stretch; aromatics also show bands in the regions


of-plane).



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Show NMR answer

Again, the integral values need to be multiplied by two to give the right

number of Hs.

http://orgchem.colorado.edu/Spectroscopy/Spectroscopy.html
http://orgchem.colorado.edu/Spectroscopy/Spectroscopy.htmlhttp://orgchem.colorado.edu/Spectroscopy/Spectroscopy.htmlhttp://orgchem.colorado.edu/Spectroscopy/Spectroscopy.html


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CHEM 39

Answers to Combined Spectral Problems:

Unknown StructureAlternative

structureStudy questions

A

How do you know that we

haveparadisubstituted benzene?

How can you recognize presence of the

ethyl group?

Can the carbonyl groups of an ester and

amide be easily distinguished?



B

Does the presence of the two carbonyl

peaks in IR mean that we have two

nonequivalent carbonyl groups? Does the MS allows one to distinguish the

two possibilities?

Could you propose a structure just on the

basis of the molecular formula and 13C

NMR spectrum?



C

What kinds of carbons are present based on the

proton and carbon NMR spectra? If the clue about carbon connectivity was not given,

could you distinguished between the two structures? What is the spectral evidence for the presence of

chlorine?


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D

What is the evidence for the presence of bromine? How do you know that that there is a CH2group

present?

Could you decide the position of the Br on thedouble bond based on the

1H NMR spectrum?

Could you tell theEandZisomers apart for the

alternative structure?



E

What does the

13C spectrum tell you about symmetry

in this molecule?

What type of functional group is present based on the

1735 cm-1

band in the IR spectrum? Both the

1H and off-resonance decoupled

13C spectra

indicate the presence of a -CH2- group. How do they

do that? How can you distinguish between the alternative

structures based on1H spectrum?



F none

The molecule has 9 carbons. Why there are onlythree

13C resonances?

Show how this substitution pattern is consistent

with1H spectrum.

Is there another way to arrange three methyls on

an aromatic ring to give a compound that willhave only three carbon resonances?



G

How can you tell the difference between the phenyl

and alkenyl carbons? How can you tell the difference between the phenyl

and alkenyl protons? How do you know that the phenyl ring is

monosubstituted? If the alternative structure were right, could you

distinguish between theE andZisomers?


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H

What MS pattern tell you about the presence of two

bromine atoms? How can you tell that the phenyl ring

isparadisubstituted (and not ortho,as suggested in

the alternative structure)? List all the spectral evidence that is inconsistent with

the alternative structure.



I

Can the presence of the nitro group be deduced from

the IR spectrum? Can you deduce presence of the propyl group

from1H spectrum alone?

Would an isopropyl group produce the same1H

spectrum? Do you know a way to distinguish between the

alternative structures for the functional group (-

NO2vs. -ONO)?



J

What would the13

C spectrum for the alternative

structure look like? What would the

1H spectrum for the alternative

structure look like?

What condition would have to be satisfied for bothcompounds to show identical13

C and1H spectra?


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Solving Spectral Problems November 21, 1997

The following is an example of the typicalsort of approach that one uses to solve

spectral problems. What you should learn from it is a sense of the logicthat'sinvolved, not the order in which you consider the data (especially that in the nmr

spectrum). It is most important that you work on developing yourowngeneralapproach to solving a problem - as usual, this can only be

accomplished through practice. There are lots of opportunities for this between the

problems in Chapter 11 of Ege, Problem Set 9, and Assignment 3.

First, qualitativelyexamine allthe data given, and extract any information that

"jumps out" at you. Then proceed to a more detailed inspection of the data.

The infrared and 1H nmr spectra of a compound with molecular formula

C10H12O2are shown below. Deduce the structure of the compound from these data.

Note that in the nmr spectrum, the integration has been done for you; the numberof protons responsible for each of the "signals" is indicated right above it in the

spectrum.


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1.

Molecular

Formula

C10H12O2-> 5 sites of unsaturation. For a molecule this size, (i.e., only ten

carbons), such a large number of sites of unsaturation is usually due to the

presence of a phenyl ring (which provides 4sites on its own). For a quick proof,

take a quick glance at the 1H nmr spectrum, and look for signals in the 7-8 ppm

(aromatic proton) region. No question, they're there! This leaves only one

siteunaccounted for.

2.

Infrared

Spectrum

- strong peak at ~1700 cm- -> C=O (this accounts for the other site of

unsaturation!)

- nothing at ~3500 cm-1-> no OH, so a carboxylic acid or alcohol can be ruled

out.

- weak absorption in the 3000-3100 cm-1region indicative of sp2C-H stretches,

in addition to prominent bands in the 2800-3000 cm-1range due to sp3C-H.

There is lots more information here, but this is the easiest to pull out, and all we

need for the present.

H NMR

spectrum- 6 types of H's:

- singlet at 9.8 ppm (1H)-> aldehyde or carboxylic acid proton. The ir spectrum

rules out a carboxylic acid, so this must be an aldehyde. The absence of couplingindicates that the -CHO group is bonded to an atom with no protons attached to

it.

- two doublets at ~6.9 and ~7.7 ppm (4H)-> aromatic protons; symmetrical

pattern of two doublets suggests a 1,4-disubstituted benzene. Each doublet

corresponds to 2 equivalent protons, each of which has one nearest neighbour:


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Note that the two doublets are skewed towards one another and that the

magnitude of the splitting is the same for both - this verifies that the two

doublets "belong together"; i.e., they represent a pair of protons coupled to each

other.

We've accomplished alot; check the molecular formula to see what's left! We've

accounted for a -CHO group and a -C6H4-. All that's left is three carbons

bearing 7 hydrogens and an oxygen. Since all sites of unsaturation are

accounted for, the seven H's must all be bound to sp3carbons.

- triplet at ~4 ppm (2H)-> two equivalent protons with 2 nearest neighbours.

With only 3 carbons left, this must be due to a CH2attached to a CH2!

- triplet at 1 ppm (3H)-> three equivalent protons with 2 nearest neighbours.

This must be due to a methyl group attached to a CH2.

The last two bits of information, coupled with the fact that we have only

C3H7left, are onlyconsistent with one structure - CH2CH2CH3!

- sextet at ~1.8 ppm (2H)-> two equivalent protons with 5 nearest neighbours.

This is the middle CH2in the -CH2CH2CH3chain.

Let's take stock of what we have:


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There are onlyTWOpossible ways that these fragments can be put together:

To choose between the two, we need to take a closer look at the nmr spectrum andask "how will the spectra of these two compounds differ?" Oneset of protons will

have a significantly different chemical shift in the two structures. (Sometimes,

more than one set can be identified as being significantly different; sometimesnone can, in which case we would need more data). This is the terminal

CH2groupin the CH2CH2CH3chain. In A, it's attached to oxygen and shouldcome at around 4 ppm (look at the table of "representative chemical shifts" in Ege).In B, it's attached to an sp2carbon and should come at 2-2.5 ppm. Clearly,

structure A is the one that fits the spectrum the best.

Go to:Instructions for Printing this Document

Chem2O6 Problem Sets & AnswersChem2O6 Home Page.

21nov97; wjl

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