7th solutions manual for spectrometric identification of organic compounds

Answers for Student Exercises 1.6 to 1.11 9Base Peak Molecular Ion Fragement Ion Fragment Loss Rearrangment... Answers for Student Exercises 1.6 to 1.11 10Base Peak Molecular Ion Frage

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Answers for Student Exercises 1.1 to 1.5 1

Atom Type # of Atoms Mass Total Mass Hydrogen 22 1.00783 22.17226Carbon 10 12.00000 120.00000 Nitrogen 0 14.0031 0.00000

CnHmXxNyOz IHD Bromine 0 78.9183 0.00000Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 0 Exact Mass 142.17226

a

3-ethyloctane

C10H22

- (C2H5 )Rule 3

- (C5H11 )Rule 3

- (CH2=CH2)Rule 9H

or

- (C4H9 )Rule 5 and 6

- (CH2=CH2)Rule 6

- (C3H7 )Rule 3

Atom Type # of Atoms Mass Total Mass Hydrogen 18 1.00783 18.14094 Carbon 10 12.00000 120.00000 Nitrogen 0 14.0031 0.00000

C n H m X x N y O z IHD Bromine 0 78.9183 0.00000Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 2 Exact Mass 138.14094

b

3-butylcyclohex-1-ene

C10H18

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- (C6H13)Rule 5H

- (C7H14)Rule 9

O c

2-methyl-1-phenylhexan-1-one

C13H18O

O

- (C6H13 )Rule 8

O

- (C6H5 )Rule 8

O

4H8)Rule 9

OH

O

- (CO)Rule 8+9

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e

1-(hexan-2-yl)benzene

C12H18

- (CH3 )Rule 7

- (C4H9 )Rule 7

- (C4H8 )Rule 9

H

OHf

- (CH3 )Rule 8

- (H )Rule 8

OH

OHH

OH

2O)Rule 9

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COOHg

- (C9H18)Rule 9

CO

OH

O

COH

O

COHO

OHh

- (C5H11 )Rule 8

- (H )Rule 8

H2COH

O

HH

OH

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OHi

Sulfur 1 31.9721 31.97210

Sj

butyl(propyl)sulfane

C7H16S

S

- (C3H7 )Rule 8

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OHk

(2-hexylphenyl)methanol

C13H20O

CH2OH

CH2

- (C5H11 )Rule 7

- (H2O)Rule 7

- (OH )

Rule 9

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O m

H O

Br

- (H )Rule 8

Br

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H

o

C8H14O2

OOcyclohexyl acetate

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Base Peak

Molecular Ion Fragement Ion Fragment Loss Rearrangment

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Base Peak

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Base Peak

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Base Peak

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Base Peak

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Base Peak

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Answers for Student Exercise 2.1 – 2.9 1

c diphenyl sulfone

OO

C11H23

OHOLauric acid

Na+

O

OSodium propionate Spectrum E Spectrum H Spectrum I Spectrum F Spectrum G

-OAllyl phenyl ether

O

BenzaldehydeH

HOo-Cresol

OHO

m-Toluic acid

2.1 Since the indicated carbon of phenylacetonitrile is sp3 hybridized, it is reasonable for this compound to show C−H stretching at less than 3000 cm-1 (2960-2940 cm-1) Where as benzonitrile has only aromatic C−H stretching which is typically between 3100 - 3000 cm-1 (page 87)

1,3-Cyclohexadiene Diphenylacetylene 1-Octene 2-Pentene

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H2N

Aniline

NNAzobenzene

NHO

Benzophenone oxime

H2N

Benzylamine

NH2 ClDimethylamine hydrochloride Spectrum Q Spectrum O Spectrum P Spectrum N Spectrum R

enol

2.6

2.7 Methyl isothiocyanate CH3−N=C=S

2.8 This diketone exists primarily (~ 90%) in enol form (in CCl4)

Enols such as this display broad, shallow O−H stretching bands (here from 3400-2500 cm-1) The strong band at ~1600 cm-1 is the enolic coupled C=C−OH band (See pages 80, 94, and 98)

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H2NButylamine

2-Phenoxylethanol

OPhenetole

OOMethyl Butyrate

HOO

OMethyl salicylate

O

HN

Caprolactam

NN

2-methylpyrazine

N+O

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Answers for Student Exercise 3.1 to 3.3 1

3.1(a) 1,2,3,4 are a spin system Pople notation is A3M2S2X2 1,2, and 3 are enantiotopic

3.2 (a)

Spin Appr δ Table Coupled with Multiplet (n+1) Integration

3.3 (a)

Br

(a) 1-bromobutane

1 2

3 4

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3.1 (b) 1,2,3,4,5,6,7 are one spin system, and 9 is the second 4,4’ ; 5,5’ ; 6,6’and 7,7’ are

diastereotopic

3.2 (b)

Spin Appr δ Table or Calculation Coupled with Multiplet (n+1) Integration

* Coupling constants may not be the same, but make the assumption that they are

**These are diastereotopic, and are coupled to diastereotopic protons, they would not be first order

We will assume they are first order for the drawn spectrum

8 9(b) (cyclohex-2-enyl)methyl acetate

2

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3.1 (c) 2,3,4,5,6 are a spin system, and 8,9,10 another spin system 3,3’ ; 4,4’ ; 9,9’ are diastereotopic 3.2 (c)

** These are diastereotopic, and are coupled to diastereotopic protons; they would not be first order

We will assume they are first order for the drawn spectrum

3.3 (c)

OH

O1

2 3 4 5 6

7 8

(c) hydroxy-2-methylpropan-1-one

1-(cyclohexa-1,5-dienyl)-3-9

10

2 6 5 9 8 3,4 OH 10

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3.1 (d) 5,4,3 are a spin system and 1 is second 5,4,3 is Pople notation is A3M2X2; may have long rang coupling 3+4 are enantiotopic

3.2 (d)

3.3 (d)

H(d) pent-1-yne

1

3 2

4 5

3 1 4 5

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3.1 (e) 3,4,5,6 are a spin system, and 1 is another spin system Pople notation is A3G2MX 5 is

enantiotopic

3.2 (e)

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3.1 (f) 3,4,5,6 are a spin system, and 1 is another spin system Pople notation is A3G2MX 4 is

enantiotopic

3.2 (f)

3.3 (f)

2 3 1 4 5

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*See Section 3.6.2 for coupling between 1 and N-H

3.3 (g)

NH 3 1 4 5

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3.3 (h)

3 2 1

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HO(i) 2-methylpentan-2-ol

1

2 3 4 56

1 and 6 Large singlet

of 6 H

3.1 (i) 3,4 and 5 is a spin system 1 and 6 are enantiotopic to each other and 3 and 4 are also

enantiotopic Pople notation is A3M2X2

3.2 (i)

3.3 (i)

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S

S(j) 1,4-bis(methylthio)butane

3.1 (j) Not First order (no Pople) See Section 3.11.2 Protons 2 and 3 are enantiotopic Protons on 1,2,3 are Chemical Shift Equivalent

3.2 (j)

* Would not be first order We will assume they are first order for the drawn spectrum

3.3 (j)

Actual spectrum would look like this!

0 1

2

PPM

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NO

(k) N,N,4-trimethylbenzamide

1 2 3 4

5 6

7 8

2' 3'

3.1 (k) 2,2’; 3,3’ are CSE but not ME 6 and7 are diastereotopic Pople notation for ring is AA’ XX’ 3.2 (k)

2, 2’ δ 7.80 Appendix D, Chart D1 none Doublet** 2

* There will be two singlets because of the restricted rotation of the NR2

** These can not be first order Not magnetically equivalent

3.3 (k)

2,2’ 3,3’ 6,7 8

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O

OH(l) 1-(2-hydroxyphenyl)ethanone

1 2 3 4 5

6

7 8

3.1 (i) N.A

3.2 (i)

* Will have long range coupling

3.3 (i)

OH 6 5 4 3 8

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O(m) 2-chloroacetaldehyde

1 2

3.1(m) 1 and 2 is a spin system Pople notation is A2X 2 protons are enantiotopic

3.2 (m)

Spin Appr δ Table Coupled with Multiplet (n+1) Integration

1 δ 4.95 Appendix A, Chart A.1

Plus Appendix B, Table B.1*

2 Triplet 1

2 δ 9.80 Appendix D, Chart D.6

extrapolation

1 Doublet 2 Calc is based on Chart A1 3.45 for ClCH2 plus Table B.1 value of 1.50 for HC=O

3.3 (m)

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H

F(n) fluoroethene

1

2 3

Protocol of the H-1 NMR Prediction:

Node Shift Base + Inc Comment (ppm rel to TMS)

2 3

4 5

6 7

PPM

3.1(n) Charts are not available in text for this problem Pople notation is AGMX, where X is F

3.2 (n)

3.3 (n)

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O

(o) cyclohexyl acetate

O1

2 3 4

5 6

2' 3'

3.1 (o) 1,2,2’,3,3’, and 4 is a spin system, and 6 is another spin system 2,2,;2’,2’;3,3;3’,3’;4,4 protons are diastereotopic 2 and 2’ are CSE, also 3 and 3’ are CSE

3.2 (o)

2,2’ δ 1.60 Appendix A, Chart A.2 1,(3or3’) Quartet* 4 3,3’ δ 1.44 Appendix C, Table C.1 (2 or 2’),4 Quintet* 4

* These are diastereotopic, and are coupled to diastereotopic protons; they would not be first order We will assume they are first order for the drawn spectrum

3.3 (o)

1 6 2,2’ 3,3’, and 4

4

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Answers for Student Exercise 3.4 16

0.80.91.01.11.21.31.41.51.61.71.81.92.02.12.22.3

2.4

1.01.5

2.02.5

3.03.5

2.53.0

3.54.0

4.55.0

5.56.0

6.57.0

1.01.5

2.02.5

1

2 3 4 5

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0.80.91.01.11.21.31.41.51.61.71.81.92.02.12.22.32.4

2.02.22.42.62.83.03.23.43.63.84.04.2

0.80.91.01.11.21.31.41.51.61.71.81.92.02.12.22.32.42.5

2.6

1.52.0

2.53.0

2-Hexanone

1 2 3

OHOPropionic Acid

1 2 3 4

H2NButylamine

1 2

3OH

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1.01.5

2.02.5

3.03.5

4.0

1.01.5

2.02.5

3.0

1.52.0

2.53.0

3.54.0

4.55.0

5.56.0

6.57.0

2.02.5

3.03.5

4.04.5

5.05.5

6.06.5

1 2 3

4

6OOH

2-Phenoxylethanol 1

2 3

5

4 5

6OPhenetole

1 23

5

OOMethyl Butyrate

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1.52.0

2.53.0

3.54.0

4.55.0

5.56.0

6.5

7.37.4

7.57.6

7.77.8

7.98.0

8.1

4.04.5

5.05.5

6.06.5

7.0

3.03.5

4.04.5

5.05.5

6.06.5

7.07.5

8.0

2490 2500

OHN

Caprolactam

1 2 3 4

6

7NN

2-methylpyrazine 1

2

5

6 7

HOO

OMethyl salicylate

1 2

3 4

5 8

N+O

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1.52.0

2.53.0

3.54.0

1.01.2

1.41.6

1.82.0

2.22.4

2.6

1.52.0

2.53.0

3.54.0

4.5

1.61.8

2.02.2

2.42.6

2.83.0

3.23.4

1500 1510

11

1 proton (x32)

Problem 4.3 R C6H10O Unsaturated alcohol

Problem 4.3 Q C7H13BrO2 Acid

Problem 4.3 S C8H14O Unsaturated ketone

5 2

3 41

6

7

8O

6-Methyl-5-hepten-2-one

2 3 4

OHOHexanoic acid

3 42

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5.75.85.96.06.16.26.36.46.56.66.76.86.97.07.17.27.3

7.4

2.02.5

3.03.5

4.04.5

5.05.5

6.06.5

2.53.0

3.54.0

4.55.0

5.56.0

2,6-Dichlorophenol

1 3 4 2

5HO

2,6-Dimethylphenol

3 4

1 2

O

2-Cyclohexen-1-one

3 4

1 2 5

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3.5A (1.58(300) – 0.90(300))/7 = about 29; (2.36(300) – 1.58(300))/7 = about 33

3.5E (2.38(300) – 1.52(300))/7 = about 37; (1.52(300) – 1.38(300))/7 = about 6

(1.38(300) – 0.87(300))/7 = about 22

3.5F (2.38(300) – 1.13(300))/7 = about 54

3.5G (2.61(300) – 1.35(300))/7 = about 54; (1.35(300) – 1.29(300))/7 = about 2 to 3

(1.29(300) – 0.84(300))/7 = about 19

3.5H Long range coupling; can’t measure coupling constant

3.5I (4.33(300) – 2.01(300))/7 = about 99; (2.01(300) – 0.99(300))/7 = about 44

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Answers for Student Exercises 3.6 23

C

1

2 3 4 5

Two spin systems Pople Notation AA’XX’ 2 and 2’ are CSE but not magnetically equivalent as are

F

A3X2 Protons on two are enantiotopic

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1 2 3 4

H2NButylamine

G

Pople Notation – A3G2M2X2One spin system 1, 2, and 3 are enantiotopic

1 2

3OH

2-Phenoxylethanol

J

1 2 3

5 2'

3'

Two spins systems; Pople Notation AA’MM’X and A2X2 2 and 2’are CSE but not magnetically equivalent as are 3 and 3’ 5 and 6 are enantiotopic

4 5

6OPhenetole

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5 6

OHN

Caprolactam

M

1 2 3 4

One spin system 2,2’; 3,3’; 4,4’; 5,5’ and 6,6’ are diastereotopic

6

7NN

HOO

OMethyl salicylate

O

1 2

3 4

5 8

Two spin systems Pople Notation – AGMX

N+O

4 3' 2'

One spin system 2 and 2’ are CSE but not magnetically equivalent as are 3 and 3’

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