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Problem of spectroscopy

Students: Group 7

Pham Van Trung

Duong Thi Hai Hoa

Teacher: Nguyen Thien Thao

Contents

Mass spectroscopy

Problem 1

Problem 2

Problem 3

Problem 4

Summary

11/15/2010 2 Spectroscopy Methods I

Mass spectrum

Molecular ion peak (M+) m/z corresponds to molecule weight of singly-charged molecule.

In general the following group of compounds will, in order of decreasing ability, give prominent molecular ion peaks: aromatic compounds> conjugated alkenes > cyclic compounds > organic sulfides > short, normal alkanes > mercaptans.

Fragment peak m/z less than molecule weight of singly-charged molecule.

Base peak most intense m/z11/15/2010 3 Spectroscopy Methods I

Mass spectrum


Mass spectrum

Isotopes: present in their usual abundance. Hydrocarbons contain 1.1% C-13, so there will be a

small M+1 peak. If Br is present, M+2 is equal to M+. If Cl is present, M+2 is one-third of M+. If iodine is present, peak at 127, large gap. If N is present, M+ will be an odd number (The

nitrogen rule) If S is present, M+2 will be 4% of M+.


Mass spectrum


Problem 1

A organic compound consists 68.5% C, 4.92% H and 26.23% O.

Following are MS,1H-NMR, 13C-NMR, and IR spectrum for a compound.

Determine the structural formula of organic compound.


Solution

Using for CxHyOzNt

A organic compound consists 68.5% C, 4.92% H and 26.23% O have

We have ratio 3.5: 3:1 and the simple formula is (C3.5H3O)n

% % % %: : : : : : : : : : : :

12 1 16 14C H O N

C H O Nx y z t n n n n a b c d

68.5 4.93 26.23: : 5.7 : 4.93:1.64

12 1 16


Mass spectrumBase peak

M+ +2

M+ +1

M+


-CHO

-OH

Solution

Most intense m/z at 121 is base peak. The presence fragment peak m/z at 77 aromatic

compounds the molecular-ion peak M is strong. We can determine molecular-ion peak m/z at 122

because it is usually the peak of highest mass number except for the isotope peaks.

Molecule weight is 122 (C3.5H3O)n = 122 n = 2

Molecular formula is C7H6O2


Solution

Degree of unsaturated a compound C7H6O2

Look for an aromatic ring in IR spectrum. Also look for a carbonyl, a double bond, or a ring because

the degree of unsaturated is 5

(2.7 2) 65

2S


IR spectrum

O-H stretch

C-H stretch of Aldeheyde

C-H stretch of Aromatic C=O C- C in ring C-O

A compound have Aromatic ring, O-H group and H-C=O group


1H-NMR of spectrum

Aromatic C-H

RCHO

OHPhenol


13C-NMR of spectrum

RCHO

A compound have seven type of carbon one of C=O and six carbon in aromatic ring.

OH

CHO

2-hydroxybenzaldehyde


Problem 2 C10H15N

A organic compound have a molecule formula C10H15N .

Following are IR, 1H-NMR, 13C-NMR, and DEPT spectrum for a compound.

Determine the structural formula of organic compound.


Solution

Degree of unsaturated a compound C10H15N

Look for an aromatic ring in IR spectrum because the degree of unsaturated is 4.

(2.9 2) 124

2S


IR spectrum of C10H15N

C-H stretch of Aromatic

C-H stretch of Alkanes

C- C in ring

C-H bend of Alkanes

C-N stretch of Aromatic amine

C-H oop


Solution

A compound have N and C-N stretch aromatic amines but hasn’t the absorption at 3400 - 3250cm-1 corresponds to a N-H stretch of primary, secondary amines, amide.

A compound is a tertiary amine and have one aromatic ring.

N


1H-NMR of spectrum

A compound have 15 H. From the integration, the hydrogens in the ratio 2:3:4:6

6H

4H

2H3H

AromaticAlkanes


13C-NMR of spectrum

In 13C-NMR of spectrum have 6 types of two in region alkanes and 4 in Aromatic.

Alkanes

Aromatic

abc

d

e

f


DEPT spectrum

CH2


Structure formula

N

N,N-diethylbenzenamine

aa

bb

cc

d ee

f


Problem 3

A organic compound consists 90% C, 10% H. Following are 1H-NMR, 13C-NMR, IR and mass

spectrum for a compound. Determine the structural formula of organic

compound.


Solution

A organic compound consists 90% C, 10% H and have

We have ratio 1:4/3 and the simple formula is (CH4/3)n

90 10: 7.5 :10

12 1


Problem 3: C9H12

Base peak

M+

M++1M+- 15- CH3

M+- 29- C2H5

C6H+5


Solution for test

Most intense m/z at 91 is base peak. The presence fragment peak m/z at 77 aromatic

compounds the molecular-ion peak M is strong. We can determine molecular-ion peak m/z at 120 have

an isotopic peak at (M+ + 1) is 121 Molecular ion peak (M+) m/z = 120 corresponds to molecule weight is 120. Using for(CH4/3)n = 120 n = 9

Molecular formula is C9H12


Solution for test

Degree of unsaturated a compound C9H12

Look for an aromatic ring in IR spectrum because the degree of unsaturated is 4.

(2.9 2) 124

2S


IR spectrum of C9H12

C-H stretch of Aromatic

C-H stretch of Alkanes C-H rock of

Alkanes or C-H oop aromatic

C-H bend of Alkanes

C-C stretch in ring aromatic

A compound have aromatic and anlkanes ( C-H oop aromatic)11/15/2010 28 Spectroscopy Methods I

1H-NMR of spectrum

Alkanes

Aromatic


1H-NMR of spectrum

CH3

CH2-CH2 -CH3Ar-CH2

Aromatic


13C-NMR of spectrum

Alkanes

Aromatic

A compound have seven type of carbon.

1-propylbenzene


Problem 4

A compound: %C = 55.17, %H = 8.06, %O = 36.78 Following are H-NMR, C-NMR and MS for a

compound. Determine the structural formula of organic

compound.


Problem 4

Using for CxHyOz

Ratio 2: 4: 1 the simple formula is

(C2H4O)n

55.17 8.06 36.78: : 4.6 :8.06 : 2.3

12 1 16


Mass of spectrum


M+

M++1

Mass spectrum

Most intense m/z at 129 is base peak. Don’t presence fragment peak m/z at 77 Don’t

have aromatic compounds. We can determine molecular-ion peak m/z at 174 have

an isotopic peak at (M+ + 1) is 175 Molecular ion peak (M+) m/z = 174 corresponds to molecule weight is 174. Using for (C2H4O)n = 174 n = 4

Molecular formula is C8H14O4


Solution

Degree of unsaturated a compound C8H14O4

A compound can has two total carbonyl, and/or double bond, and/or triple bond

(2.8 2) 142

2S


IR C8H14O4

C-H (alkane)

C=O C-O

C-H bend

Alkanes


A compound have C=O and C-H alkanes

There are four types of hydrogen

1H-NMR of spectrum C8H14O4


C8H14O4

R-CH2-CH3


R-CH-CH3

C8H14O4


R-CH2-CH3

C8H14O4


1H NMR show four signals: 6H: triplet 3H: doublet 2H: multiplet 1H: tetraplet 1.21 – 1.26 ppm: RCH2CH3

1.36 – 1.4 ppm: RCHCH3

3.36 – 3.42 ppm: -OOCCHCH3

4.1 – 4.22 ppm: -OCH2CH3

C8H14O4


13C-NMR of spectrum C8H14O4

RCH3

RCH2RR3CH

RCOOR


C8H14O4

13C NMR shows four signals: 13.5, 13.9: RCH3

46.1: RCH2R 61.2: R3CH 170.1: RCOOR


Propose a structure:

C8H14O4


CH3CH2O C

O

CH

CH3

C O

O

CH2CH3

Summary

Six step Analyze the spectra Step 1: Determine the simple formula; Step 2: Using mass spectroscopy determine the molecule formula

of organic compound; Step 3: Calculate the degree of unsaturated a organic compound; Step 4: Using IR spectroscopy detecting certain functional groups

in a molecule; Step 5: Using NMR (1H-NMR and 13C-NMR spectroscopy,

DEPT) gives valuable data about the carbon chain. Step 6: Put this information together to arrive at the following

structural formula.

11/15/2010 Spectroscopy Methods I47

References [1]. Các phương pháp phân tích quang phổ trong Hóa học Hữu cơ. Tiến sĩ

Nguyễn Thanh Hồng, Nhà xuất bản khoa học và kỹ thuật. [2]. GS-TS Khoa học Nguyễn Đình Triệu, Các phương pháp phân tích vật

lý và hóa lý- tập 1 [3]. A Handbook of Spectroscopic Data CHEMISTRY (UV, IR, PMR,

13CNMR and Mass Spectroscopy) B.D. Mistry B.K.M. Science College.Valsad - (Gujarat)

[4]. Brooks Cole - Organic Chemistry 5edition by Brown; Foote; Iverson and Anslyn. Chapter 12, 13 and 14. page 456-540

[5] Dr. Walt Volland, Bellevue Community College All rights reserved 1999, Bellevue, Washington

http://orgchem.colorado.edu/hndbksupport/spectprob/problems.html http://www.chem.ucla.edu/~webspectra/


http://orgchem.colorado.edu/hndbksupport/spectprob/problems.html

http://www.chem.ucla.edu/~webspectra/

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