CH. 12 - STRUCTURE DETERMINATION: MASS SPECT AND...

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CH. 12 - STRUCTURE DETERMINATION: MASS SPECT AND INFRARED SPECTROMETRY

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CONCEPT: PURPOSE OF ANALYTICAL TECHNIQUES

Classical Methods (Wet Chemistry): Chemists needed to run dozens of chemical reactions to determine the type of

molecules in a compound.

EXAMPLE: Tollen’s Test

Instrumental Methods (Dry Chemistry): Expensive scientific instruments investigate the properties of molecules.

EXAMPLE: 1H NMR

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CONCEPT: IR SPECTROSCOPY- GENERAL FEATURES

● IR Spectroscopy is a chemical analytical method that uses differing frequencies of infrared light to detect predictable

types of chemical bonds in molecules.

□ The frequencies will cause certain bonds to _________________

□ Stretching, Twisting, Wagging, Scissoring, etc.

□ If the molecule is symmetrical, e.g. N2, the band is not observed in the IR spectrum.

Major regions of absorption Common IR Ranges

3200 – 3600 -OH Strong, Broad

3300 -NH Peaks = H’s

SP3 = 2900 – 3000

SP2 = 3000 – 3150

SP = 3150 - 3300

-CH Choppy

2200-2300 C≡C

C≡N

Medium, Sharp

1700 C=O Very Strong, Sharp

1650 C=C Medium, Sharp

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CONCEPT: IR SPECTROSCOPY- FREQUENCIES

● There are specific absorption frequencies in the functional group region that we should be familiar with

EXAMPLE: What are the major IR absorptions for each compounds?

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PRACTICE: Answer each of the following questions based on the images below.

O

A

O

B

O

C

OH

H

D

OF3C O

CF3

E

a) Which compounds show an intense peak ~ 1700 cm-1?

b) Which compound shows an intense, broad peak at ~ 3400 cm-1?

c) Which compound has a peak at ~1700 cm-1, but no peaks at 2700 cm-1?

d) Which compound has no signal beyond the fingerprint region?

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PRACTICE: The following compound contains two carbonyl groups. Identify which carbonyl group will exhibit a signal at a lower wavenumber.

O

O

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CONCEPT: IR SPECTROSCOPY- DRAWING HYDROCARBONS

Alkanes:

Alkenes:

Terminal Alkynes:

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CONCEPT: IR SPECTROSCOPY- DRAWING ALCOHOLS AND AMINES

Alcohols:

1° Amines:

2° Amines:

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CONCEPT: IR SPECTROSCOPY- DRAWING SIMPLE CARBONYLS

Ketones:

Esters:

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CONCEPT: IR SPECTROSCOPY- DRAWING COMPLEX CARBONYLS

Aldehydes:

Carboxylic Acids:

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CONCEPT: IR SPECTROSCOPY- DRAWING CONCEALED FUNCTIONAL GROUPS

Alkyl Haldies: Ethers: 3° Amines:

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PRACTICE: Based on IR data given determine the structure of the unknown. Unknown compound A has molecular formula C4H11N. It shows a peak at 2900 cm-1 and peaks in the fingerprint region.

PRACTICE: Based on IR data given determine the structure of the unknown. Unknown compound B has molecular formula C4H11N. It shows a single peak at approximately 3400 cm-1 as well as peaks at 2900 cm-1 and in the fingerprint region. Compound B also possesses a branched alkyl group.

PRACTICE: Based on IR data given determine the structure of the unknown. Unknown compound C has molecular formula C6H10O3. It shows peaks at 2900, 1850 , 1740 cm-1 and in the fingerprint region.

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PRACTICE: Match the following functional group choices with the supplied infrared spectra data

A) Ether B) Ketone C) Alcohol D) Alkene E) Nitrile

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PRACTICE: Match the following functional group choices with the supplied infrared spectra data. A) Alkyl Halide B) Alkyne C) Carboxylic Acid D) Alkene E) Ketone

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PRACTICE: Match the following functional group choices with the supplied infrared spectra data.

A) Aldehyde B) Alkane C) Carboxylic Acid D) Ester E) Ether PRACTICE: Match the following functional group choices with the supplied infrared spectra data. A) Ketone B) Alkyne C) Alkene D) Alkyl Halide E) Amine

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CONCEPT: MASS SPECT- INTRODUCTION

Mass Spectrometry is usually accomplished through a technique called electron impact ionization (EI)

Electrons are beamed at molecules, generating high energy intermediates called radical cations

● This is known as the molecular ion _______ or as the parent ion.

● Only fragment cations are deflected by the magnetic field, the smaller ones more than the bigger ones.

□ Detects the mass-to-charge ratio ______, which means that it detects the MW of cationic fragments

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CONCEPT: MASS SPECT- FRAGMENTATION

Ionization Potentials:

Some electrons require less energy to ionize than others.

Simple Fragmentation Mechanisms:

The molecular ion will often fragment into smaller, sometimes more stable ion fragments.

● The stability of the cation fragment usually determines the relative amounts of fragments observed

● Radicals tend to form on the less stable side of the fragment

Common Splitting Fragments:

EXAMPLE: Fragmentation of Butane

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PRACTICE: Draw the most likely ion fragment for the following molecules

a.

b.

PRACTICE: Propose the molecular ion and likely fragmentation mechanism for the following molecule. What would be the

value of the base peak?

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CONCEPT: MASS SPECT- COMMON ISOTOPES

Isotopes are often visible on a mass spectrum, due to their differing weights. They can be used for structure determination.

Understanding the (M + 1) Peak

1.1% of all carbon is found as 13C, adding a small but distinctive (M + 1) peak proportional in size to the number of carbons.

● This proportion is fairly consistent, so it gives rise to two helpful equations

Understanding the (M + 2) Peak

The halogens –Cl and –Br give distinctive (M + 2) peaks due to their unusual patterns of isotopic abundance

● 35Cl = 75.8% and 37Cl = 24.2%, yielding an approximate 3:1 ratio at (M + 2)

● 79Br = 50.7% and 81Br = 49.3%, yielding an approximate 1:1 ratio at (M + 2)

The Nitrogen Rule

Unlike carbon, nitrogen forms 3 bonds. We can use this information to determine the number of nitrogens in a molecule.

● Even or odd molecular weight of parent ions usually indicates and even or odd number of nitrogens present

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PRACTICE: Propose the number of carbons for a compound that exhibits the following peak in its mass spectrum:

a. (M)+• at m/z = 72, relative height = 38.3% of base peak

(M+1)+• at m/z = 73, relative height = 1.7% of base peak

b. Predict the approximate height of the (M + 1) peak for the molecule icosane, molecular formula C20H42.

c. Draw the expected isotope pattern that would be observed in the mass spectrum of CH2Br2. In other words, predict

the relative heights of the peaks at M, (M + 2), and (M + 4) peaks.

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