CH. 15 - ANALYTICAL TECHNIQUES: IR, NMR, MASS...

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CH. 15 - ANALYTICAL TECHNIQUES: IR, NMR, MASS SPECT

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

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.

OF3C O

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.

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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: 1H NUCLEAR MAGNETIC RESONANCE- GENERAL FEATURES

1H (Proton) NMR is a powerful instrumental method that identifies protons in slightly different electronic environments by

inducing tiny magnetic fields in the electrons around the nucleus.

General Spectrum:

● ____________ is the standard reference point for NMR

● Electrons __________ protons from the effects of NMR

● The further downfield, the more __________ the proton

● There are 4 types of information we can gain from

NMR spectra.

Four Types of Information

1. Total Number of Signals

● Describes how many different types of hydrogens are present

2. Chemical Shift

● Describes how shielded or deshieldied the hydrogens are

3. Height of Signals (Integration)

● Describes the relative ratios of each type of hydrogen

4. Spin-Splitting (Multiplicity)

● Describes how close or far the different hydrogens are to each other

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CONCEPT: 1H NMR – TOTAL NUMBER OF SIGNALS

There are as many signals on each spectrum as there are unique, non-equivalent protons.

● Equivalent protons are defined as protons that have the same prospective on the molecule

● For now, let’s assume that hydrogens bound to the __________ ________________ are equivalent

□ Symmetry will reduce the total number of signals

EXAMPLE: How many different types of protons (signals) are there on each molecule?

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PRACTICE: How many types of electrically unique protons (peaks) are there in the following molecule?

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CONCEPT: 1H NMR – PROTON RELATIONSHIPS

Hydrogens attached to the same carbon actually do have different relationships based on their chirality.

□ The Q-Test is used to determine the specific type of chirality of each hydrogen.

a. Homotopic Protons

Q-Test DOES NOT yield new chiral center

● Protons are always homotopic and are considered ______________________ (They share a signal)

● In general, the three hydrogens on -CH3 groups will always be homotopic

b. Enantiotopic Protons

Q-Test DOES yield new chiral center

● No original chiral centers = protons are still _____________________________ (They share a signal)

c. Diastereotopic Protons

Q-Test DOES yield new chiral center

● 1+ original chiral centers = protons are now __________________________ (Each proton gets its own signal)

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EXAMPLE: How many signals will each molecule possess in 1H NMR?

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PRACTICE: Identify the indicated set of protons as unrelated, homotopic, enantiotopic, or diastereotopic.

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CONCEPT: 1H NMR – PROTON RELATIONSHIPS

d. E / Z Diastereoisomerism

Q-Test DOES yield new trigonal center on terminal double bonds

● Protons are always diastereotopic and are ____________________________ (Each proton gets its own signal)

EXAMPLE: How many peaks will each molecule possess in 1H NMR?

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CONCEPT: 1H NMR – CHEMICAL SHIFTS

The chemical shift indicates the exact electrochemical environment that each proton is experiencing.

● In general, electronegative groups will pull electrons away from nuclei, deshielding them

● Shifts increase (move downfield) as protons become more deshielded

C – H 1 – 2 C = C 4.5 – 6

C ≡ C 2.5 Benzene 6 – 8

Z – C – H 2 – 4 Aldehyde, -CHO 9-10

OH, NH 1 – 5 Carboxylic Acid , -COOH 10-13

Your professor will determine how many chemical shifts you should memorize. We’ll go over them just in case.

EXAMPLE: Order the following five protons from most deshielded to most shielded

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PRACTICE: Which of the labeled protons absorbs energy most upfield in the 1H NMR?

PRACTICE: Which of the labeled hydrogens will be most de-shielded?

A B C D E

PRACTICE: Which compound possesses a hydrogen with the highest chemical shift for its 1H NMR signal?

A B C D

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CONCEPT: 1H NMR – SPIN-SPLITTING WITHOUT J-VALUES

Also known as spin-spin coupling, or J-coupling, this describes the distances between different protons.

Note: This topic can be taught with or without J-values. Check with your professor to determine how much detail you should learn. For now, we will start with the simplest explanation, (should suffice for 90% of professors), which is without J-values.

● Adjacent, _______ - ____________________ protons will split each other’s magnetic response to the NMR

□ We use the ______________ rule to determine how many splits we will achieve

□ Pascal’s Triangle predicts the shape of the splits we will get

EXAMPLE: How will the following protons be split?

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PRACTICE: Predict the splitting pattern (multiplicity) for the following molecule:

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PRACTICE: Which of the following compounds gives a 1H NMR spectrum consisting of only a singlet, a triplet, and a pentet?

a) CH3OCH2CH2CH2CH2OH

b) CH3OCH2CH2OCH2CH3

c) CH3OCH2CH2CH2OCH3

d) CH3OCH2CH2OCH3

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CONCEPT: 1H NMR – SPIN-SPLITTING WITH J-VALUES AND TREE DIAGRAMS

Coupling-Constants, also known as J-values, describe the amount of interaction that a proton will have on another.

Here are some examples of common coupling-constants (measured in Hz):

Pascal’s Triangle only helps to predict the shapes of splits when all of the J-values are assumed to be the same.

● When multiple J-values are involved, tree diagrams are needed to predict the shapes of the splits.

Drawing Simple Tree Diagrams:

First, let’s use tree diagrams to help us understand why Pascal’s Triangle and the n + 1 Rule make sense.

● Each split represents the J-value in Hz of a single proton. What does n + 1 predict here? ________________ ANSWER

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CONCEPT: 1H NMR – SPIN-SPLITTING WITH J-VALUES AND TREE DIAGRAMS

Drawing Complex Tree Diagrams:

Now let’s use an example where multiple J-values are involved. Always split in order of highest to lowest values.

● Before starting, what does the n + 1 Rule predict here? ___________________ ANSWER

EXAMPLE: Use a tree diagram to predict the splitting pattern of the bolded proton.

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PRACTICE: Draw a tree diagram for H* in the structure below.

F2CH*CH(CH3)2 JH*-F = 50 Hz JH*-H = 7 Hz

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CONCEPT: 1H NMR SPIN-SPLITTING – COMMON PATTERNS

Some splitting patterns are highly indicative of certain structures. We can get ahead by memorizing them.

EXAMPLE: Which common 1H NMR splitting pattern seen below could help us determine the molecular structure?

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CONCEPT: 1H NMR – INTEGRATION

Integration describes how many of each type of hydrogen are present, expressing this information as relative ratios.

● Uses the Area Under the Curve (AUC) to visually demonstrate which hydrogens are most prevalent.

EXAMPLE: Draw the complete NMR spectrum:

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PRACTICE: Which of the following molecules gives a 1H NMR spectrum consisting of three peaks with integral ratio of 3:1:6?

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PRACTICE: Draw the approximate positions that the following compound might show in its 1H NMR absorptions?

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PRACTICE: Draw the approximate positions that the following compound might show in its 1H NMR absorptions?

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CONCEPT: 13C NMR – GENERAL FEATURES

13C NMR is a more limited type of nuclear magnetic resonance that identifies 13C instead of 1H.

● Due to low natural incidence of the 13C isotope, ______________ is NOT observed. ( ---------) (---------) =

● All of the other principles from 1H NMR apply, except that we must learn new shift values:

C – H 5 - 45 C = C 100 - 140

C ≡ C 65 - 100 Benzene 120 - 150

Z – C – H 30 - 80 Carbonyl 160 - 210

EXAMPLE: How many 13C signals would ethylbenzene give?

EXAMPLE: Which compound(s) will give only one peak in both its 1H and 13C spectra?

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CONCEPT: STRUCTURE DETERMINATION – MOLECULAR SENTENCES

The holy grail of this section is structure determination.

● You may be asked to produce a structure from scratch given only a MF, NMR Spectrum and IR Spectrum.

● Our goal is to build a strong “molecular sentence” by gathering clues, then propose drawings.

How to build a molecular sentence:

1. Determine IHD.

2. Analyze NMR, IR and splitting patterns, integrations for major clues (i.e.).

● NMR = 9.1 ppm __________________

● IR = 1710 cm-1 __________________

● Triplet/Quartet __________________

● 9.1 ppm (2H) __________________

3. Calculate 1H NMR Signal : Carbon Ratio.

● Ratio < ½ suggests symmetrical, whereas ratio > ½ suggests asymmetrical

□ Never rule out a structure based on symmetry (you may not be able to visualize it)

4. State the number of 1H NMR signals needed.

--- DRAW POSSIBLE STRUCTURES ---

5. Use a combination of Shifts, Integrations, and Splitting to confirm which structure is correct.

EXAMPLE: Build a strong molecular sentence using the following data.

MF: C4H6O2 IR: peak at 2950 cm-1 1H NMR

peak at 2700 cm-1 - 2.2 (doublet, 4H)

peak at 1720 cm-1 - 9.4 (triplet, 2H)

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PRACTICE: Propose a structure for the following compound that fits the following 1H NMR data:

Formula: C3H8O2 1H NMR: 3.36 δ (6H, singlet)

4.57 δ (2H, singlet)

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Formula: C2H4O2 1H NMR: 2.1 δ (singlet, 1.2 cm)

11.5 δ (0.5 cm, D2O exchange)

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Formula: C10H14 1H NMR: 1.2 ppm (6H, doublet)

2.3 ppm (3H, singlet) 2.9 ppm (1H, septet) 7.0 ppm (4H, doublet)

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PRACTICE: Propose a structure for the following compound, C7H12O2 with the given 13C NMR spectral data:

Broadband decoupled 13C NMR: 19.1, 28.0, 70.5, 129.0, 129.8, 165.78 δ DEPT-90: 28.0, 129.8 δ DEPT-135: 19.1 δ (↑ ), 28.0 (↑ ) , 129.8 δ (↑ ) , 70.5 δ (↓) & 129.0 δ (↓)

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PRACTICE: Propose a structure for the following compound, C5H10O with the given 13C NMR spectral data:

Fully Broadband decoupled 13C NMR and DEPT: 206.0 δ (↑ ); 55.0 δ (↑ ); 21.0 δ (↓) & 11.0 δ (↑ ).

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PRACTICE: Provide the structure of the unknown compound from the given information.

Formula: C4H10O IR: 3200-3600 cm-1 1H NMR: 0.9 ppm (6H, doublet)

1.8 ppm (1H, nonatet) 2.4 ppm (1H, singlet) 3.3 ppm (2H, doublet)

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PRACTICE: Provide the structure of the unknown compound from the given information.

Formula: C4H9N IR: 2950 cm-1, 3400 cm-1 1H NMR: 1.0 ppm (4H, triplet)

2.1 ppm (4H, triplet)

3.2 ppm (1H, singlet)

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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

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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CH. 15 - ANALYTICAL TECHNIQUES: IR, NMR, MASS...

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