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Analysis Flashcards

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1
Q

What is nuclear magnetic resonance

A
  1. The nucleus has a property called spin, which is significant if there is an odd number of nucleons (protons and neutrons).
  2. For organic chemistry NMR is relevant for 1H 13C - the isotopes of carbon and hydrogen with an odd number of neutrons
  3. It can also be used to detect isotopes of other elements e.g 19F and 31P
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2
Q

What does a NMR spectrum graph look like

A
  1. chemical shift on the x-axis
  2. Energy on the y-axis
  3. Peaks at different frequencies
  4. The chemical shift increases from right to left
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3
Q

What is chemical shift

A
  1. In an organic molecule every carbon and hydrogen atom is bonded to other atoms.
  2. All atoms have electrons surrounding the nucleus which shifts the energy and radio frequency needed for nuclear magnetic resonance to take place
  3. The frequency shift is measured on a scale called chemical shift in units parts per million (ppm)
  4. Standardises the NMR
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4
Q

What is used as the standard for chemical shift measurements

A
  1. Tetramethylsilane (TMS) (CH3)4Si

2. It has the chemical shift value of 0 ppm

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5
Q

What does the amount of chemical shift depend on

A
  1. It is determined by chemical environments especially the presence of nearby electronegative atoms
  2. So depending on the chemical environment, nuclear magnetic resonance requires a different energy and frequency, producing absorption peaks at chemical shifts
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6
Q

How do you run an NMR spectra

A
  1. The sample is dissolved in a solvent and placed in a narrow NMR sample tube, together with a small amount of TMS
  2. The tube is placed inside the NMR spectrometer where it is spun to even out any imperfections in the magnetic field within the sample
  3. THe spectrometer is zeroed against the TMS standard and the sample is given a pulse of radiation containing a range of radio frequencies whilst maintaining a constant magnetic field
  4. Any absorptions of energy resulting from resonance are detected and displayed on a computer screen
  5. After the analysis the sample can be recovered by evaporation of the solvent
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7
Q

What type of solvents are used in NMR and why

A
  1. Molecules of most common solvents contain carbon and hydrogen atoms, which will produce a signal in both 13C and 1H NMR spectra
  2. A deuterated solvent is usually used in which 1H atoms have been replaced by 2H atoms (deuterium)
  3. Deuterium produced no NMR signal in the frequency ranges used in 1H and 13C NMR spectroscopy.
  4. Deuterated trichloromethane CDCl3, is commonly used as a solvent in NMR spectroscopy but this will produce a peak in a carbon-13 NMR spectrum. But the computer usually filters out this peak before displaying the spectrum.
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8
Q

What information can you get from a carbon-13 NMR spectrum

A
  1. The number of different carbon environments - from the number of peaks
  2. The types of carbon environment present - from the chemical shift
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9
Q

What determines the chemical environment of a carbon atom

A
  1. Carbon atoms that are bonded to different atoms or groups of atoms have different environments and will absorb different chemical shifts
  2. If two carbon atoms are positioned symmetrically within a molecule, then they are equivalent and have the same chemical environment. They will then absorb radiation at the same chemical shift and contribute to the same peak
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10
Q

Identify the number of chemical environments in the following molecules:

  1. CH3CH2CH2CH3
  2. CH3CH(CH3)CH(OH)CH2CH3
  3. 1-methylcyclohexane
  4. 2,5-methylhexane
A
  1. 2
  2. 5
  3. 5
  4. 3
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11
Q

How do you interpret a carbon NMR spectra

A
  1. Number of peaks= number of different carbon environments
  2. Position of peaks= chemical shift
  3. Height of the peak = means nothing
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12
Q

What information can you get from a Proton NMR spectrum

A
  1. The number of different proton environments- from the number of peaks
  2. The types of proton environment present- from the chemical shift
  3. The relative numbers of each type of proton- from integration traces or ratio numbers of the relative peak areas
  4. The number of non-equivalent protons adjacent to a given proton- from the spin-spin splitting pattern
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13
Q

What determines the chemical environment of a hydrogen atom

A
  1. If two or more protons are equivalent, they will absorb at the same chemical shift, increasing the size of the peak
  2. Protons of different types have different chemical environments and are non-equivalent- they absorb at different chemical shifts
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14
Q

Identify the number of hydrogen environments and the ratio of the areas under each peak in the following molecules:

  1. (CH3)2CHCH2CH3
  2. Benzene
  3. 2,4-dibromopentane
  4. CH3CH(OH)CH=CH2
A
  1. 4 6:1:2:3
  2. 1 6
  3. 3 6:2:2
  4. 5 3:1:1:1:2
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15
Q

What are splitting patterns caused by

A
  1. The proton’s spin interacting with the spin states of nearby protons that are in different environments
  2. This provides information about the number of protons bonded to adjacent carbon atoms
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16
Q

What is the n+1 rule

A
  1. The splitting of a main peak into sub-peaks is called spin-spin coupling or spin-spin splitting
  2. THe number of sub-peaks is one greater than the number of adjacent protons causing the splitting
  3. For a proton with n protons attached to an adjacent carbon atom, the number of sub-peaks ina splitting pattern = n+1
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17
Q

What are the possible splitting pattern for different sequences of protons in an organic molecule

A
  1. 0 adjacent hydrogens- singlet- 1 peak
  2. 1 adjacent hydrogen- doublet- 2 peaks
  3. 2 adjacent hydrogens- triplet- 3 peaks
  4. 3 adjacent hydrogens- quartet- 4 peaks
  5. > 3 adjacent hydrogens- multiplet- 5+ peaks
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18
Q

Show some common spin-spin coupling pairs

A
  1. Quartet+triplet = CH2-CH3
  2. Triplet+doublet = CH-CH2
  3. Triplet+triplet = CH2-CH2
  4. Doublet+doublet = CH-CH
  5. Quartet+doublet = CH-CH3
  6. Multiplet+doublet = CH-(CH3)2
19
Q

What do chemical shifts look like for -OH and -NH groups

A
  1. The NMR peaks are often broad and of variable chemical shift. Difficult to identify
  2. The broadening of the peaks means that OH and NH are not involved in spin-spin coupling
20
Q

How are -OH and -NH groups identified

A
  1. Technique called proton exchange
  2. A proton NMR is run as normal
  3. A small volume of deuterium oxide D2O is added, the mixture is shaken and a second spectrum is run
  4. Deuterium exchanges and replaced the OH and nh protons in the sample with deuterium atoms e.g with methanol
    CH3OH + D2O ↔ CH3OD + HOD
  5. As deuterium does not absorb in this chemical shift range the -OH/NH peak disappears
21
Q

Describe the steps used to identify a compound with using combined spectral techniques

A
  1. Elemental analysis- Use of percentage composition by mass to determine the empirical formula of a compound
  2. Mass spectra- Use of the molecular ion peak to determine the molecular mass and fragment ions to determine parts of the molecules. The molecular formula can then be determined from the empirical formula and molecular mass.
  3. Infrared spectra- Use of absorption peaks to identify bonds present and functional groups
  4. NMR spectra- To determine the number and types of carbon and hydrogen atoms from the chemical shifts of peaks and the order of atoms within molecules from splitting patterns
22
Q

How do you test for an alkene

A
  1. Add bromine water drop-wise

2. Bromine water decolourises from orange to colourless

23
Q

How do you test for a haloalkane

A
  1. Add silver nitrate and ethanol and warm to 50 degrees in a water bath
  2. Chloroalkane- white precipitate
  3. Bromoalkane- cream precipitate
  4. Iodoalkane- yellow precipitate
24
Q

How do you test for a carbonyl

A
  1. Add 2,4-dinitrophenylhydrazine/ 2,4-DNPH

2. Orange precipitate forms

25
Q

How do you test for an aldehyde

A
  1. Add Tollens’ reagent and warm

2. Silver mirror forms

26
Q

How do you test for primary and secondary alcohol, and aldehyde

A
  1. Add acidified potassium dichromate and warm in a water bath
  2. Colour change from orange to green
27
Q

How do you test for a carboxylic acid

A
  1. Add aqueous sodium carbonate

2. Effervescence is produced

28
Q

What is chromatography used for

A

Chromatography is used to separate individual components from a mixture of substances

29
Q

What do all forms of chromatography have

A
  1. Stationary phase

2. Mobile phase

30
Q

What is the stationary phase in chromatography

A
  1. The phase does not move

2. It is normally solid or a liquid supported on a solid

31
Q

What is the mobile phase in chromatography

A
  1. The mobile phase does move

2. It is normally a liquid or a gas

32
Q

What is thin layer chromatography used for

A
  1. It is a quick and inexpensive analytical technique

2. It indicated how many components are in a mixture

33
Q

Describe what is used in thin layer chromatography

A
  1. Uses a TLC plate which is usually a plastic sheet or glass, coated with a thin layer of a solid adsorbent substance- usually silica.
  2. In TLC, the adsorbent is the stationary phase.
  3. The different components in the mixture have different affinities for the absorbent and bind with differing strengths to its surface.
  4. Adsorption is the process by which the solid silica holds different substances in the mixture to its surface
  5. Separation is achieved by the relative adsorptions of substances with the stationary phase
34
Q

Describe how to carry out TLC

A
  1. Take a TLC plate, using a pencil draw a line across the plate about 1 cm from one end of the plate. This is the base line
  2. Using a capillary tube, spot a small amount of a solution of the sample onto the base line on the plate
  3. Prepare a chromatography tank for the TLC plate. This can be made from a small beaker with a watch glass placed on the top. Pour some solvent into the beaker to a depth of about 0.5 cm
  4. Place the prepared TLC plate in the beaker, making sure that the solvent does not cover the spot. Cover the beaker with the watch glass and leave it undisturbed on the bench. The solvent will rise up the TLC plate
  5. Allow the solvent to rise up the plate until it is baou 1cm below the top of the plate. Remove the plate from the beaker and immediately mark the solvent front with a pencil. Allow the plate to dry.
  6. If there are any visible spots circle them with a pencil. Or hold a UV lamp over the plate and circle any spots that you can see.
  7. Sometimes the plate is sprayed with a chemical or locating agent e.g iodine, to show the position of spots that aren’t visible to the naked eye.
35
Q

What do you calculate to analyse thin layer chromatograms

A
  1. The retention factor Rf for each component.
36
Q

How do you calculate the retention factor

A
  1. Retention factor= distance moved by the component/ distance moved by the solvent front
37
Q

What can you do that means you don’t need to calculate any Rf values

A
  1. Run a TLC of a sample alongside pure samples of compounds that may be present
  2. It is then easy to identify the compounds present visually without needing to calculate Rf values
38
Q

What is gas chromatography used for

A
  1. It is useful for separating and identifying volatile organic compounds present in a mixture
39
Q

Describe what the sationary and mobile phasees are for gas chromatography

A
  1. Stationary phase is a high boiling liquid adsorbed onto an inert solid support.
  2. Mobile phase is an inert carrier gas such as helium of neon
40
Q

Describe how gas chromatography works

A
  1. A small amount of the volatile mixture is injected into the apparatus, called a gas chromatograph.
  2. The mobile carrier gas carries the components in the sample through the capillary column which contains the liquid stationary phase adsorbed onto the solid support.
  3. The components slow down as they interact with the liquid stationary phase inside the column. The more soluble the component is in the liquid stationary phase, the slower it moves through the capillary column
  4. The components of the mixture are separated depending on their solubility in the liquid stationary phase.
  5. The compounds in the mixture reach the detector at different times depending on their interactions with the stationary phase in the column.
  6. The compounds retained in the column for the shortest time has the lowest retention time and is detected first.
41
Q

What is the retention time

A

The time taken for each component to travel through the column.

42
Q

What are the two pieces of information that can be obtained from a gas chromatogram

A
  1. Retention time can be used to identify the components present in the sample by comparing these to retention times for known components.
  2. Peak integrations (area under each peak) can be used to determine the concentrations of components in the sample
43
Q

How can you determine the concentration of a component in a sample

A
  1. By comparing its peak integration with values obtained from standard solutions of the component:
  2. Prepare standard solutions of known concentrations of the compounds being investigated
  3. Obtain gas chromatograms for each standard solution.
  4. Plot a calibration curve of peak area against concentration. This is called external calibration and offers a method for converting a peak area into concentration.
  5. Obtain a gas chromatogram of the compound being investigated under the same conditions.
  6. Use the calibration curve to measure the concentration of the compund

Chemistry (25 decks)

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