A2.12 Chromatography And Qualititaive Analysis Flashcards

1
Q

Chromatography

A

Used to separate and identify the different components of a mixture

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2
Q

TLC

A

Thin layer chromatography is used mainly to follow progress of a reaction or identify the purity of a product

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3
Q

Stationary phase - TLC

A

Does not move

Is usually a solid

Adsorbs

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4
Q

Mobile phase in TLC

A

Moves over stationary phase and is a liquid

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5
Q

TLC plate

A

Spot a sample onto the base line using capillary tube

Base line is drawn in pencil 1cm up

TLC plate is placed in shallow level of solvent (must be below baseline)

Travels up TLC plate
No of spots = no of parts to mixture

Elongated spots could be due to 2 parts with very similar structures

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6
Q

Rf

A

Distance spot travels
——————————-
Distance to solvent front

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7
Q

How far a substance travels in TLC depends on……

A

1) how strongly the substance adsorbs onto stationary phase

2) how soluble substance is in mobile phase

If it travels far - weakly adsorbed and very soluble

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8
Q

GC

A

Gas chromatography is useful in separating and analysing the amounts of different organic compounds in a mixture

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9
Q

Retention time

A

Time taken for a component of a mixture to reach the detector

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10
Q

Stationary phase in GC

A

Normally a liquid coating a tube or beads in a tube

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11
Q

Mobile phase in GC

A

Inert gas e.g. He or N

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12
Q

Retention time depends on…….

A

How soluble substance is in stationary phase

More soluble = longer retention time

Peaks = components

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13
Q

External calibration curve

A

Obtain several GC of same substance in different concentrations

Plot graph of relative peak area vs conc

Use GC for unknown concentration and use graph

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14
Q

Peak furthest to right on NMR

A

TMS - Si(CH3)4

Reference peak

Lots of equivalent H to give single environment

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15
Q

13C NMR

A

Each peak = 1 environment

Tells us functional group in environment (furthest to left if multiple)

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16
Q

1H NMR

A

Each peak is an environment

Number above peak - no of H in environment

Peak splits into patterns showing how many H in neighbouring environment

Split is n + 1
So if 4 in neighbouring then splits into triplet

17
Q

Solvents in NMR

A

CDCl3 for polar

CCl4 for non polar

Don’t have H atoms so won’t give peak

18
Q

D

A

Isotope of hydrogen - 2H (has 1 neutron)

19
Q

D2O in NMR

A

Confirms presence of OH or NH

Peak will be removed as H is removed and exchanged for D