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Chemistry > Topic 18 Organic Chemistry III > Flashcards

Topic 18 Organic Chemistry III Flashcards

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1
Q

Aromatic

A

Delocalised electrons forming pi bonding in a hydrocarbon ring.

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2
Q

Aliphatic compounds

A

All compounds that are not aromatic.

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3
Q

Physical properties of benzene

A

Colourless liquid. Boiling point: 80 °C. Insoluble in water. Toxic carcinogen. Found in crude oil and the derived fuels.

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4
Q

Molecular formula of benzene

A

C6H6

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5
Q

If the Kekule structure for benzene were correct, what would happen when shaken with bromine water?

A

Bromine water would decolourise in an addition reaction.

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6
Q

Kekule structure

A

Ring/cyclic structure with 3 double bonds.

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7
Q

What does happen when benzene is shaken with bromine water?

A

A substitution reaction occurs. There are no individual C=C bonds, so there is no addition reaction with Br2.

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8
Q

What suggests that the bonds between the carbon atoms in the benzene ring are the same?

A

There should be 4 isomers with the molecular formula C6H4Br2 (dibromobenzene), if the Kekule structure were correct. Instead, there were three indicating that the two isomers with the bromine on adjacent carbon atoms were identical– not joined by C-C in one and C=C in another.

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9
Q

What suggests the carbon-carbon bonds in benzene are all the same?

A

C-C bond length in benzene is in-between the bond lengths of C=C (shorter) in cyclohexene and C-C (longer) in cyclohexene.

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10
Q

Units for bond length

A

pm

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11
Q

IUPAC name for the Kekule structure

A

Cyclohexa-1,3,5-triene.

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12
Q

How much lower is the enthalpy change of hydrogenation for actual benzene than the theoretical structure with 3 C=C bonds?

A

152 kJ mol-1.

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13
Q

What was the expected value for the enthalpy of hydrogenation of benzene?

A

Triple that of cyclohexene, so -360 kJ mol-1.

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14
Q

How does the delocalised structure of benzene form?

A

P-orbitals overlap sideways forming a pi-electron cloud above and below the plane of carbon atoms. Each carbon atom contributes 1 electron to the cloud, the delocalised system has 6 electrons in total.

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15
Q

Why does the actual structure of benzene mean that there are only 3 isomers of C6H4Br2?

A

When the Br atoms are on adjacent carbon atoms, there is no difference in the arrangement of electrons between these atoms.

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16
Q

Why is benzene 152 kJ mol-1 more stable compared to cyclohexa-1,3,5-triene?

A

The charge is spread around in the species.

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17
Q

Explain why benzene does not undergo addition reactions with Br2, but it does undergo substitution instead?

A

Substitution preserves the stability of the delocalised electrons in the pi bond. The addition reaction would instead produce a compound with 2 C=C double bonds, which would lack the extra stability of the substitution product.

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18
Q

Hydrogenation of benzene

A

C6H6 + 3H2 –> C6H12
Heat under pressure with a nickel catalyst– the same as with alkanes.

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19
Q

Bromination

A

Putting bromine into a molecule. Either by addition or substitution, but substitution in the context of benzene.

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20
Q

Halogen carrier

A

A catalyst that helps to introduce a halogen atom into a benzene ring.

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21
Q

Why does benzene combust in air with a smoky flame?

A

High carbon-to-hydrogen ratio, so incomplete combustion.

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22
Q

Equation for the combustion of benzene.

A

C6H6 + 7.5O2 –> 6CO2 + 3H2O.

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23
Q

Bromination: a type of substitution. Equation.

A

C6H6 + Br2 –> Bromobenzene + HBr

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24
Q

Reagents & conditions for the bromination of benzene.

A

Heat under reflux with a halogen carrier catalyst. This is FeBr3 here.

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25
Q

Nitration

A

The substitution of a hydrogen atom by a nitro group. NO2 is the source of the NO2 group, and sulfuric acid is the catalyst.

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26
Q

Equation for nitration

A

Benzene + HNO3 –> Nitrobenzene + H2O.
Put H2SO4 above the arrow.

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27
Q

Reagents & conditions for nitration.

A

Warm benzene with a mixture of concentrated sulfuric and concentrated nitric acid.

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28
Q

Friedel-Crafts Reactions

A

Alkylation & Acylation.

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29
Q

What 3 features do the Friedel-Crafts reactions have in common?

A
  • Use a reagent represented by XY, one of the H atoms in benzene is substituted by Y and the other product is HX.
  • A catalyst is needed. It is AlCl3.
  • Anhydrous conditions.
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30
Q

Why do Friedel-Crafts reactions require anhydrous conditions?

A

Water would react with the catalyst and sometimes also with the organic product.

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31
Q

Equation for alkylation:

A

Benzene + CH3Cl –> Methylbenzene + HCl. Where chloromethane is the halogenoalkane, but this could be any R-group.
Write the AlCl3 catalyst above the arrow.

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32
Q

What happens in alkylation?

A

One of the H atoms is substituted by an alkyl group.
The reagent: halogenoalkane.
The product: alkylbenzene + HCl.

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33
Q

What happens in acylation?

A

One of the hydrogen atoms is substituted by an acyl group.
Reagent: acyl chloride.
Product: ketone + HCl.

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34
Q

Equation for acylation:

A

Benzene + CH3OCl –> Phenylethanone + HCl
Write the AlCl3 above the arrow.

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35
Q

What is the name used to represent benzene when it is attached to another functional group?

A

Phen or phenyl. All other naming rules apply.

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36
Q

How does the benzene ring attract electrophiles?

A

Although it is electrically neutral, the delocalised electrons in the pi bond above and below the ring mean it is electron-rich.

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37
Q

In all electrophilic substitution reactions, what happens in the…
- 1st step.
- 2nd & 3rd steps.
- Last step.

A

1: forming the electrophile.
2 & 3: the actual substitution in the benzene ring.
4: what happens to the replaced H+.

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38
Q

Electrophilic substitution of benzene: a general mechanism.

A

The electrophile approaches the delocalised pi bond and forms a covalent bond to it by attracting 2 electrons. This gives a species with a positive charge. The 4 remaining delocalised electrons are represented by 2/3 of a circle. This intermediate is less stable, so the H leaves as H+. The 2 electrons in the C-H bond join with the remaining 4 to restore its stability.

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39
Q

Where is the positive charge shown on the intermediate in the electrophilic substitution of benzene?

A

The centre of the circle.

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40
Q

Bromination: formation of the electrophile.

A

AlCl3 + Br2 –> Br+ + AlCl3Br-
Bromine reacts with the catalyst to form Br+.

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41
Q

Step 2 in the bromination of benzene.

A

Curly arrow from the ring to Br+ –> Br & H joined to the same carbon with a + in the centre of the 2/3 circle.

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42
Q

Step 3 in the bromination of benzene.

A

Curly arrow from the C-H bond to the centre of the ring to form bromobenzene + H+.

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43
Q

Step 4 in the bromination of benzene.

A

AlCl3Br- + H+ –> AlCl3 + HBr
The catalyst is regenerated & HBr is also formed.

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44
Q

Nitration: formation of the electrophile.

A

HNO3 + H2SO4 –> NO2+ + HSO4- + H2O
Conc. nitric + conc. sulfuric acids react to form NO2+, the nitronium ion,.

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45
Q

Nitration: step 2, electrophilic attack.

A

Curly arrow form the ring to the NO2+. –> NO2 & H joined to the same carbon of a benzene ring with a 2/3 circle & a + in the centre.

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46
Q

Nitration: step 3, forming the aromatic product.

A

Curly arrow from the C-H bond to the centre of the ring. –> Nitrobenzene + H+

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47
Q

Nitration: step 4, forming the inorganic products & regenerating the catalyst.

A

HSO4- + H+ –> H2SO4

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48
Q

Alkylation: step1, forming the electrophile.

A

AlCl3 + CH3Cl –> CH3+ + AlCl4-
The halogenoalkane reacts with the catalyst to form the R+ electrophile.

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49
Q

Alkylation: step 2, electrophilic attack.

A

Curly arrow from the ring to the CH+. The H & CH3 are now both joined to the same carbon.

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50
Q

Alkylation: step 3, forming the aromatic product.

A

Curly arrow from the C-H bond to the + in the centre of the 2/3 ring. –> methylbenzene + H+.

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51
Q

Alkylation: step 4, forming the inorganic products.

A

AlCl4- + H+ –> AlCl3 + HCl

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52
Q

Phenol

A

At least 1 hydroxyl group attached to a a benzene ring.

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53
Q

Under what conditions does the bromination of phenols occur?

A

RTP without a catalyst. Uses bromine water.

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54
Q

What is observed during the bromination of phenols?

A

The bromine water decolourises, and the organic product forms a white precipitate.

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55
Q

Equation for the bromination of phenols.

A

Phenol + 3Br2 –> 2,4,6-tribromophenol + 3HBr
The organic product has OH on carbon 1 and Br on carbons 2, 4 & 6.

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56
Q

Activation (phenols)

A

The electrons in the p-orbitals of the oxygen atom become part of the delocalised electron system. This increases the electron density above and below the ring, which makes the molecule more attractive towards electrophiles.

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57
Q

Why does the bromination of phenols occur more easily than the bromination of benzene?

A

The oxygen in the OH group has lone pairs of electrons, which can merge with the electrons in the delocalised pi bond. Br2 molecules are polarised as they approach the ring. Eventually, the Br-Br bond breaks and the Br+ electrophile attacks the benzene ring.

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58
Q

2,4,6-trichlorophenol

A

The product of chlorination of phenols. TCP antiseptic.

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59
Q

Uses of phenols

A

Antiseptic, manufacture of polymers and pharmaceuticals.

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60
Q

Phenolphthalein

A

The structure changes, so that in acidic conditions it is protonated, but in basic conditions, it is deprotonated and there are no OH groups in it.

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61
Q

Shape of amines

A

Trigonal pyramidal

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62
Q

Describe the basic structure of amines.

A

N with a lone pair and 3 bonds. If just one of these bonds is to an alkyl group, it’s primary. If 2 of these bonds are to alkyl groups, it’s secondary. If all 3 of these bonds are to alkyl groups, it’s tertiary.

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63
Q

The simplest aromatic amine

A

Phenylamine C6H5NH2

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64
Q

What are the two possible starting materials for making primary aliphatic amines?

A

Halogenoalkanes & nitriles.

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65
Q

Preparation of primary aliphatic amines from halogenoalkanes: reagents & conditions.

A

Heat with gaseous ammonia in a sealed tube under pressure or with concentrated aqueous ammonia. The lone pair on the N of ammonia means this is a nucleophilic substitution reaction.

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66
Q

Preparation of primary aliphatic amines from halogenoalkanes: equation.

A

CH3Cl + NH3 –> CH3NH2 + HCl
OR
CH3Cl + 2NH3 –> CH3NH2 + NH4Cl

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67
Q

Why do unwanted side reactions, such as the forming of secondary amines, occur?

A

The primary amine formed has a lone pair on the N too. This means it can act as a nucleophile and compete with NH3 to attack the halogenoalkane.
CH3Cl + CH3NH2 –> (CH3)2NH + HCl

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68
Q

How do we prevent unwanted side reactions from occurring when making our primary amine from a halogenoalkane?

A

Add the NH3 in excess, so it outnumbers the molecules of primary amine formed. Some of the excess NH3 will react with the HCl formed.

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69
Q

How can primary aliphatic amines be prepared from nitriles?

A

Reduction using LiAlH4 in dry ether.
CH3CN + 4[H] –> CH3CH2NH2

70
Q

What is phenylamine made from?

A

The reduction of nitrobenzene.

71
Q

How are aromatic amines prepared?

A

Reducing agent: tin mixed with concentrated HCl.
Heat under reflux.
C6H5NO2 + 6[H] –> C6H5NH2 + 2H2O

72
Q

How is the reduction involved in preparing aromatic amines achieved?

A

Partly through the oxidation of tin to tin(II) ions & tin(IV) ions.
Partly through the hydrogen produced between the reaction between tin & the acid.

73
Q

Just like other amines, phenylamine is basic, so reacts with the acid present to form a phenylammonium ion. How can this be converted to phenylamine?

A

Add alkali such as NaOH (aq). C6H5NH3^+ + OH- –> C6H5NH2 + H2O

74
Q

Solubility of amines in water

A

The first few primary aliphatic amines are completely miscible in water, but solubility decreases as the hydrocarbon part of the molecule increases. They dissolve by forming H bonds with the water. Phenylamine is only slightly soluble in water.

75
Q

Amines react slightly with water to form…

A

…alkaline solutions.

76
Q

Equation for the reaction of methylamine with water:

A

CH3NH2 + H2O <–> CH3NH3^+ + OH^-
The N atom uses its lone pair to form a dative bond with the water.
This is almost the same as when ammonia reacts with water, except one of the H atoms on the ammonium ion has been replaced by the methyl group.

77
Q

Basicity

A

The extent to which a base can donate a lone pair of electrons to the hydrogen atom of a water molecule.

78
Q

How can we measure basicity?

A

Ka or pKa

79
Q

pKa of water

A

7.00 Water acts equally as a base & an acid.

80
Q

How does increasing the carbon chain of an aliphatic amine affect pKa?

A

It increases slightly pKa, indicating greater basicity, as the electron-releasing effect increases slightly.

81
Q

Why are aliphatic amines stronger bases than ammonia?

A

The alkyl chain is electron-releasing, so increases the electron density on nitrogen compared to NH3.

82
Q

Why is phenylamine less basic than water? (pKa of 4.62.)

A

The lone pair of electrons on nitrogen joins the delocalised electrons in the benzene ring. This makes the nitrogen less electron-rich and the lone pair less available to donate to the H of a water molecule.

83
Q

Amine + strong acid –>

A

Ionic salt!
E.g., Methylamine + HNO3 –> Methylammonium nitrate

84
Q

Addition-elimination reaction

A

Occurs when 2 molecules join together followed by the loss of a small molecule.

85
Q

Amide functional group

A

Carbonyl next to a NH group/an amino group.

86
Q

Amine + acyl chloride –>

A

Amide + HCl

87
Q

General formula for the reactions of amines with halogenoalkanes

A

R1NH2 + R2X –> R1NHR2 + HX, where R1 is the alkyl group in the amine & R2 is the alkyl group in the halogenoalkane. Substitution.
Produces a secondary amine + a hydrogen halide.

88
Q

What happens in the substitution reaction between amines & halogenoalkanes when the halogenoalkane is in EXCESS?

A

Ammonium salt formation is more likely.
The initial product of a 1° amine + halogenoalkane is a 2° amine with an electron-rich nitrogen atom + HX.
The product then reacts with more halogenoalkane to form a 3° amine + HX.
The 3° amine then reacts with even more halogenoalkane to form a quaternary ammonium salt. E.g., CH3CH2N(CH3)2 + CH3Cl –> CH3CH2N+(CH3)3 + Cl-

89
Q

1° amine + hexaaquacopper(II) ions e.g., with CH3CH2NH2:
Note: similar to when NH3 reacts with hexaaquacopper(II) ions.

A

First, a pale blue precipitate forms. Then, with excess amine, the precipitate dissolves to give a deep blue solution.
[Cu(H2O)6]^2+ + 2CH3CH2NH2 –> [Cu(H2O)4(OH)2] + 2CH3CH2NH3+
With excess:
[Cu(H2O)4(OH)2] + 4CH3CH2NH2 –> [Cu(CH3CH2NH2)4(H2O)2]^2+ + 2H2O + 2OH^-

90
Q

Physical properties of amides

A

Solid (except for methanamide, which is liquid).
Lower aliphatic amides are water soluble due to their polar bonds & 2 electronegative atoms meaning they can form H bonds with water.

91
Q

How can we make an amide?

A

Acyl chloride + concentrated NH3 (aq). Misty fumes of HCl are also produced.
The lone pair on N of NH3 is attracted to the electron-deficient C of the acyl chloride.
The HCl produced then reacts with the NH3.

92
Q

How can we make a polyamide?

A

Condensation reaction:
Dicarboxylic acid + diamine (releases H2O).
OR: Dioyl chloride + diamine (releases HCl).

93
Q

Polyamide formation example: HOOCCOOH + H2NCH2NH2

A

One has acidic groups, one has basic groups. They react to form a CONH group & eliminate H2O.

94
Q

Nylon 6,6

A

A polyamide formed by the reaction between hexanedioic acid + hexane-1,6-diamine

95
Q

Kevlar

A

A polyamide formed by the reaction between a benzene ring with 2 acyl chloride groups on it + a benzene ring with 2 NH2 groups attached.

96
Q

Isoelectric point

A

The pH of an aqueous solution of amino acid in which it is neutral.

97
Q

Zwitterion

A

A molecule containing positive & negative charges, but which has no overall charge.

98
Q

The zwitterion exists in aqueous solution at the…

A

Isoelectric point.

99
Q

Low isoelectric point

A

The molecule is predominantly acidic.

100
Q

High isoelectric point

A

The molecule is predominantly basic.

101
Q

All amino acids except glycine contain a chiral centre.

A

So they are optically active! But if the amino acid is synthesised in a lab, a racemic mixture forms.

102
Q

All amino acids react with acids & bases to form salts– example: alanine.

A

Alanine + acid –> this protonated structure:
CH3N+H3CHCOOH.

103
Q

Glutamic acid reacts with NaOH to form 3 possible salts: how?

A

It has 2 COOH groups. Either one or both can react.
H2N–CH–COO-Na+
|
CH2–CH2–COOH

104
Q

What happens when 2 amino acids react together?

A

An acid-base condensation reaction. Forms a dipeptide via peptide bond formation. The OH of the COOH reacts with the H of the NH2 to form water + an amide group (CO–NH).

105
Q

Peptide bond

A

The bond formed by a condensation reaction between the carbonyl group of one amino acid + the amino group of another amino acid.

106
Q

Two different amino acids. How many possible dipeptides can form?

A

2, but if one or more of the 2 amino acids has more than 1 NH2 or COOH group, then more possible dipeptides can form.

107
Q

When one molecule of each of 3 different amino acids reacts, how many tripeptide possibilities are there?

A

6 (unless 2 or more of the same molecule can react to form a tripeptide in which case many more)!

108
Q

Protein

A

Very long polypeptide with 2° 3° and quaternary structures as well.

109
Q

2 steps in protein structure discovery:

A
  1. Find which amino acids are present.
  2. Find out the order in which they occur in the polypeptide chain.
110
Q

Hydrolysing a protein

A

Prolonged heating with H2O + concentrated HCl breaks the peptide bond. Due to the acidic conditions, all the amino acids are in their protonated (NH3+) form.

111
Q

How can the amino acids from a hydrolysed protein be identified?

A

Chromatography. Spray the chromatogram with developing agent, e.g., ninhydrin, so the positions of the normally colourless amino acids can be observed. Identify the amino acids from their Rf values.

112
Q

4 different ways of extending the carbon chain

A
  1. Halogenoalkane + CN- –> nitrile with one more carbon than the halogenoalkane.
  2. The addition of HCN to a carbonyl group.
  3. The alkylation of benzene, which introduces an alkyl group into a benzene ring.
  4. Grignard reagents.
113
Q

How can we make a Grignard reagent?

A

Mg in dry ether + a halogenoalkane or haloarene (e.g., bromobenzene). Heat under reflux.

114
Q

Grignard reagents react with water. What is the consequence of this?

A

Grignard reagents are both made & used in dry ether.

115
Q

General equation for the formation of a Grignard reagent:

A

R–Br + Mg –> R–Mg–Br

116
Q

Grignard + CO2 –>

A

Carboxylic acid

117
Q

Grignard + methanal –>

A

1° alcohol

118
Q

Grignard + an aldehyde –>

A

2° alcohol

119
Q

Grignard + ketone –>

A

3° alcohol

120
Q

After the Grignard reagent has reacted with the CO2 or carbonyl, what must be added to obtain the desired product?

A

Hydrolyse with dilute acid
H2O/H+

121
Q

How can we form 3-methylbutanoic acid from a Grignard?

A
  1. Form the Grignard:
    (CH3)2CHCH2Br + Mg –> (CH3)2CHCH2MgBr
  2. React with CO2:
    (CH3)2CHCH2MgBr + CO2 –> (CH3)2CHCH2COOMgBr
  3. Hydrolysis with a dilute acid:
    (CH3)2CHCH2COOMgBr + H2O –> (CH3)2CHCH2COOH + Mg(OH)Br
122
Q

How can we make propan-1-ol from a Grignard? (Abbreviated reaction scheme.)

A

CH3CH2Br + Mg/ether –> CH3CH2MgBr+ HCHO then H2O/H+ –> CH3CH2CH2OH

123
Q

Hazard

A

Property of a substance that could cause harm to a user.

124
Q

Risk

A

Possible effect that a substance may cause to a user, depending on factors such as concentration & apparatus. The level of risk is controlled via control measures.

125
Q

Heating under reflux.

A
  • Heat the reaction mixture in a flask fitted with a reflux condenser.
  • Heat using hot water or oil in a beaker heated by a Bunsen burner or by using an electric mantle.
  • Anti-bumping granules make the boiling smooth.
  • Cold water enters the bottom of the condenser.
  • All the vapours rising from the reaction mixture during heating enter the condense, condenser back into liquids and return to the flask, so unreacted compounds can react.
  • Volatile organic compounds cannot escape.
126
Q

Simple distillation

A

Heat an impure liquid in a flask fitted to a condenser. The liquid with the lowest boiling point evaporates and passes into the condenser first, so can be collected in the receiver separately from any other liquids.

127
Q

What is the thermometer used for in simple distillation?

A

To monitor the temperature of the vapour as it passes into the condenser. If the temperature remains steady, one compound is being distilled. If after a while the temperature begins to rise, another compound is being distilled.

128
Q

Pros of simple distillation over fractional distillation.

A

Easier to set up and quicker.

129
Q

Cons of simple distillation over fractional distillation.

A

Does not separate the liquids as effectively. It should only be used if the boiling temperature of the liquid being purified is very different from the other liquids in the mixture, ideally by more than 25 °C.

130
Q

What is steam distillation used to separate?

A

An insoluble liquid from an aqueous solution.

131
Q

Fitting of the condenser in practical techniques:

A

Vertically in reflux.
Horizontally in all types of distillation.

132
Q

How does steam distillation work?

A

Pass steam into a reaction mixture that contains an aqueous solution and a liquid that forms a separate layer.
The agitation of the liquid due to the steam bubbling through the mixture ensures both the insoluble liquid and the aqueous solution are on the surface, so can form part of the liquid that evaporates.

133
Q

Pro of steam distillation:

A

The insoluble liquid is removed from the reaction mixture at a temperature below its boiling point, so there’s less chance of decomposition.

134
Q

Pros & cons of fractional distillation

A

Takes longer than simple distillation.
Used when the difference in boiling temperatures is small and when there are several compounds to be separated from the mixture.

135
Q

How does fractional distillation work?

A

The fractionating column between the heating flask & still head is filled with glass beads, which act as surfaces on which the vapour leaving the column can condense and the be evaporated again as more hot vapour passes up the column. This means that the vapour undergoes several repeated distillations as it passes up the column, hence the better separation.

136
Q

What sort of solvent should be added in solvent extraction?

A
  • The solvent should be immiscible with the solvent containing the desired product.
  • The desired product should be much more soluble in the added solvent than in the reaction mixture.
137
Q

How does solvent extraction work?

A
  1. Place the reaction mixture in a separating funnel, then add the chosen solvent. It should form a separate layer.
  2. Place the stopper in the neck of the funnel. Gently agitate the contents of the funnel: invert, open the tap, agitate in a circular motion, close the tap & allow the product to return to its original position.
  3. Allow the contents to settle into 2 layers.
  4. Remove the stopper, and open the tap to allow the lower layer to drain into a flask and the upper layer drains into a different flask.
  5. Use simple/fractional distillation to separate the organic product from the solvent used.
138
Q

In solvent extraction, why is better to use the solvent in small portions with the same total volume rather than a single volume?

A

More efficient as more of the desired organic product is removed.

139
Q

Washing

A

Uses water or an organic solvent to remove impurities from a solid or liquid by dissolving impurities, but as little of the substance being purified as possible.

140
Q

Washing: solid

A

Stir the impure solid in some of the solvent. Filter the mixture.

141
Q

Washing: liquid

A

Mix with the impure liquid. Shake the mixture in a separating funnel. Allow the 2 layers to separate. Open the tap to allow each layer to drain into a separate container.

142
Q

Drying an organic solid

A

Leave in a warm place or in a desiccator with a drying agent.

143
Q

Drying organic liquids

A

The drying agent must not react with the organic liquid. Add it to the organic liquid. Shake the mixture. Leave for a period of time. The drying agent is removed by either decantation or filtration.

144
Q

What are type of compounds are drying agents generally?

A

Anhydrous metal salts. These can form hydrated salts, so when they come into contact with the water in an organic liquid, they absorb it as water of crystallisation.

145
Q

3 common drying agents

A

Calcium chloride.
Magnesium sulfate.
Sodium sulfate.
All anhydrous.

146
Q

How do we know when a liquid is dry?

A

The drying agent initially appears powdery, then absorbs water and looks more crystalline. If more drying agent is added and remains powdery, the liquid is dry.
The liquid also changes from cloudy to clear when the water is removed.

147
Q

Recrystallisation

A
  1. Add the impure solid to a conical flask.
  2. Add the chosen solvent, and warm until the mixture nears boiling.
  3. If there is still undissolved solid, warm until the mixture boils again.
  4. Continue adding further solvent, and heating until all of the soluble solid has dissolved.
  5. If insoluble impurities are present, hot filtration could be performed using fluted filter paper in a heated funnel.
  6. Allow the liquid to cool until crystals of the organic solid have formed.
  7. More crystals can be obtained by cooling the solution below room temperature in an ice bath.
  8. The mixture is then filtered under reduced pressure to remove soluble impurities using a Buchner or a Hirsch funnel, then dried in a desiccator or an oven.
148
Q

Filtration under reduced pressure.

A

Uses a vacuum pump. Involves a Hirsch or a Buchner funnel.

149
Q

Testing the purity of a solid

A

Impurities reduce the melting point of a solid.
Attach a capillary tube containing the solid to a thermometer bulb.
Place this assembly in a beaker containing a liquid (with a boiling point higher than the melting point of the solid).

150
Q

Testing the purity of a liquid

A

Impurities increase the boiling point of a liquid. Usually, use simple distillation to measure the boiling point, though the thermometer may not always measure accurately. Compare the value obtained with the value in the databook.
Remember: some organic compounds may have the same boiling point by coincidence!

151
Q

Why do we need a lid in simple chromatography?

A

Prevent evaporation of the solvent.

152
Q

Stationary phase

A

In paper chromatography, the liquid or solid that does not move.

153
Q

Mobile phase

A

The liquid that moves through the stationary phase and transports the components.

154
Q

What is simple chromatography used for?

A
  • To check whether an organic compound is pure or not, and if not, what impurities are present.
  • To separate a mixture into its individual components.
  • To identify components of the mixture by their Rf values/comparison to known samples.
155
Q

What is the stationary phase in TLC?

A

A sheet of glass or plastic coated in a thin layer of solid such as silica or alumina.

156
Q

What is the advantage of column chromatography?

A

Much larger quantities of material can be separated than with paper chromatography.

157
Q

How does column chromatography work?

A

Alumina or silica is packed into a tube and soaked in solvent. The mixture is added. More solvent (the mobile phase) is added. When the tap is opened, the solvent drips through the tip and the components of the mixture begin to move down the tube and separate.

158
Q

When giving equations for the reactions of amino acids with acids & alkalis, what form should the reacting amino acid be in?

A

Zwitterion.

159
Q

Why does benzene undergo substitution not addition reactions?

A

The delocalised pi electrons of the ring make benzene more stable. Substitution reactions retain this stable arrangement.

160
Q

What is a Grignard reagent?

A

A nucleophilic reducing agent.

161
Q

In an organic synthesis pathway, how would you add benzene rings together?

A

Brominate 1 benzene and turn it into a Grignard. Turn the the Grignard into an alcohol, then an acyl chloride. Then use AlCl3 to add another benzene to it (acylation).

162
Q

Why do amino acids have relatively high boiling points compared to alkanes?

A

Amino acids exist as zwitterions, so the strongest intermolecular forces they possess are ionic.

163
Q

Why do some amino acids have longer retention times than others in column chromatography?

A

When a polar stationary phase is used, amino acids with a higher charge will be more strongly attracted to the stationary phase.

164
Q

Why is filtering under reduced pressure used when recrystallising a solid?

A

It removes more of the soluble impurities faster & produces a drier product.

165
Q

What conditions are required to produce an amine from a nitrile?

A

Reduce using LiAlH4 in dry ether. Then add dilute acid (aq).

166
Q

What conditions are required to reduce nitrobenzene to phenylamine?

A

Sn, conc. HCl & heat under reflux.

167
Q

How does having nitro groups very close together on benzene affect boiling point?

A

If they are too close, the nitrobenzene forms intramolecular H bonds instead of intermolecular. If nitro groups are at opposite ends of the molecule, molecules are joined in a straight chain by H bonds.

168
Q

Benzene burns with a smoky flame. What else does this?

A

Alkenes.

169
Q

Why is phenylamine a weaker base than an aliphatic amine?

A

The lone pair of electrons on the N atom delocalises into the benzene ring, so it’s less able to accept a proton. Whereas, the alkyl group pushes electrons towards the lone pair on N, so it’s more able to accept an electron.

170
Q

What does it mean when the window in a fume cupboard is above the safety line?

A

The exhaust system is not strong enough to draw in the fumes, so toxic fumes will escape into the lab.

171
Q

Bromination of phenol vs. bromination of benzene.

A

Bromination of benzene requires heating under reflux & a halogen carrier catalyst. Bromination of phenols needs no catalyst & occurs at RTP, forming 2,4,6-tribromophenol. This is because the lone pair on the O interacts with the benzene ring of delocalised electrons, so electrophilic attack is more likely. Both occur by electrophilic substitution. Phenol requires Br2(aq), whereas benzene requires Br2(l).

Chemistry (23 decks)

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