Amines and Amides Flashcards

(74 cards)

1
Q

primary aliphatic amines

A

Primary aliphatic amines can be prepared by the reaction of ammonia with halogenoalkanes and by the reduction of nitriles.

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2
Q

aromatic amines

A

Aromatic amines, prepared by the reduction of nitro compounds, are used in the manufacture of dyes.

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3
Q

amines =

A

weak bases = lone electron pair on nitrogen atom can accept protons

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4
Q

primary aliphatic amines

A

Primary aliphatic amines act as Bronsted-Lowry bases because the
lone pair of electrons on the nitrogen is readily available for forming a
dative covalent bond with a H+
and so accepting a proton. They are
weak bases as only a low concentration of hydroxide ions is produced.

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5
Q

formation of ammonium ions

A

NH3 (aq) +H2O (l) ⇌NH4+(aq)+OH-(aq)

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6
Q

weak to strong base order

A

Aromatic amines < ammonia< primary amines< tertiary amines< secondary amines

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7
Q

amine + acid =

A

ammonium salts

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8
Q

amines

A

mine + acyl chloride or acid anhydride = amide

Naming amines:

e.g 4 carbon chain = butylamine 

Amine attached to benzene ring = phenylamine 

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9
Q

what are aromatic amines

A

Aromatic amines = only when nitrogen is direcetly bonded to benzene ring

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10
Q

content

A

Tertiary amines and amides do not experience hydrogen bonding

Amines can use their lone pairs to act as a nucelophile =electron pair donor

Amines convert CO2 and hydrogen sulfuide into less harmful products

Aliphatic amines –> ammonia –> aromatic amines = decreasing levels of pH

The more avaliable a lone pair is the stronger the base (more likely to accept a proton)

Aliphatic amine: An amine in which nitrogen is bonded only to alkyl groups

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11
Q

nucelophilic addition elimination

A

In nucelophilic addition elimination reactions the amine acts as a nucelophile

Chance of nucelophilic subsitution reactions occuring happens when excess halogenoalkane is used

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12
Q

cattionic surfaces

A

Quaternary ammonium ion salts can be used as cationic sufactants (soaps)

-surfactants = polar end and non-polar end

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13
Q

bonding in amines

A

-H2N – CH3 = methylamine (primary amine)

-hydrogen bonding is present between the molecules of methylamine

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14
Q

aromatic amines

A

-in aromatic amines the nitrogen lone pair is partially delocalised into the benzene ring, lowering base strength

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15
Q

naming amines

A

Amines = count the number of carbons directly attached to the nitrogen atom

-quaternary ammonium salts are not amines

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16
Q

phenylammonium chloride production

A

Phenylamine + HCl –> phenylammonium chloride (Water soluble ionic salt)

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17
Q

6 strong bases

A

-LiOH = lithium hydroxide

-NaOH = sodium hydroxide

-KOH = potassium hydroxide

-Ca(OH)2 = calcium hydroxide

-Sr(OH)2 = strontium hydroxide

-Ba(OH)2 = barium hydroxide

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18
Q

strength of bases

A

Dimethylamine is a stronger base than methylamine as it is secondary amine. Has a greater electron density on the nitrogen lone pair and therefore a greater ability to accept a proton (H+)

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19
Q

comparing strengths

A

-alkyl group release electrons away from alkyl group and towards the nitrogen atom (positive inductive effect)

-this inductive effect increases the electron density on the nitrogen atom (better electron pair donor_

-primary amine is stronger than NH3

-secondary amine is stronger than primary amine

-teritary amine is not a stronger base (less soluble in water)

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20
Q

teritary amines

A

-not stronger bases than secondary amines bc they are less soluble in wtaer

-teritary amines don’t have N-H bonds

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21
Q

Whats stronger propylamine or phenylamine:

A

Whats stronger propylamine or phenylamine:

-propylamine = lone pair on phenylamine is partially delocalised into the benzene ring

-propylamine has a greater eletron density on the nitrogen lone pair

-propylamine has a greater ability to accept H+

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22
Q

more alkyl groups

A

The more alkly groups attached to the amine the more strongly proton is attracted to the amine. The strength of bases is determined by how readily it will accept a proton

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23
Q

nucleophilic subsitution

A

Nucelophilic subsitution:

-haloalkane + ammonia = primary amine and ammonium salt

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24
Q

reduction of nitriles

A

-reducing nitriles via hydrogenation can produce amines

-requires LiAlH4 (reducing agent)

-R-C=N + reducing agent) –> R-CH2-NH2

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25
production of aromatic amines
Production of aromatic amines: -produced from the reduction of nitrobenzene using HCl and a tin catalyst
26
CH3Br --> C2H3N
KCN alcohol
27
C2H3N --> C2H7N nitrile --> amine
H2 Ni catalyst
28
Butylamine can also be prepared in a 2 step synthesis starting from 1-bromopropane
step 1 = CH3CH2CH2Br + KCN --> CH3CH2CH2CN + KBr step 2 = CH3CH2CH2CN + 2H2 --> CH3CH2CH2CH2NH2
29
why is butylamine a stronger base than ammonia
-more alkly groups = lone pair on nitrogen increases electron density due to R group -positive inductive effect
30
identify a substance that could be added to aqueous butylamine to produce a basic buffer solution
HCl
31
teritary amine isomer of butylamine
CH3CH2N(CH3)2
32
ionic salt =
more soluble in the body so can be used in medicinal treatments
33
give the reagents and equations needed for the nitration of benzene
Reagents = conc HNO3 + conc H2SO4 equation = HNO3 + 2H2SO4 --> NO2+ + H3O+ + 2HSO4-
34
reagents required for the reduction of nitrobeznene to phenylamine
Ni catalyst H2
35
Give a use for J =
cationic surfatant / making dyes
36
why is J a weaker base than K
-lone pair on nitrogen -delocalised into ring -less avalable for protonation
37
name the compound (CH3)2NH
dimethylamine
38
(CH3)2NH can be formed by the reaction of excess CH3NH3 with CH3Br. Name and outline the mechanism
nucelophilic subsitution
39
suggest how the reaction conditions could be modified o minimise this contamination
use a large excess of NH3
40
give a reagent that could be used in a test-tube reaction to distinguish between benzene and cyclohexene
bromine water benzene = no reaction cyclohexene = brown to colourles
41
C6H5NH2 =
phenylamine
42
C6H5 =
phenyl
43
nitrogen =
lewis base (electron donor) bronstey lowrd base = proton acceptor
44
amines =
higher boiling point due to hydrogen bonding heavier molecules = less solubility
45
ammonia = base and acid
CH3NH2 + H2O --> CH3NH3+ + OH- C6H5NH2 + HCl --> C6H5NH3+Cl-
46
C6H5NH3+Cl-
phenylammonium chloride
47
R group =
increases electron density on the nitrogen
48
impurity of propylamine
(C3H7)2NH
49
nitrile to amine =
reduction H2/Ni catalyst
50
why is phenylammonium chloride soluble in water
ionic salt = soluble in water
51
1,6dibromohexane
Br(CH2)6Br
52
1,6,diaminohexane
H2N(CH2)6NH2
53
nitrile to amine
reduction reaction
54
true or false = amines are weak bases
true = lone pair electron on nitrogen can accept protons
55
alkyl groups =
positive inductive effect push electron density towards nitrogen making it more available
56
reversible equations
CH3NH2 +H2O ⇌CH3NH3++OH- NH3 (aq) +H2O (l) ⇌NH4+(aq)+OH-(aq)
57
bromoethane + ammonia equation
CH3CH2Br + 2NH3 ---> CH3CH2NH2 + NH4Br
58
making buffer from an amine
Basic buffers can be made from combining a weak base with a salt of that weak basee.g. Ammonia and ammonium chloride Methylamine and methylammonium chloride Ethylamine and ethylammonium chloride
59
nitrile
-nitrile = C=N -order of strength = phenylamine --< ammonia -> dimethylamine -nucelophilic subsitution = introduce functional groups into a molecule
60
preparation of primary aliphatic amines
-nuceloohilic subsitution -CH3Br + KCN --> CH3C=N + KBr -CH3C=N + 4H2 --> CH3CH3NH2
61
preparation of aromatic amines
-reduction of nitro compounds using tin (Sn) and concentrated HCl -nitrobenzene + 6H2 --> phenylamine + 2H2O -using a large amount of excess ammonia increases the yield of the primary amine as there is a better chance that the halogenoalkane is formed
62
forming secondary amines
-halagoenoalkane + primary amines via an nucelophilic subsitution mechanism -formation of diamines e.g dimethylamines + NH4X
63
forming tertiary amines
-diamines + halogenoalkane = triamines
64
quaternary ammonium salts
Quaternary ammonium salts = cationic surfactant -forms salt N+ with 4 R groups
65
nitriles
-primary aliphatic amines can be prepared by the reduction of ammonia and halogenoalkanes + reduction of nitriles -aromatic amines can be prepared by the reduction of nitro compounds Halogenoalkanes react with cyanide in aqueous ethanol to form a nitrile RBr + CN- --> RCN + Br- RCN + 2H2 --> RCH2NH2
66
reduction of nitrile compounds
Reduction of nitrile compounds = more favourable = doesn’t continue to produce secondary amines. Nitriles = high atom economy Production of primary aliphatic amines -use H2 and Ni catalyst Cyanide = toxic side effects
67
production of aromatic compounds
1) nitration of benzene 2) nitrobenzene to phenylamine HNO3 + H2SO4 Use Sn and HCl as a reducing agent C6H4N2O4 + 12H --> C6H8N2 + 4H2O
68
naming amides
secondary amides are named using an N to designate the alkyl group is on the nitrogen atom
69
CH3-CH2-C=O-NH-CH3 name
N-methylpropananmide
70
state one simple method of distinguishing the 2 solutions
pH meter methylamine would have higher pH than ammonia as it is a stronger base
71
true or false - tertiary amines dont have N-H bonds
true
72
disadvantage of forming a nitrile (CN)
toxic/poisinious
73
disadvantage of using one route excess NH3
further subsitution occurs = not most efficient
74
equation for dibromohexane with excess ammonia
Br(CH2)6Br + 4NH3 --> H2N(CH2)6NH2 + 2NH4Br