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A Level Chemistry > 4.1 Core organic chem 11.1-13.5 > Flashcards

4.1 Core organic chem 11.1-13.5 Flashcards

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1
Q

Saturated hydrocarbons

A

Has single bonds only

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2
Q

Unsaturated hydrocarbons

A

Contains carbon to carbon multiple bonds

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3
Q

Homologous series

A

A family of components with similar chemical properties whose successive members differ by the addition of a – CH2 – group

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4
Q

Functional groups

A

The part of the organic molecule that is largely responsible for the molecule’s chemical properties

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5
Q

Hydrocarbons can be classified as:

A

Aliphatic
Alicyclic
Aromatic

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6
Q

Aliphatic

A

Carbon atoms are joined to each other in unbranched (straight) or branched chains, or non-aromatic rings

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7
Q

Alicyclic

A

Carbon atoms are joined to each other in ring structures with or without branches

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8
Q

Aromatic

A

Some or all of the carbon atoms are found in a benzene ring

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9
Q

Three homologous series of aliphatic hydrocarbons:

A

Alkanes
Alkenes
Alkynes

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10
Q

Alkynes

A

Containing at least one triple carbon to carbon bond

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11
Q

Stem of the name

A

Indicates the number of carbon atoms in the longest continuous chain in the molecule

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12
Q

Prefix of the name

A

Can be added before the stem, often to indicate the presence of sidechains or a functional group

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13
Q

Suffix of the Name

A

Added after the stem to indicate functional groups

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14
Q

Aldehyde

A

– CHO
–al
(End carbon atom of a branch, double bond with oxygen and single bond with hydrogen)

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15
Q

Ketone

A

-C(CO)C-
-one
(Middle carbon atom Double bonded with oxygen)

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16
Q

Carboxylic acid

A

– COOH
– oic acid
(End carbon atom double bonded to oxygen and single bonded to OH)

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17
Q

Molecular formula

A

Shows the number and type of atoms of each element present in a molecule

Ethanol is C2H60

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18
Q

Empirical formula

A

The simplest whole number ratio of the atoms of each element present in a compound

Glucose has the molecular formula C6H1206 and therefore the empirical formula CH20

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19
Q

General formula

A

The simplest algebraic formula for any member of the homologous series

Alkanes – CnH2n +2

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20
Q

Displayed formula

A

Shows the relative positioning of all the atoms in the molecule and the bonds between them

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21
Q

Structural formula

A

Uses the smallest amount of detail necessary to show the arrangement of the atoms in a molecule

Butane – CH3CH2CH2CH3

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22
Q

Skeletal formula

A

A simplified organic formula

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23
Q

Structural isomerism

A

Compounds with the same molecular formula but different structural formulae

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24
Q

Types of bond fission

A

Homolytic fission and heterolytic fission

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25
Q

Homolytic fission

A

When a covalent bond breaks by homolytic fission each of the bonded atoms takes one of the shared pair of electrons from the bond

Each atom now has a single unpaired electron
An atom or group of atoms with an unpaired electron is called a radical

H3C – CH3 —> H3C+ CH3
radicals

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26
Q

Heterolytic fission

A

When a covalent bond breaks by heterolytic fission, one of the bonded atoms takes both of the electrons from the bond

The atom that takes both electrons becomes a negative ion
The atom does not take the electrons becomes a positive ion

H3C – Cl—> H3C+ + Cl-

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27
Q

Curly arrows

A

Used to show the movement of electron pairs when bonds are being broken or made

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28
Q

Types of reaction

A

Addition
Substitution
Elimination

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29
Q

Addition reaction

A

Two reactants join together to form one product

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30
Q

Substitution reaction

A

An atom or group of atoms is replaced by a different atom or group of atoms

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31
Q

Elimination reaction

A

Involves the removal of a small molecule with a larger one. In an elimination reaction, one reactant molecule forms two products

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32
Q

Alkanes

A

CnH2n+2

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33
Q

The bonding in alkanes

A

Alkanes are saturated hydrocarbons, containing only carbon and hydrogen atoms joined together by single covalent bonds
Each carbon atom in an alkane is joined to 4 other atoms by single covalent bonds. These are a type of covalent bond are called Sigma bonds

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35
Q

The shape of alkanes

A

Each carbon atom is surrounded by four electron pairs in four Sigma bonds. Repulsion between these electron pairs results in a 3-D tetrahedral arrangement around each carbon atom. Each bond angle is approximately 109.5°

36
Q

Variations in the boiling point of alkanes

A

Boiling point increases with a larger amount of carbon atoms per Alkane

37
Q

Effects of chain length on boiling point

A

London forces act between molecules that are in close surface contact. As the chain length increases, the molecules have a larger surface area, so more surface contact is possible between molecules. The London forces between the molecules will be greater and so more energy is required to overcome the forces

38
Q

Effects of branching on boiling point

A

Isomers of alkanes have the same molecular mass. If you compare the boiling points of branched isomers with straight chain isomers, you find that the branched isomers have lower boiling points
This is because there are fewer surface points of contact between molecules of the branched alkanes, giving fewer London forces. Another factor lies with the shape of the molecules. The branches get in the way and prevent the branched molecules getting as close together as straight-chain molecules, decreasing the intermolecular forces further.

39
Q

Reactivity of alkanes

A

Alkanes do not react with most common reagents. The reasons for the lack of reactivity are:
C – C and C – H Sigma bonds are strong
C – C bonds are nonpolar
Electronegativity of carbon and hydrogen is so similar that the C – H bonds can be considered to be non-polar

40
Q

Combustion of alkanes

A

Despite their low activity, all alkanes react with a plentiful supply of oxygen to produce carbon dioxide and water
All combustion processes give out heat and alkanes are used as fuels because they are readily available, easy to transport and burn in a plentiful supply of oxygen without releasing toxic products

41
Q

Complete combustion of alkanes

A

In a plentiful supply of oxygen, alkanes burn completely to produce carbon dioxide and water

42
Q

Incomplete combustion of alkanes

A

When oxygen is limited during combustion, the hydrogen atoms in the alkane are always oxidised to water, but combustion of carbon may be incomplete, forming the toxic gas carbon monoxide or even a carbon itself as soot.

43
Q

Reactions of alkanes with halogens

A

In the presence of sunlight, alkanes react with halogens. The high energy ultraviolet radiation present in sunlight provides the initial energy for a reaction to take place.
For example, methane react with bromine as shown below
CH4+ Br2 –> CH3Br + HBr

This is a substitution reaction, as a hydrogen atom in the alkane has been substituted by a halogen atom

44
Q

Mechanism for bromination of alkanes

A

A chemical reaction can often be represented by a series of steps showing how electrons are thought to move during the reaction. The mechanism for the bromination of methane is an example of radical substitution
The mechanism takes place in three stages, called initiation, propagation, and termination

45
Q

Step one: initiation (radical substitution)

A

The reaction is started when the covalent bonds in a bromine molecule is broken by homolytic fission . Each bromine atom takes one electron from the pair, forming two highly reactive bromine radicals. The energy for this bond fission is provided by UV radiation.
Br– Br–>Br+BR

46
Q

Alkyl group

A

CnH2n+ 1
An alkyl group has a hydrogen atom removed from an alkane parent chain.
e.g. methyl

47
Q

Step two: propagation (radical substitution)

A

Reaction propagates through to propagation steps, chain reaction
Propagation step one – CH4+ Br* —> *CH3+ HBr
Propagation step two – CH3+ Br2 – CH3Br+ Br
In the first step a bromine radical reacts with a C – H bond in the methane, forming a methyl radical and a molecule of hydrogen bromide
In the second step each methyl radical reacts with another bromine molecule, forming the organic product bromomethane, together with a new bromine radical

48
Q

Step three: termination (radical substitution)

A

To radicals collide, forming a molecule with all electrons paired. There are a number of possible termination steps with different radicals in the reaction mixture

Br+Br—>Br2
*CH3+ CH3 —> C2H6
CH3+Br—>CH3Br
When two radicals collide and react, both radicals are removed from the reaction mixture, stopping the reaction

49
Q

Limitations of radical substitution in organic synthesis

A

Although radical substitution gives us a way of making haloalkanes, this reaction has problems that limit its importance for synthesis of just one organic compound

50
Q

Structure of alkenes

A

Unsaturated hydrocarbons
Contain at least one carbon to carbon double bond in the structure
Aliphatic alkenes that contain more than one double bond have the general formula CnH2n

51
Q

alkenes double bond

A

For each carbon atom of the double bond, three of the four electrons are used in three Sigma bonds, one to the other carbon atom of the double bond and the other two electrons to 2 other atoms
This leaves one electron on each carbon atom of the double bond not involved in Sigma bonds. This electron is in a P orbital. A pi bond is formed by the sideways overlap to P orbitals, one from each carbon atom of the double bond. Each carbon atom contributes one electron to be electron pair in the pi bond. The pi electron density is concentrated above and below the line through the nuclei of the bonded atoms. The P bond locks the two carbon atoms in positioning and prevents them from rotating around the double bond.

52
Q

The shape around a double bond

A

The shape around each of the carbon atoms in the double bond is trigonal planar, because:
There are three regions of electron density around each of the carbon atoms
The three regions repel each other as far apart as possible, so the bond angle around each carbon atom is 120°
All of the atoms are in the same plane

53
Q

Stereoisomers

A

Have the same structural formula but a different arrangement of atoms in space

54
Q

Two types of stereoisomerism

A

E/Zisomerism and optical isomerism

55
Q

E/Z isomerism

A

Stereoisomers around double bonds arises because rotation about the double bond is restricted and the groups attached to each carbon atom are therefore fixed relative to each other. The reason for the rigidity is the position of the pipe on the electron density above and below the plane of the significant
If a molecule satisfies both of the following conditions it will have a E/Z isomerism:
A C = C double bond
Different groups attached to each carbon atom of the double bonds

56
Q

What conditions must a molecule satisfy to have E/Z isomerism?

A

A C = C double bond

Different groups attached to each carbon atom of the double bond

57
Q

Cis- trans isomerism

A

The name commonly used to describe a special case of E/Z isomerism. Molecules must have a C = C double bond and each carbon in the double bond must be attached to 2 different groups. Also in cis – trans isomerism, one of the attached groups of each carbon atom of the double bond must be hydrogen
In cis– trans isomerism:
The cis isomer is the Z isomer
The trans isomer is the E isomer

58
Q

Using the Cahn-Ingold-Prelog rules

A

In this system the atoms attached to each carbon atom in a double bond are given a priority based upon their atomic number
If the groups of higher priority are on the same side of the double bond, the compound is the Z isomer
If the groups of higher priority are diagonally placed across the double bond, the compound is the E isomer

59
Q

The reactivity of alkenes

A

They are more reactive than alkanes because of the presence of the pi bond
The C = C double bond is made up of a Sigma bond and a pi bond, and the electron density is concentrated above and below the plane of the Sigma bonds
Being on the outside of the double bond, the pi electrons are more exposed than the electrons in the sigma bond. A pi bond readily breaks and alkenes undergo addition reactions relatively easily

60
Q

Addition reactions of the alkenes

A

Alkenes undergo many addition reactions for example, with:
Hydrogen in the presence of a nickel catalyst
Halogens
Hydrogen halides
Steam in the presence of an acid catalyst
Each of these reactions involve the addition of a small molecule across the double bond, causing the pi bond to break and for new bonds to form

61
Q

Hydrogenation of alkenes

A

When an alkene, such as propene, is mixed with hydrogen and passed over nickel catalyst at 423 Kelvin, and addition reaction takes place to form an alkane. This addition reaction, in which hydrogen is added across the double bond, is known as hydrogenation
All C=C bonds react with hydrogen in this way

62
Q

Halogenation of alkenes

A

Alkenes undergo a rapid addition reaction with the halogens chlorine and bromine at room temp.
Addition of bromine across the double bond of an alkene. Form alkane w 2 bromine

63
Q

Testing for unsaturation

A

The reaction of alkenes with bromine can be used to identify if there is a C=C bond present and the organic compound is unsaturated
When bromine water (Orange solution) is added dropwise to a sample of an alkene, bromine adds across the double bong. The orange colour disappears, indicating the presence of a C=C bond

64
Q

Addition reactions of alkenes with hydrogen halides

A

Alkenes react with gaseous hydrogen halides at room temp to form haloalkanes
If the alkene is a gas, like ethene, then the reaction takes place when the two gases are mixed
If the alkene is a liquid, then the hydrogen halides is bubbled through it.
Alkenes also react with concentrated hydrochloric or concentrated hydrobromic acid, which are solutions of the hydrogen halides in water

65
Q

Hydration of alkenes

A

Alcohols are formed when alkenes react with steam, H2O, in the presence of a phosphoric acid catalyst, H3PO4
Steam adds across the double bond
This addition reaction is used widely in industry to produce ethanol from ethene. As with the addition of with hydrogen halides, there are two possible products

66
Q

Electrophilic addition reactions

A

Alkenes usually take part in addition reactions to form saturated compounds. The mechanism for this reaction is called electrophilic addition
The double bond in an alkene represents a region of high electron density because of the presence of the pi electrons
The high electron density of the pi electrons attracts electrophiles
An electrophile is an atom or group of atoms that is attracted to an electron rich centre and accepts an electron pair
An electrophile is usually a positive ion or a molecule containing an atom with a partial positive charge (delta positive)

67
Q

The mechanism for a electrophilic reaction

E.g. addition reaction between but-2-ene and hydrogen bromide

A
  1. Bromine is more electronegative than hydrogen, so HBr is polar and contains dipole H(delta plus)— Br(delta minus)
  2. The electron pair in the pi bond is attracted to the partially positive hydrogen atom, causing the double bond to break
  3. A bond forms between the hydrogen atom of the H-Br molecule and a carbon atom that was part of the double bond
  4. The H –Br bond breaks by heterolytic fission, with the electron pair going to the bromine atom
  5. a bromide ion and a Carbocation are formed. A carbocation contains a positively charged carbon atom
  6. in the final step the Br ion reacts with the carbocation to form the addition product
68
Q

Carbocation stability

A

Carbocations are classified by the number of alkyl groups attached to the positively charged carbon atom. An alkyl group is normally represented with the symbol –R. Tertiary carbocations (with three R groups) are the most stable, and primary Carbocations are the least stable
Carbocation stability is linked to the electron – donating ability of alkyl groups. Each alkyl group donates and pushes electrons towards the positive charge of the Carbocation. The positive charge is spread over the alkyl groups. The more alkyl groups attached to the positively carbon atom, the more the charge is spread out, making the ion more stable, therefore tertiary carbocations are more stable than secondary ones, which are more stable than primary ones

69
Q

Addition polymerisation

A

Unsaturated alkene molecules undergo addition polymerisation to produce long saturated chains containing no double bonds
They have high molecular masses
Synthetic polymers are usually named after the monomer that reacts to form their giant molecules, prefixed by poly

70
Q

Poly(ethene)

A

Made by heating a large number of ethene monomers at a high pressure

71
Q

Poly(chloroethene)

A

PVC

Can be prepared to make a polymer that is flexible or rigid

72
Q

Disposing of waste polymers

A

Bruh I cba

73
Q

Addition reaction example

A

But-2-ene +H2O —> butan-2-ol

74
Q

Substitution reaction example

A

1-bromopropane + OH- —> propan-2-ol +Br-

75
Q

Elimination reaction example

A

Propan-2-ol acid—>catalyst propene + H2O

76
Q

Processing waste polymers

A

Disposing of plastics is a big environmental concern. Processing waste polymers more sustainably would benefit the environment.

77
Q

Disposing of waste polymers

A

The unreactivity of plastics is a useful property while they are being used, but it makes their disposal more of a challenge.
Landfill sites are a common way of disposing of plastics, but conventional polymers will not break down for hundreds of years. This is not a sustainable solution because there is limited space for landfill sites.
Finding a way to process polymers more sustainably would be of great benefit to the environment.

78
Q

Combusting waste polymers

A

Combustion of waste polymers in an incinerator disposes of the polymers, but it also produces energy that can be used.
This would lower our reliance on fossil fuels.

79
Q

Organic feedstock

A

Waste polymers could be used as organic feedstock to produce plastics and other organic compounds.
Feedstocks are the starting materials in a manufacturing process, and organic feedstocks are made of organic material.

80
Q

Removing toxic waste products

A

Combusting halogenated plastics (like PVC) produce toxic waste products such as HCl gas.
Removing the toxic products by neutralising them can be expensive, but is necessary to ensure that the combustion is safe.

81
Q

Bio and photodegradable polymers

A

Conventional polymers take hundreds of years to break down. Biodegradable and Photodegradable polymers are designed to break down more quickly, reducing the burden on the environment.

82
Q

Biodegradable polymers

A

Biodegradable polymers can be broken down by microorganisms.
Biodegradable bags are used to collect food waste in households, and they can be put into the compost along with the contents, there is not need to separate the two.
In the right conditions, microorganisms can break down the polymers completely into carbon dioxide and water.
Burying plastics at landfill sites can create an anaerobic environment, which reduces the capacity of microorganisms to decompose the polymer.

83
Q

Photodegradable polymers

A

Photodegradable polymers can be broken down into smaller pieces by sunlight.
The idea is that smaller pieces will be able to biodegrade more easily.
Sometimes, the smaller pieces just accumulate in the environment without getting broken down further. This causes problems, particularly in oceans, where animals ingest plastics and they build up in the food chain.
If photodegradable plastic is buried at a landfill site, it is unlikely to be exposed to sufficient sunlight to break up into small pieces.

84
Q

What is the name of the addition polymer used in bulletproof armour?

A

Kevlar

A Level Chemistry (16 decks)

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