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Unit F212: Molecules, Biodiversity, Food and Health

Module 1:

Biological Molecules

(a) Describe how hydrogen bonding occurs between water

molecules, and relate this, and other properties of water, to the roles of

water in living organisms (HSW1);

Hydrogen

bond is a weak interaction that can occur wherever molecules contain a slightly

negatively charged atom bonded to a slightly positively charged hydrogen. The

bonds between oxygen and hydrogen are covalent, with electrons shared between

them. They exist, as oxygen in water is more electronegative than the hydrogen

so a permanent dipole forms. Water is polar as oxygen atom pulls the shared

electrons towards it.

Property

Key Points

Role of water

Good

solvent for charged & uncharged substances

Water molecules are attracted to ions and polar molecules e.g.

glucose

Transport

in blood, xylem and phloem

High

specific heat capacity

4.2kJ are necessary to increase the temperature of 1kg of water

by 1oC. The thermal energy absorbed is used to break hydrogen bonds.

Helps

prevent rapid body temperature changes allowing them to keep their

temperature fairly stable

High

latent heat of vaporization

Much thermal energy is used to cause water molecules to change

to water vapour – this happens in transpiration in plants, and in sweating

and panting in mammals.

Coolant

– water is used efficiently as a small amount of water absorbs and uses as

much thermal energy when water evaporates

High

cohesion

Hydrogen bonds stick water molecules together as they’re polar

Helps

draw up water in xylem, helps water flow making it great for transporting

substances

Can

be reactive/Metabolic

Water reacts with other substances

Involved

in hydrolysis reactions and in photosynthesis

Incompressibility

Outside pressure cannot force water into a smaller space

Hydrostatic

skeleton for some animals e.g. earthworms and provides turgidity in plant

cells

(b) Describe, with the aid of diagrams, the structure of an

amino acid;

H2N-CHR-COOH. Cells

polymerise amino acids into polypeptides by forming peptide bonds. A peptide bond forms between an amine

group and a carboxylic acid group.

(c) Describe, with the aid of diagrams, the formation and

breakage of peptide bonds in the synthesis and hydrolysis of dipeptides and

polypeptides;

Amino acids are linked

together by peptide bonds to form dipeptides (2) and polypeptides (2+). A

molecule of water is eliminated in a condensation reaction. The chemical

addition of water breaks the bond is hydrolysis. Properties of proteins depend

on R group, some are charged (polar) and interact with water hydrophilic and

some aren’t – hydrophobic, and can interact with phospholipids within

membranes. R groups determine active site of an enzyme.

(d) Explain, with the aid of

diagrams, the term primary structure

The sequence of amino acids that make up the protein.

e) Explain,

with the aid of diagrams, the term secondary structure with reference to

hydrogen

bonding

The coiling or pleating of

parts of the polypeptide molecule stabilised by the numerous hydrogen bonds. Formed when the chain of amino

acid coils to form an α-helix or coils to form a

β-pleated sheet. Hydrogen bonds between the amino acids hold the coils in

place.

f) Explain, with the aid of diagrams, the term tertiary

structure, with reference to hydrophobic and hydrophilic interactions,

disulfide bonds and ionic interactions;

Refers to the overall 3D structure of the

final polypeptide or protein molecule. It’s the further coiling and folding of

a polypeptide to give a more complex 3D shape specific to the function. 3D

shape is held in place by a number of different types of bonds and

interactions.

·

Hydrogen bonds between polar

groups. Wherever slightly positively charged groups are found close to slightly

negatively charged groups.

·

Disulfide bonds is when the

amino acid cysteine contains sulfur and 2 cysteines are found close to each

other a covalent bond can form.

·

Ionic bonds between R groups

which ionize to form positively and negatively charged groups that attract each

other.

·

Hydrophobic interactions

between non-polar R groups. Hydrophobic amino acids will be most stable if they

are held together with water excluded. Hydrophilic amino acids tend to be found

on the outside in globular proteins, with hydrophobic amino acids in the

centre.

(Some bonds break when proteins are heated

up or treated with acids and alkalis, the tertiary structure changes and

protein does not function. Destruction of shape and function is denaturation)

g)

Explain, with the aid of diagrams, the term quaternary structure, with

reference to the structure of hemoglobin.

Protein structure where a protein consists

of more than one polypeptide chain. Hemoglobin and insulin both have a

quaternary structure. Hb is made of 4 polypeptide chains bonded together. For

proteins made from 1+ polypeptide chain, the quaternary structure is the

protein’s final 3D structure.

h)

Describe, with the aid of diagrams, the structure of a collagen molecule

Made up of three polypeptide chains, each

about 1000 amino acids that are tightly coiled into a strong triple helix. The

chains are interlinked by strong covalent bonds. Minerals can bind to the

triple helix to increase its rigidity.

(i)

Compare the structure and

function of haemoglobin (as an example of a globular protein) and collagen (as

an example of a fibrous protein);

Haemoglobin

Collagen

Globular transport protein transports

oxygen and carbon dioxide

Fibrous structural protein strengthens

tissues

Soluble in water (good for transport in

the blood)

Insoluble in water

Wide range of amino acid constituents in

primary structure

Approximately 35% of the molecule’s

primary structure is one type of amino acid (glycine)

Contains a prosthetic group – haem

Doesn’t contain prosthetic group

Much of the molecule is wound into alpha

helix structures

Much of the molecule consists of

left-handed helix structures

Complex 3D shape

Triple helix

In red blood cells other globular proteins

e.g. enzymes found in all organisms and plasma proteins and antibodies found

in the blood of mammals

Found in skin, bone, cartilage, teeth,

tendons, muscles, ligaments and walls of blood vessels other fibrous protein

e.g. keratin which is skin, nails and hair and elastin which is in blood

vessel walls and alveoli

(ii)

Describe, with the aid of

diagrams, the molecular structure of alpha-glucose as an example of a

monosaccharide carbohydrate;

Glucose is a source of energy in organisms and is built up into

macromolecules. It is a monosaccharide with six carbon atoms in each molecule.

It is polar so it interacts with water – soluble and suitable for transport in

the blood.

k)

State the structural difference between alpha-

and beta- glucose;

In α-glucose the –OH on carbon 1 is below the plane of the ring.

In β-glucose it is above the chain of the ring. Since respiration is powered by

enzymes, which need to be complementary to the shape of the substrate- glucose,

one enzyme which can bind with α-glucose may not be able to bind with β-glucose

meaning that plants and animals are unable to use β-glucose for respiration.

l)

Describe, with the aid of diagrams, the formation and breakage of glycosidic

bonds in the synthesis and hydrolysis of a disaccharide (maltose) and a

polysaccharide (amylose);

Two monosaccharide molecules can be joined

together in a condensation reaction forming a disaccharide. A new covalent bond

called a 1,4-glycosidic bond forms between C1 on one glucose and C4 on

another and water is eliminated (released) and the

reverse hydrolysis reaction uses a water molecule to break the glycosidic bond.

Two α-glucose molecules are joined

together by a glycosidic bond to form maltose

Lots of α-glucose molecules are joined together by 1,4-glycosidic

bonds to form amylose

m)

Compare and contrast the structure and functions of starch (amylose) and

cellulose;

Amylose

Cellulose

Monomer α-glucose

Monomer β-glucose

Unbranched helix so compact

Unbranched straight chain

Angles of 1,4-glycosidic bonds give it a

coiled structure

β-glycosidic bond can only be broken down by cellulose enzymes,

which humans do not. Because the H and OH groups on the C1 and C4 atoms are

on opposite sides of the ring, each alternating molecule is rotated through

180°

Energy storage in plants

Forms plant cell walls

Insoluble in water

Many microfibrils are held together by hydrogen bonds to form

macrofibrils.

n) Describe, with the aid of diagrams, the structure of

glycogen;

Made up of α-glucose subunits. Similar to amylopectin but 1-4 linked glucose chains in

glycogen is shorter and have more branches extending from chain. Forms glycogen

granules in animal cells, in liver and muscle cells, it is the energy store in

animals, they have higher energy demand than plants so needs to be able to

break down glycogen quickly into many glucose molecules. It is a compact

molecule. They can easily be broken off from ends to provide glucose for

respiration when required.

o) Explain how the structures of glucose,

starch (amylose), glycogen and cellulose molecules relate to their functions in

living organisms;

Glucose

Simplest sugar,

small carbohydrates, soluble and crystalline. Used in respiration.

Starch (Amylose)

Amylose in starch

is unbranched and coiled à compact àgood energy storage to break the

glycosidic bonds and release glucose quickly. It’s insoluble so It can be

stored in cells without causing water to enter by osmosis which would cause them

to swell, doesn’t affect water potential of the cell.

Glycogen

Highly branched

so can be broken down into many glucose molecules quickly

Cellulose

Macrofibrils have

great mechanical strength due to large number of hydrogen bonds helps cell

walls provide strength to each cell, allows water to easily move through and

along macrofibrils and prevents cell from bursting in turgid conditions.

p)

Compare, with the aid of diagrams, the structure of a triglyceride and a

phospholipid

Glycerol is a three-carbon compound, each

carbon atom ha a hydroxyl group (-OH) which can react with a fatty acid to form

an ester bond. It has a glycerol + 3 fatty acids. Phospholipid has a glycerol +

2 fatty acids + a phosphate group.

q)

Explain how the structures of triglycerides, phospholipid and cholesterol

molecules relate to their functions in living organisms;

Triglycerides

Phospholipid

Cholesterol

Long hydrocarbon

fatty acid tails contain lots of chemical energy so releases lots of energy

when respired.

Insoluble so

doesn’t affect water potential of cell

Hydrophilic

phosphate head and hydrophobic tail repelled forms stable bilayers when in

water – an ideal structure for cell membranes as it creates barrier, as

water-soluble substances can’t easily pass through.

Small size and flattened

shape allows cholesterol to fit in between phospholipid molecules. Bind to

hydrophobic tails causing them to pack more closely.

Thermal

insulators, buoyancy, protecting vital organs, energy store

Maintain membrane

fluidity even in low temperatures, no. of unsaturated fatty acid tails is

increased

Controls membrane

fluidity. Gives membrane strength and stability.

r)

Describe how to carry out chemical tests to identify the presence of the

following molecules:

·

Protein (biuret test) – if

present turns from pale blue à lilac/violet/purple

·

Reducing sugars (Benedict’s

test) – Add Benedict’s solution (Cu2+), heat to 80oC from blue à orange-red with a precipitate (of red copper (l) oxide). If not the

blue colour will have no change.

·

Non-reducing sugars (Benedict’s

test) – If reducing sugars test is negative boil with hydrochloric acid, cool

and neutralise with sodium hydrogencarbonate/alkali before doing Benedict’s

test. Repeat Benedict’s test.

·

Starch (Iodine solution) –

turns from yellow à blue black

·

Lipids (emulsion test) – Add

ethanol to dissolve lipid and pour alcohol into water in another test tube. White

emulsion forms near top of water if lipid is present.

s)

Describe how the concentration of glucose in a solution may be determined using

colorimetry

Colorimeter measures the degree of

‘blueness’ of the solution remaining after the reducing sugar has reacted with

Benedict’s solution. After boiling a small sample of test solution with a

larger volume of Benedict’s solution, filter the test solution to remove the

precipitate. Solution that remains blue has no reducing sugar in it and gives a

high reading for absorbance in colorimeter. A solution that has no blue colour

has a high concentration of reducing sugar and gives a low reading for

absorbance.

Nucleic

Acids

a)

State that deoxyribonucleic acid (DNA) is a polynucleotide, usually double

stranded, made up of nucleotides containing the bases adenine (A), thymine (T),

cytosine (C) and guanine (G);

b)

State that ribonucleic acid (RNA) is a polynucleotide, usually single stranded,

made up of nucleotides containing the bases adenine (A), uracil (U), cytosine

(C) and guanine (G);

Each nucleotide consists of pentose sugar

(with 5 carbon atoms), phosphate and nitrogen-containing (nitrogenous) base.

c) Describe,

with the aid of diagrams,

How

hydrogen bonding between complementary base pairs (A to T, G to C) on two

antiparallel DNA polynucleotides leads to the formation of a DNA molecule,

There

are two types of nucleotide bases – pyrimidines and purines. The bases always

arranged so that a purine is opposite pyrimidine: A-T, G-C. There are 2

hydrogen bonds between A and T and three between G and C. The bases are

complementary in size and shape so that only the pairings A-T and G-C fit into

the space between the sugar-phosphate backbone of DNA.

And

how the twisting of DNA produces its ‘double-helix’ shape;

The antiparallel chains twist like a rope

ladder to form the final structure

d)

Outline, with the aid of diagrams, how DNA replicates semi-conservatively, with

reference to the role of DNA polymerase;

·

Double helix unwinds and DNA

‘unzips’ as hydrogen bonds between the polynucleotides are broken.

·

Existing polynucleotides act as

templates for assembly of nucleotides

·

Free nucleotides, which have

been made in the cytoplasm, move towards the exposed bases of DNA

·

Base pairing occurs between

free nucleotides and exposed bases. A matches T and C matches G. Hydrogen bonds

form between complementary bases.

·

Enzyme DNA polymerase forms

covalent bonds between free nucleotides attached to each template

·

Two daughter DNA molecules form

separate double helices.

Proofreading enzyme DNA polymerase prevents

mistakes, the ‘wrong’ nucleotide is cut out and the correct one is inserted.

e)

State that a gene is a sequence of DNA nucleotides that codes of a polypeptide;

f) Outline the roles of DNA and RNA in living

organisms (the concept of protein synthesis must be considered in outline

only).

Protein synthesis

Vital for living

organisms to produce proteins in order to grow and develop. DNA carries

instructions to make proteins found in the nucleus. Organelles that make

proteins are found in cytoplasm but DNA molecules are too large to move out of

nucleus so sections of DNA are copied into RNA. RNA leaves nucleus and joins

with a ribosome used to synthesise a protein.

·

(Transcription) Sequence coding

for a particular protein (a gene) can be exposed by splitting the hydrogen

bonds that hold the double helix together in that region

·

RNA nucleotides form a

complementary strand (mRNA). This mRNA molecule is a copy of the DNA coding

strand or gene

·

mRNA peels away from DNA and

leaves the nucleus through a nuclear pore

·

mRNA attaches to a ribosome

·

Then tRNA molecules bring amino

acids to the ribosome in the correct order, according to the base sequence on

the mRNA

·

(Translation) Amino acids are

joined together by peptide bonds to give a protein with a specific primary

structure (which then gives rise to the secondary and tertiary structures)

Types of RNA

Structure of polynucleotide

Function

Messenger RNA (mRNA)

Variable length, no base pairing

Transfers genetic information from DNA to ribosomes,

after which it is broken down

Transfer RNA (tRNA)

Clover-leaf structure with some base pairing

Carries amino acids to ribosomes

Ribosomal RNA (rRNA)

Folded and attached to proteins to make ribosomes

Provides site for assembly of amino acids to make

proteins

Enzymes

a) State that enzymes are globular proteins, with a

specific tertiary structure, which catalyse metabolic reactions in living

organisms;

b) State that enzyme action may be intracellular or

extracellular;

c) Describe, with the aid of diagrams, the mechanism

of action of enzyme molecules, with reference to

specificity: The active

site of an enzyme is a specific shape, depending on the reaction that it catalyses,

meaning that other molecules won’t fit into the active site.

Active site: The

area on an enzyme to which the substrate binds

Lock and key

hypothesis: The active site has a complementary shape to the substrate molecule

so that binding is much easier as substrate fits into enzyme, this is described

as being like a lock and key. The substrate is then held in place so that the

reaction can occur.

Induced-fit

hypothesis:

AS Unit F322: Chains, Energy and Resources

Basic Concepts

a)

Interpret and use the terms:

(i)

Empirical formula is the

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

(ii)

Molecular formula is the actual

number of atoms of each element in a molecule

Relative molecular mass/Relative empirical mass=n then

empirical formula xn

(iii)

General formula is the simplest

algebraic formula of a member of a homologous series, i.e. for alkanes: CnH2n

+ 2 , Alkenes: CnH2n , Alcohols: CnH2n+1OH

(iv)

Structural formula is the minimal detail that shows the

arrangement of atoms in a molecule. E.g. for butane: CH3CH2CH2CH3 or

CH3(CH2)2CH3

(v)

Displayed formula is the relative positioning of atoms and

the bonds between them.

(vi)

Skeletal formula is the simplified organic formula, shown

by removing hydrogen atoms from alkyl chains, leaving just a carbon skeleton

and associated functional groups.

b)

Interpret, and use, the terms:

(i)

Homologous series is a series

of organic compounds having the same functional group but with each successive

member differing by CH2.

(ii)

Functional group as a group of

atoms responsible for the characteristic reactions of a compound.

c)

Use the general formula of a

homologous series to predict the formula of any member of the series.

They contain same

functional group, have similar properties but each successive member of a

homologous series differs by one carbon and two hydrogen atoms: a –CH2-

group.

d)

State the names of the first

ten members of the alkanes homologous series.

No. Of carbons

Name

Formula

1

Methane

CH4

2

Ethane

C2H6

3

Propane

C3H8

4

Butane

C4H10

5

Pentane

C5H12

6

Hexane

C6H14

7

Heptane

C7H16

8

Octane

C8H18

9

Nonane

C9H20

10

Decane

C10H22

e)

Use IUPAC rules of nomenclature

for systematically naming organic compounds:

Nomenclature is a system

of naming compounds.

-Identifying parent chain

(longest) and assign appropriate stem

-Add a suffix or prefix to

stem to indicate the functional group

-Add a prefix to show the

identity of any alkyl groups and side chains and use multipliers such as di,

tri, tetra when more than one of the same type of side chain is used

-Number the position of

all side chains and functional group, if required

If the suffix starts with a

vowel, remove the final e from the alkane suffix e.g. methaneal à methanal

A stem tells us the longest

no. of carbon atoms in a chain.

A prefix or suffix is used

for different functional groups, alkyl groups are also given stems.

An Alkyl group is an alkane

with a hydrogen atom removed.

No. Of carbons

Stem

Alkyl

group

1

meth-

methyl

2

eth-

ethyl

3

prop-

propyl

4

but-

butyl

5

pent-

pentyl

6

hex-

hexyl

7

hept-

heptyl

8

oct-

octyl

9

non-

nonyl

10

dec-

decyl

Type of compound

Functional group

Prefix

Suffix

Comments

Alkane

C-C

-ane

Alkene

C=C

-ene

Halogenoalkane

-F

-CL

-Br

-I

Fluoro-

Chloro-

Bromo-

Iodo-

Alcohol

-OH

Hydroxyl-

-ol

Hydroxyl used if there are other

functional groups also present

Aldehyde

-CHO

-al

No no. ever required as the functional

groups is always on carbon 1

Ketone

C-CO-C

-one

Carboxylic acid

-COOH

-oic acid

f)

Describe and explain the terms:

(i)

Structural isomers as compounds

with the same molecular formula but different structural formulae

(ii)

Stereoisomers as compounds with

the same structural formula but with a different arrangement in space

(iii)

E/Z isomerism

as an example of stereoisomerism, in terms of restricted rotation about a

double bond and the requirement for two different groups to be attached to each

carbon atom of the C=C group

(iv)

Cis-trans isomerism as a

special case of E/Z isomerism

in which two of the substituent groups are the same;

g)

Determine the possible

structural formulae and/or stereoisomers of an organic molecule, given its

molecular formula;

Example:

Molecular formula

Structural isomers

C4H10

CH3CH2CH2CH3

CH3CH(CH3)CH3

butane

2-methylpropane

C3H7Cl

CH3CH2CH2Cl

CH3CHClCH3

1-Chloropropane

2-chloropropane

C4H9OH

CH3CH2CH2CH2OH

CH3CH2CH(OH)CH3

(CH3) 3COH

(CH3) 2CHCH2OH

butan-1-ol

butan-2-ol

2-methylpropan-2-ol

2-methylpropan-1-ol

h)

Describe the different types of

covalent bond fission:

(i)

Homolytic fission forming two

radicals:

Each bonded atom takes one of the shared pair of

electrons. Each atom has a radical. Two species of the same type are

produced. X-Y à X + Y (2 radicals)

Homolytic fission is the breaking of a covalent bond, with one of

the bonded electrons going to each atom, forming 2 radicals.

(ii)

Heterolytic fission forming a

cation and an anion;

Heterolytic fission is the breaking of a covalent bond

with both of the bonded electrons going to one of the atoms, forming a cation

(+ion) and an anion (-ion).

One of the bonded atoms takes both of the shared paid of electrons.

Two ions are produced: Atom that takes both shared electrons becomes negatively

charged ion (anion), atom that doesn’t take the shared electrons becomes a

positively chared ion (cation). X-Y à X+ + Y- (2 ions)

(j)

Describe a ‘curly arrow’ as the movement

of an electron pair, showing either breaking or

formation of a covalent bond

A curly arrow will be used in both addition and

substitution mechanisms, arrow goes to the electrophile in addition reactions

and arrow goes away from nucleophile in substitution reactions.

An electrophile is an atom (or group of atoms) that is

attracted to an electron-rich centre or atom, where it accepts a pair of

electrons to form a new covalent bond. An atom with a δ+ partial charge. E.g. Br2,

HBr, NO2

An

addition reaction is where a reactant is added to an unsaturated molecule to

make a saturated molecule. 2 reactants à 1 product

E.g.

Ethene + Bromine (electrophile) à 1,2-dibromoethane

A

nucleophile is an atom (or group of atoms) that is attracted to an

electron-deficient centre of atom, where it donates a pair of electrons to form

a new covalent bond. δ- partial

charge. E.g. :Br-, :OH-, H2O:, :NH3

A substitution reaction is a reaction in which an atom

or group of atoms is replaced with a different atom or group of atoms. 2

reactants à 2 products.

Elimination reactions is the removal of a molecule from

a saturated molecule to make an unsaturated molecule. 1 reactant à 2 products.

Oxidation is the gain of oxygen or loss of hydrogen.

Also the loss of electrons from the organic compound. [O] is used to show an

oxidizing agent.

E.g. C2H5OH + [O] à CH3CHO + H2O

Reduction is the loss of oxygen or gain of hydrogen.

Also gain of electrons to the organic compound.

Neutralisation: Acid + base à salt + water

(k)

Carry out calculations to determine the

percentage yield of a reaction.

%Yield = (Actual

amount, in mol, of product/theoretical amount in mol, of product) x 100

A 100% percentage yield is when all of the reactants

turn into products (rare). Lower atom economies result because of:

· Equilibrium reactions preventing the

reaction reaching completion

· Side reactions can occur, forming by-products

· Reactants are not pure

· Reactants and products left behind on the

apparatus

· Separation and purification results in loss

of product

Other e.g.: When distilling, not all the distillate was

evaporated from the mixture, when recrystallising, not all the crystals

crystallised from solution, the reaction did not go to completion, the receiver

was not cooled so not all the distillate condensed.

(l)

Explain the atom economy of a

reaction as:

(Molecular mass of the

desired products/sum of molecular masses of all products) x 100%

A reaction may produce by-products,

as well as the product we want to make. The by-products may be useless waste

and disposed of – costly and wasting resources posing environmental problems.

By-products may also be

sold on or used elsewhere in the chemical plant. (minimising waste)

Atom economy describes the

efficiency of a reaction in terms of all the atoms involved

(m) Explain that addition reactions have an atom economy of 100% whereas

substitution reactions are less efficient.

·

Addition reactions will

theoretically have a 100% atom economy.

·

Substitution and elimination

reactions have lower atom economies.

(n)

Carry out calculations to

determine the atom economy of a reaction;

(o)

Describe the benefits of

developing chemical processes with a high atom economy in terms of fewer waste

materials;

A higher atom economy

reduces the waste of the Earth’s resources.

(p)

Explain that a reaction may

have a high percentage yield but a low atom economy.

A reaction can have a high

percentage yield but a low atom economy.

Hydrocarbons from crude oil

(a)

Explain that a hydrocarbon is a

compound of hydrogen and carbon only;

They are extracted from

crude oil

(b)

Explain the use of crude oil as

a source of hydrocarbons, separated as fractions with different boiling points

by fractional distillation, which can be used as fuels or for processing into

petrochemicals;

It is a fossil fuel made

naturally from decayed plant & animal material. Takes millions of years to

form. Mixture of about 150 hydrocarbons, most are straight chained and

unbranched.

· Takes

place in a fractionating column. Crude oil is heated and passed into the column

· The column is hottest at the bottom and

coolest at the top

· The

crude oil gasses pass through bubble caps into a series of compartments

· Eventually,

each hydrocarbon reaches a compartment with a temperature below its boiling

point

· The

particular hydrocarbon then condenses in that compartment

· The

liquid fractions are tapped off into storage compartments

· Short-chained

hydrocarbons with lower boiling points condense near the top of the column

· Long-chained

hydrocarbons with higher boiling points condense near to the bottom

· Gasses

do not condense and pass through an outlet at the top of the column as

‘petroleum gas’

· The

residue from the process is bitumen, which is removed from the bottom of the

column. It can be used in road surfacing

· Fractions

obtained can be used as fuels or may be further separated in a dedicated

fractionating column, to obtain pure hydrocarbons. Processed further to make

petrochemicals – chemicals made from natural gas and oil, distilled further to

form pure liquids in other fractionating columns, operating over a narrower

range of temperatures.

(c)

State that alkanes and

cycloalkanes are saturated hydrocarbons

Alkanes are a homologous

series of hydrocarbons. Alkanes and cycloalkanes have no multiple bonds in

their molecules and are therefore saturated hydrocarbons.

(d)

State and explain the

tetrahedral shape around each carbon atom in alkanes

All bon angles are 109.5o

- tetrahedral shape gives the

maximum separation of bond-pair electrons. Carbon atom in centre with the other

atoms at each of the four corners.

The only bonds present in

the molecules are either C-C or C-H bonds that are non-polar making alkanes and

cycloalkanes unreactive – they have few reactions. Since they are non-polar

molecules, the forces between molecules are weak van der Waals’ forces.

Uses: fuels and feedstock

for production of other chemicals. Longer-chain alkanes are cracked to give

alkenes and shorter alkanes that are used as fuels.

(e)

Explain, in terms of van der

Waals’ forces, the variations in the boiling points of alkanes with different

carbon-chain length and branching;

·

The longer the carbon chain

length, the greater the surface area of the molecule that can take part in

intermolecular bonding, and the greater the van der Waals’ forces.

·

When alkanes are branched, the

surface area is less and consequently the van der Waals’ forces are weaker.

Hydrocarbons as fuels

(f)

Describe the combustion of

alkanes, leading to their use as fuels in industry, in the home and in

transport;

(g)

Explain, using equations, the

incomplete combustion of alkanes in a limited supply of oxygen and outline the

potential dangers arising from production of CO in the home and from car use;

AS Unit G482: Electrons, Waves and

Photons

Electric current

Candidates should be able to:

(a) Explain that electric current is a

net flow of charged particles;

An electric current is a

flow of charge. Current is measured in amps. One amp is one coulomb per second.

Charge cannot get used up therefore current is conserved.

(b) Explain that electric current in a

metal is due to the movement of electrons, whereas in an electrolyte the

current is due to the movement of ions;

In metals charge is carried by electrons, which are negatively

charged. In an electrolyte, the mobile charges are the positive and negative

ions.

(c) Explain what is meant by

conventional current and electron flow;

Conventional current is from positive to negative, but in fact

electrons flow around the circuit from negative to the positive terminal.

Electron flow is in the opposite direction to conventional current.

(d) Select and use the equation △Q = I△t;

Charge = Current x time

Q = I x t

(e) Define the coulomb;

One coulomb is the total charge supplied by a current of one

ampere in a time of one second.

(f) Describe how an ammeter may be used

to measure the current in a circuit;

To measure current, an ammeter is placed in series in a circuit.

Connected in series as the current it is measuring must flow through it)

(g) Recall and use the elementary

charge e=1.6 x10-19 C;

(h) Describe Kirchhoff’s first law and

appreciate that this is a consequence of

conservation of charge;

The sum of the currents entering any junction is always equal to

the sum of the currents leaving the junction.

(i) State what is meant by the term mean

drift velocity of charge carriers;

Drift velocity (v) is the average velocity of an electron as it

travels through a wire due to a p.d across it.

The conduction of electrons in a metal move around rapidly in

random directions. When a voltage is applied, electrons gain an additional

velocity along the wire. The greater the current (I), the greater the value of

v. Current also depends on cross-sectional area A of the conductor and on the

number density n of electron sin the metal.

(j) Select and use the equation I = Anev;

Current = Cross-sectional area x number density of electrons in

the metal per m3 x electron charge x mean drift velocity

(k) Describe the difference between conductors, semiconductors and

insulators in terms of the number density n.

Metals have high number densities of

electrons so drift velocities are low – of the order of mm s-1v.

Semiconductors have far fewer free electrons (n is much smaller) so drift

velocities are much higher. For insulators, n is close to zero.

Resistance

(a) Recall and use appropriate circuit symbols as set out in SI

Units, Signs, Symbols and Abbreviations (ASE, 1981) and Signs, Symbols and

Systematics (ASE, 1995);

(b) Interpret and draw circuit diagrams using these symbols.

Candidates should be able to:

a) Define potential difference (p.d.);

The p.d. (V) across a

component is the energy transferred per unit charge when electrical energy is

converted into some other form of energy.

b) Select and use the equation W = VQ;

Energy transferred (J) = Voltage x Charge

c) define the volt;

Volt is the unit of potential difference and e.m.f (V)

One volt is one joule per coulomb.

1 V = 1 J C-1

d) Describe how a voltmeter may be used

to determine the p.d. across a component;

A voltmeter must be connected across the terminals of a component

to measure the p.d. across it/it must be connected in parallel with it. Recall:

Voltmeter is measuring the p.d. across the component.

e) Define electromotive force (e.m.f.)

of a source such as a cell or a power supply;

E.m.f is the energy transferred per unit charge when one other

type of energy is converted into electrical energy. Chemical energy to

electrical energy in a cell. E.m.f=electrical energy transferred/charge

W=VQ

(Think as Work, J)

Combing e.m.f.s – when two or more

cells or other sources of e.m.f. are connected in series, their voltages add

up. They must be connected +-to- or the e.m.f.s subtract.

f) Describe the difference between e.m.f. and p.d. in terms of

energy transfer.

Potential difference is the electrical

energy transferred per unit charge when electrical energy is converted into

another form e.g. electrical energy to light energy whereas e.m.f is the energy

transferred when a type of energy is being converted into electrical energy.

(a) Define resistance;

A

property of a component that regulates the electric current through it.

(b) Select and use the equation for

resistance

Resistance = Voltage / Current R(Ω) V / I

(c)

Define the ohm;

Ohm is the unit of resistance 1 Ω = 1VA-1 One volt per amp. Kilo-ohm is 103,

mega-ohm is x106

(d) State and use Ohm’s law;

Ohm’s law states that the current through a conductor is

proportional to the potential difference across it, provided physical

conditions, such as temperature, remain constant. Greater the p.d., the greater

the current. Ohmic conductor is one in which the current that flows is

proportional to the p.d. that pushes it, it obeys ohm’s law. V=IR so resistance

of conductor is constant, doesn’t depend on the p.d. across it.

(e) Describe the I–V characteristics of

a resistor at constant temperature, filament lamp and light-emitting diode

(LED);

For a metallic conductor at constant temperature, an ohmic

conductor gives a straight line through the origin. Metals obey Ohm’s law –

provided the temperature remains constant.

Non-ohmic conductors has an I-V characteristic graph with no

straight line.

Filament lamp (non-ohmic conductor): As metal filament gets hotter with increasing current and

voltage, its resistance increases. The current is less than it would be if it

remained proportional to the voltage.

Semiconductor diode: In the forward direction allow current to flow when the p.d.

is above 0.7V. In the reverse direction, only a very small current flows. A

light-emitting diode (LED) shows similar pattern, it emits light when a

sufficiently large current flows through it.

(f) Describe an experiment to obtain

the I–V characteristics of a resistor at constant temperature,

filament lamp and light-emitting diode (LED);

· An ammeter connected in series measures the current through

the resistor

· A voltmeter connected in parallel measures the p.d. across

the resistor

· Reversing the connection to the resistor makes the current

· Use long thing wire to keep current low so that heating

effect is negligible

· Take readings from voltmeter and ammeter and increase the

no. of cells and take readings for each added cell

· Plot a graph of current against potential difference/I-V

characteristics

(g) Describe the uses and benefits of

using light- emitting diodes (LEDs);

Advantages of LEDs as light sources:

· Requires only low p.d.

· Have a long working life

· Are very robust

· Can be used singly as indicators or in large arrays to

illuminate

· Switch on instantly

Candidates should be able to:

(a) Define resistivity of a

material;

ρ of a wire of length, l, resistance

R and area of cross-section A is given by ρ=RA/l

Resistivity

is the ratio of the product of resistance and cross-sectional area of a

component and its length. Unit of resistivity is Ωm, ohm metres. Resistivity is constant for all materials at a

set temperature.

A

typical value for a good conductor: resistivity of copper = 1.6 x 10-8

Ωm

(b) Select and use the equation Rpl/A

R= resistance p= resistivity l=length

A= cross-sectional area.

R ∝ l : The longer the wire, the greater its

resistance

R ∝1/A : The fatter the wire, the less its resistance

So doubling the length of a wire doubles its resistance;

doubling its cross-sectional area halves its resistance.

(c) Describe how the resistivities of

metals and semiconductors are affected by temperature;

A semiconductor has many fewer conduction electrons than a metal

(number density is less), when a semiconductor is heated, more electrons break

free from their atoms increasing the no. of free electrons that can carry the

current (the number density increases rapidly). So resistivity of the

semiconductor decreases as it is heated – the opposite of a metal.

(d) Describe how the resistance of a

pure metal wire and of a negative temperature coefficient (NTC) thermistor is

affected by temperature.

When a metal is heated, its resistivity increases as temperature

rises atoms of the metal vibrate with increasing amplitude making it more

likely that conduction electrons will collide with the atoms as they try to

travel through the metal. Electrons lose energy when they collide and so we

observe an increase in the metal’s resistivity. Metal’s resistance increases

gradually as temperature is increased.

Negative Temperature Coefficient (NTC) thermistors - Resistance

decreases rapidly over a narrow range of temperature. Resistance goes down as

temperature goes up.

Candidates should be able to:

(a) Describe power as the rate of

energy transfer;

The rate of doing work, measured in Watts, W.

Power (W) = Energy transferred (J)/time

taken (s) P = W/t W = Pt

Energy transferred = power x time

1 watt = 1Js-1

1 kilowatt = 103W = 1000W

1 megawatt = 1MW = 106W = 1000000W

1 gigawatt = 1GW = 109W = 1000000000W

(b) Select and use power equations P = VI, P=I2R

and P=V2/R

The greater the current and p.d V that

is flowing through, the greater the power P.

Power = current x p.d. P=IV combing this with V = IR gives

P=I2R and P=V2/R the equation for energy

transferred W becomes W=current x p.d. x time.

W=IVt

(c) Explain how a fuse works as a

safety device (HSW 6a);

Protects electrical appliances from dangerous overloading by fuse

in plug. Fuse is made of thin copper wire in a ceramic casing. Above a certain

value of current, wire becomes so hot that it melts breaking circuit. Fuse

should be chosen so that its rating is a little higher than the maximum current

drawn by the device when it is operating correctly.

(d) Determine the correct fuse for an

electrical device;

Most household appliances uses fuses of 3A, 5A or 13A and household

mains uses 230V. To work out correct fuse needed use P=IV so I=P/V e.g. if

current worked out is 4.3A, so a 5A fuse is needed.

(e) Select and use the equation W = IVt; (f) define the kilowatt-hour (kW h) as

a unit of

energy;

Unit of energy (kWh) used by

electricity companies when charging for electricity. 1kWh=1000W for

3600s=3.6MJ. 1kWh= 3.6x106J (as 1hour = 3600s and 1kW=1000Js-1)

(g) Calculate energy in kW h and the cost of this energy when

solving problems (HSW 6a).

Energy transferred (kWh) = power (kW) x

time (h)

DC Circuits

(a) State Kirchhoff’s second law and

appreciate that this is a consequence of conservation of energy;

The sum of the e.m.f.s around any circuit loop is equal to the sum

of the p.d.s: ΣE=ΣIR

(E.g. by the time the charge has completed the circuit, it has

lost as much energy as it gained – example of conservation of energy as energy

cannot be created or destroyed in a closed loop)

(b) Apply Kirchhoff’s first and second

laws to circuits;

By applying Kirchhoff’s first law, we know that each resistor has

the same current flowing through it, current is always conserved.

(c) Select and use the equation for the

total resistance of two or more resistors in series;

Rtotal=R1 + R2 +R3 + ….

Resistors in series must have the same current through them and

the p.d. of the supply is shared between them.

So Rtotal = R + r and hence E=I(R+r) where I is current

supplied.

This value is given by V=IR. The voltmeter reading is less than

the e.m.f. of the battery so

E=I(R + r) = IR + Ir

E=V + Ir

(d) Select and use the equation for the

total resistance of two or more resistors in parallel;

I=I1 + I2 + I3 = E/R1

+ E/R1 + E/R3 since I=E/R where R is the total

resistance, we get:

E/R=E/R1 + E/R1 +E/3 and

cancelling throughout by E gives 1/R=1/R1 + 1/R1 + 1/R3

The battery has negligible internal resistance using Kirchhoff’s

second law for the complete loop from the battery to R1 and back again, we find

that the p.d. across the resistor equals the e.m.f. of the battery. This will

be true for each resistor.

Resistors in parallel have the same p.d. across them but the

current flowing from supply is shared between them.

(e) Solve circuit problems involving

series and parallel circuits with one or more sources of e.m.f.;

(f) Explain that all sources of e.m.f.

have an internal resistance;

Same current flows through the interior of the supply (from – to

+). Interior of a supply is made up of chemicals or metal wire and must have

resistance – this is the internal resistance r of the supple.

(g) Explain the meaning of the term terminal

p.d.;

Terminal p.d. is the p.d, across the load (external) resistance.

Terminal p.d. is the energy transferred when 1 coulomb of charge flows through

the load resistance. If there was no internal resistance the terminal p.d.

would be the same as the e.m.f.

(h) Select and use the equations

e.m.f. = I (R + r),

and e.m.f.

= V + Ir .

(a) draw a simple potential divider circuit;

(b) explain how a potential divider circuit can be used to produce

a variable p.d.;

(c) selectandusethepotentialdividerequation Vout R2 Vin ;

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