Sheet 1 of 11 RI 2010 Year 4 Chemistry Speed Revision Notes v7: A Learner-Outcome/Reference-based Approach in 7495 words and 11 sheets

Drawing (that’s right) courtesy of Daniel Siew (RI4H’10)

RI 2010 Year 4 Chemistry Speed Revision Notes v7: A Learner-Outcome/Reference-based Approach in 7495 words and 11 sheets best used as a supplement alongside your Chemistry file (there are things which I’ve omitted because they’re relatively simple or easy to derive. I’ve tried to cover everything in the EOY SoW though. I’m pretty sure reading these 11 pages will be sufficient to jog your memory and help you with rote-memorising stuff like QA! Brevity: Wit’s soul.)

CC BY-NC-SA 3.0 (SIN) DISCLAIMER: Use these notes at your own risk. Questions? Comments? Errata? Send to:

ADVANCE GRATITUDE to: Mr Ong, for the help and guidance Friends and batchmates who liked the MYE/Y3 EOY’10 incarnation and helped with content/proofreading Daniel Siew for the dove drawing Sources, (hopefully) credited where appropriate (most content is paraphrased from the worksheets) The Gentle Reader, for reading this. My apologies if the notes sound a little self-indulgent ._. List of Topics A. Periodic Table B. Salt Preparation C. Qualitative Analysis D. Metal Reactivity E. Metal Extraction F. Electrolysis G. Chemical Energetics H. Reaction Kinetics I. Chemical Equilibria J. Organic Chemistry K. Others

A. Periodic Table A0. Summary of Periodic Trends (adapted from http://www.seab.gov.sg/aLevel/20102011Syllabus/9647_2011.pdf)

A-1. Redox rules that you absolutely must not forget Oxidation Reduction

OIL (e-) RIG (e-) Loss of H Gain of H Gain of O Loss of O

A1. Definitions Actual Nuclear Charge (Z)

Proton number of atom

Effective Nuclear Charge (Zeff)

Resultant +ve charge acting on a valence shell electron Zeff = Z – S

Shielding Electrons (S)

Electrons not in the valence shell of the atom they shield the outer shell electrons from the attraction force of +ve nuclear charge

1st I.E. Measure of energy (usually in kJ/mol) required to remove one electron from one mole of X (g)

A2. Periodic Trends Across a period

(L R) Down a group

Metallic character

Decreases -

Atomic radius

Decreases (higher Zeff greater electrostatic attraction between e- and +ve nucleus)

Increases (more electron shells)

Ionic radius

Decreases (for isoelectronic ions as proton number increases)1

Increases (more electron shells)

1st I.E.2 Increases (higher Zeff more energy required to ionise)

Decreases (increasing distance of valence e- from nucleus)

Z Increases (higher proton number)

Increases (higher proton number)

S Constant Increases (more e-) Zeff Increases (since

Z increases but S remains constant)

Constant (but distance prevails down group, e- lost more readily

1 Isoelectronic = same electronic configuration. P3- with electronic configuration 2.8.8 obviously isn’t going to be smaller than Al3+ with 2.8 2Smaller I.E. easy e- loss OIL easily oxidised

A3. Oxides (hydroxides behave the same way) Basic Amphoteric Acidic Neutral

Metals Group I and II, transition metals

ZAP: Zn, Al, Pb

Non-metals CO2, SO2, SO3, NO2, SiO2, P4O10

Other non-metals CO, NO, O2 ^^

Reacts with converse Yields H+ (aq)/OH- (aq) Basic CaO (s) + 2HCl (aq) CaCl2 (aq) + H2O (l)

CaO (s) + H2O (l) Ca(OH)2 (aq)

Amphoteric ZnO (s) + 2HCl (aq) ZnCl2 (aq) + H2O (l) ZnO (s) + 2NaOH + H2O Na2Zn(OH)4 (aq)

-

Acidic SO2 (g) + 2NaOH (aq) Na2SO3 (aq) + H2O (l) SiO2 (s) + hot conc.1 2NaOH (aq) Na2SiO3 (aq) + H2O (l)

CO2 (g) + H2O (l) H2CO3 (aq) SO2 (g) + H2O (l) H2SO3 (aq) SO3 (g) + H2O (l) H2SO4 (aq) NO2 (g) + H2O (l) HNO3 (aq) P4O10 (s)2 + 6H2O (l) 4H3PO4 (aq)

1 Else your flasks will start to dissolve in NaOH. 2 Phosphorous pentoxide can react with carboxylic acids: P4O10 + RCO2H → P4O9(OH)2 + [RC(O)]2O. AARGH Summary: Acidic/basic oxides (1) react with alkalis/bases to form a salt, or (2) form an acid/alkali in water.

Sheet 2 of 11 RI 2010 Year 4 Chemistry Speed Revision Notes v7: A Learner-Outcome/Reference-based Approach in 7495 words and 11 sheets

A4. Group Characteristics Group I

Alkali Metals Transition Metals Group VII

Halogens Group VIII

Noble Gases M.p./b.p. Decreases down group (more

electron shells greater distance between positively charged nuclei and negatively charged sea of delocalised electrons weaker electrostatic forces of attraction less thermal energy needed to overcome forces decrease in m.p./b.p.)

- Increases down group (more electron shells larger surface area larger IM forces more thermal energy needed to overcome increase in m.p./b.p.)

Increases down group (more electron shells larger surface area larger IM forces more thermal energy needed to overcome increase in m.p./b.p.)

Oxidation State

+1 Variable -1 (usually) 0

Reactivity Reactive (reactivity increases down group; more electron shells

greater distance between nucleus and valence electrons lower I.E. higher reactivity) Good reducing agents; easily oxidized (OIL, low I.Es) see A2

Moderately reactive; see reactivity series!

Reactive (reactivity decreases down group; more electron shells greater distance between nucleus and outer electron shells lower E.A. lower reactivity) Good oxidizing agents; easily reduced (RIG, high electron affinity - E.A.)

Mostly unreactive due to stable octet structures [NOT completely filled valence shells! Look at Kr; its 4th shell isn’t full (2.8.18.8), and neither is Xe’s 5th shell (2.8.18.18.8)]

Catalytic Property

None MnO2 (s): 2H2O2 2H2O (l) + O2 (g) Fe (s): Haber Process (NH3 (g)) Ni (s): Hydrogenation Cu2+ (aq): Acid + zinc Pt (s): Petroleum cracking

None None

Appearance and Density

Low density solids Appears as soft, silvery metals

High density solids Appears as hard, coloured metals

Range from gases (F2, Cl2) to liquids (Br2) to solids (I2) Colour darkens down group (F2: yellow; Cl2: greenish; Br2: brown; I2: black/purple)

Colourless gases

With water Forms corresponding hydroxide (alkali)

No reaction (but an aqueous solution of its ions is usually coloured)

Dissolves (this is different from formation of halide ions X- (aq))

-

Bonding Metallic (positively charged nuclei in a sea of negatively charged delocalized electrons)

Metallic (positively charged nuclei in a sea of negatively charged delocalized electrons)

Intramolecular: Covalent; diatomic molecules Intermolecular: van der Waals IM forces of attraction (instantaneous dipole-induced dipole interaction)

Monoatomic molecules “Intermolecular”: van der Waals IM forces of attraction (instantaneous dipole-induced dipole interaction)

A5. Halogens More reactive halogens will displace less reactive halogens from their aqueous halide solutions. In terms of reactivity, F > Cl > Br> I, e.g.: Cl2 (g) + 2Br- (aq) 2Cl- (aq) + Br2 (aq) By the way…

Colour as (aq)

Colour in organic solvent

Cl2 Colourless Colourless Br2 Brown Orange I2 Brown Violet

A6. Miscellaneous stuff Same group

same number of valence electrons same charge as ions (mostly…)

Same period same number of electron shells

Sheet 3 of 11 RI 2010 Year 4 Chemistry Speed Revision Notes v7: A Learner-Outcome/Reference-based Approach in 7495 words and 11 sheets

B. Salt Preparation B1. Solubility Rules

Soluble Insoluble Nitrates Oxides

Chlorides Hydroxides Sulphates Carbonates

Group I salts Ammonium salts

CaO, SrO, BaO… PbCl2, AgCl Ca(OH)2, Sr(OH)2… CaSO4, BaSO4, PbSO4

Solubility of Group II oxides and hydroxides increases down the group. Mg is stubborn. B2. Methods of Salt Preparation

Precipitation 1. Mix two relevant aqueous solutions

2. Filter to obtain residue salt 3. Wash residue with distilled

water 4. Dry residue between sheets of

filter paper Titration 1. Titrate with suitable indicator till

endpoint is reached 2. Repeat titration without

indicator 3. Heat obtained solution to 1/3

volume (for saturation) 4. Leave to solution to cool 5. Filter to obtain residue salt 6. DO NOT WASH (else it will

dissolve) 7. Dry residue between sheets of

filter paper Excess Base 1. Add excess base/carbonate

into acid, stirring and heating until no more can dissolve

2. Filter off excess base to obtain filtrate

3. Heat filtrate to 1/3 volume (for saturation)

4. Leave to solution to cool 5. Filter to obtain residue salt 6. Dry residue between sheets of

filter paper

B3. Decision Flowchart (visual version of B5) Given that we need to produce salt XY (where X is the cation and Y is the anion):

B4. Caveats You can try reacting acids with metals to create certain Group II (Mg, Zn) salts but this is not recommended to prepare the following:

Insoluble salts

A thin layer of the insoluble salt will form on the metal, hence preventing further reactions between the metal and the acid. Reaction thus ends prematurely.

Salts of unreactive metals (Pb, Cu, Ag)

Being unreactive, they will not react with the acid.

Group I salts Group I metals react with both the acid and the water in the acid, thus giving you the relevant soluble Group I hydroxide mixed with the relevant salt

B5. Summary of Salt Preparation Approach (textual version of B3) Grey boxes denote desired products. Latter two cases consider the scenario where you have to obtain a salt XZ given a related salt XY.

Method Anion (-) Contributor Cation (+)

Contributor Product Byproduct

Titration acid + alkali soluble salt + water (aq) (aq) (aq) H2O (l)

Excess Base acid + base soluble salt + excess base (aq) (s) (aq) (s)

Precipitation solution + solution insoluble salt + solution (aq) (aq) (s) (aq)

Soluble salt XZ from soluble compound XY

[Na2CO3, Acid]

sodium carbonate + soluble compound insoluble carbonate + sodium salt solution Na2CO3 (aq) XY (aq) XCO3 (s) NaY (aq)

acid + insoluble carbonate soluble salt + carbon dioxide

+ water HZ (aq) XCO3 (s) XZ (aq) CO2 (g), H2O (l)

Insoluble salt XZ from base XY

[HNO3, Precipitate]

nitric acid + base soluble nitrate + acid reaction byproduct

HNO3 (aq) XY (s) XNO3 (aq) CO2 (g), H2O (l) or H2 (g)

sodium salt solution + soluble nitrate soluble salt + sodium nitrate

NaZ (aq) XNO3 (aq) XZ (aq) NaNO3 (aq)

Is XY soluble?

Yes No

Is X2O or X2CO3 soluble?

Yes No

Precipitation by XNO3 (aq) + NaY (aq)

Excess base: add excess X2O (s) or X2CO3 (s) with HY

(aq)

Titration of XOH (aq) with HY (aq)

Sheet 4 of 11 RI 2010 Year 4 Chemistry Speed Revision Notes v7: A Learner-Outcome/Reference-based Approach in 7495 words and 11 sheets

C. Qualitative Analysis C1. Cation Tests • To distinguish between Al3+ and Pb2+, KI should be added. Al3+ cations

form AlI3 which dissolves; yielding no ppt. Pb2+ ions form yellow ppt PbI2. • Ppts marked with (soluble) dissolve when corresponding excess base is

added. • Solutions are colourless save for Cu2+ + NH3 (aq) (a dark blue solution).

Cation NaOH (aq) NH3 (aq) NH4+ Δ Gas evolved: NH3 Cu2+ Light blue ppt (insoluble) Light blue ppt (soluble, dark

blue soln) Fe2+ Dirty green ppt (insoluble) Fe3+ Reddish brown ppt (insoluble) Zn2+ White ppt (soluble) Ca2+ White ppt (insoluble) No ppt Pb2+ White ppt (soluble) White ppt (insoluble) Al3+ White ppt (soluble) White ppt (insoluble)

C2. Anion Tests For all tests, acidify the sample by adding HNO3 first in order to remove any CO32-, UNLESS the anion to be tested for is NO3- (adding HNO3 invalidates the test by introducing NO3- anions).

Anion Reagents to add Result CO32- HNO3 Effervescence of CO2

Cl- AgNO3 White ppt AgCl I- Pb(NO3)2 Yellow ppt PbI2

NO3- NaOH (aq), Al (s), Δ Gas evolved: NH3 SO42- Ba(NO3)2 White ppt BaSO4

Standard carbonate ion reaction: 2H+ (aq) + CO32- (aq) CO2 (g) + H2O (l) For the intrepid, here’s the reaction for the identification of NO3- (aq): 3NO3-(aq) + 8Al(s) + 18H2O(l) + 21OH-(aq) 8[Al(OH)6]3-(aq) + 3NH3(g) C3. Gas Tests

Gas Test and Results NH3 Turns damp red litmus paper blue CO2 (1) Forms white ppt when bubbled into limewater

(2) Extinguishes lighted splint Cl2 Bleaches damp litmus paper H2 Extinguishes lighted splint with a pop O2 Relights glowing splint

SO2 (1) Reduces acidified orange Cr2O72- (aq) to green Cr3+ (2) Reduces acidified purple MnO4- (aq) to pink Mn2+

C4. (extra) Notes on Gas Tests This can be skipped; it’s for understanding…(since when did that become sidelined? Oh, right. Mugging.) NH3 CO2 NH3 (g) + H2O (l) NH4+ (aq) + OH‐ (aq) This would explain why it turns damp red litmus paper blue.

CO2 (g) + Ca(OH)2 (aq) white ppt CaCO3 (s) + H2O (l) With prolonged bubbling: CO2 (g) + CaCO3 (s) + H2O (l) Ca(HCO3)2 (aq) Ca(HCO3)2 is not a compound that you can actually extract by evaporation. It’s just free‐floating Ca2+ (aq), CO2 (aq), CO32‐ (aq) and HCO3‐ (aq). Either way, it’s colourless, so prolonged bubbling makes the white ppt (CaCO3) disappear.

Cl2 SO2 Cl2 (l) + H2O (l) HCl (aq) + HClO (aq) This explains why damp blue litmus paper turns red first. Cl2 is a very strong oxidising agent and thus oxidises the litmus dyes in litmus paper, hence effectively bleaching it.

An example of the oxidation of SO2 (g) and reduction of Cr2O72- (aq) in acidic media: Cr2O72- (aq) + 2H+ (aq) + 3SO2 (g) 2Cr3+ (aq) + 3SO42- (aq) + H2O (l) Note: Acidified KMnO4 (aq) and K2Cr2O7 (aq) test for reducing agents. Just because a gas makes orange acidified K2Cr2O7 (aq) solution turn green doesn’t mean that it MUST be SO2.

C5. Explanations of (less intuitive results in) C1 (there actually isn’t anything left to say…most stuff here probably isn’t tested.) Explanations paraphrased from http://home.clara.net/rod.beavon/cations.htm (a very good site IMHO) Some aqueous metal cations exist as their aqua complexes e.g. Cu2+ (aq) can be written as [Cu(H2O)6]2+ (aq). Hooray for d orbital awesomeness! Then again, please just stop at the simple equations (e.g. Cu2+ (aq) + 2OH- (aq) Cu(OH)2 (s)) for the examinations, or else the examiner will kill you for acting smart (there’s A-levels for that to happen, la la)

NH4+ Consider aqueous ammonia’s dissociation reaction: NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) By LCP, the addition of NaOH (aq) and the subsequent increase in [OH- (aq)] will cause the equilibrium to shift to the left, hence increasing [NH3 (aq)] which is then liberated from the solution upon heating.

Cu2+ + NaOH (aq): light blue ppt insoluble in xs [Cu(H2O)6]2+ (aq) + 2OH- (aq) Cu(OH)2 (s) + 6H2O (l) 2[Cu(H2O)6]2+(aq) + 2OH-(aq) + SO42- (aq) Cu(OH)2•CuSO4(s) + 12H2O (l) + NH3 (aq): light blue ppt, soluble in xs to give dark blue solution [Cu(H2O)6]2+ (aq) + 2OH- (aq) Cu(OH)2 (s) + 6H2O (l) Cu(OH)2 (s) + 4NH3(aq) + 2H2O(l) [Cu(NH3)4(H2O)2]2+(aq) +2OH-(aq) then again the number of H2O varies…FYI only

Fe2+/ Fe3+

+ NaOH (aq) or NH3 (aq): dirty green (II) / brown (III) ppt insoluble in xs. Fe doesn’t form complexes with NH3 (aq). (II): [Fe(H2O)6]2+(aq) + 2OH-(aq) Fe(OH)2(s) + 6H2O(l) (III): Apparently, aqueous iron (III) ions exist as [Fe(H2O)5OH]2+ after protonating water: [Fe(H2O)6]3+ (aq) + H2O (l) [Fe(H2O)5OH]2+ (aq) + H3O+ (aq). [Fe(H2O)6]3+(aq) + 3OH-(aq) Fe(OH)3 (s) + 6H2O(l)

Zn2+ + NaOH (aq): white ppt, soluble in xs to give colourless solution [Zn(H2O)6]2+(aq) + 2OH-(aq) Zn(OH)2(s) + 6H2O(l) Zn(OH)2(s) + 2OH-(aq) Zn(OH)42-(aq) + NH3 (aq): white ppt, soluble in xs to give colourless solution [Zn(H2O)6]2+(aq) + 2OH-(aq) Zn(OH)2(s) + 6H2O(l) Zn(OH)2(s) + 4NH3(aq) [Zn(NH3)4]2+(aq) + 2OH-(aq) soluble! note: distinct from the NaOH (aq) reaction

Ca2+ + NaOH (aq): white ppt, insoluble in xs Ca2+ (aq) + 2OH- (aq) Ca(OH)2 (s) [OH-(aq)] is too low in NH3 (aq) to make Ca(OH)2 precipitate (it isn’t too soluble; waiting will get you limewater: Ca(OH)2 (aq).)

Pb2+/Al3+

+ NaOH (aq): white ppt, soluble in xs to give colourless solution Pb2+(aq) + 2OH-(aq) Pb(OH)2(s) Pb(OH)2(s) + 2OH-(aq) [Pb(OH)4]2-(aq) + NH3 (aq): white ppt, insoluble in xs Pb2+(aq) + 2OH-(aq) Pb(OH)2(s) Pb doesn’t form complexes with NH3(aq), and NH3(aq) is not strong enough a base to react with amphoteric Pb(OH)2(s)

+ NaOH (aq): white ppt, soluble in xs to give colourless solution [Al(H2O)6] 3+(aq) + 3OH-(aq) Al(OH)3 (s) + 6H2O(l) Al(OH)3(s) + 3OH- (aq) [Al(OH)6]3-(aq) + NH3 (aq): white ppt, insoluble in xs Al3+(aq) + 3OH-(aq) Al(OH)3(s) Al does not form complexes with NH3 (aq), and NH3 is not strong enough a base to react with amphoteric Al(OH)3(s)

Sheet 5 of 11 RI 2010 Year 4 Chemistry Speed Revision Notes v7: A Learner-Outcome/Reference-based Approach in 7495 words and 11 sheets

D. Metal Reactivity D1. Reactivity Series (Reactivity w/)

- Metal Reactivity w/ Cl2, Δ Cold

water Steam HCl/

H2SO4 O2, Δ

P K MCl MOH MO MCl/

MSO4 M2O (lim. O2), M2O2 (xs O2) S Na

C Ca Burn MO M Mg A Al - C C Z Zn I Fe T Sn - L Pb Surface MO H H - C Cu M Hg S Ag - G Au P Pt

D2. Reactivity Series (Thermal Stability/Solubility) - Metal Thermal Stability Solubility

MO MOH MCO3 MNO3 MO MCl P K MO MOH MCO3 MNO2

+ O2

MOH Sol.

(except PbCl2,

dissolves in hot water)

S Na C Ca MO MO

+ CO2 MO

+ NO2 + O2

Insol. M Mg A Al C C Z Zn I Fe T Sn L Pb H H C Cu M Hg M M

+ CO2 + O2

M + NO2 + O2

Insol. S Ag G Au P Pt

D3. General Trends Down the reactivity series:

I.E. Increases Ease of oxidation (OIL) Decreases Reactivity Decreases Strength as reducing agent Decreases

More reactive metals will displace the cations of less reactive metals from their aqueous solution (see A5) e.g. Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s) D4. On oxide layers

Porous Non-porous Na Mg Fe Al

Oxide layer falls off and exposes more metal to

oxidation/corrosion

Oxide layer is non-porous and protects

from further corrosion (disallows penetration of water and oxygen)

Al oxide layers can be anodized to make them thicker and hence better protect the metal. Anodizing involves the introduction of dyes, hence also colouring it.

D1 (alt). Reactivity credits to Theophi (4P’10) for the mnemonic!

D5. Electric Cells (a.k.a. Galvanic cells) An electric (or galvanic cell) comprises two electrodes (metals of different reactivities) immersed in an electrolyte containing free mobile ions (doesn’t matter what kind). Due to the differing reactivity (and hence ease of electron gain/loss) of the two electrodes, a potential difference is set up.

More reactive metal (M)

Less reactive metal (Y)

Anode (since ox.) Cathode (since red.) Oxidised (OIL)

M (s) M+ (aq) + e- Reduced (RIG)

Y+ (aq) + e- Y (s) M+ enters electrolyte

e- enters electrode, flows into wire

e- flows from wire out of electrode to reduce Y+ (aq) in electrolyte

Negative electrode Positive electrode Substance produced

(reduced form of cation in electrolyte;

e.g. H+ (aq) produces H2 (g))

D2 (alt). Thermal Stability

D6. Examples of Galvanic Cells • Seriously, any mobile

ions in solution will do for an electrolyte

• Salt bridges (typically filter paper dipped in KNO3 or) maintain electrical neutrality that was previously caused by e- flow (ions of salt bridge flow to areas of charge imbalance)

Using electrolyte containing metal cations of one of the electrodes

Using acid electrolyte

Using two electrolytes and a salt bridge

Sheet 6 of 11 RI 2010 Year 4 Chemistry Speed Revision Notes v7: A Learner-Outcome/Reference-based Approach in 7495 words and 11 sheets

E. Metal Extraction E1. Alloys

Adapted from image found at

http://www.spaceflight.esa.int/impress/text/education/Solidification/Question_Solidification_Phase_Diagram_13.html

Alloy Constituents Bronze Copper, tin Brass Copper, zinc Steel Iron, carbon

Pure metals tend to be soft and malleable, since the metals are arranged in neat regular rows that slide across each other easily when a force is applied. Hence, alloying a metal with another substance (another metal, or carbon) disrupts these regular rows, preventing them from sliding easily and hence increasing the resultant mixture (alloy)’s strength and hardness.

E4. Iron Extraction & the Blast Furnace diagram adapted from worksheet

List of substances

Coke C (s) Used for CO generation Haematite (iron ore)

Fe2O3 (s) iron (III) oxide

“Substrate”

Hot air - Provides a supply of oxygen

Lime CaO (s) Reacts with and removes acidic impurities (e.g. SiO2)

Limestone CaCO3 (s) Produces lime after thermal decomposition

Slag1 CaSiO3 (s) calcium silicate

Byproduct; used to make roads

1 Floats on top of Fe (l) so it can be easily collected Why do we need…?

CO (g) C (s) + O2 (g) CO2 (g) CO2 (g) + C (s) 2CO (g) Fe2O3 (s) + 3CO 2Fe + 3CO2 (g)

CaO (s) CaCO3 (s) CaO (s) + CO2 (g) CaO (s) + SiO2 (s) CaSiO3 (s)

The obtained iron (termed pig iron) has 5% C by weight and is too brittle to be of industrial use – it can be processed to produce cast iron (low tensile strength applications e.g. stoves) and wrought iron (0.1% C; soft, tough, malleable), or steel (by oxidising pig iron’s impurities with oxygen)

E5. Rusting and Corrosion Rusting Corrosion of iron Corrosion Gradual destruction of any metal due to reaction

with air, water or other chemicals For iron to rust, it needs (1) water and (2) oxygen. Mechanics Red. (RIG): 0.5O2 (g) + H2O (l) + 2e- 2OH- (aq) Ox. (OIL): Fe (s) Fe2+ (aq) + 2e- -------- [1] Redox: Fe (s) + 0.5O2 (g) + H2O (l) Fe(OH)2 (s) Further Ox. (OIL): Fe2+ (aq) Fe3+ (aq) + e- Fe3+ (aq) + 3OH- (aq) Fe (OH)3 (s) Conditions that speed up Rusting (1) Water contains dissolved ionic substances (e.g. seawater, acid) Rusting is like a galvanic cell; water’s the electrolyte XD

Right: Presence of free mobile ions

in water (i.e. “electrolyte”)

speeds up electron transfer

(2) Iron is in contact with less reactive metal This will speed up reaction [1] earmarked above under Mechanics. Recall that higher reactivity greater ease of electron loss Hence, iron will lose electrons to copper: Fe + Cu2+ Fe2+ + Cu (The formation of Cu2+ is a mystery we will solve later. Possibly corrosion? Meh)

Rust Prevention can be achieved by… (1) Applying a protective coating (paint/oil/grease/tin/chromium) Problems: Coating will be ineffective even if a tiny bit is scratched off (and a tin coating will make iron rust even faster; recall that tin is less reactive than iron) (2) Sacrificial corrosion (i.e. connecting a more reactive metal to iron) In this case, the more reactive metal is more likely to lose electrons in the oxidising environment created by the presence of water and oxygen, and the iron atoms are hence not oxidised. Since both substances are metals, electrons flow readily within and between them and a connection between iron and the more reactive meal is sufficient (i.e. no need to coat the whole thing). Example: underwater pipes with magnesium, iron cans with galvanized zinc coating

E2. Steel The hardness and brittleness of steel increases as the carbon content increases. Lower carbon steels are softer and more easily shaped.

Alloy Composition Features Uses Mild Steel Fe: 99 -99.5 %

C: 0.15 -0.25%

Hard, strong, malleable, ductile; can withstand great stress and strain

Car bodies, machinery, steel rods to reinforce concrete

Stainless Steel

Fe: 90-95% Cr/Ni: 5-10% C: variable

Hard; resistant to corrosion; attractive appearance

Cutlery, surgical instruments, chemical plants

E3. Methods of Metal Extraction P S C M A C Z I T L H C M S G P

Electrolysis Smelting with coke

Heat/Physical Extraction

Electrolysis: Reactive metals form stable compounds with strong bonds. Thus, extracting reactive metals from their compounds would require a lot of energy, and electrolysis is thus used. Smelting with coke: Less reactive metals can be extracted by heating the relevant metal oxide with coke (carbon is “more reactive” and can hence “displace” the metal from oxygen), which will reduce the metal.

Sheet 7 of 11 RI 2010 Year 4 Chemistry Speed Revision Notes v7: A Learner-Outcome/Reference-based Approach in 7495 words and 11 sheets

F. Electrolysis F1. Definitions and Basics Electrolysis is “the conduction of electricity by an ionic compound (electrolyte), when molten or dissolved in water, leading to the decomposition of the electrolyte”. Since stuffing carbon electrodes into a huge crystal of NaCl and passing current through the electrodes won’t get you anything, electrolysis shows that ions are held immobile in an ionic lattice, and are mobile only when molten or in solution. That’s just what the scope wanted me to tell you. Caveat: Oxidation occurs at the anode, and reduction occurs at the cathode! Forget this not!

F3. Summary Table (well, supposedly) Mat. = Material of electrode, Pdt. = Product, R = Reason (see F2) adapted from the worksheet

Electrolyte Cathode (negative electrode) Site of Reduction (of cations)

Anode (positive electrode) Site of Oxidation (of anions) Effect on electrolyte

Cations Mat. Pdt. R Anions Mat. Pdt. R PbBr2 (l) Pb2+ (l) C Pb (l) - Br- (l) C Br2 (g) - - H2O (l)5 H+ (aq) Pt H2 (g) - OH- (aq) Pt O2 (g) - - dil. H2SO4 (aq) H+ (aq) Pt H2 (g) - SO42- (aq), OH- (aq) Pt O2 (g) 1 More acidic1 conc. HCl (aq) H+ (aq) C H2 (g) - Cl- (aq) , OH- (aq) C Cl2 (g) 2 More dilute2 conc. NaCl (aq) Na+ (aq), H+ (aq) C H2 (g) 1 Cl- (aq) , OH- (aq) C Cl2 (g) 2 More alkaline3

Na+ (aq), H+ (aq) Hg Na/Hg 3 Cl- (aq) , OH- (aq) C Cl2 (g) 2 More dilute4 CuSO4 (aq) Cu2+ (aq), H+ (aq) C Cu (s) 1 SO42- (aq), OH- (aq) C O2 (g)6 1 More acidic1

Cu2+ (aq), H+ (aq) Cu Cu (s) 1 SO42- (aq), OH- (aq) Cu (anode dissolves) 3 Concentration unchanged

1 [H+ (aq)] increases since [H2O(l)] decreases as OH- (aq) is discharged at anode 2 [HCl (aq)] decreases since Cl- (aq) is discharged at anode 3 Water usually has only a small [OH-(aq)] from auto-ionisation, but as H+ (aq) and Cl- (aq) are discharged at the cathode and anode respectively, [Na+ (aq)] and [OH- (aq)] increase electrolyte becomes more alkaline. 4 [NaCl (aq)] decreases since Na+ (aq) and Cl- (aq) are discharged at the cathode/anode respectively 5 Catalytic amounts of H2SO4 (aq) usually added to increase the poor conductivity of water (a weak electrolyte) 6 This O2 (g) can cause the C anode to burn up (more prevalent at high temperatures)

F2. Determining substances preferentially discharged Three things come into play: (1) Reactivity Less reactive species usually get preferentially discharged. Consider potassium and gold: potassium is reactive and loves to lose an electron so it would rather stay as K+ (i.e. not liberated), whereas gold is unreactive and would rather stay as Au rather than lose an electron to become Au+ (i.e. Au+ discharged! Au+ reduced at cathode) (2) Concentration (and the eponymous concentration effect, or CE) Species in high concentration usually get preferentially discharged, though reactivity is a more important factor. This is usually relevant only when the two ions we want to compare are of similar reactivity. Consider the electrolyte NaCl (aq) (ions present: Na+, H+, Cl-, OH-) When CE works: Cl-, OH- are of similar reactivity, [OH-(aq)] > [Cl-(aq)]

OH- (aq) discharged as O2 (g) at anode (+ve electrode) When CE fails: [Na+ (aq)] > [H+(aq)] , but H+ is less reactive

H+ (aq) discharged as H2 (g) at cathode (-ve electrode) (3) Nature of electrode Inert electrodes (e.g. platinum and carbon, which are attacked only by liberated chlorine and oxygen respectively) don’t affect discharge, but certain reactive electrodes do. Example: Cu electrodes in CuSO4 (aq): Instead of considering SO42- (aq) or OH- (aq) to lose electrons and get oxidised (OIL) and discharged at the +ve electrode (anode), Cu at the anode undergoes the reaction Cu (s) Cu2+ (aq) + 2e- since Cu loses electrons more readily than SO42- (aq) or OH- (aq). Example: Hg cathode in NaCl (aq): Usually H+ gets preferentially discharged, but Na and Hg tend to amalgamate and hence less energy is needed to discharge Na+ over H+, therefore Na+ is reduced and discharged.

F4. Preferential Discharge Chart Species at the bottom tend to get preferentially discharged first as per F2.

K+ Ca2+ Na+ Mg2+ Al3+ Zn2+ Fe2+ Sn2+ Pb2+ H+

Cu2+ Ag+ Au+

F- SO42- NO3- Cl- Br- I-

OH-

F6. Electroplating Object to be electroplated is made the cathode (where cations of electroplating material are discharged). Metal used for electroplating is made the anode. Consider the use of silver for electroplating: Cathode: Ag+ (aq) + e- Ag (s) Anode: Ag (s) Ag+ (aq) + e- The discharged electroplating material is deposited as a layer on the electroplated object.

From worksheet

F7. Extraction of Aluminium Carbon cathode (reduction, RIG): Al3+ (l) + 3e- Al (l) Carbon anode (oxidation, OIL): 2O2- (l) O2 (g) + 4e- Overall: 4Al3+ (l) + 6O2- (l) 4Al (l) + 3O2 (g) Process Bauxite (aluminium ore) ↓Purification1 Alumina (Al2O3 (s)) ↓Melting, solvation in cryolite (Na3AlF6) Al2O3 – Na3AlF6 electrolyte mixture ↓Electrolysis Aluminium (and O2(g) liberated)

1 Bauxite + hot NaOH (aq), filter filtrate: sodium aluminate; residue: Fe2O3 (s) Sodium aluminate + dil. Acid Al(OH)3 (s) Al(OH)3 (s) + heat Al2O3 (s)

Rationale for adding cryolite M.p. of Al2O3 = 2065oC (too high to be cost-effective/safe) M.p. of Na3AlF6 = 950oC

Al2O3 – Na3AlF6 electrolyte mixture can remain liquid (i.e. contain free mobile ions) at a temperature lower than 2065oC Problem of carbon anodes (as per F3 fn.6) Oxygen gas discharged attacks carbon anode at high temperature: C (s) + O2 (g) CO2 (g) Hence, carbon anodes must be frequently replaced.

F5. Electrolysis for Refining Impure metals (e.g. copper) are made the anode and electrolysis is carried out. Pure metal accumulates on the cathode (which itself is the desired pure metal). Cathode: Cu2+ (aq) + 2e- Cu (s) Anode: Cu (s) Cu2+ (aq) + 2e- Impurities collect below the anode as anodic sludge.

Sheet 8 of 11 RI 2010 Year 4 Chemistry Speed Revision Notes v7: A Learner-Outcome/Reference-based Approach in 7495 words and 11 sheets

G. Chemical Energetics G1. Exothermic and Endothermic Reactions diagrams adapted from http://schools-wikipedia.org/images/58/5853.png.htm

Exothermic Reaction Endothermic Reaction

Ereactants > Eproducts Ereactants < Eproducts ΔH = negative ΔH = positive

Energy released to environment Energy absorbed from environment Bond making Bond breaking

Think about it…if you need to add energy to break bonds (bond breaking is endothermic), then the logical converse is that bond making is exothermic. Hurrah! Expressions for enthalpy change ΔH = Eproducts – Ereactants ΔH = Ebond breaking – Ebond making

G4. Hydrogen as a fuel if this wasn’t in the scope, no one would have known that this had to do with Chemical Energetics XD Pros and cons (+) Burns cleanly in air, forming H2O (g) (+) High energy released per gram: H2 125kJ/g; CH4 61kJ/g (-) Low b.p. (-252oC), hard to transport/store Production (1) Cracking (see X) (2) Steam Reforming H2O (g) + CH4 (g) cat. CO (g) + 3H2 (g) H2O (g) + CO (g) cat. CO2 (g) + H2 (g) H2 (g) separated from CO2 (g) by passing mixture through alkali (reacts with acidic CO2 (g)) Uses of hydrogen (1) As a fuel in rockets or fuel cells (2) Manufacture of NH3 (g) in Haber process, hydrogenation of margarine (vegetable oil) G5. Hydrogen-Oxygen Fuel Cells diagram adapted from worksheet Definitions NOTE: Fuel cells are electric/galvanic cells that produce current, and are NOT electrolytic cells! Fuel cells use a fuel to react with atmospheric oxygen to directly produce electrical energy. Reactions Anode (oxidation, OIL): 2H2 (g) 4H+ (aq) + 4e- Cathode (reduction, RIG): O2 (g) + 4H+ (aq) + 4e- 2H2O (g) Overall: 2H2 (g) + O2 (g) 2H2O (g) Mechanism Electrodes are usually carbon (cheap, and the O2 (g) produced isn’t hot) or platinum (amazing efficiency but costly). Fuel cell works like a battery (produces current), but since electrodes are inert, they do not deteriorate chemically constant fuel supply = constant electrical energy!

G2. Useful algebraic manipulation of standard enthalpy changes Reversing PQ (g) P (g) + Q (g) ∆HΘ = +500kJ/mol P (g) + Q (g) PQ (g) ∆HΘ = -500kJ/mol

Multiplying A (l) + B (l) AB (l) ∆HΘ = -500kJ/mol 2A (l) + 2B (l) 2AB ∆HΘ = -1000kJ/mol

G3. Some standard enthalpy changes (SECo = Standard Enthalpy Change of…) Assume all reactions occur at standard state/conditions 1atm, 298K ∆HrΘ SECoReaction “1mol” of reactions occurs Variable ∆HfΘ SECoFormation 1mol of substance formed in standard

state from constituent elements in their standard states

Variable (sometimes for elements in standard state bonds must be broken first, so this isn’t always –ve as per bond making)

∆HcΘ SECoCombustion 1mol of substance completely combusted

Always –ve (burning releases energy)

∆HmΘ SECoFusion 1mol of substance from (s) to (l) Always –ve (fusion, the opposite of melting, releases energy)

∆HvapΘ SECoVapourisation 1mol of substance from (l) to (g) Always +ve (requires energy) ∆HatΘ SECoAtomisation 1mol of substance from standard state

to individual (g) atoms Always +ve (requires energy)

∆HlatΘ Lattice enthalpy ∆HfΘ of 1mol of ionic compound from gaseous ions

Always –ve (bond making release energy)

∆Hi.e. Θ First IE 1mol of e- removed from 1mol of (g) atoms

Always +ve (requires energy)

∆HsolΘ Enthalpy of solution 1mol of substance dissolved in water Variable (-ve for acids)

Sheet 9 of 11 RI 2010 Year 4 Chemistry Speed Revision Notes v7: A Learner-Outcome/Reference-based Approach in 7495 words and 11 sheets

H. Reaction Kinetics H0. Preamble Frequency of effective collisions (right orientation, above minimum energy) ∝reaction rate Minimum energy needed is known as activation energy (Ea).

high Ea = slow reaction rate (less molecules have sufficient energy to react, reaction proceeds slower) low Ea = fast reaction rate (more molecules have sufficient energy to react, reaction proceeds faster)

The Maxwell-Boltzmann distribution curve (Graph A) shows the number of gas molecules that posses a certain amount of energy, over a range of possible energies. As you can see, most of the molecules have an intermediate amount of energy. The area under a section of the curve gives the number of gas molecules with energies falling within that range (think integration!)

Diagram adapted from

http://www.webchem.net/notes/how_far/kinetics/maxwell_boltzmann.htm Diagram from http://www.physchem.co.za/OB12-che/catalysis.htm

Diagram from http://www.physchem.co.za/OB12-che/catalysis.htm Graph A Graph B Graph C

H2. Catalysts Definition

Catalysts lower the Ea of a reaction and hence enable the reaction to proceed to equilibrium more quickly. Catalysts are not consumed or used up during the reaction. Biological catalysts are called enzymes.

Catalysts can be “poisoned” by molecules other than the intended reactant molecules that are preferentially adsorbed/permanently attached to the catalyst surface, hence reducing the number of available active sites and catalytic activity.

Catalyst Definition Example Homogeneous1 Same state as reactants H2SO4 (aq) in water electrolyte Heterogeneous1 Different state as reactants MnO2 (s) in H2O2 (l) decomposition

1 Not to be spelled as “homogenous” or “heterogenous”; those have different meanings. Examples (not exhaustive! Includes our variable oxidation state transition metal friends from A4)

Catalyst Reaction Product Ti/Al compound (s) Low-P polymerisation polyethene

Fe (s) Haber Process NH3 (g) V2O5 (s) Contact Process H2SO4 (aq)

Zeolite1 or Pt (s) Catalytic cracking gasoline Ni (s) Hydrogenation Margarine

Zymase enzyme (yeast) Fermentation Wine MnO2 (s) H2O2 (l) decomposition O2 (g) Cu2+ (aq) Acid + zinc Zinc salt + H2 (g)

1 Microporous, aluminosilicate minerals (thanks, Wikipedia!) Basically, very large surface area.

Mechanism (not strictly relevant; still…) Catalysts provide a different pathway of reaction. For example, reactant molecules can collide with them to form an intermediate which breaks down into the (restored) catalyst and the product. Diagram from the worksheet (AARGH NOTEMAKING BURNOUT)

H1. Factors affecting rate of reaction: to increase the rate of reaction, we can: Increase T (see Graph B)

Higher temperature particles have higher KE particles move faster, collide harder more particles have E > Ea

increased frequency of effective collisions Usually, increasing the temperature by 10K will double the reaction rate

Increase concentration [ ]

Higher concentration more particles/unit volume particles closer, increased frequency of collision increased frequency of effective collisions

Increase P (for gaseous reactants)

Higher pressure particles closer, increased frequency of collision increased frequency of effective collisions

Increase surface area of reactants

Increased surface area Increased area of reactant Increased frequency of collision/unit time increased frequency of effective collisions

Use of catalyst (see Graph C)

Catalyst Lowers Ea of reaction (usually by providing collision site; see H2) increased frequency of effective collisions

Increasing catalyst amount will increase reaction rate up to a point (after which catalyst > substrate)

H3. “Investigating the effect of a given variable on the speed of a reaction” -.- To monitor the rate of reaction, measure: • Δ mass of reaction mixture • Δ volume of gas evolved • Δ amount of product/reactant titrating samples of reaction mixture removed at fixed

intervals • Δ colour intensity w/ colorimeter (one of the reactants/products must be coloured) • Δ pressure in closed reaction vessel (when amount of gaseous products ≠ amount of gaseous

products) • Δ temperature

Sheet 10 of 11 RI 2010 Year 4 Chemistry Speed Revision Notes v7: A Learner-Outcome/Reference-based Approach in 7495 words and 11 sheets

I. Chemical Equilibria I1. Definitions

Reversible Reaction

Chemical reactions that can proceed both forward and backward1. Look out for “ ” as opposed to “ ” Example: NH4Cl (s) NH3 (g) + HCl (g)

Equilibrium A state in which the concentration of the reactants and products are constant (and the forward and backward reactions are still occurring).

1 Strictly speaking, all chemical reactions are “reversible”; just that the equilibrium constant Kc is so high for the forward reaction that the reaction virtually goes to completion…but hey, that’s beyond the scope of this test.

I3. Contact Process Reactions (1) S (s) + O2 (g) SO2 (g)

(2) 2SO2 (g) + O2 (g) 2SO3 (g) (3) SO3 (g) + conc. H2SO4 (aq) H2S2O7 (aq)* (4) H2S2O7 (aq)* + H2O (l)2 conc. 2H2SO4 (aq)

Conditions (bolded step)

450oC Catalyst: V2O5 (s) vanadium (V) oxide

*Oleum, more accurately described as nSO3•H2O or H2SO4•nSO3. H2S2O7 is actually its empirical formula – it merely describes SO3 (g) dissolved in conc. H2SO4 (aq). Dissolving SO3 (g) in water gives a fine mist of H2SO4 (g) that is “difficult to manage” (thanks, Wikipedia!) 2 I’m assuming T is a lot lower than 100oC at this point.

I4. Haber Process Reactions 3H2 (g) + N2 (g) 2NH3 (g) ∆HrΘ = -92kJ/mol Conditions 450oC

200atm Catalyst: Fe (s)

N2 (g) comes from the air (78% N2, oh yeah!). H2 (g) comes mainly from petroleum cracking. Elucidating the rationale of said conditions for the Haber Process (5 syllables! Score!) Given the reaction: 3H2 (g) + N2 (g) 2NH3 (g) ∆HrΘ = -92kJ/mol and applying the concepts presented in I2 (2), we rewrite this as such: 3H2 (g) + N2 (g) 2NH3 (g) + heat Δ temperature LCP tells us that we should carry out the reaction at low temperatures to maximize NH3 (g) concentration Δ pressure Looking at the amount of gas on each side, we see that LHS > RHS (4mol > 2mol). Hence, LCP tells us that we should carry out the reaction at high pressure to maximize NH3 (g) concentration. Kinetic considerations Since a higher temperature generally connotes a faster rate of reaction, it would be reasonable to assert that a low temperature entails a slower rate of reaction. (At this juncture it is worth noting that equilibria position and reaction rate are two unrelated parameters of a reaction). Based on this alone, we should carry out the reaction at high temperatures. Economic considerations It is very expensive to maintain high pressure systems; hence low pressure industry setups are preferred. In the interests of cost, as well as considerations in terms of kinetics and equilibria, scientists/engineers have settled on the 450oC/200atm configuration. (not something too slow/expensive like 60oC/1000atm) This is what could have been:

From http://www.ausetute.com.au/haberpro.html

I2. Le Chatelier’s Principle (LCP; a.k.a >) our UG friends are not amused. If a system at equilibrium is subjected to a small change, the equilibrium’s response is to shift so as to minimize the effect of the change. Well, that probably isn’t intuitive enough, so a few case studies would come in handy! (1) Δ concentration Consider A + B C + D By LCP, removing A or B will cause the equilibrium to shift left – the system converts some C and D to replenish A and B. By LCP, adding A or B will cause the equilibrium to shift right – the system converts some A and B to C and D to maintain the ratios of the amounts of each species. (2) Δ temperature (now is a good time to be familiar with Section G) Consider A + B C ∆HrΘ = +500kJ/mol This reaction is endothermic (takes in energy), so we can rewrite it as: A + B + heat C Now, by treating heat as a “reactant”, we have generalized the Δ temperature case to the Δ concentration case! By LCP, removing heat (i.e. cooling the system) will cause the equilibrium to shift left, and adding heat (i.e. warming the system) will cause the equilibrium to shift right. (Of course, you can rewrite A + B C ∆HrΘ = -500kJ/mol as A + B C + heat) (3) Δ pressure For this to make sense, at least one of the products/reactants should be a gas. Consider A (g) + 2B (g) 4C (g) + 8D (g) As you can see, the RHS has a greater amount of gas (12mol) as compared to the LHS (3mol). When a closed reaction vessel is under pressure, it tries to dissipate this excess pressure. This is normally done by reducing the amount of gas in the system. By LCP, increasing pressure will cause the equilibrium to shift to the left, and decreasing pressure will cause the equilibrium to shift to the right.

Sheet 11 of 11 RI 2010 Year 4 Chemistry Speed Revision Notes v7: A Learner-Outcome/Reference-based Approach in 7495 words and 11 sheets

J. Organic Chemistry! J1. Fractional Distillation of Petroleum Fractions that condense higher up the column are displayed higher.

Fraction C B.p. Uses Petroleum gases 1-4 Below r.t. Bottled gas for gas cookers Petrol/gasoline 5-10 35-75 Motor vehicle fuel

Naphtha 8-12 70-170 Petrochemical feedstock Kerosene/Paraffin 10-14 170-250 Jet aircraft fuel, oil stove fuel

Diesel oil 15-25 250-340 Diesel engine fuel Lubricating oil 19-35 340-500 Engine lubricant, wax/polish making

Bitumen >70 >500 Road surfacing Higher up the column

(vice versa lower down the column)

Smaller molecules/Mr Low m.p./b.p.

Volatile Less viscous

Flammable

Explanation of Trends Smaller molecules (and Mr) Smaller surface area

Weaker van der Waals IM forces of attraction Less thermal energy needed to overcome IM forces

Low m.p./b.p./lower viscosity volatile (evaporate more easily) more flammable

J6. Summary of tested chemical transformations For the purposes of demonstration, I’ve used specific molecules to illustrate the related reactions. I’ve also omitted byproducts. Meh.

- Oxidation reactions can be written as CH3CH2OH (aq) + 2[O] CH3COOH (aq) + H2O (l). - Depending on how long you let the reaction proceed, you can get multi-substituted chloroalkanes. * As a strong oxidising agent, KMnO4 (aq) will oxidise ROH to RCOOH directly. K2Cr2O7 (aq) will oxidise ROH to RCHO (aldehyde) first; upon prolonged oxidation, the RCOOH will be obtained. Of course, atmospheric oxygen will also do the job (albeit slowly). For this examination, treat KMnO4 (aq) and K2Cr2O7 (aq) as equals. * Bromine water (Br2 (aq)) reacts with alkenes to give bromohydrins as a major product and disubstituted alkanes as a minor product (for ethene, it will give 2-bromoethanol). Solvent is important! For this examination, take it that bromine water + alkene

1,2-dibromoalkene. *The oxidation of alkenes isn’t actually tested. You can get cool stuff like the epoxide, ethylene oxide (C2H4O), from Ag catalysed oxidation of ethene. Depending on the substituents of an alkene (2x -H, -CH3 & -H, 2x -CH3), the (two; not necessarily similar) products of oxidative cleavage are also different (CO2 (g) and H2O (l), carboxylic acid and ketone respectively)

J2. Definitions Homologous series

A group of compounds with a general formula, similar chemical properties and a gradation in physical properties as a result of increase in the size and mass of molecules

General formula e.g. alkynes: CnH2n-2 Molecular formula e.g. ethanol: CH3CH2OH

alternatively (though I don’t like this one): C2H6O Displayed formula

e.g. butanol:

J3. Saturation, unsaturation and degrees of unsaturation* (*not in syllabus) Alkanes are saturated hydrocarbons, whereas alkenes and alkynes are unsaturated hydrocarbons. Given a random hydrocarbon like C20H38 and further information that all the bonds in that molecule are either single or double bonds, we can actually figure out the number of double bonds in the molecule. Using the general formula for alkanes (CnH2n+2), we would expect 2x20+2 = 42 H atoms. However, we are short of 42-38 = 4 H atoms. This tells us that there are 4/2 = 2 C=C double bonds, because: For every 2 less H atoms as would be expected from an alkane of that number of carbons , there is one degree of unsaturation (1 degree a C=C bond, 2 degree

C≡C bond or 2 C=C bonds). J7. Miscellaneous noteworthy (wow, first pun! And we’re on the last page!) information - Meth-, eth- (= acet-), prop-, but-… (after this it’s easy) - Polymerisation: Stuff like ethene can polymerise to become poly(ethene)/polythene - Reaction of alkenes with bromine water: Alkenes decolourise bromine water; e.g. ethene with Br2 (aq) yields 1,2-dibromoethane (major product; just state this in the examination) and 2-bromoethanol (minor product) - Esters: can be used as solvents/flavouring! Wow! (ethanol can be used as a solvent too) BTW ethers (R-O-R’) ≠ esters. - Hydrogenation: of unsaturated vegetable oils to saturate them is done for the manufacture of margarine. (removal of double bonds including bent cis- isomers less “kinked” molecules stronger van der Waals IM forces of attraction more thermal energy needed to overcome higher m.p. solid state at r.t.p.!)

J4. Cracking of alkanes always yields at least one alkene molecule. Possible products are alkenes (always), alkanes, H2 (g) and C (s). Alkane alkene + alkane Alkane alkene + H2 (g) Alkane alkene + alkane + C (s)

J5. cis-trans isomerism* since double bonds aren’t rotatable like single bonds…

cis- same side

trans- opposite side

of double bond