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TOPIC 16 & 17 CHEMISTRY (LAST MINUTE REVISION)

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TOPIC 16 & 17 CHEMISTRY (LAST MINUTE REVISION)

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Here are the satanic things you have to relearn for these 2 topics, to hopefully increase your survivability for the test in 6th Feb 2025.

Oh yeah btw, the bunch of "subtopic" is actually "Sub-subtopic", i'm just too lazy to change it

remember, simply reading the book isn't enough cus past papers practice is crucial. (PPQ practice is useless if your brain is large enough, which is something that i 100% don't have)

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A. Standard Electrode potential I

B. Redox in Action

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Oxidation is the gain of electrons of a chemical species.

Reduction is the loss of electrons of a chemical species.

For the s, p, and d block, electrons are lost first at the highest energy level.

1) Example (Oxidation):

Copper electron config:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰

So, when copper loses 2 electrons, the cation created would have the electron config of:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁸

See? 3d energy level that loses the electrons first, because the electrons from this energy requires the leadt amount of energy to ionise.

Another example:

Aluminium (Al) has the electron config of:

1s² 2s² 2p⁶ 3s² 3p¹

When it stabilizes into its ionic state. It becomes:

1s² 2s² 2p⁶ 3s¹

Which is isoelectronic to Sodium (Na)

Al loses electrons from 3p first.

2) Example (reduction):

Bromine electron config:

1s² 2s² 2p⁶ 3s² 3p⁵

Chlorine has to gain one electron to reach its noble gas electronic config (isoelectronic as Argon). This electron is accepted at the unpaired P orbital. Thereby:

1s² 2s² 2p⁶ 3s² 3p⁶

Another Example (reduction):

Arsenic electron config:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p³

Arsenic must gain 3 more electrons to reach its noble gas electronic configuration (isoelectronic as Krypton). These electrons are accepted at the unpaired p orbitals.

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

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This concept is interlinked with equilibria and synonymous with standard redox equilibria. As you can already tell, this reaction is a redox reaction.

Definition:

Standard electrode potential is the emf measured in a half-cell that contains 1.00 mol/dm³ of its ions, connected to a standard hydrogen electrode under standard conditions, the conditions applies for both half-cells.

Standard conditions: 298K and 100kPa (or 1 Bar).

>But why do you need to connect it to standard hydrogen potential?

-First, it is not possible to measure the absolute potential difference between a metal electrode and its own solution. This metal electrode must either gain/lose electrons for emf to be established; the only way to get this done is to connect it to another electrode of a different metal.

-Second, the standard hydrogen potential (SHE) is used as a reference electrode, which is used with another electrode to measure its relative potential to the hydrogen half-cell.

So, those reversible reactions with varying E(0) values measured in Volts, are the standard electrode potential with reference that metal, which is evident on the reference booklet that'll be given during the exam.

I'm pretty sure the definition is already given in the reference booklet.

IMPORTANT INFO:

The more negative the Standard electrode potential E(0), the more the equilibrium position lies towards the left and the species on the right-hand side of the equation is more readily oxidized.

For example:

The standard electrode potential of zinc

Zn(2+) (aq) + 2e- -> Zn (s) E(0) = -0.76 V

This means that the equilibrium lies towards to the left, thereby Zinc Zn(s) is more readily oxidised to give out 2 electrons and a zinc ion Zn(aq).

And vice versa for the positive value of E(0)

I think you already understand this, but i just want to make sure

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This is an electrode that consists of hydrogen gas at a pressure of 100kPa bubbling over a piece of platinum foil dipped into a solution of HCl with a concentration of 1 molar (1 mol/dm³) at a temperature of 298K.

Also, the Pt electrode is covered in porous platinum to increase that sweet sweet... (you know what it is)

Why is it so important?

These are used as reference electrodes used to measure the relative potential of an electrode to that of a hydrogen half-cell. As it is used as a reference, its value is declared to be zero.

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i. For a metal, the standard electrode potential is set up using the SHE connected with a half cell of another metal. The electrode of this metal half-cell will be purely made out of a specific element that is complementary with the solution.

For example:

When investigating the standard electrode potential of copper using the SHE. The electrode in the copper solution will need to be made out of pure copper. This is because copper can either be oxidised (dissolved into the solution, which will need a supply of copper) or reduced (copper is deposited on the electrode). Which is only possible when the electrode is made out of copper.

ii. For a non-metal, the standard electrode potential is set up using the SHE too, why? (reread Subtopic 2). For gaseous non-metal. The same apparatus is used as for the SHE. (re-see Standard hydrogen electrode to see the full apparatus).

Now, for the ions of the same element with varying oxidation states.

This will indicate a colour change in the solution cus same elements with different oxidation states tend to have different colour on their aqueous forms.

Some VERY IMPORTANT info:

A salt bridge is needed to complete the electrical circuit; this allows the ions to move and maintains electrical neutrality between the 2 solutions.

The salt bridge will contain any ionic salt, but the ions should not interact (chemically) with both of the 2 solutions.

A high-resistance voltmeter is used to measure the emf of standard electrode potentials. WHY?

This is to ensure that the flow of electrons (current) is minimal around the external circuit. This is because a high current would disrupt the equilibrium in both half-cells, which would lead to inaccurate values.

emf is the measured potential difference of a cell when NO current is flowing.

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This involves some calculations (requires some big brains)

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E(cell) = E(reduction) - E(oxidation)

yeah as simple as that. Example:

Will copper produce H2 gas when reacts with sulfuric acid?

Cu(s) + 2H+ -> Cu(2+) (aq) + H2(g)

Standard electrode potential of copper = -0.34 V

Standard electrode potential of H2 =

0.00 V

Thereby, what you need to do is:

-0.34V + 0.00V = -0.34V

Due to the negative sign of the E(cell), the reaction isn't thermodynamically feasible.

Ok, now another example:

Would Zn displaces Cu in a copper(II) solution?

E(0) Zinc = -0.76 V

E(0) Copper = +0.34 V

Remember, E(0) will always be written in its reduction format

First, you have to write the overall reaction in according to the question (unless the equation is already given).

You then have to flip the equation to match it with the equation of the reaction.

Zn(s) + Cu(2+) -> Zn(2+) + Cu(s)

Then, -0.76 -> +0.76 V

Also, +0.34 -> -0.34 V

So, E(cell) = E(oxidation) - E(reduction)

(+0.76) - (-0.34) = +1.10 V

As the value is positive, then the reaction is thermodynamically feasible.

Disproportionation reactions.

This will involve E(0) of the half-cells having both negative/positive. Thereby, from this. You just need to see which E(0) is more positive than the other.

More positive than the other half-cell: The more the equilibrium position lies towards the right, (despite majorly to the left), thereby this species is more likely to be reduced.

No example cus I'm too lazy and you already understood it by now.

mmkay, some IMPORTANT infos:

>Thermodynamic feasible: it means that the reaction will take place spontaneously in standard conditions, no change is required to make the reaction to occur.

>Kinetically stable: this means that the reaction does not take place or is very slow because the activation energy for the reaction is too high.

>Disproportionation reaction: These are reaction where it involves the reduction and oxidation of the same chemical species at the same time.

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Apparently, E(cell) has some relations with entropy (WHYYYYYY SO MUCH THINGS TO LEARN)

RInK = [nFE(cell)]/T

Where:

R = 8.31 J-1K-1

K = equilibrium constant

n = moles of electrons involved in the cell reaction.

F = Faraday's constant

E(cell) = self-explanatory

T= Temperature (in Kelvins)

So, from the equation itself, it can be seen that. InK is directly proportional to the E(cell).

So, the higher the K, the higher the E(cell) value.

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There is some limitations when using the standard electrode potential when used for predictions.

First, standard electrode potential ONLY determines the chemical reaction's thermodynamic feasibility.

A reaction may be thermodynamically feasible. But they can be kinetically stable, meaning that the reaction is energetically favorable but the rate of reaction can be very slow.

This is evident on the decomposition of H2O2. This chemical compound decomposes at standard conditions, but they are VERY VERY SLOW. So, a catalyst is needed to speed up the reaction by providing an alternative reaction pathway that is more energetically favorable.

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These calculations will take too long to be put into this section of padlet (also I'm too lazy and time is expensive for me).

So, I'm just going to list the things you need to do:

1. Calculate the mean titre from your concordant results.

2. Calculate the concentration of the analyte

3. Calculate the mass of the analyte.

4. Whatever the questions is asking you to do next

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Some uncertaintis when measuring the standard electrode potentials

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1. The standard electrode potentials are measured in standard conditions, meaning that in reality, these conditions: 1 Bar, 298K, 1 M for all aqueous solution

Cannot always be precisely maintained, which will lead to deviations in the values of the E(cell).

2. Metal electrodes must be pure and free from their oxide layer; this is because surface contamination can affect electron transfer, which will alter the measured electrode potential.

3. When using the SHE. The SHE must have constant supply of pure hydrogen gas at 1 atm. Small variations in the pressure can affect the potential of the reference electrode leading to systematic errors

4. High-resistance volt meter must minimize the current draw, otherwise, too much flow of current will lead to disruption in the equilibrium of both half-cell. Leading to minor uncertainties in the E(0) value.

5. Calculation error. If the value of one of the E(0) half-cell contains a small error, this can be passed through the E(cell) which will lead to false predictions.

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Fuel cells involving hydrogen and oxygen to produce water.

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In the hydrogen-oxygen fuel cell, there are 2 conditions that the reaction can hold:

Acidic electrolytes:

negative electrode = H2 (g) -> 2H+ (aq) + 2e-

positive electrode = (1/2)O2 (g) + 2H+ (aq) + 2e- -> H2O (l)

Basic electrolyte:

negative electrode = H2 (g) + OH- (aq) -> 2H2O (l) + 2e-

positive electrode =(1/2)O2 (g) + H2O (l) + 2e- -> 2OH- (aq)

advantages:

>No greenhouse gas produced, ONLY water

>Offers an alternative to the direct use of fossil fuels such as petrol and diesel

They are lighter and more efficient than he engines that uses fossil fuels

Disadvantages:

>Compressing the gas

>Absorbing it onto the surface of a suitable solid material.

>Acquiring hydrogen gas will take a lot of energy and effort also cost

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Transition metals and their chemistry Topic 17

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Transition metals and their chemistry Topic 17

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A. Principles of transitional metal chemistry I

B. Transition of metal reactions.

C. Transition metals as catalysts

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These are d-block elements that form one or more stable ions with incompletely filled d-orbital.

>This is the edexcel spec definition.

To get more marks, just list their characteristics.

>They are hard solids

>Have high melting and boiling point

>Can be catalytically active

>Formed colored ions and compounds

>Forms ions with varying oxidation states

>forms ions with incompletely filled d-orbital.

Zinc and scandium are d-block elements that shares similar characteristics as transition metals but they are not transition metals because:

>Only have 1 oxidation states.

>Do not form colored ions/compounds

>the d-orbitals (zinc has full 3d orbitals) where as scandium only has 1 which is not enough.

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Reread Subtopic 1 redox equilibrium Topic 16

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Why does transition metal show variations in oxidation states?

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>Transition metal exhibits variable oxidation states due to the presence of incompletely filled d-orbitals; this allows them to either gain/lose electrons at different quantities.

>The energy difference between the 4s and the 3d orbitals are small, thereby electrons from both 4s and 3d orbitals are able to participate in chemical bonding, this allows transition metals to lose/gain varying number of electrons, leading to different oxidation states.

>When a metal loses an electron, the successive ionization energy becomes higher, however this is compensated by:

1. The lattice energy. This is the energy released when oppositely charged ions came together to form a solid crystal lattice. So, when a metal forms a high oxidation states, it produces highly charged small cations which will attract the anion very strongly, this increases the lattice energy. Thereby, when the lattice energy is high enough then the energy released can compensate for the high sucessive ionisation energy. This allows for better tendency for the metal to lose electrons, which will show variations in its oxidation states.

2. The hydration energy is the energy released when water surrounds and stabilizes a gaseous ion. So, small cations with very high charge, such as Ti(4+) will have a very strong electrostatic attraction with water, which will lead to high hydration energy. If the hydration energy is higher than the ionization energy required, a higher oxidation state of that metal can be stabilized in the solution.

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Ligands are species that use lone pair of electrons to form a dative covalent bond with a metal ion.

But how do transition metals bond with LIGANDS?

Transition metals can form a high oxidation states with smaller ionic radius, this means that transition metal ions have a high charge density. This allows transition metals to attract electron-rich species more strongly.

In fact the attraction is so strong that these LIGANDS forms a specific number of dative bonds. For transition metal chemistry, you will address them by coordinate bonds.

In case if you need a refresher, dative bonds are bonds formed when an electron-rich species donates their lone pair of electrons to an electron-deficient species.

But a (co-ordinate) bond is a bond formed between the central metal ion and a ligand in which both of the bonding electrons are supplied by the LIGANDS. When referring to a LIGAND, the metal ion is referred as central metal ion.

So... metal ions forms coordinate bonds with ligands to form a complex ions. If the complex has no charge, then just address them by complex.

Transition metals will exhibit different coordination number. Which is really, the number of dative bonds around the central metal ion. Yes as simple as that.

Ligands will have specific prefixes added for the name.

Water - aqua - no charge

Hydroxide - hydroxo - -1 charge

Ammonia - ammine - no charge

Chloride - Cl- -1 charge

For example:

[Fe(H2O)6]2+ -> hexaaquairon(II)

[Ti(OH)6]2- -> hexahydroxotitanate(IV)

The roman numeral is just to show it's oxidation state.

So, when there is 6 ligands forming coordinate bonds with the central transition metal ion, these coordination can also be called as six-fold coordination, where the shape of the molecule would be octahedral, where the bond angle on ANY side is perfectly 90*.

Ligands can be monodentate, bidentate, hexadentate and multidentate.

monodentate = These are molecules/ions that donates ONE lone pair of electrons to the central transition metal ion.

bidentate = These are molecules/ions that donates 2 lone pair of electrons to the central metal ion.

hexadentate = These are molecules/ions that donates 6 lone pairs of electrons to the central metal ionm.

multidentate = These are molecules that can donate more than 2 lone pair of electrons.

BIG IMPORTANT TIP:

If a metal complex is in it's ionic form, aka having a charge, within a solution. This means that the metal complex is dissolved in the solution.

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Transition metals will form colored compounds, especially when it is dissolved, aka aqueous.

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When a transition metal forms a complex, its 3d energy level splits into 2 energy levels with slightly different energies.

This slight difference in the energy allows electrons from the lower energy level to move by excitation or promotion to a higher energy level of an unpaired/empty orbital by absorbing energy from the visible spectrum. When the electrons moves to a high energy level, the amount of energy it absorbs is dependent on the difference in the energy of the 2 levels. The energy absorbed by the electron is inversely proportional to the wavelength of light (complementary colors).

so when there is less energy absorbed, the wavelength emitted will have a higher wavelength of light.

To identify the part of the visible spectrum absorbed/emitted, then you will need to use a color wheel. (just look at textbook at pg. 190).

However, metals such as scandium and zinc won't create colored compounds but why?

Scandium only has 1 electron in the 3d energy levels. So, when the metal forms a metal ion, all of the electrons from 3d and 4s are removed/ionised. preventing the opportunity for electron excitations, as there is no electrons present in its 3d orbitals. This is evident in its colourless color.

Zinc only loses 2 electrons from it's 4s orbitals, thereby the 3d orbital is remained unchanged. Due to the completely filled d-orbitals, no electrons can be excited within the d-d transitions, thereby zinc will possess no color when dissolved onto a solution.

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There are 4 types of reactions involving transition metals.

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Different types of reactions involving LIGANDS

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Different types of reactions involving LIGANDS

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These reactions are:

Redox - which are just the changes in the oxidation number of the metal ions.

For example:

CuSO4 + I- -> CuI + SO4(2-).

In here, the oxidation states of copper moves from (2+) into (2-) ions.

Deprotonation - This is when one or more ligand gains or loses a hydrogen ion (proton)

For example:

[Fe(H2O)6](2+) + 2NH3 -> [Fe(H2O)4(OH)2)] + NH4+

Coordination number change - The number of ligands changes.

For example,

[Cu(H2O)6](2+) + 4Cl- -> CuCl4(2-) + 6H2O

In here, the coordination number changes from 6 to 4.

Ligand substitution, replaces the ligand of something with another

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pg. 210-211 for the ultimate summary

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Iron(II) complexes with aqueous alkalis.

[Fe(H2O)6](2+) (aq) + 2OH- (aq) -> [Fe(H2O)4(OH)2] (s) + 2H2O(l)

with ammonia (NH3) aqueous

[Fe(H2O)6](2+) (aq) + 2NH3 (aq) -> [Fe(H2O)4(OH)2] (s) + 2NH4+ (aq)

Observation: pale green solution -> green precipitate.

Upon standing (left for a bunch of time), the green precipitate gradually turns into brown as oxygen from the atmosphere causes oxidation, which oxidizes iron(II) to Iron(III), which forms [Fe(H2O)3(OH)3] (s).

This is known as triaquatrihydroxoiron(III) complex. (remember, the IUPAC must be in alphabetical order)

Iron(III) complexes with aqueous alkalis.

[Fe(H2O)6](2+) (aq) + 3OH- (aq) -> [Fe(H2O)3(OH)3] (s) + 3H2O(l)

With NH3 aquoeus

[Fe(H2O)6](3+) (aq) +3NH3 (aq) -> [Fe(H2O)3(OH)3] (s) + 3NH4+ (aq)

Observation: yellow-brown solution -> brown precipitate

For this compound, there will be NO change in color when left standing.

If concentrated alkali or ammonia is used, then all of the ligands are replaced by OH- or NH3

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Chromium(III) complexes with alkalis.

[Cr(H2O)6)](3+) (aq) + 3OH- (aq) -> [Cr(H2O)3(OH)3] (s) + 3H2O (l)

This is generally what happens when aqueous alkali is added.

When using NH3

[Cr(H2O)6](2+) (aq) + 3NH3 (aq) -> [Cr(H2O)3(OH)3] (s) + 3NH4+ (aq)

This reaction occurs when added aqueous ammonia.

Observation: green solutions form a green precipitate.

There are no further changes when left standing.

When excess NaOH is added

[Cr(H2O)3(OH)3 (s) + 3OH- (aq) -> [Cr(OH)6](3-) (aq) + 3H2O (l)

Observation: The green precipitate dissolves to form a green solution.

When an excess of aqueous ammonia is added on

[Cr(H2O)3(OH)3] (s), the precipitate will be slow to dissolve.

[Cr(H2O)3(OH)3] (s) + 6NH3 (aq) -> [Cr(NH3)6](3+) (aq) + 3H2O (l) + 3OH- (aq)

Observation: Green precipitate dissolves to form a purple solution.

Further oxidation of [Cr(OH)6](3-) (aq) can occur with the addition of an oxidizing agent, hydrogen peroxide.

2[Cr(OH)6](3-) (aq) + 3H2O2 (aq) -> 2CrO4(2-) (aq) + 2OH- + 8H2O (l)

Observation: green solution changes colour to yellow solution. Despite CrO4(2-) is a complex, it is not enclosed in the square brackets. The chromium here are Chromium(VI)

Chromate(VI) ions are stable in an alkaline solution. But in a more acidic solution, dichromate (VI) is more stable.

2CrO4(2-) (aq) + 2H+ (aq) -> Cr2O7(2-) (aq) + H2O (l)

Chromate(VI) always exists in equilibrium.

Dichromate = orange colour

Chromate = yellow colour

When a zinc metal is added to a solution of dichromate (VI), there will be a reduction of chromium from +3 to +2.

First stage

Zn (s) + Cr2O7(2-) (aq) + 14H+ (aq) -> 2Cr(3+) (aq) + 7H2O

+ 3Zn(2+)

Observation: orange solution changes color to green solution, Zinc dissolves into the solution.

second stage

Zn (s) + 2Cr(3+) (aq) -> Zn(2+) (aq) + 2Cr(2+) (aq)

Observation: green solution changes color to blue solution.

go to pg 206 for more info.

In conclusion.

Cr exhibits +2, +3, +3.5, +6

Based on textbook

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when aqueous NaOH is added to hexaaquamanganese(II) ion.

[Mn(H2O)6](2+) (aq) + 2OH- (aq) -> [Mn(H2O)4(OH)2] (s) + 2H2O (l)

When aquoeus NH3 is added:

[Mn(H2O)6](2+) (aq) + 2NH3 (aq) -> [Mn(H2O)4(OH)2] (s) + 2NH4+ (aq)

These are deprotonation reaction.

Observation: Pale pink solution forms a pale brown precipitate.

When left standing, the brown precipitate turns into a darker brown precipitate [Mn(H2O)3(OH)3] (s) as it is oxidized by the air, which then turns into very dark brown precipitate to form a hydrated manganese(IV) oxide MnO2xH2O

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Vanadium is a transtion metal that forms ions with several oxidation states

Here are the colors:

V(2+) - purple

V(3+) - green

VO(2+) - blue

VO2(+) - yellow

Reactions - +5 to +2

We will need to use zinc, which will be added into the solution

2VO2(+) (aq) + 8H+ (aq) + 3Zn(s) -> 3Zn(2+) (aq) + 4H2O (l) + 2V(2+) (aq)

Observation: yellow solution changes color to purple solution

Reactions - +5 to +4

2VO2(+) (aq) + 4H+ (aq) + Zn (s) -> 2VO(2+) (aq) + Zn(2+) (aq) + 2H2O (l)\

Observation: yellow solution changes color to blue solution.

Reactions - +4 to +3

2VO(2+) (aq) + 4H+ (aq) + Zn (s) -> 2V(3+) (aq) + Zn(2+) (aq) + 2H2O (l)

Observation: blue solution changes color to green solution.

Reactions - +3 to +2

2V(3+) (aq) + Zn(s) -> 2V(2+) (aq) + Zn(2+) (aq)

Observation: green solution changes color to purple solution.

Reactions - +2 to 0

Zn(s) + V(2+) (aq) -> Zn(2+) (aq) + V (s)

We dont need to know this cus the reaction isn't therodynamically feasible.

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[Ni(H2O)6](2+) (aq) + 2OH- (aq) -> [Ni(H2O)4(OH)2] (aq) + 2H2O (l)

From this reaction, aqueous NaOH is added for this to occur. This deprotonation reaction can also be done by adding aqueous NH3.

[Ni(H2O)6](2+) (aq) + 2NH3 (aq) -> [Ni(H2O)4(OH)2] (aq) + 2NH4+ (aq)

Both reactions observation: green solution changes colour to green precipitate.

But, when adding aqueous ammonia is added in excess, then a deep blue solution is formed.

[Ni(H2O)6](2+) (aq) + 6NH3 (aq) -> [Ni(NH3)6](2+) (aq) + 6H2O (l)

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THis metal is very simple luckily. They only form white precipitate and colorless solutions.

They dissolve in excess NaOH to form colorless solution

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What's a catalyst?

This is a chemical compound that speeds the rate of reaction by lowering down the energy needed to be obtained by the colliding reactant particles to reach the energy level of the transition state by providing an alternate chemical pathway that is more efficient and has a lower energy requirement. These chemical compounds are NOT used up in the chemical reaction.

How do they really work?

Catalyst will speed up the rate of reaction, as within a system, they lower the activation energy of a chemical reaction by providing an alternate reaction pathway that is more efficient. As there are greater proportions of reactant particles possessing lower energy. Lower activation energ means that the reactant particles will have enough energy to surpass the activation energy. Leading to an increase in the rate of reaction.

To make a catalyst to be most effective:

Heterogenous catalyst

Very large surface area to volume ratio

DOES NOT REACT with reactant particles

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 H E T E R O G E N O U  S   C A T A L Y S I S

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 H E T E R O G E N O U  S   C A T A L Y S I S

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This is just a catalyst that is in a different phase from the reactants. It's usually solid catalysts.

The contact process

This will use V2O5 as the catalyst

Decomposition of H2O2

This will use manganese (IV) oxide

Catalytic converters

Uses palladium

okay but how does solid catalysts really work?

1. Adsorption - The reactant particles becomes attached to the surface of the catalyst

2. Reaction - the adsorption is followed by the weakening of the bonds in the adsorbed reactants.

3. Desorption - reactant products becomes completely detached from the catalysts surface

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 H O M O G E N O U S   C A T A L Y S I S 

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 H O M O G E N O U S   C A T A L Y S I S 

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This is just a catalyst that has the same state as the reactant particles. This is not really commonly used commercially, because it's going to be quite difficult to separate the catalyst from the solution.

On our edexcel spec, homogenous catalysts are present in these reactions:

The reaction of S2O8(2-) ion

General equation:

S2O8(2-) (aq) + 2I- (aq) -> 2SO4(2-) (aq) + I2 (s)

The reaction is very slow at step.

This is because both of the reactants are negatively charged, so they will repel each other leading to lower collision frequency and energy.

The reaction will occur much faster upon the addition of Fe(2+) ion.

Mechanism:

Step 1:

S2O8(2-) (aq) + Fe(2+) (aq) -> 2SO4(2-) (aq) + 2Fe(3+) (aq)

Step 2:

2Fe(3+) (aq) + 2I- (aq) -> 2Fe(2+) (aq) + I2 (s)

See? Fe(2+) is regenerated again. The amount of Fe(2+) in the solution remains fairly constant throughout the reaction. Thereby, it is a catalyst.

Oxidations of ethanedioate ions

This reaction will involve Potassium manganate (VII)

2MnO4- (aq) + 5C2O4(2-) (aq) + 16H+ (aq) -> 2Mn(2+) (aq) + 5CO2 (g) + 8H2O (l)

K is just a spectator ion, so it won't be written!

From this reaction, Mn(2+) is the catalyst.

The formation of the reaction product itself, catalyzes the reaction, increasing its rate of reaction further. This effect is known as autocatalysis.

Evidence???

go to pg.215,

Here you can see that the graph shows the reactant products against time.

The graph shows a slow decrease in the reactant concentration, as time passes. The concentration of the reactants decreases more rapidly. This is due to the increasing in the concentration of the Mn(2+), there will be greater autocatalysis. The graph line then decreases more slowly as there is little amount of reactant left.

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This is another satanic things you will have to learn about topic 18 chemistry to increase your survivability on the MOCK EXAM on 23rd March.

remember again guys, simply reading the book isn't enough cus PPQ practice is importante!!!

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It was on 6th Feb

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Please don't confuse yourself with the usage of either K's or Lonsdale. ALWAYS use Lonsdale unless you're told to draw K's struture!

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Here is the list of things why Kekule is objectively WRONG!

Thermochemical data

This data used is the enthalpy change of hydrogenation.

(The change in energy where one mole of an unsaturated compound is fully hydrogenated into a saturated compound)

Evidence:

As there are 3 C=C bonds. Kekelu's model suggests it would need 3 mols of H2 to fully hydrogenate benzene to an aliphatic hydrocarbon

1 mol of H2 would release around -120 kJmol-1, so Kekule's structure would release around -360 kJmol-1 so benzene is fully hydrogenated, if a certain data values is given in the question. USE THOSE.

While the actual value of the enthalpy change of hydrogenation is -208 kJmol-1, this difference in value suggests that his model is WRONG!

But, why is this?

Benzene has 6 fully delocalised π electrons.

The 6 delocalised π electrons spread evenly within the molecule, creating equal C-C bond length

This delocalization lowers benzene's energy making it more stable.

More energy is needed to break the π electron system.

More energy is absorbed --> Less energy is released --> Lower enthalpy change of hydrogenation

X-ray diffraction

The data from the X-ray diffraction suggests that C-C bonds in benzene are the same length.

Whereas in Kekule's (K) structure, it would suggest that there would be different bond length due to the alternating C=C bonds.

C-C in cyclohexane = 154 nm

C=C in cyclohexane = 135 nm

K's structure would suggest there would be alternating 154 nm and 135 nm bond lengths on its ring.

While evidence suggests that the bond length in benzene ring are the same length.

Infrared data

Cyclohexene has an absorption at around 1650 cm-1 due to its isolated alkene stretch.

But evidence suggests that benzene has a strong absorption at around 1500 cm-1 and also absorption at 1580 cm-1 and 1450 cm-1, this is typical for an aromatic C=C stretching.

The chemistry spec only suggests these 3 as evidence for the Lonsdale structure. But I'm pretty sure there'd be more evidences.

Bromine water

If benzene did indeed contain 3 C=C bonds, it should readily decolorize bromine water due to an addition reaction.

Instead, a substitution reaction occurs, which doesn't decolorize the bromine water.

Isomerism

If K's structure were correct, then there'd be 4 possible isomers of dibromobenzene, where instead it only shows 3 isomerism.

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The modern structure of benzene.

This model suggests:

Each carbon in a benzene molecule has 3 σ bonds

2 σ with adjacent carbon

1 σ with adjacent hydrogen atom

From this generation of 3 bonds, each carbon from benzene has one bonding electron left in its p orbital.

(Carbon forms 4 bonds, remember.)

This electron from the p-orbital overlaps sideways, sideways overlap is contributed by all 6 carbons.

All of the sideways overlaps combine/overlap to create a delocalized π electron cloud. Where 6 electrons are

shared evenly within the 6 carbons on benzene

This lowers the energy for benzene, making it more stable. Resulting in lower ΔH.

Benzene's resistance to bromination (compared to alkenes)

Due to the existence of a delocalized π electron density, they are more resistant to bromination by:

Polarization

Benzene has a delocalized π electron density which is less electron-rich than a localized π electron density in alkenes e.g. C=C.

benzene does not polarize bromine as easily compared to an alkene

benzene is less susceptible to electrophilic attacks

to deal with this, Benzene would require a stronger electrophile, which are Br+ generated by AlBr3

Aromatic stability

Benzene's aromaticity has a high stabilization energy compared to an localized π system (C=C bonds in alkenes) When benzene undergoes an addition reaction with bromine:

Benzene will have to lose its aromaticity to form cyclohexene which is far less stable and requires large input of energy

The reaction will not proceed due to the high energy requirements, thereby the addition reaction of benzene with bromine is thermodynamically unfeasible.

Benzene will only undergo substitution reaction to preserve its aromaticity.

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Benzene will have a much higher chance to react if the reaction preserves benzene's aromaticity. Benzene will participate ina reaction of the following:

oxygen in air (combustion to form a smoky flame)

Benzene is a hydrocarbon so it is able to participate in a combustion reaction. When benzene burns in air, it produces a smoky flame since it has a high carbon : hydrogen ratio, which therefore can produce 6CO2 per molecule of benzene.

C₆H6 + 7½O2 --> 7CO₂ + 3H₂O

Bromine in a presence of a catalyst

Benzene molecule is very stable due to its aromaticity therefore bromine will need a halogen carrier.

Benzene will react with bromine under electrophilic substitution. This reaction uses bromine and aluminium tribromide.

Bromination of benzene requires heating under reflux

This reaction substitutes hydrogen with a Bromine group There are 4 steps!

Step 1: generation of electrophile (Br⁺)

AlBr₃ + Br2 --> [AlBr₄]⁻ + Br⁺

You can replace Al with Fe if u want... (depends on the question)

The generation of Br+ is important as benzene is a weak nucleophile, so a stronger electrophile is needed to initiate the reaction

Step 2: The electrophillic attack

You must ensure to draw the ⅔ on the benzene itself and it must face to the area where electrophillic substution has taken place.

Step 3: formation of the product

Ensure that you have drew every species of this step.

Step 4: regeneration of ctalayst

[AlBr₄]⁻ + H⁺ --> AlBr₃ + HBr

Acylation!

You will have to use Ethanoyl chloride and Aluminium trichloride to synthesize the product. This reaction substitutes hydrogen with methyl ketone group.

Step 1: Generation of electrophile

AlCl₃ + CH₃COCl CH₃CO⁺ + [AlCl₄]⁻

Step 2: electrophilic attack

Step 3: formation of the aromatic product

Step 4: regeneration of catalyst

[AlCl₄]⁻ + H⁺ --> AlCl₃ + HCl

Alkylation!

You will have to use Chloromethane and Aluminium trichloride to produce the product.

This reaction substitutes hydrogen with methyl group

4 steps as usual

Step 1: Formation of electrophile (CH₃⁺)

CH₃Cl + AlCl₃ --> [AlCl₄]⁻ + CH₃+

Step 2: Electrophilic attack

Step 3: Formation of the aromatic product

Step 4: regeneration of catalyst

[AlCl₄]⁻ + H⁺ --> AlCl₃ + HCl

Fuming sulfuric acid (sulfonation)

This is the replacement of a hydrogen atom with a sulfonic acid group.

This reaction is carried out by warming benzene in a fuming sulfuric acid at 40*C for 20-30 mins.

Sulfonation can also be written by using SO ₃ but it really depends on the question given.

Nitrations!

Same goes with this, benzene will need a strong electrophile to attack benzene due to its very stable aromatic conformation. You will have to use Nitric acid and sulfuric acid, BOTH CONCENTRATED!

The reaction is carried out between 50*C and 60*C

Temperature too low = no reaction (slow as heck)

Temperature too high = side product forming (dinitrobenzene and trinitrobenzene)

This reaction substitutes hydrogen with nitro group

There are also 4 steps!

Step 1: The formation of electrophile (NO₂⁺)

HNO₃ + H₂SO4 --> NO₂⁺ + HSO4⁻ + H₂O

Step 2: The electrophilic attack

Step 3: Formation of the aromatic product

Step 4: regeneration of catalyst

H⁺ + HSO₄- --> H₂SO₄

Hydrogenation!

Basically, creating cyclohexane.

This is done by mixing benzene with hydrogen and heating under pressure with a nickel catalyst.

C₃H6 + 3H₂ C₆H₁₂

This extreme conditions is required cus benzene\s electronic conformation is very stable.

Phenol

This thing consists of a hydroxyl group joined to a benzene ring. They will undergo electrophilic substitution to preserve its aromaticity, but the difference is that Phenol is much more reactive due to its π electron system. Which is evident from its bromination reactions.

Bromination of Phenol

The bromination of Phenol occurs at room temperature without a catalyst, decolorizes bromine water to form a white precipitate.

Benzene undergoes bromination but it needs a catalyst and reaction mixture must be heated under reflux.

So. how is it more reactive than benzene?

This is the result of the lone pair of electrons in the p orbital on the oxygen atom interacts with the delocalised π electrons in the ring

Therefore, the electron density on the delocalised π electron is increased, the molecule became much more reactive towards electrophiles

This will lead to the greater polarization of the Br molecule, where the Br-Br bond can be broken and Br+ electrophile can be formed to attack the benzene ring.

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Pls ignore the "sp2 hybridized orbitals" word there, cus it would explode ur brain

Pi system = delocalized pi electrons

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Step 3: The formation of aromatic product

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Step 3: The formation of aromatic product

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Step 3: The formation of the aromatic product

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Step 3: The formation of the aromatic product

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Step 3: The formation of the aromatic product

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Step 3: The formation of the aromatic product

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Here is the list of things you have to relearn for this heinous topic

Topic 18 & 19 will be essesential for the last topic Organic Synthesis

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There are 3 types of nitrogen compounds you have to relearn... AGAIN

Amides

Structural formulae rule

RCONH2 for primary amides

RCONHR' for a N-subtituted amides

RCONR'R" for a double N-subtituted amides

General rule:

(Main alkyl groups naming) + (amide)

Suffix: (-amide)

Prefix: (Main alkyl groups naming)

Methan - 1 C

Ethan - 2 C

Propan - 3 C

Butan - 4C

Pentan - 5C

-You get the point, there is no need to give more names.

For example, an amide with 3 C

This would be Propan amide

You may also be given N-substituted amides

This will have a similar general rule as before but a bit different.

Structural formula naming: RCONHR'

It would be:

N-(Side alkyl group naming directly connected to NH group) + (Main alkyl groups naming directly connected to CO group) + (amide)

Methyl - 1 C

Ethyl - 2 C

Propyl - 3 C

Butyl - 4 C

Pentyl - 5 C

For example: CH₃CH₂CH₂CONHCH₂CH₃

N-ethylbutanamide

Another one:

CH₃CONHC₆H₅

N-phenylethanamide

You may also be given double substituted amide

The naming is identical to the previous one, but you will have to prioritize one group over another in alphabetical order.

Due to 2 alkyl groups directly connected to NH you will have to use 2 N prefixes.

If it contains 2 side alkyl groups connected to NH group, you will have to use the classic numbering system.

For example:

CH₃CH₂CH₂CON(C₆H₅)CH₂CH₃

N-ethyl-N-Phenylbutanamide

Another example:

C₆H₅CON(CH₃)CH2 CH₂CH₂CH₃

N-methyl-N-Butylbenzamide

Another one:

HCON(CH₂CH3 )CH2 CH₃

N,N-Diethylmethanamide

Amines

Structural formula rule:

RCNH2 For a primary amines

RCNHR' for a secondary amines

RCNHR'R" for a tertiary mines

Quaternary we dont need to learn.

General Rule

Suffix: (-amine)

Prefix: (Side alkyl group naming)

Methyl - 1 C

Ethyl - 2 C

Propyl - 3 C

Butyl - 4 C

Pentyl - 5 C

In amines, the carbon chain is a side group

It would be:

(Side alkyl groups)(-amine)

For example:

CH₃CH₂NH(CH₂CH₃)

There is 2 ethyl group, thereby:

Di ethylamine

Another example:

CH₃CH₂CH₂CH₂NH₂

There is 4 Carbon so it would be:

Butyl amine

Another example:

C₆H₅NH₂

There is a benzene group here, and it is a side chain so it would be:

Phenyl amine

Amino acids

Structural formula rule:

RCONHR"

General rule:

Prefix: (x-amino), then (Main alkyl group naming)

Suffix: (-oic acid)

naming:

Methyl - 1 C

Ethyl - 2 C

Propyl - 3 C

Butyl - 4 C

Pentyl - 5 C

So it would be:

(x-amino)(Main alkyl group naming)(-oic acid)

Example:

H₂NCH₂COOH

There is carbon here, so it would be:

2-aminoethan oic acid

Another one

CH₃CH(NH₂)COOH

it may look like there is addition methyl group, but what's changing is the position of the amine group. This would be:

2-aminopropan oic acid

Example again

C₆H₅CH(NH₂)COOH

This time the addition of alkyl group is replaced with a benzene ring, which is treated as a side group. The naming would be:

2-aminophenylethanoic acid

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Amines have similar properties to ammonia itself. So, there are 6 reactions that revolve around amine compounds.

Reactions with halogenoalkanes

This will produce secondary amines if the alipathic amines is a primary amine.

Where the general formulae is given by (for a primary amine):

R'NH2 + R''X --> R'NHR'' + HX

Example: (ethylamine and chloroethane)

CH₃CH₂NH2 + CH₃CH₂Cl --> CH₃CH₂NHCH₂CH₃ + HCl

Whenever they react together, they will form hydrogen halides

Another example: (phenylamine and iodopropane)

C₆H₅NH₂ + CH₃CH₂CH₂I --> C₆H₅NHCH₂CH₂CH₃ + HI

From these reactions the product formed are secondary amines.

These secondary amines can react with remaining halogenoalkane to form a tertiary amines. Which also can react again, to form quarternary ammonium salts.

To limit the possibilities of these side reactions, excess ammonia is used to limit the side reactions.

Reactions with water to form an alkaline solution

Amines are similar to ammonia where both are weak bases, they accept a proton. Creating an alkylammonium ion and a hydroxide ion. Thereby mkaing the solution more basic.

From our previous topics on pH we should know that more OH- = more basicity.

For example:

NH3 + H₂O ⇌ NH₄₊ + OH⁻

Which is similar to:

CH₃CH₂NH₂ + H₂O ⇌ CH₃CH₂NH₃ + OH⁻

Amines are stronger than ammonia due to the presence of an alkyl group attached to it.

Alkyl groups are electron-releasing, which increases the electron density of the nitrogen on the amine functional group.

The increase in electron density increases the tendency of amines to attract a proton. The attraction for proton is stronger than ammonia. which makes it a stronger base.

Reactions with acids to form a salt

Amines vary in their basicity as it is heavily dependent on the groups they are attached to. They react with acids to form a salt.

For example:

CH₃CH₂NH2 + HNO3 C --> H₃CH₂NH₃⁺NO₃⁻

Ethylamine reacts with nitric acid to form Ethylammonium nitrate

C₆H₅NH₂ + HBr --> C₆H₅NH₃⁺ Br⁻

Phenylamine reacts with hydrobromic acid to create

Phenylammonium Bromide

Reactions with ethanoyl chloride

This reaction will produce an N-substituted amide. Where it willk undergo a addition-elimination reaction.

For example:

CH₃COCl + CH₃CH₂CH₂NHCH₂CH₃ --> CH₃CONH(CH₂CH₂CH₃)CH₂CH3 + HCl

Reactions with Cu(2+)

Amines just like ammonia can form a complex. Which is caused by 2 steps of reaction. In this spec, amines will react with [Cu(H₂O)₆]²⁺

Deprotonation

Before partaking in ligand exchange reaction, amines deprotonate [Cu(H₂O)₆]2+ to form [Cu(H₂O)₄(OH)₂] and an alkylammonium ion

Which can be seen here:

[Cu(H₂O)₆]2+ + 2CH₃(CH₂)₅NH₂ à [Cu(H₂O)₄(OH)₂] + 2CH₃(CH₂)₅NH₃⁺

Observation:

Pale blue precipitate forms within the solution

Ligand exchange

When the amines are present in excess then ligand exchange can occur. Which again can be seen here:

[Cu(H₂O)₄(OH)₂] + 4CH₃(CH₂)₅NH₂ --> [Cu(2CH₃(CH₂)₅NH₂)₄(H₂O)₂]²⁺ + 2H₂O + 2OH⁻

If its a phenylammonium, then there can only be 4 coordination with Cu²⁺ ion as they're bulky and also VSEPR theory too

[Cu(H₂O)₄(OH)₂] + 4C₆H₅NH₂ --> [Cu(C₆H₅NH₂)₄]²⁺ + 4H₂O + 2OH⁻

Reaction of phenylamine with HNO2 to form a benzenediazonium ion

To prepare this particular compound, you will have to create HNO2. This particular compound doesn't exist at stp, thereby we have to isolate it.

NaNO₂ + HCl --> HNO₂ + NaCl

Then it can finally react with phenylamine.

C₆H₅NH₂ + HCl + HNO₂ --> C₆H₅N₂⁺ + Cl⁻ + 2H₂O

Conditions:

Reaction vessel must be kept in ice-cold mixture which is about 5*C or lower

The concentration of HCl must be dilute

If there is a sudden increase in temperature, Phenol is formed.

It then can react further with phenol to form (4-hydroxyphenyl)azobenzene.

C₆H₅N₂⁺ + C₆H₅O⁻ + OH⁻ --> C₆H₅N=NC₆H₅O⁻

+ H₂O

This reacts further with a proton

C₆H₅N=NC₆H₅O- + H⁺ --> C₆H₅N=NC₆H₅OH

There (4-hydroxyphenyl)azobenzene is created which has a similar dyeing property as methyl orange

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The solubility of amines

Key points to remember:

-Hydrogen bonding with water (why and how?)

-The predominance of intermolecular forces

-solute-solute and solute-solvent interactions

-Enthalpy change of dissolution (depends on question)

-London force interaction with organic solvents.

The presence of a lone pair of electron on a highly electronegative atom Nitrogen on the -NH2 group, allowing to form hydrogen bonds with the hydrogen atoms with water.

This solute-solvent interactions is energetically favorable which helps to overcome the solute-solute interactions, causing the few members of primary amines to be highly miscible with water.

However, as the length of the alkyl groups increases, the influence of london dispersion force becomes more predominant, this disrupts the hydrogen bond interactions of these primary amines with water, this makes solute-solvent interactions much less favorable through the increasing enthalpy change of dissolution, making it less soluble in water.

However, again, involving organic solvents, amines becomes increasingly soluble as the alkyl length increases, this increases the quantity of electrons in the compound, which strengthens the london force intermolecular force, not only that, it also increases their points of contacts and their surface area with the organic solvent, making the solute-solvent interactions much more favorable, this increases the enthalpy change of dissolution, making it more soluble in organic solvents.

Amines can form strong hydrogen bonding with water which easily overcome the solute-solute intermolecular interaction, thereby solute-solvent interactions is more favorabler, making it miscible with water.

However, as the chain length increases, London forces becomes more predominant; this disrupts the hydrogen bonding interactions between -NH2 and water, increasing the ΔH and decreases it's solubility.

It also becomes increasingly soluble in organic solvents, the increase in the predominance of london forces, makes the solute-solvent interactions more favorable due to better points of contacts with other molecules, increasing it's solubility in organic solvent and decreases its ΔH

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Amines can be prepared by 3 ways.

Halogenoalkanes and NH3

First method: This method involves heating a halogenoalkane with ammonia in a closed container and under pressure.

Second method: Halogenoalkane is mixed with concentrated aqueous ammonia.

Reaction equation: (bromopropane with some NH3)

CH₃CH₂CH₂Br + 2NH₃ --> CH₃CH₂CH₂NH₂ + NH₄Br

The reaction mechanism is not in spec.

Preparation from nitriles

Nitriles can be reduced to primary amines by using the LiAlH4 which are mixed with dry ether, this is important cus H2O can interfere with the reaction.

Reaction equation: (2-methylbutanenitrile)

CH₃CH₂C(CH₃)CN + 4[H] CH₃CH₂C(CH₃)CH₂NH₂

if u are going to study chem-related subjects then u have to learn the side reactions

Preparation of aromatic amines from nitrobenzene

Tin and concentrated HCl is mixed to form Tin(II) and Tin(IV) ions, this reduces nitrobenzene into phenylamine.

The reaction is carried on heating under reflux

Reaction equation:

C₆H₅NO₂ + 6[H] --> C₆H₅NH₂ + 2H₂O

nitrobenzene is reduced to phenylamine

C₆H₅NH₃⁺ + OH⁻ --> C₆H₅NH2 + H₂O

some of the phenylamine exist in a form of an ammonium salt as it reacts with the HCl, so NaOH is added which forms the phenylamine

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The AMIDES!!!

Amides are formed by the reaction of amines with acyl chlorides. (Check subtopic 2)

in this subtopic we are much more focused on polyamides.

The formation of polyamides.

This polymer is formed by 2 monomers which are dicarboxylic acid/dioyl chlorides and diamine. These 2 reacts together to form an amide. Here are the polymers of amides:

-Nylon

Monomers needed:

hexanedioic acid

1,6-diaminohexane

Both these react together to form nylon-6,6 by condensation polymerisation.

Forms strong polymer that can be wounded

-Kevlar

Monomers needed:

Benzene-1,4-dicarbonyl dichloride

Phenylene-1,6-diamine

Both these react together to form a strong polymer kevlar by condensation polymerisation.

Properties: Forms strong polymers that is stress-resistant

-Poly(propenamide)

This is made by the addition polymeristion of 2-propenamide

-Polyethenol

This is an addition polymer made by ethenyl ethanoate where there are 2 stages of the polymerisation,

Stage 1: formation of poly(ethenyl ethanoate)

Stage 2: Ester exchange with methanol to form poly(ethenol) and methyl methanoate

Properties: The more the -OH group, the more soluble it is on water.

Uses: can be used for hospital laundry bag which removes the need for hospital workers to touch contaminated clothes as during washing, the bag completely dissolves and clothes are washed clean.

It is also used to make liquitabs which dissolves in water, releasing a quantified amount of detergent to the water.

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These are compounds that contain both -NH2 group and -COOH group.

Here are the properties of amino acids:

Isoelectric points

This is the pH of the (aq) solution when the amino acid dissove to exist as a zwitterion, the amino acid is roughly electrically neutral due to the presence of both -COOH and -NH3.

This measures how acidic/basic an amino acid is.

Formation of salts

All amino acids forms salt with both acids and bases. -NH2 accepts a proton to become -NH3+ forming an ammonium salt. -COOH dissociates to form -COO(-) to form a carboxylate salt.

Optical activity

Almost all amino acids contains a chiral centre due to their molecular structure.

Natural amino acids will rotate a plan polarised monochromatic light meaning they're optically active.

Artificial amino acids WON'T have an optical activity as both dextro . and levo enantiomers exist on the products.

Formation of polypeptides

This Involves the hydrolysis reaction between the -NH2 group and the -COOH group to form a peptide bond.

polypeptide combine together by the folding of polypeptides to form a complex 3d structure known as a protein.

Polypeptides can be returned back to its amino acids constituent though hydrolysis. This is done by prolonged heating of the polypeptides with concentrated HCl.

This breaks the peptide bond by the protonstion of the amine group.

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This is the final boss among them all where it links almost all of the organic chemistry we learnt from AS and IAL.

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Functional group testing (Extra but IMPORTANT)

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Functional group testing (Extra but IMPORTANT)

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Alkene

1) Decolorises Bromine water.

Hydroxy group (any)

1) PCl5 forms misty fumes when mixed with the compound.

Primary alcohol

1) Turns acidified potassium dichromate(VII) from orange to green.

2) Produces silver mirror with Tollens' reagent

Secondary Alcohol

1) Turns acidified potassium dichromate(VII) from orange to green.

2) DOES NOT produce Silver mirror with tollen's reagent.

Tertiary Alcohol

1) Won't react with both acidified potassium dichromate(VII) and tollen's reagent.

Carbonyl group (any)

1) produces an orange precipitate with 2,4-dinitrophenylhydrazine.

Aldehyde

1) Turns Fehling's/Benedict's solution from blue to red

2) Produces silver mirror with tollen's reagent.

Ketones

1) Doesn't react to any chemical compounds listed on aldehydes. but they do form orange precipitate

Methyl ketone/methyl alcohol

1) Forms a yellow precipitate on an alkaline iodine solution

Carboxylic acid

1) Reacts with NaHCO3/Na2CO3 to form effervescence

2) Reacts with PCl5 to form misty fumes.

Phenol

1) decolorises bromine water and forms a white precipitate

2) Soluble in NaOH (aq) but insoluble in HCl

Halogenoalkanes

1) warm with NaOH and add dilute acidified AgNO3

2) Precipitates forms where:

White ppt = chlorine

Cream ppt = Bromine

Yellow ppt = Iodine

3) These precipitates can be further tested using NH3s where

White ppt = both soluble in dilute and conc NH3

Cream ppt = soluble in excess NH3 but insoluble in dilute NH3

Yellow ppt = insoluble in both conc and cilute NH3

Phenylamine

1) decolorises bromine water and forms a white precipitate

2) Insoluble in NaOH (aq) but soluble in HCl

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Empirical formulas

1) Calculate the mass of each element from their compound counterpart.

Element relative mass/(relative mass of compound) x (mass of compound) = mass of element

2) Calculate the moles of each element

(Mass of element)/(relative mass of element) = moles of element.

3) Divide each moles of the elements calculated by the smallest number of moles

4) The empirical formula is now found. Multiply with a common factor, so that all numbers are integers.

Tip: In combustion analysis questions, where oxygen is incorporated into the fuel and the mass of it, is unknown. DONT subtract the number of oxygen present as they're present as an excess.

Mass spectra

This analytical technique allows us to determine the accurate relative Mr of a compound

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1) The last peak toward the right of the spectra is called the molecular ion peak. This shows the Mr of the organic compound.

2) There will be lots of fragments created which are recorded on the mass spectra. Fragments are created on each carbon-carbon single bond, which have a specific mass that will be recorded on the mass spectra.

3) The abundance of the fragments explains a lot about the presence of the fragments, so by using the relative abundance of the compound masses, you can predict what would be the molecular mass of the compound.

When asked regarding mass spectra. Ensure that you write the structural/molecular formula of the fragments and molecular ion.

Visit topic 2 AS chem textbook if unsure.

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IR Spectra

This modern analytical technique will allow us to identify the particular functional group and features in an organic compound.

On this technique, you will have to use the data booklet to answer the question.

1) Match the reading of the IR spectra with the data booklet

2) Write down the functional group you think is present like this:

"1720 cm-1 absorbance which indicates there is a C=O group present, which could either be a ketone or an aldehyde"

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NMR spectra of carbons

C13 NMR spectra will allow us to identify the number of different chemical environments for carbon of an organic compound.

On this technique, you will have to use the data booklet to answer the question.

1) Match the reading of the C-13 spectra with the data booklet, compare them.

2) The number of peaks in a C13 spectra will show the number of carbon environment present on the organic compound.

3) Then write down the functional group you think is present which would be responsible for each peaks, such as:

"The peak at 8 = 203 ppm is due to an aldehyde or a ketone"

"The peak in the range 8= 75-55 ppm corresponds to the C-OH bond present"

"The peaks between 37-6 ppm has a higher absorption by a magnitude of 2x, than other peaks, meaning that there is 2 of the same carbon environment"

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NMR Spectra of protons (high resolution)

This will allow us to identify groups of hydrogen atoms in a compound.

Proton NMR spectra will allow us to identify the number of different chemical environments for carbon of an organic compound.

There will be splitting patterns in the proton NMR spectra, this is due to the carbon atoms have an influence on protons on adjacent/neighboring carbon atoms.

The Spectra will show the height of the absorbance this will show the number of protons existed on the organic compound. The height and the range of the chemical shift will indicate the functional group of the organic compound.

The number of sub-peaks (smaller peaks) will indicate the number of protons that is adjacent to them.

Whenever we work with an NMR spectra of a compound, we have to use the...

(N+1) Rule

This is the rule to calculate the number of peaks + subpeaks each proton will have.

Where N = total number of protons in all of the adjacent carbon to the proton.

+ 1 = the peak of absorption of the desired proton.

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The Grignard reagent is used to:

Extend the carbon chain of a carbonyl compound.

Making carboxylic acids.

Grignard reagent has to be prepared in order to be used. This can be prepared by mixing a halogenoalkane of choice with magnesium in a solvent of dry ether, this is done by heating under reflux. There Girgnard reagent is ready to be consumed.

It is important that the preparation NEEDS to be in dry ether because water can react with reagent to form an alkane and hydroxides of magnesium.

Gignard reagent's general formulae is as follows:

R-Br + Mg --> R-Mg-Br

Girgnard reagent's will react with the 2nd reagent of choice to extend their carbon chain (or the halogenoalkane depends on ur perspective). here are the common reaction for Girgnard.

1) Reaction with carbon dioxide CO2

RMgBr + CO2 + H2O --> RCOOH + Mg(OH)Br

This produces a carboxylic acid

2) Reaction with methanal HCHO

RMgBr + CH2O + H2O --> RCH2OH + Mg(OH)Br

This produces a primary alcohol

3) Reactions with aldehydes RCH2O

RMgBr + R'CHO + H2O--> RCH2OHR'

This produces a secondary alcohol

4) Reactions with ketones RCOR'

RMgBr + R"COR' +H2O --> R"R'COR + Mg(OH)Br

This produces a tertiary alcohol

Acid is added to protonate the formally negative charged oxygen on the C-O bond. To achieve the final desired product.