CM1401 NotesThanks to Dr Linda Sellou, NUS Department of Chemistry. For her physical chemistry lectures, slides and tutorials from which this note is based on.

Also thanks to lecturer Fun Man for this videos and slides on organic chemistry

Accuracy of notes not guaranteed, so refer to textbook to be sure of concepts.

Miscellaneous

1. Always change eV to J before calculations2. Accuracy of answer depends on the decimal place of the data.

a. Do not give more accurate answers, unless it is unreasonable to do so.

Equations

1. Energy of photon: E=hv=hcλ

a. Used to find wavelength of light absorbed or emitted

2. Electron volt: eV= Joules of objectCharge of anelectron

3. Bhor’s equation: n=−13.6( Z2n2 )eV=−2.18×10−18(Z2n2 )Ja. Used to find energy level of a shellb. Only applicable when only 1 electron is present

4. Bond order = Number of electrons∈bonding orbitals−number of electrons∈antibondingorbitals

2a. Used to determine stability of molecule formed and bond strength

5. Equilibrium constant and reaction quotient = aProduct 1Stoichiometryof product 1×aProduct 2

Stoichiometry of product 2

aReactant1Stoichiometryof reactant 1×aReactant 2

Stoichiometry of reactant 2

a. a is the activity of the compound. How much it contributes to the reaction.b. K and Q shows the ratio of products and reactants at specific times of a reactionc. Equilibrium is when time = infinity

6. Ideal Gas Law: pV=nRTa. Where R = 8.314 J/K.mol

7. Equations to find pHa. pH = −lg ¿¿

b. pH = p Ka+lgconjugate base

acidc. pH = p Kw−pOH

8. Change in internal energy: ΔU=w+qa. w is work, q is heat

9. Energy flow due to work: w=−pext ΔV10. Enthalpy: H=U+ pV=q+V Δp

a. At constant pressure: ΔH=q p=C pΔT=mc p ΔT=nc p , m ΔTb. The subscript p indicates constant pressure

c. The subscript m stands for molar, to show that the c is used for molar heat capacity

11. Heat capacity of an object: C p=qΔT

a. Used to find energy needed to cause a certain change in temperature

12. Enthalpy change of a reaction: ΔH r=∑i

v i ΔH f , i( products)−∑i

v i ΔH f , i(reactants)

a. v i is the stoichiometric coefficient

13. Kirchhoff’s Law: ΔH ro (T ' )=ΔH r

o (T )+ΔCp , ro ×(T '−T )

a. Used to find enthalpy change of reaction at a temperature different from given datai. New temperature T’ from given data at T

ii. ΔC p , ro =∑

i

v i ΔCp , io (products )−∑

i

v iΔCp , io (reactants)

b. Only possible when heat capacity is given, else use assumption that enthalpy is constant over a range of temperatures

14. Entropy change of phase transition: Δ S transO =

ΔH transO

T trans

a. Usable only for phase transition at constant pressure at transition temperatureb. Δ S trans

O =−Δ SsurroundingO only at phase transition

15. Entropy change of surroundings: Δ SsurrO =

−ΔH rO

T surr

16. Entropy change of a reaction: Δ SrO=∑

i

v i ΔS iO( products)−∑

i

v i Δ SiO(reactants)

17. Change in Gibbs Free energy (ΔG r)a. From enthalpy and entropy: ΔG r=ΔH r−T ΔSr

b. From formation: ΔG r=∑i

v i ΔGf , i(products )−∑i

v iΔGf , i(reactants)

c. From standard conditions and Q: ΔG r=ΔGrO+RT lnQ

i. R = 8.315 J/K.mold. At equilibrium, Q = K, ΔG r=0: ΔG r

O=−RT ln Ke. From cell potential difference: ΔG r=−nF Ecell

f. From reverse reaction: ΔG forward=−ΔGbackward

18. Transition temperature for Gibbs free energy: T transition=ΔH r

O

ΔSrO

a. The temperature where ΔG rO changes sign

19. Standard potential for voltaic cell: EcellO =Ecathode

O −EanodeO

a. Data depends on temperatureb. Data given is always from reduction perspective, so need change sign

20. Nernst equation: Ecell=EcellO −RT

nFlnQ

a. F = 9.6485 x 104 C/mol21. Integrated rate law

a. Zero order: [A ]t=−kt+[A ]0a. First order: ln [ A ]t=−kt+ln [A ]0

b. Second order: 1

[A ]t=kt+ 1

[A ]0

22. Arrhenius equation: ln k=

−Ea

R∗1

T+ ln A

a. Where A is the collision frequency factor that is affected by temperature, and Ea is the activation energy

23. To substitute intermediate with original reactant in the rate law:

rate=k intemediate∗kreactant fwd

kreactant bwd[B ]n

a. Where B is the original reactant, and n is its order of reaction

b. Basically sub in [A ]m=kreactant fwdk reactant bwd

[B ]n

i. Where kreactant fwdkreactant bwd

is the net formation of A if the formation of A is

reversibleii. Else just [A ]m=kreactant [B ]n

Physical Chemistry

Electrons in an Atom

1. A diffraction pattern is formed when a wave splits up after passing through a crystal2. Electron are shown to behave like waves when passed through an Aluminium atom, as there

is the formation of a diffraction pattern, much like one produced by X-raysa. As in there is clearly wave interference, proving the wave-particle duality

3. The Bhors model of electron shells is useful in explaining why the light emitted by Hydrogen is different from the light emitted by Helium

a. Electrons absorb different wavelength of light when excited and produce a different colour upon release

b. Electrons can only be excited and change shells by absorbing or emitting a specific quanta of energy (photon)

i. Quantum is the minimum amount of any physical entity involved in an interaction

ii. Photon can refer to light and any other electromagnetic radiation4. Bhors model is limited as it cannot predict the light spectrum of other atoms, working for

only 1 electron atoms or ions.a. Bhors assumes all the sublevels in a shell have the same energyb. There are additional electron forces that complicate the prediction

i. Electron-electron repulsionii. Effective nuclear charge

5. Energy of a particle can be written in Joules or electronvoltsa. eV is the amount of energy gained by the charge of a single electron moved across

an electric potential difference of one voltb. Remember to write the correct units!

6. Schordinger says that we do not need to know the exact position of the electron, just use the wave properties of the electron to find the atomic orbital (psi), and this the probability of finding the electron (psi squared)

7. An electron has 4 numbersa. Principle is the shell it belongs inb. Angular is the shape of its orbitalc. Magnetic is the orientation of the shaped. Spin is the direction of electron spin

8. Pauli exclusion principlea. No two electrons can have the same set of quantum numbers

i. Electrons in the same orbital have the same energy but opposite spins9. Probability density (psi squared) does not equal the probability of electron position.

a. Psi squared is: Probability / Volume10. R2(r) = Psi squared11. Radial distribution function P(r) is the probability of finding the electron on the surface of

circle with radius r.a. P(r) is the sum of all the Psi squared at all points on the circle

12. The boundary of space of an electron is where it spends 90% of its time13. The graph of P(r) shows that they highest probability of finding the electron is not at or very

near the nucleus

14. With the 2s and 3s orbitals, there are nodes where there is a 0 probability of finding an electron

15. The bigger an orbital, the further away the electron spends most of its time.16. The energy is negative in an energy level diagram, as it is convention

a. 0 is taken as a point infinitely away from the atom17. When an electron drops down from the excited state, to a lower shell it can drop in two

waysa. It can drop directly back to ground state, release the same wavelength it absorbedb. It can drop by steps, to an intermediate electron shell, releasing the corresponding

wavelength for the energy gap. Thus this produces visible light (or not).18. There is a splitting of energy level within the shell, which gives rise to the p orbital having

more energy than the s orbital.a. Energy level can differ between different elements, due to nuclear chargeb. Or within the shell, due to shielding effect (repulsion) and penetration effect (orbital

shape)i. Penetration effect is where the electrons in the s orbital have a probability

of being close to the nucleus, closer than the s orbital below it, effectively penetrating the shielding effect.

Properties of atoms

1. Atomic radiusa. Decreases along the period

i. Additional proton increases nuclear chargeii. Increase is greater than loss from electron-electron repulsion

Even when electron is added to another subshelliii. Thus effective nuclear charge increases, drawing the electrons closer to the

nucleusiv. Exceptions of period 4 and below need not be known

2. Ionisation energya. Increases along the period

i. As effective nuclear charge decrease as more protons in nucleusii. More energy needed to remove electron that is closer to the nucleus

b. Except when electron is added to new subleveli. Shielding effect of the lower sublevel decreases effective nuclear charge

ii. Valence electron is easier to remove due to reduced attraction.c. Except when first electron pair is made by adding to an unpaired electron

i. Electron-electron repulsion within the orbital decreases energy required to remove electron

d. To answer questions about ionisation energy,i. Use energy level diagram

ii. Or write out the electronic configuration3. Electron affinity

a. Decreases along the period, becoming more negativei. Additional protons increases nuclear charge

ii. Thus more energy is released when an electron is added to the valence orbital

b. Except when adding to an orbital with an unpaired electroni. Electron-electron repulsion needs to be overcome when adding the electron

ii. Thus less energy is releasedc. Except when adding to a new sublevel

i. The electron is added to an orbital further away from the nucleusii. Thus less energy is released.

d. Except when adding to an already half-filled subleveli. The half-filled sublevel is stable as all orbitals contain one electron

ii. Energy is required to disrupt this stability, thus less energy is released The EA of nitrogen is positive because of this

e. Electron affinity is negative as it is the energy difference of the electron shell from shell infinity

4. For group numbers, follow the number given in the periodic table

Electron orbitals in covalent bonds

1. Covalent bonds arises from attractive and repulsive forcesa. The atoms are most stable at a specific distance from each other

i. Bond length determined by attraction between nucleus and electrons, and nucleus-nucleus repulsion and electron-electron repulsion

2. Lewis structure (aka dot-cross diagram)a. Shows electrons involved in bonding as well as long pairs of electrons

3. Valence-Shell Electron Pair Repulsion (VSEPR) theorya. Model that minimises electron-electron repulsion by arranging electrons are far

apart as possibleb. Structure is based on number of lone pairs and bond pairs

i. Can be written with VSEPR notation: A,X and E A is the central atom, X represents a bond pair, E is a lone pair Use subscript to represent multiple pairs of the same type

c. Shape differs between electron group arrangement and molecule shapei. Electron group arrangement counts both lone and bonding pairs

ii. Molecule shape removes lone pairs from the shape.4. Valence bond theory is one way to explain covalent bonding

a. Valence bond theory explains the VSEPR structure with orbital hybridizationi. Each electron pair (bond or lone) requires an orbital, so the orbitals

participating in the hybridization equals the number of pairsb. Hybridization is needed as the stable configuration of 8 electron requires 4 pairs of

electrons to be involved in bonding (or lone), and the p sublevel can hold only 3 pairs

c. Covalent bonds are formed by the overlap of the hybridised orbitals i. Double bonds are considered as only one bond pair when calculating

number of electrons pairsd. Useful to show molecular shape as obtained with VSEPR theorye. Limitations

i. Electrons are not localised in a specific regionii. Bond energy cannot be evaluated

5. Molecular orbital theory is another way to explain covalent bondinga. The theory does not hold the electrons within a specific orbital, but consider them

delocalised over the whole moleculei. Uses Schrodinger equation to calculate probability of the electron position

ii. The molecules does not have the shape described by VSEPRb. When molecules bond, the orbitals interact

i. There is the same number of new orbitals as the number of component orbitals

Two s orbitals interacting will produce two resultant orbitalsii. Only the same shell and shape of orbitals can merge

The results depends on the waves of the electrons and their interaction

iii. The bonding orbital has a lower energy than the original orbitals The atoms are stabilized as electron density between the atoms is

increased Formed when the orbitals are in-phase and constructively interfere

iv. The anti-bonding orbital has a higher energy than the original orbitals The atoms are destabilized as the electron density between the

atoms is reduced, as they intensify nearer the nucleus, facing away from the bond.

Formed when the orbitals are out-of-phase and destructively interfere

v. A sigma bond is formed when the orbitals interact in the same plane as the bond axis

For the first bond between atomsvi. A pi bond is formed when there is a side to side overlap of orbitals

For the second and higher bonds between atomsvii. There is also electronic configuration for molecules, much like that of atoms,

which shows where the electrons are in each mixed orbitalc. Bond order determines the stability of the molecule

i. The molecule is stable when the bond order is greater than 0 The higher the bond order, the stronger the bond

ii. When bond order = 0, the molecule is not stableiii. Negative bond order is not possible as the electrons fill the lower energy

bonding orbitals firstd. When drawing the energy diagram

i. Optional to count the total number of valence electrons and write the electronic configuration of the molecule

ii. First draw all the orbitals of the two atoms and label the element for both Note the slight energy difference between the orbitals of different

elements due to nuclear charge The difference between p orbitals is smaller than that

between s orbitals Draw the electrons present in the orbitals

iii. Draw the bonding and anti-bonding orbitals and label them There should be only one pair of sigma bond orbitals, with the rest

being pi bond orbitals For each orbital interaction, like s with s, and p with p

The anti-bonding orbitals have a higher energy than the bonding ones

The anti-bonding sigma bond orbitals always have more energy than the anti-bonding pi bonds

The position of the bonding sigma bond orbitals depends on the elements

Due to s-p orbital mixing, making the s orbitals more stable, while destabilising the p orbital, increasing its energy

For Boron to Nitrogen, the bonding sigma bond orbitals are above the pi bond ones

For Oxygen and Fluorine, the bonding sigma bond orbitals are below the pi bond ones

The presence of a less electronegative atoms will cause s-p orbital mixing, regardless of the presence of a more electronegative one

iv. Exception when bonding with Hydrogen

Since Hydrogen only has 1s, when bonding with Period 2 elements, it will interact with the orbital having a close energy with it

Calculate the energy of the orbital with Bhor’s equation Usually 2s will have too little energy, so it interacts with 2p

instead When bonding, 1s can only interact with the orbital that is

symmetrical around the bonding axis By lecture’s definition, it is the 2pz orbital The other 2 orbitals do not interact and remain at the same

energy level Draw the bonding and anti-bonding orbital and label

Do not label with subscript since different orbitals are interacting, just the sigma and sigma star.

e. The molecules formed can be either paramagnetic or diamagnetici. Paramagnetic molecules are attracted to a magnetic field

Due to unpaired electrons in the resultant orbitalsii. Diamagnetic molecules are repelled from a magnetic field

Due to paired electrons in the resultant orbitalsf. Useful to explain bond energy, magnetism and bond polarity

Chemical Equilibrium

1. Equilibrium reactions are dynamic, reversible and equilibrium is obtained from either direction

2. Equilibrium constant, Ka. Ratio of activity of products over activity of reactants at equilibrium

i. Only affected by temperatureb. Based on a (activity)

i. It is the propensity for a material to contribute to a reactionii. It is unitless

Thus K is also unitlessiii. For pure solids, liquids and solvents, a = 1iv. For dilute solutions, ideal gases, their activity coefficient is approximately 1v. Otherwise the activity is pressure or concentration

c. It indicates the ratio of products over reactants at equilibriumi. Shows if the reaction is product or reactant favoured, or both are equal

3. Reactant quotient, Qa. Ratio of activity of products over activity of reactants at any time during the reaction

i. Ratio of products and reactants will change till Q = K When Q < K, more net products will form When Q > K, more net reactants will form

4. Le Chatelier’s principlea. States that if a system at equilibrium is disturbed, the system will tend to shift its

equilibrium position to counter the effect of the disturbance5. Strength of acid and bases is the inversely proportional to the strength of their conjugate

base and acida. Strong acids have a very weak conjugate base, allowing complete dissociation

i. K >> 103

b. Weak acid will produce a weak conjugate base, creating an equilibrium reactionc. The reactivity of the acid and conjugate base are related

i. pKw = pKa + pKb

ii. pKw = 14 at 25oC6. Amphiprotic (slightly different from amphoteric) compounds have both acidic and basic

properties and can accept or donate a protona. Difference with amphoteric lies with the Lewis definition based on exchange of

electronsi. Reaction can occur where electrons are exchanged without movement of

protonsb. All amphiprotic compounds are also amphotericc. Zwitterions are amphiprotic as they have both negative and positive charges

7. Polyprotic acids can donate 2 or more protons8. Acid strength depends on several factors

a. Bond strength decreases down a group, allowing the proton to be released more easily, as the bond requires less energy to break

i. Bond strength decrease due to increasing atomic radii increasing bond length

b. Electronegativity increases across a period, allowing the proton to be released more easily, as the electrons on Hydrogen are pulled away from it

i. Electronegativity increases due to increasing effective nuclear chargeii. Electronegative groups also affect the dissociation of hydroxyl groups

Quantity and quality of the electronegative groups play a part.9. Buffers resist changes in pH

a. Le Chatelier’s principle shifts the equilibrium of the acid dissociation10. Question solving skills

a. Always write out the equation of the reaction and equation for Ki. Sometimes the inverse of K is needed for the reaction

ii. When weak acid and weak base react, sub in Ka, Kb and Kw to find the reaction’s K

b. Strong acid and bases dissociate completely, thus for their reaction, the ions can be ignored, leaving H+ and OH- to react

c. Finding pH when weak acid is dissolvedi. Write out the equation

ii. Calculate the initial and final composition of the reactioniii. Write out the K for each reaction involvediv. Assume that the acid is weak, thus the change in initial concentration is

negligible Always do this, or at least address this step, as it is part of the

markingv. Calculate pH

vi. Verify assumption is plausible, as the dissociated amount is less than 5% of the initial concentration

Else solve the quadratic equation instead using x=−b±√b2−4ac2a

d. Finding pH when strong acid reacts with weak basei. Write out the equation

ii. Calculate the initial and final composition of the reactioniii. If equal amounts are used

pH depends on the conjugate acidiv. If more acid than base

Assume pH depends only on the strong acid left, rather than the weak conjugate acid

Find pH using concentration of H+ producedv. If more base than acid

pH depends on pKa and ratio of conjugate acid and base lefte. Finding pH when multiple acid and/ or bases are added

i. Identify the strongest base and acid from their dissociation constantii. Assume all other acids and bases have a negligible effect

iii. Write the equation between the strongest acid and baseiv. Find the initial and final composition of the solutionv. After the reaction between the strongest acid and bases has occurred,

assume that further dissociation is negligible and assume the compound with the highest K determines the pH

vi. Calculate pHf. Preparing buffer of certain pH

i. Choose the weak acid/base with a pKa closest to the pH

ii. Find the concentration of H+ at the target pH and calculate ratio of conjugate and original from there.

Thermodynamics

1. Internal energy (U) is the sum of all kinetic and potential energies of all particles in a systema. Dependent on temperature, pressure, volume, and molar quantityb. Path-independentc. Only change in internal energy can be calculated, not absolute

2. There are two ways to transfer energy into a closed system. ∆U=w+qa. By doing work, wb. Or by heating, qc. Only in an open system does adding or removing matter changes the internal energy

3. Work is the amount of energy flowing across the system-surroundings boundary, accompanying a change in the volume of the system

a. It is path dependent4. Heat is the amount of energy flowing across the system-surroundings boundary, due to

differences in temperaturea. Heat flows spontaneously from high temperature regions to low temperature onesb. It is path dependent

5. Enthalpy is the thermodynamic potential of the system. H=U+ pVa. At constant pressure (and therefore volume), only the heat flow changes the

internal energy, and thus enthalpy6. Enthalpy change in the system is taken from the surroundings: ΔH system=−ΔH surroudning

7. Heat capacity is the energy needed to raise the temperature of an object by 1 Ka. At constant pressure, q p=C p ΔTb. Specific heat capacity is the energy needed to raise 1 g of an object by 1K

i. C p=mc pc. Molar heat capacity is the energy needed to raise 1 mole of an object by 1K

i. C p=nc p , m m stands for molar, that the c used is for molar heat capacity

instead of specific heat capacity. It does not mean a variable is fixed8. Standard condition fixes the pressure at 1 bar, and concentration at 1 M

a. Temperature is not fixedb. The compound is in its most stable form at 1 barc. Represented by the naught superscript

9. Reference state of an element is its most stable forma. Instead of using gaseous molecules in an equation, the element’s crystal can be used

10. The enthalpy change of formation is when 1 mole of a product is formed from pure elements in their standard states.

a. Can be used to find the enthalpy change of reaction11. Using Hess’s Law, the enthalpy change of reaction can be found through the addition of

enthalpy changes of intermediate steps12. Kirchhoff’s Law is used to find enthalpy change of reaction at a temperature different from

given dataa. Heat capacity of products and reactants are neededb. Else use assumption that enthalpy change is constant over a range of temperatures

13. Take note of the different units between the data given for enthalpy, entropy and heat capacity

14. Entropy is how much (total energy) and how widely energy (number of possible microstates) is dispersed

a. A pure crystal only has one arrangement of atoms at 0K, thus it is the reference point

15. Entropy changes when there is a change in state, or change in number of moles16. The standard entropy of a substance is the increase in entropy when it is converted from its

perfect 0K crystal, to its standard state conditions17. Spontaneous reactions move towards equilibrium without continuous external aid

a. Catalyst is not external aid, since it does not affect equilibrium, only speed18. Gibbs Free energy is the change in entropy of the universe, which by the second law of

thermodynamics, should always increasea. Gibbs free energy is negative for spontaneous reactions as

ΔG r=−T ΔS (universe)b. The change in Gibbs free energy is the maximum energy available to do work

19. Gibbs free energy under standard conditions is related to the equilibrium constanta. When the reaction is spontaneous, ΔG r

O<0, the reaction is product favoured, K > 1

b. When the ΔG rO>0, the reaction is reactant favoured, K < 1

c. When ΔG rO=0, the reaction favours neither side, K = 1

i. Also occurs at temperature of transition.20. Gibbs free energy is related to the reaction quotient

a. When ΔG r<0 ,Q<K , so the reaction occurs spontaneously to form more productsb. When ΔG r>0 ,Q>K , so the reaction is spontaneous in the reverse directionc. When ΔG r=0 ,Q=K , so no net change in entropy as the reaction is in equilibrium

d. When, ΔG r=ΔGrO ,Q=1, with the

productsreactants

=1

i. Because ln(1) = 021. Gibbs free energy of a forward reaction is the negative of that of the backward reaction

a. ΔG forward=−ΔGbackward

Electrochemistry

1. Two types of cellsa. Voltaic/ Galvanic cells are spontaneous and produces an electric currentb. Electrolytic cells are non-spontaneous, using electric current for chemical reactions

2. Similarities between the two cell typesa. Both involve redox reactionsb. Both have oxidation at the anode, and reduction at the cathode

i. Electrons flow from anode to cathode3. Differences between the two cell types

a. The cathode in a galvanic cell is positive, while it is negative in an electrolytic celli. Electrolytic cell has a negative cathode to attract the positive ions

b. A galvanic cell requires two half cells separated by a salt bridge, an electrolytic cells doesn’t need the two electrodes to be separated

c. A galvanic cell produces a voltage equal to the cell potential difference between the cells, an electrolytic cell requires a voltage greater than the cell potential difference.

4. Cell notationa. Anode electrode | Pre-oxidised form | Post-oxidised form || Pre-reduced form |

Post-reduced form | Cathode electrodei. Shows how electrons flow.

ii. Pre-oxidised form may be the same as electrode5. Standard cell potential for half-cell reactions needs to include the phase when writing the

equation6. Since half-cell potentials cannot be measured directly, the reference electrode is they

hydrogen half cella. Pt | H2 (1 bar) | H+ (1 M) || = EH 2

O = 0V7. Half-cell potentials shows their relative ability to act as oxidising agents (get reduced)

a. The more positive the potential values, the more likely they are to get reduced8. The cell potential difference is related to the Gibbs free energy of the reaction

a. The cell potential difference has to be positive for the reaction to be spontaneous, else the reaction will happen in the opposite direction

9. The cell potential difference at any point during the reaction can be found using the NERNST equation

Kinetics

2. Reaction rate is the change in concentration of reactants/ products per unit timea. Product concentration will increase, while reactant will decrease over time until

equilibrium

b.Rate of reaction=

−1aΔ [A ]

Δti. Where A is the reactant, and a is the stoichiometric ratio of A

ii. Negative since the concentration of A is decreasingiii. Unit is concentration over time

M s−1

mol L−1s−1

c. Rate of reaction is determined from experimental datai. Such as change in light absorption amount by reaction vessel

ii. Such as change in conductivity if ionic products are formediii. Such as change in pressure if gaseous compounds are involved

3. Rate Law is an equation that shows the dependence of the reaction rate on the concentration of each reactant

a. Rate=k [ A ]m [B ]n

i. k is the rate constant and is only affected by temperature (and catalyst) k is not affected by catalyst if catalyst concentration is included in

the rate law equation Unit depends on the order of reaction

Zero order is mol L−1s−1

First order is s−1

Second order is Lmol−1s−1

Third order is L2mol−2 s−1

Remember to use the correct units for kii. m and n are the order with respect to the reactant

b. The order with respect to the reactant is determined experimentallyi. It is independent of the stoichiometry of the reactions

ii. Done by changing the initial concentrations of reactants and measuring the initial rate

4. To find reactant at time t , with known k (or vice-versa) and final and initial concentrations, from integrated rate law

a. For zero order reaction with respect to the reactanti. [A ]t=−kt+[A ]0

Thus −k is the slope, −k=[A ]1− [A ]0t1−t0

b. For first order reaction with respect to the reactant

i. ln ([A ] t[ A ]0

)=−kt

Convert known concentrations to ln [A ]ii. ln [ A ]t=−kt+ln [A ]0 for y=mx+c graph

Thus – k is the slope, −k=ln [A ]1−ln [A ]0

t 1−t0c. For second order reaction with respect to the reactant

i.1

[A ]t=kt+ 1

[A ]0

Thus k is the slope, k=

1[A ]1

− 1[ A ]0

t 1−t0 Plotting a graph of ln [A ] against t will not give a straight line

d. Unknown order of reaction can be tested by putting in known t and k to see which equation fits

i. Or plot the graphs to see which kind of x and y gives a straight linee. The equations apply only to reactants, since initial concentration of product is 0

i. Find product concentration by first finding reactant’s5. Half-life is the time required for the reactant concentrations to drop to half of its initial value

a. Substitute [A ] 12

=[A ]02

into the integrated rate law equation to find t (which is t 12)

b. For zero order reaction

i. t 12

=[A ]02k

c. For first order reaction

i. t 12

=ln 2k

The half-life is independent of the initial reactant concentration Each half-life measured from any reactant concentration is an equal

period of time Thus different first order reactions with the same k gives the

same half lifed. For second order reaction

i. t 12

= 1k [A ]0 Each half-life depends on the initial reactant concentration Thus each half-life is twice as long as the preceding one

6. Reaction mechanism is a sequence of reaction steps that describes the pathway from reactants to products

a. An elementary reaction is a reaction where reactants react directly to form the products in a single reaction step with a single transition state

i. Each ‘reaction step’ is an elementary reactionb. The rate law for elementary reaction follows its molecularity

i. Molecularity classifies elementary reactions based on the number of molecules on the reactant side of the equation

c. The rate determining step is the slowest step in the reaction mechanism as it limits the rate of the entire reaction

i. The overall reaction’s rate equation follows the molecularity of the slow step

7. Catalyst is a substance that increases the rate of reaction without itself being consumed at the end of the reaction

a. It is used in one step and regenerated in another.b. The concentration of catalyst can be included in the rate law equation too

Organic Chemistry

Nomenclature

1. Find the parent chaina. The longest carbon chain that contains the functional group

2. Start numbering the carbons from the end that gives the lowest first functional group or substituent number

a. Lowest first functional group takes priority.b. In cycloalkane, make the carbon-carbon double bonds between carbon 1 and 2c. Compare second substituent number only if the first one is tied from both directionsd. If all the substituents can receive the same number, number by alphabetical priority

3. Name substituents in alphabetical order with their locanta. Prefixes indicating identical group quantity is not considered

4. Use comma to separate numbers, dashes to separate branches, and brackets when there is secondary branching

a. No dash between branch and parent at the end5. Prefixes

a. Functional groupsi. Cyclo-___-ane

ii. Cyclo-___-yliii. Phenyl

When benzene is a substituentiv. Nitro

-NO2

v. ab. Numbers

i. Meth-ii. Eth-

iii. Prop-iv. But-v. Pent-

vi. Hex-vii. Hept-

viii. Octix. Non-x. Dec-

xi. Undec-xii. Dodec-

xiii. Tridec-xiv. Tetradec-xv. Add an ‘a’, when there are 2 or more substituents. For example,

Metha- Buta- Trideca-

xvi. But not for alcohols since the substituent numbers are added after the ‘ane’ Hexane-1,4-diol

6. Suffix

a. Alkanei. –ane

b. Alkane branchi. –yl

c. Alkenei. –ene

d. Alkene branchi. –enyl

e. Alkynei. –yne

f. Alkynes branchi. –ynyl

g. Alcoholi. –anol

ii. –ane-X,X-diol There is an e if more than one alcohol group

h. Alcohol alkenei. –X-en-X-ol

i. Etheri. ___-yl ____-yl ether

j. Ether as substituenti. –oxy

k. Aldehydei. –anal

l. Ketonei. –an-X-one

ii. –ane-X,X-dionem. Ketone alkene

i. –X-en-X-onen. Ketone as substituent

i. –oylo. Carboxylic acid

i. –anoic acidii. –anedioic acid

Since the acid function group must be at the terminalsp. Carboxylic acid in a cyclo___

i. –carboxylic acid Like cyclopent-1-enecarboxylic acid

q. Nitrilei. -anenitrile

General formula

1. Alkane parent chaina. CnH 2n+2

2. Alkane brancha. CnH 2n+1

3. Cycloalkane

a. CnH 2n

4. Alkenea. CnH 2n

5. Alkynea. CnH 2n−2

6. Benzene

Drawing organic molecules

1. Newman projectiona. Two carbon atoms are aligned and viewed end-on

i. Carbon atoms represented by a dot and circle R-groups attached to the first carbon are joined to the dot R-groups attached to the second carbon are joined to the circle

b. All R groups are shown with spatial orientationi. As in the direction of bond matters

ii. Torsional strain can be shown Staggered projection where all bonds are evenly spaced out Eclipsed projection where all bond are overlapping Thus the rotation is not perfectly free

iii. Steric strain can be shown When two bulky groups are 60 degrees to each other (Gauche

conformation) instead of being 180 degrees apart (anti conformation)

2. Sawhorsea. Shows a carbon-carbon bond from an angleb. All R groups are shown with spatial orientationc. Chair conformation for cyclohexane

i. Draw two parallel linesii. Have a carbon situated at ¾ of the distance between the lines at both sides

iii. Join all the six carbons together in a ringiv. Fill in the 6 equatorial hydrogen atoms

4 by following the zig-zag lines 2 by drawing lines parallel to the original two lines at the middle

carbonsv. Fill in the 6 axial hydrogen atoms

3 above the plane to make a triangle formation 3 below the plane to make a triangle formation

d. Ring flip can occur for cyclohexanei. Draw the chair conformation with the parallel lines facing a mirrored

directionii. Carbon 1 goes from bottom left to top left

iii. Carbon 4 goes from top right to bottom rightiv. All the equatorial and axial R-group swap orientation.

Reactions

1. Alkanesa. Combustion with oxygen

i. Uses oxygen and heatii. Produces carbon dioxide and water

b. Halogenationi. Uses halogen and UV light

ii. Free-radical substitution occurs Initiation Propagation Termination

iii. Mixture of halogenated products are formed2. Alkenes

a. Electrophilic addition of wateri. Uses H+/H2O and organic solvent (ether) to dissolve organic compound

ii. Carbocation is formed on the carbon with more substituents, after addition of one hydrogen atom

Due to hyperconjugation stabilising the charge This is Markovnikov’s rule

iii. Hydride shift can occur to form a more stable carbocation Note that there is no quaternary carbocation

iv. Nucleophile, H2O, is then added to the carboncation to give -OHb. Electrophilic addition of hydrogen halide

i. Uses HF and organic solvent (ether) to dissolve organic compoundii. Similar to above, but nucleophile is halogen to give –X

c. Electrophilic addition of halogeni. Uses X2 and organic solvent (ether) to dissolve organic compound

ii. The halogen approaches the double bond and undergoes induced polarisation

iii. Halogen bonds breaks, giving a halogen ion, and a halogen with a positive charge bound to both carbons

Forms a 3-membered ring, that has high ring strain and is unstableiv. Nucleophile, X-, attacks the electrophile joined to the carbons, X+

Nucleophile is added to the carbon that is more stable with a partial positive charge

d. Electrophilic addition of hypohalous acidi. Uses X2 and H2O, to give HOX, and organic solvent (ether) to dissolve organic

compoundii. Similar to above, where X+ is the electrophile that joins to both carbons

iii. While OH- is the nucleophilee. Electrophilic addition of Hg(OAc)2

i. Uses Hg(OAc)2/H2O, and organic solvent (ether) to dissolve organic compound

ii. Similar to above, but with HgOAc+ as electrophile that binds to both carboniii. OH- is the nucleophile

f. Electrophilic addition of BH3

i. Uses BH3 and organic solvent (ether) to dissolve organic compoundii. H-BH2 approaches the double bond from the same plane

It is orientated to avoid steric hindrance of BH2 Follows anti-Markovnikov’s rule

Since Hydrogen is added to the carbon with more substituents, to avoid steric hindrance

g. Electrophilic addition of hydrogeni. Uses H2, PtO2 catalyst and organic solvent (CH3COOH) to dissolve organic

compound Or Pd/C and CH2CH2OH

ii. The hydrogen molecule is adsorbed onto the catalyst, allowing it to attack the alkene from the same plane

h. Reduction of R-groups to produce alcoholi. -HgOAc can be reduced by NaBH4, to give –OH

ii. –BH2 can be reduced by H2O2/OH- to give –OH i. 2 step addition of diol

i. Using mCPBA and CH2Cl2 as solvent Gives epoxide, where an oxygen is bonded to both carbons Use H3O+/H2O to add –OH to both carbons

ii. Using OsO4 and Pyridine as solvent Gives cyclic osmate intermediate Use NaHSO3 and H2O to add –OH to both carbons

j. Oxidation reaction to split the bondi. Use O3 to give ozonide

ii. Use Zn and H3O+ to add double-bond O to both carbons3. Cycloalkenes

a. Similar to alkenes but there is anti- and syn-additioni. All reactions that involve electrophiles and nucleophile will be anti-addition

Since nucleophile will attack from the opposite planeii. Hydroboration (H-BH2) and hryogenation will be syn-addition, since they

approach from the same planeb. Subsequent reduction would give the syn- or anti-addition of water, and on different

carbonsc. mCPBA gives anti-addition as it is a two step reactiond. OsO4 gives syn-addition as it is a one step reaction

4. Alkynesa. Electrophilic addition similar to alkenes

i. Turns the triple bond into a double bondii. Can undergo an further reaction to give single bond

b. Nucleophilic substitution to add carbonsi. Use terminal alkyne and strong base (NaNH2) to produce acetylide anion and

cation (salt)ii. Add Halogenoalkane that contains the desired carbon chain

Halogen reacts with nucleophile (anion) and transfers the carbon chain to the alkyne

The triple bond is still preserved5. Benzene

a. Electrophilic substitution of halogen (Halogenation)i. Uses X2/FeBr3

The X2 reacts with FeBr3 to produce an electrophile, X+

The electrophile attacks the double bond, causing a carbocation This is stabilised by resonance

Re-aromatization occurs to eject the hydrogen, keeping the halogenb. Electrophilic substitution of halogenoalkane (Friedel-Crafts Alkylation)

i. Uses halogenoalkane and AlCl3

The halogenoalkane reacts with AlCl3 to produce an electrophile Acyl cation is stabilised by resonance

Take note of hydride shift Attack is similar to above

ii. As alkyl is an activating group, it causes further alkylationiii. Does not occur if a strong electron-withdrawing group is present

c. Electrophilic substitution of nitrogen dioxide (Nitration)i. Uses concentrated HNO3/H2SO4

The reacts HNO3 with H2SO4 to produce NO2+

Attack is similar to aboved. Electrophilic substitution of SO3H (sulfonation)

i. Uses fuming H2SO4 (fuming means contains SO3) The reacts SO3 with H2SO4 to produce SO3H+

Attack is similar to abovee. Oxidation of alkyl substituent

i. Uses KMnO4, H2O and heat OR Na2CrO7, H2SO4/H2O and heat Requires the first carbon to not be quaternary as hydrogen is

neededii. Converts the alkyl chain into –COOH

f. Hydrogenation of benzene requires a highly effective catalysti. Uses H2, Rh/C, ethanol at 1 atm and 25 degrees Celsius

Syn-addition of hydrogen6. Halogenoalkane

a. Elimination of hydrogen halidei. Uses KOH and organic solvent

ii. Gives alkene, KX and water X goes to K, H goes to OH.

7. Alcohola. Elimination of water

i. Uses H2SO4, H2O and THF (organic solvent) at 50 degrees Celsius ii. Gives alkene and water

8. Step up reactionsa. Terminal alkynes through 2-step

i. Use strong base to give salt NaNH2/NH3

ii. Use RX, where R is the desired carbon chainiii. The triple bond is still intact

b. Friedel-Craft Alkylationi. Only for benzene through electrophilic substitution

ii. Use RX and FeBr3

c. Friedel-Craft Acylationi. Only for benzene through electrophilic substitution

ii. Use RCOX and FeBr3

iii. Forms a ketoned. Grignard Reagent with carbonyl compound (nucleophilic addition)

i. R from R-Mg-X gets added to the R’-CR”=O, forming R’-CRR”-OH e. SN2 or SN1 with nucleophile containing R

i. Make nucleophile strong Make –OH strong by reacting with NaH to produce salt that has

negative charge9. Quick reference reagents

a. Alkene and Alkynei. H3O+ + H2O

Electrophilic addition of water (carbocation)ii. HX

Electrophilic addition of HX (carbocation)iii. X2 +H2O (excess)

Electrophilic addition of HOX (three-membered ring)iv. X2

Electrophilic addition of X2 (three-membered ring)v. Hg(OAc)2 + H2O + NaBH4

Electrophilic anti-addition of H & OH (three-membered ring)vi. BH3 + H2O2,OH-

Electrophilic syn-addition of H & OH (parallel)vii. H2,Pd

Electrophilic syn-addition of H2 (parallel)viii. mCPBA + H3O+ +H2O

Electrophilic anti-addition of OH & OH (three-membered ring)ix. OsO4 + NaHSO3

Electrophilic syn-addition of OH & OH (parallel)x. O3 + Zn,H3O+

Breaking double bond into ketone/aldehydexi. NaNH2,NH3 + RX

For alkyne, substitution with R Step up reaction

xii. NBS, uv Bromination of allylic position (note free radical shift)

b. Benzenei. X2,FeBr3

Electrophilic substitution with X (halogenation)ii. Concentrated HNO3, concentrated H2SO4

Electrophilic substitution with NO2 (nitration)iii. Fuming H2SO4

Electrophilic substitution with SO3H (sulfonation)iv. RX,FeBr3

Electrophilic substitution with R (Friedel-Crafts alkylation)v. RCOX,FeBr3

Electrophilic substitution with RCO (Friedel-Crafts acylation)vi. KMnO4 + H3O+

Oxidation of alky substitute (not tertiary) into COOHvii. H2, Rh on C

Electrophilic addition of H2 to convert benzene to cycloalkanec. Alkyl halides

i. Mg, THF Produces Grignard Reagent Reacts with weak acid (H2O, ROH. RCO2H, RNH2) to give alkane

Forms a salt with the acid as by-product Reacts with carbonyl compounds to give alcohol in a step up

reaction Reacts with dry ice to give carboxylic acid (in ion form)

ii. Nucleophile with lone pair of electrons SN2 reaction SN1 reaction E2 reaction E1 reaction

d. Alcoholi. Nucleophile for tertiary alcohol only

SN1 reaction Use HX to produce alkyl halide

ii. SOCl2 SN2 Reaction for primary and secondary alcohol to produce alkyl

chlorideiii. PBr3

SN2 Reaction for primary and secondary alcohol to produce alkyl bromide

iv. TosCl To make better leaving group for SN2, SN1, E2, E1

v. Grignard Reagent Gives H to R in R-Mg-X, while making a salt

vi. H3O+, THF Acid-catalysed dehydration to give alkene More stable alkene is the major product

vii. Carboxylic acid or acid halide Forms ester

viii. Dess-Martin periodinane Oxidises primary alcohol to aldehyde only

ix. CrO3 + H3O+

Oxidises primary alcohol to carboxylic acid Oxidises secondary alcohol to ketone

x. K2Cr2O7 + H3O+

Oxidises primary alcohol to carboxylic acid Oxidises secondary alcohol to ketone

xi. NaH,THF + RX Forms ether (R-O-R’)

e. Esteri. Grignard Reagent + H3O+

Removes previous alcohol R, and replace with 2 R’ from 2 R’-Mg-X, while producing a tertiary alcohol

Thus two Grignard Reagent is needed for every one ester By-product is the ROH, the original alcohol that formed the

ester

ii. LiAlH4 + H3O+

Reduction to give acid and alcohol, but the acid is further reduced to alcohol

iii. DIBAH + H3O+

Reduction to give acid and alcohol, but the acid is further reduced to aldehyde

f. Thiolsi. Br2 or I2 (H2O2 in cells)

Oxidised to yield disulfides (RSSR’) By-product is HX

Reversed with Zn, H3O+ (FADH2 in cells)ii. NaH,THF + RX

Forms sulphide (R-S-R’), faster rate than ether formation in alcoholg. Sulphide

i. H2O2 + H2O Oxidised to form sulphoxide (R2SO)

Further oxidation with CH3CO3H gives sulphone (R2SO2)h. Carbonyl compounds

i. NaBH4 + H3O+

Reduces aldehydes and ketones to alcoholii. LiAlH4 + H3O+

Reduces aldehydes and ketones to alcohol Also strong enough for acids and ester

iii. Grignard Reagent + H3O+

Nucleophilic addition forms an alcohol while stepping up by the R in R-Mg-X

iv. CrO3+ H3O+

Oxidise aldehyde to carboxylic acidv. RNH2

The C=O becomes C=N-R By-product is water

vi. H2O, H3O+

Acid-catalysed hydration to give diolvii. H2O, OH-

Base-catalysed hydration to give diolviii. ROH, H3O+

Acid-catalysed nucleophilic addition to give acetals [R2C(OR’)2] Two units of alcohol for each unit of carbonyl

i. Carboxylic acidsi. ROH

Esterification Step up reaction

j. Nitrilei. OH-, H2O

Base-catalysed hydrolysis to give an amide (in ion form) first, further hydrolysis to give carboxylic acid (in ion form)

ii. H3O+, H2O

Acid-catalysed hydrolysis to give an amide first, further hydrolysis to give carboxylic acid

iii. LiAlH4 Forms amine

iv. Grignard Reagent + H2O Forms an imine anion that hydrolyses to ketone

Alkanes

1. Alkanes are saturated hydrocarbons consisting of only carbon and hydrogena. Each carbon is sp3 hybridised, forming a bond angle of 120 degrees

2. Normal alkanes are unbranched3. Isomers are compounds with the same molecular formula, but different structures

a. Different structures like presence and location of branching4. The degree of a carbon depends on how many other carbons it is attached to

a. Primary carbon is only bonded to one other carbon, with the other 3 bonds being attached to Hydrogen or other elements

b. Secondary carbon is bonded to 2 other carbons. Tertiary to 3, quaternary to 4.5. Cycloalkanes are aliphatic cyclic hydrocarbons

a. They are saturated non-benzene rings6. Melting and boiling point increases with molecular weight due to greater London forces

a. The shape of the molecule also affects surface area for intermolecular forcesb. Relatively low compared to other organics as VDW forces are easily broken

Cycloalkanes

1. Cycloalkanes are aliphatic cyclic hydrocarbonsa. They are saturated non-benzene ringsb. There is no free rotation around the carbon-carbon bond without breaking the ring

i. This allows for cis-trans isomerism2. There are angle strains in cycloalkanes, as bonds are forced into an angle differing from the

120 degrees expected from an sp3 hybridised carbona. Greater in small rings (up till 5) that have little flexibility

3. There are three types on strains present in cycloalkanesa. Angle strain

i. Due to expansion or compression of bond anglesb. Torsional strain

i. Due to eclipsing, adjacent bondsc. Steric strain

i. Due to repulsion of atoms in close proximityd. Cyclopropane has the greatest strain, followed by cyclobutane and cyclopentane

i. Angle strain decreases as deviation is reduced, but torsional strain variesii. Cyclopentane has the least torsional strain due to non-planar arrangement

One carbon rises above the plaine. Cyclohexane has no strain due to chair conformation which is conformationally

mobilei. No angle strain as bond angle is 111.5 degrees

ii. No torsional strain as carbon-hydrogen bonds are staggerediii. No steric strain, as substitutes are hydrogen that have small electron clouds

iv. Ring flip can occur to change the axial and equatorial R-groups4. Substituted cyclohexane experiences 1,3-diaxial steric interactions when the substituted R-

group is in the axial positiona. Steric strain between the 3 R-groups in the same plane

i. Total steric strain equals twice the energy of the 1,3-diaxial steric interactionb. Can be avoided by ring flip so that the bulkier R-group faces outwards into empty

space5. Polycyclic molecules cannot undergo ring flip, as the fused rings are rigid

Alkenes

1. Alkenes are unsaturated hydrocarbons that contain at least one carbon-carbon double bond2. Alkenes that are more substituted are more stable than those less substituted

a. Due to hyperconjugation, where there is interaction between the pi carbon-carbon double bond, and the sigma carbon-hydrogen bonds on adjacent atoms.

Alkynes

1. Alkynes are unsaturated hydrocarbons that contain at least one carbon-carbon triple bond2. Terminal alkynes are slightly acidic, and can react with strong bases (NaNH2)

a. Terminal hydrogen is removed to form an acetylide anion

Benzene

1. Benzene is a cyclohexa-1,3,5-triene, with all the 6 pi electrons delocalised into an electron ring around both sides of the plane.

a. The molecule is planar as all carbons are sp2 hybridisedb. The delocalisation makes the benzene stable

i. Thus less reactive and does not undergo electrophilic addition2. Inductive effects are the withdrawing of electrons towards a pi bond3. Resonance effects are the movements of electrons that shift a charge around by bond

formation and breakinga. Positive charge is most stabilised on a nitrogen atom, followed by oxygen, then the

halogensi. Since they have a lone pair to form a bond, to shift the charge

ii. Stabilisation decreases since they increase in electronegativity4. Arenes are alkyl-substituted benzenes5. Positions of di- or multi-substituted benzene can be named

a. Ortho- is at the relative 2nd carbonb. Meta- is at the relative 3rd carbonc. Para- is at the relative 4th carbon

6. Reactivity is determined by substituted groupsa. Electron-donating groups increase reactivity

i. By induction of alkyl substituent through hyperconjugation 2,4 directing

ii. By resonance with electronegative atoms directly connected There is a lesser inductive effect that withdraws electrons 2,4 directing

b. Electron-withdrawing groups decrease reactivityi. By halogens due to inductive effect

2,4 directing due to resonance effectii. By electronegative groups not directly connected to benzene

3 directingiii. By positive charges

Little chance of further reaction

Isomerism

1. Structural isomers have their atoms connected in a different order while having the same molecular formula

a. Branching2. Stereoisomers have their atoms connected in the same order, but different 3D geometry

a. Cis-trans isomerismi. Requires an identical R-group at both sides of the bond

ii. If the R-group are at the same side it is cis-, else it is trans-iii. Cis- isomers are less stable than trans- isomers due to steric strain

b. E,Z isomerismi. Requires two R-groups with different ‘weight’ at each side of the bond on

both sides ‘Weight’ is determined by Cahn-Ingold-Prelog sequence rules

Take the atomic number of the first atom connected to the carbon

Only consider the next atom if they are tied Treat double bonds as if bonded to two of the same atom

ii. If the high ‘weighted’ R-group is on the same side, it is (Z)-, else (E)-iii. Add a number before the letter if there is more than one restricted rotation

available (2Z,4Z,6E,8Z)-undeca-2,4,6,8-tetraene

Halo alkanes

1. SN2 reaction (second order) has inversion of stereochemistry.a. Low steric hinderance in substrateb. Strong nucleophile with lone pair of electons

i. Nucleophilicity parallels basicityii. Nucleophilicity increases down a column

iii. Nucleophilicity is greater on a negatively charged ion than neutral onec. Leaving group that stabilises the negative charge

i. Leaving group cannot be OH-, NH2-, OR- They are strong bases and will attack the carbon again Thus unlikely to break from carbon Exception for alcohol with PBr3 or SOCl2 since they activate the -OH

ii. TosO- is the best leaving group, so use TosCl to form TosOd. Aprotic solvent

i. Protic solvents would form a cage around the attacking nucleophile2. SN1 reaction (first order) produces a partial racemic mixture due to steric hindrance

a. Substrate that can stabilise the carbocationb. Nucleophile has no effectc. Leaving group should be weak base like SN2

i. Leaving group cannot be OH-, NH2-, OR- They are strong bases and will attack the carbon again Thus unlikely to break from carbon

ii. TosO- is the best leaving group, so use TosCl to form TosOd. Protic solvent

i. They stabilise the carbocation by forming a cage around it Carbocation solvation

3. Elimination reactions usually give a mixture of alkene productsa. Zaitsev’s rule states that the more highly substituted alkene is the major product

i. Due to hyperconjucation stabilising the alkeneb. Nucleophiles are also bases, so either SN or E can occur

4. E2 (second order) needs a strong base to attack the ‘acidic’ hydrogen on the carbon beside the one with the halogen

a. Substrate should be highly substituted to form a more substituted alkenei. No steric hindrance unlike SN2

ii. Thus more bulky substituents will favour E2 over SN2b. A strong nucleophile is a strong basec. Leaving group should be weak base like SN2d. Anti-elimination occurs giving only one stereoisomer product

i. For cycloalkane, the leaving group and acidic hydrogen needs to be trans diaxial

As in they are in axial position on opposite sides of the plane If one is in equatorial, it cannot occur

e. Solvent is not as important5. E1 (first order) similar to SN1

a. Substrate should be more substituted to from more stable carbocationb. Nucleophile does not affect reaction since it is not in the rate limiting stepc. Leaving group should be weak base like SN2d. Solvent should be protic to solvate the carbocatione. Halogenoalkane will give roughly equal proportions of E1 and SN1 products

Alcohol

1. Alcohols, phenols and ethers are organic derivatives of watera. R-OH is alcoholb. Ar-OH is phenolc. R-O-R’ is ether

2. Thiols and sulphides are sulphur analogues of alcohol and estersa. R-SH is thiolb. R-S-R’ is sulphide

3. Alcohol has higher boiling points than thiols since sulphur does not form hydrogen bonds4. Alcohols are both weakly basic and weakly acidic like water

a. Don’t react with weak bases (HCO3, RNH2), slightly with metal hydroxides (OH-), fully with alkali metals and strong bases (NaH, NaNH2)

b. Reacts will with Grignard Reagent from alkyl halides, to add hydrogen to the reagenti. Alcohol itself forms a salt with MgX

5. Thiols produced from adding SH- nucleophile in SN2 reaction with alkyl halide