Organic Chemistry

What are Nucleophiles?

Nucleophiles are electron-rich species that can donate electrons and attack electron deficient centre in an organic molecules.

Considered to be Lewis bases (ie. electron pair donor)

δ- : electron rich

What are Electrophiles?

Electrophiles are electron-deficient species that can accept electrons and are attacked by electron-rich centre in an organic molecule.

Considered to be Lewis acids (ie. electron pair acceptor)

δ+: electron poor

Electrophile seeking underaged electrons for love ;)

Free Radicals

Species consisting of an atom or a group of atoms with an unpaired electron

formed from homolytic fission of a covalent bond.

They are very reactive despite being electrically neutral.

Conditions for cis-trans isomerism

1) Restriction about C=C bond/ C-C single bond about ring structure

2) 2 different groups of atom on each C

cis- : same

trans- : opp

*Trigonal planar >c=c<*

Conditions for enantiomerism

1) Chiral carbon (4 different groups attached to C)

2) No internal plane of symmetry

Optically active (rotates plane-polarised light), with same magnitude but opposite direction

• Equal proportion, cancel out → racemic mixture

*Dotted line and “non-superimposable mirror images”*

Addition reactions

1 + 1 → 1

Organic molecule + Another molecule → 1 product 

Substitution reactions

One atom/ group of atoms,

• Replaced by another atom/ group of atoms

Elimination reactions

Small molecule (H₂O / HX) removed

• between 2 adjacent C

• + by product small molecule

Condensation reactions

Two molecules/ FG combined

• to form larger molecule (with new FG) 

• + by product of small molecule

Hydrolysis reactions

Molecule split into two by H₂O

Reduction reactions

Addition of H

OR

Removal of O

Oxidation reactions

Removal of H

OR

Addition of O 

Homolytic Fission

Covalent bond breaks in a way

• each of the bonded atoms 

• takes one of the two bonding electrons 

→ to form free radicals

(When 2 bonded atoms have similar e^-negativity)

Heterolytic Fission

Covalent bond breaks in a way

• only one of the bonded atoms 

• takes both bonding electrons

→ to form anion 

→ remaining molecule becomes cation

Why are π bonds electron rich?

because overlapping p-p orbitals formed,

• extend above and below the molecular axis

• creating large regions of electron density

→ that are accessible and available to react with electron-poor species

How does excitation/ promotion of electrons work in C atom?

• The energy gap between the s and p orbitals are small

• an electron from the 2s orbital is promoted to the empty 2 orbital to give 4 unpaired electrons 

Explain how sp³ hybridisation works

• Mixing of 1 s orbital with 3 p orbitals 

→ to form 4 identical sp³ hybrid orbitals 

→ Bond angle, tetrahedral, 109.5 

25% s character

75% p character

Explain how sp² hybridisation works 

• Mixing of 1 s orbital with 2 p orbitals 

→ to form 3 identical sp² hybrid orbitals 

→ Bond angle, Trigonal planar, 120 

33.3% s character

66.7% p character

Explain how sp hybridisation works

• Mixing of 1 s orbital with 1 p orbital

→ to form 2 identical sp hybrid orbitals

→ Bond angle, linear, 180

Why are alkanes generally unreactive?

1) C and H very similar in electronegativity

C-H bonds, non-polar

• unreactive towards polar reagents

2) Large amounts of energy required to break strong 

C-C and C-H bonds 

Why more exothermic ΔH_c better fuel?

Exothermic reactions release energy

• Release more energy, better fuel 

Why FRS cannot take place in aqeuous state?

Radicals cannot exist in water

• It would react with water 

Why FRS only happens when exposed to bright light?

Bright light acts as a source of energy

• To break the bonds in the X molecules

→ to produce X radicals 

FRS using limited vs excess X₂

Limited: ONLY monosubstituted products formed 

Excess: polysubstituted products formed 

→ Will carry on from the last propagation step of “Limited” framework

Explain why iodoalkanes not possible to be made in FRS

The low reactivity of I₂ with alkane

→ mainly due to weak H-I bonds formed in the reaction

Why cis-isomer higher BP than trans-isomer?

cis-isomer: polar

trans-isomer: non polar

→ Talk about pd-pd / id-id

Why cis-isomer lower MP than trans-isomer?

In solid states,

‘kinked’ shape of cis-isomer molecules does not pack as closely

than

‘straighter’ shape of trans-isomer molecules

→ Less energy required to overcome 

• less extensive id-id attraction between cis-isomer molecules 

Boiling and melting point of Straight chained Alkanes

• BP and MP increases as number of carbon atoms in alkane increases

• Increasing amount of energy is required to overcome the increasing strength of id-id attraction between molecules

• Since no. of electrons per molecule increases

Boiling and melting point of Branched Alkanes

• BP of branched alkanes are lower than the BP of straight chained alkanes

• Due to the branching of the carbon chain, branched molecules are more spherical and have smaller SFA of contact between adjacent branched molecules

• Thus less energy is required to overcome the weak id-id attraction between the branched molecules.

Solubility of Alkanes

• Alkanes are insoluble in polar solvents (eg. water) but are soluble in non-polar solvents (eg. CCl4)

• The energy released from the formation of id-id attraction between alkane molecules and water molecules is insufficient

• To compensate for the energy required to overcome the stronger hydrogen bonds between water molecules.

Density of Alkanes

• Density increases as no. of carbon atoms in the alkane increases

• Both the mass and volume increase with the number of C atoms in the molecule.

• However, the increase in mass is more significant than in volume

• As stronger id-id attraction between molecules which pulls the molecules closer to become more compact

• The greater the degree of branching in the isomer, the lower its density

How do you determine which is major product in an Elimination reaction

Using Zaitsev’s Rule, the major alkene is formed when most substituent alkyl groups attached to the C=C (Stable)

How to determine stability of carbocations?

See R groups attached to carbocation, major product will form on most stable

Tertiary > Secondary > Primary

Why is there slow and fast step in Electrophilic Addition?

Step 1: The Slow Step (Rate-Determining Step)

• The alkene attacks the electrophile (e.g., H⁺ from HBr), breaking the π bond and forming a carbocation intermediate.

• This step is slow because:

  • Breaking the π bond requires energy.

  • Formation of a high-energy, unstable carbocation takes time.

Step 2: The Fast Step

• The nucleophile (e.g., Br⁻) quickly attacks the carbocation to form the final product.

• This step is fast because:

  • Carbocations are extremely reactive.

  • The attraction between the positively charged carbocation and the negatively charged nucleophile is strong

Why is FRS not a good method of obtaining Halogenoalkane (RX)?

Substitution of H atom is random, there will be a mixture of mono-substituted and poly-substituted products formed, leading to a low-yield of desired product.

Why reaction between Alkene and aqueous X₂ produces Halohydrin as major product and Dihalogenoalkane as minor product?

H₂O has a lone pair on O and can also act as a nucleophile. Hence H₂O and X- compete to attack the electron deficient C+ of the carbocation.

Since H₂O is a solvent and is present in much larger quantity than X-, H₂O attacks C+ more readily than Br- thus halohydrin will be formed in larger proportion than dihalogenoalkane

What are the two factors that affect reactivity in R-X

1) Strength of C-X bond

• Referring to Data booklet (BE of C-X) , the weaker the C-X bond is, the more readily it breaks, the more reactive the halogenoalkane towards nucleophilic substitution

2) Stability of leaving group X^- (Use only when question asks)

• Compare pK_a values

• Lower pK_a value, implies that X^- is a weaker base and more stable leaving group than _____.

→ similar to acidity of acids

Why Halogenoarenes resistant to nucleophilic substitution

• p-p orbital overlap results in delocalisation of lone pair of electrons on X into the adjacent π-electron system of the benzene ring

• Forming partial double bond character in C-X bond and hence, strengthening C-X bond in halogenoarene

What are the uses of CFCs?

Chloroflurocarbons (CFCS) 

→ Halogenoalkanes with no hydrogen

1) Refrigerants 

→ low bp, easily liquified by pressure

→ low toxicity, odourless, inert

2) Aerosol propellants

→ volatile; vapourise easily 

3) Fire extinguishers 

→ non-flammable, volatiles, dense (high Mr) 

4) Solvents

→ cleaning/ degreasing machinery

What are problems caused by CFCs?

Due to strength of C-F & C-Cl bonds,

→ CFCs chemically unreactive

→ stable molecules under ambient conditions

• Lack of stability → CFCs lifespan more than 100 years

→ giving time to diffuse in upper stratosphere

→ where strong UV radiation cleaves relatively weak C-Cl bonds homolytically → generate chlorine free radicals 

→ catalyse decomposition of ozone into oxygen

→ depleting ozone layer 

Reactions with Na(s) with groups containing -OH

Redox reaction

Reacts:

1. R-OH 

2. Phenol

3. R-COOH 

Reactions with NaOH (aq)

Acid base reaction

Does not react:

1. R-OH

Reacts:

1. Phenol 

2. R-COOH 

Reactions with Na₂CO₃ (aq)

Acid-carbonate reaction

• Only R-COOH reacts 

Why must LiAlH₄ used in dry conditions?

As it reacts violently with water

Characteristics of NaBH₄

NaBH₄ is a weaker RA than LiAlH₄ and can only reduce carbonyl compounds (aldehydes and ketones)

Why need immediate distillation from 1° Alcohol to Aldehydes?

To remove the aldehyde immediately on formation so that it is not oxidised further into carboxylic acid

Why R-OH reacts with RCOCl don’t need heat or catalyst?

As acyl chloride RCOCl; highly reactive functional group compared to RCOOH, reaction goes to completion (full arrow)

Why phenols don’t undergo nucleophilic substitution?

• C-O bond in phenol is stronger than that in alcohols

• p-p orbital overlap results in delocalisation of the lone pair of electrons on O into the benzene ring

• leading to the formation of partial double bond character in C-O, strengthening C-O bond; making C-O bond difficult to break

Why phenols undergo electrophilic substitution?

• p-p orbital overlap results in delocalisation of lone pair of electrons on O of the -OH group into the benzene ring, making the benzene ring more electron-rich and hence more susceptible to electrophilic substitution.

• -OH group STRONGLY ring activating and 2,4,6 directing.

• Tri-substitution occurs

Why phenols need to be in NaOH(aq)

There will be an acid-base reaction

→ converted to -ve charged phenoxide ion

→ a stronger nucleophile than phenol.

What affects acid strength in alcohols and phenols?

Strength of acids depends on the relative stability of the resulting anion formed.

• Stability of anion RO^- depends on

  • EDG: Intensifies negative charge, destabilising RO^-

  • EWG: Disperses negative charge, stabilising RO^-

Why is racemic mixture produced when an aldehyde or an unsymmetrical ketone undergo N.A with HCN?

There is an equal probability of CN^- nucleophile to attack either side of the trigonal planar C of >C=O group, producing a racemic mixture.

Why do aldehydes undergo N.A more rapidly than ketones?

1) Steric Factor

• Ketone has one more bulky alkyl/aryl group attached to the carbonyl carbon, hindering the attack of Nu on the carbonyl carbon of ketones

2) Electronic Factor

• Ketone has one more EDG alkyl group which makes the carbonyl carbon in ketones less electron-deficient

Acid strength of RCOOH

Acid strength depends on the stability of the resulting anion

1) Effect of substituents groups

• EDG intensify negative charge, destabilising RCOO^- ion

• EWG disperses negative charge, stabilising RCOO^- ion

2) Position of substituent groups

• The effect of the EDG/ EWG decreases with increasing distance from the acid group

RCOOH and ROH acid strength relatively

→ p-p orbital overlap results in delocalisation of lone pair of electrons on O^- into π-electron cloud of C=O

→ the negative charge is dispersed equally between two highly electronegative O atoms

→ which can accommodate the negative charge well

→ R-COOH stronger acid than R-OH

Why is it necessary to add NaOH when reducing Nitrobenzene?

• In step 1, phenylamine produced will react with excess HCl to form the salt of phenylamine

• Phenyl group - NH₃⁺ Cl^-

• Basic NaOH (aq) added will form back the phenylamine

What are the factors that affect the basicity of amines?

Basicity of amines depends on the availability of the lone pair of electrons on N atom:

• EDG

• EWD

• Phenylamines

What is the relative ease of hydrolysis 

Acyl halides are the easiest and fastest to hydrolyse 

→ C atom in C=O is attached to two strongly electronegative O and Cl atoms

→ C atom is highly electron deficient, more susceptible to nucleophilic attack 

Why are amides neutral?

lone pair of electrons on N atom, is not available for protonation 

→ p-p orbital overlap results in the delocalisation of lone pair of electrons on N atom into the pi-electron system of C=O 

→EWG C=O further decrease availability of lone pair of electrons on N of amide group for protonation 

Why FRS continues with increasing rate after the brief exposure to the bright light has stopped

1) Describe the first 2 steps in propagation step 

→ Which creates a chain reaction 

2) Increases [Free radical], exothermic reaction 

→ Increases T, rate of reaction occurs 

What are zwitterions?

species contain both positively and negatively charged groups

• but with no overall electric charge

Why are amino acids crystalline solids?

Amino acids are crystalline solids with high MP

→ Large amount of energy required to overcome the strong electrostatic attraction

→ between oppositely charged groups of the zwitter ions

Why are amino acids soluble in polar solvents but not non-polar?

Both charged groups of the zwitterions can form ion-dipole interactions with water molecules

→ Energy released is sufficient to overcome the ionic bonds between zwitterions