Organic Chemistry- Properties and explanations

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- : identical groups on same side of double bond

trans- : identical groups on opposite side of double bond

*Trigonal planar >c=c<*

Conditions for enantiomerism

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

2) No plane of symmetry

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

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

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?

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

Since H₂O is a solvent and is present in much larger quantity than Br-, 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 _____.

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

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)

As phenol is reacted first with NaOH(aq) in an acid-base reaction, it is converted to the negatively charged penoxide ion which is 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

• EWD 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

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 wll 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:

i) Donation to an acid (Lewis acid base theory)

ii) Accept a proton/ H⁺ (Bronsted Lowry acid base theory)

• EDG, makes lone pair of electrons on N more available to accept a proton, increase basicity of amine

• EWG, makes lone pair of electrons on N less available to accept a proton, decrease basicity of amine

• Phenylamine, p-p orbital overlap results in delocalisation of the lone pair of electrons on N into the π electron system of benzene ring, making the lone pair of electrons less available to accept a proton, decrease basicity of phenylamine