Topic 20.1 nucleophile, electrophiles, and

Question 1

Electrophilic addition

Answer 1

NO₂⁺ and CH₃⁺

A reaction in which a species is attracted to an electron-rich centre, where it accepts a pair of electrons to form a new covalent bond.

CHeckk (ie alkene + halogen -> halogenalkane)

Question 2

nucleophilic substitution

Answer 2

F⁻, Cl⁻, Br⁻, I⁻, OH⁻, NH₃, CN⁻, H₂O (H₂O is weak) ₂₃₅

a nucleophile attacks an electron-deficient centre, donating a pair of electrons and forming a dative covalent bond. This replaces a leaving group.

ie C₂H₅Br(l) + NaOH(aq) -> C₂H₅OH(l) + NaBr(aq)

Question 3

nucleophilic addition

Answer 3

sodium borohydride (NaH₄) or lithium aluminium hydride (LiAlH₄) produces H⁻

and this can be used to add onto an alkene by breaking the pi bond and donating two electrons as the electrons from the pi bond are given to the other molecule attached to the carbon on the alkene.

ie aldehyde/ ketone - alcohol

Question 4

free radical substitution

Answer 4

A species with an unpaired electron is produced which initiates a chain reaction.

Question 5

electrophile

Answer 5

electron acceptor

Question 6

nucleophile

Answer 6

electron pair donator

Question 7

Types of nucleophilic substitution

Answer 7

The S៷ number is dependent on whether the compound is primary, secondary or tertiary.

S៷1 reactions ->

S៷2 reactions ->

Question 8

S៷2 reactions

Answer 8

S៷2 reactions in primary halogenoalkanes occur as there are two molecules involved in the RDS. There is a transition state where both the new molecule and molecule that is going to be the leaving group are both attached to the carbon.

During this step, the large molecule creates a steric hindrance which means the side of the large molecule (the front) cannot be attacked by the nucleophile. This means that the nucleophile attaches at 180deg to the large molecule.

As the nucleophile enters the compound, it causes an inversion of configuration (walden conversion - like an umbrella). When the nucleophile enters, both it and the leaving group have weak covalent bonding (shown as dotted line - - - - -)

rate = k(halogenoalkane)(nucleophile)

Question 9

S៷1 reactions

Answer 9

Tertiary halogenoalkanes do S៷1 reactions as there is only one molecule (unimolecular) involved in the RDS. There are two steps and only the halogenoalkane is involved in the RDS. The leaving group separates from the compound first, leaving a carbocation. This readily bonds with a nucleophile.

Rate = k(halogenoalkane)

Question 10

Carbocation

Answer 10

Ion with a C⁺ atom. Carbon normally has three bonds.

Question 11

Inductive effects

Answer 11

Carbon is slightly more electronegative than hydrogens. This causes a weak dipole and electrons to be shifted slightly towards the carbon. Alkyl groups connected to a carbocation makes it more stable. More alkyl groups - more stable.

This is why S៷1 is typically tertiary.

Question 12

Rate of nucleophilic substitution

Answer 12

Leaving group: better leaving group - higher RoR. ie more electronegative halogen, quicker halogen leaves.

Class of compound: Teriary > secondary > primary

Solvent choice:
S៷2 prefers aprotic polar solvents because there is no O-H or N-H groups so no hydrogen bonding with the nucleophile, making the nucleophile effective at forming the intermediate step.
S៷1 prefers protic polar solvents because polar bonds and O-H and N-H groups. Nucleophile solvated so it is less effective at attacking the δ⁺ carbon atom. Becomes less effective as a nucleophile, favouring S៷1 reactions.

Question 13

S៷1 vs S៷2

Answer 13

S៷1: Electrons move to leaving group, carbocation produced,
nucleophile with its free electrons to bond bonds with the carbocation. end result is nucleophilic substitution

S៷2: nucleophile attacks the partially positive carbon atom, forming a weak bond. The bond with the leaving group is weakened (shown with a dotted line). Then the leaving group leaves with an inversion of configuration.

Inversion of configuration -> final product is the 180° mirror image from the original. The intermediate is built like a star

Question 14

Effectiveness of a nucleophile

Answer 14

Depends on the charge, more negative leads to higher attraction to electrophiles.

Question 15

Markovnikov’s rule

Answer 15

Electrophiles will bond to the carbon with the most carbon constituents. Other isomers will be formed too but are a minor product.

Question 16

Electrophilic addition of hydrogen halide to an alkene

Answer 16

Hydrogen halide splits heterolytically. H⁺ electrophile attacks the double bond and bonds using the electrons. The I⁻ ion then bonds with the carbocation and then the halogenoalkane is formed.

This is halogenation of alkenes

Question 17

Electrophilic substitution of benzene

Answer 17

Produces: nitrobenzene
Conditions: 50°C, nitrating mixture of H₂SO₄ and HNO₃

Nitrate ions (NO₂⁺) are attracted to delocalised pi bonds from the benzene and these electrons are bonded to the nitrate ions, nitrating the benzene.

Although the presence of both the nitrate ion and the proton briefly eliminates the aromaticity of the arene (forming a + charge on the carbon with both attached), once the proton leaves, the aromaticity is restored.

benzene + nitric acid -> nitrobenzene + water
C₆H₆ + HNO₃ -> C₆H₅NO₂ + H₂O