2.4. Synthesis

Question 1

What is homolytic fission?

Answer 1

When a covalent bond breaks and one electron goes to each atom/ion.

Question 2

What is heterolytic fission?

Answer 2

When a covalent bond breaks and both electrons go to one atom/ion. The atom they go to is always the more electronegative one.

Question 3

What do organic chemists call electron donors?

Answer 3

Nucleophiles.

Question 4

What do organic chemists call electron acceptors?

Answer 4

Electrophiles.

Question 5

What is a _ haloalkane?

a) Primary
b) Secondary
c) Tertiary

Answer 5

A primary haloalkane is a haloalkane where the halogen is bonded to a carbon which is bonded to one other carbon.

A secondary haloalkane is a haloalkane where the halogen is bonded to a carbon which is bonded to two other carbons.

A tertiary haloalkane is a haloalkane where the halogen is bonded to a carbon which is bonded to three other carbons.

Question 6

How can monohaloalkanes undergo nucleophilic substitution reactions?

Answer 6

Due to the polar carbon to halogen bond.

Question 7

What nucleophilic substitution reactions do haloalkanes undergo?

Answer 7

Monohaloalkane + Aqueous Alkali (KOH or NaOH) -> Alcohol
The nucleophile is the OH- ion.

Monohaloalkane + Alcoholic Alkoxide (CH3OK in methanol) -> Ether
The nucleophile is the CH3O- ion.

Monohaloalkane + Potassium/Sodium Cyanide in Ethanol -> Nitrile
This reaction is a very good way of making a carbon chain longer as the nucleophile is the C to N triple bond. Nitriles can also be converted into carboxylic acids through acid hydrolysis.

Question 8

Explain the elimination reaction haloalkanes can undergo.

Answer 8

When monohaloalkanes are heated under reflux with sodium or potassium hydroxide in ethanol an elimination reaction can happen and an alkene is formed. When the double bond in the alkene is formed both a hydrogen and the halogen are eliminated from the molecule and aren’t replaced.

Question 9

Explain an SN1 mechanism.

Answer 9

SN1 stands for Substitution of a Nucleophile and only one species is involved in the rate-determining step.

Consider the nucleophilic substitution of a tertiary haloalkane and an OH- ion.
SN1 mechanisms are two step processes.
Step 1-
The C-X bond undergoes heterolytic fission and a carbocation intermediate is formed. The intermediate itself is still quite stable due to the inductive effect alkyl groups have (they push electrons onto the central carbon to stabilise it).

Step 2-
The OH- nucleophile attacks from one of the sides of the carbocation and a new molecule is formed.

Question 10

Explain an SN2 mechanism.

Answer 10

SN2 stands for Substitution of a Nucleophile and two species are involved in the rate-determining step.

Consider the nucleophilic substitution of a primary haloalkane and an OH- ion.
SN2 mechanisms are one step processes.

The hydroxide ion attacks the partially positively charged carbon atom in the primary haloalkane from the side opposite the C-X bond and begins to form a bond with it. At the same time the C-X bond begins to break. A transition state is then created where neither of these bonds are fully formed or fully broken. Once the OH- fully substitutes the halogen the reaction is complete.

Question 11

What kind of haloalkane is most likely to react with a nucleophile via an SN1 mechanism?

Answer 11

Tertiary haloalkanes due to the stability of their carbocation.

Question 12

What is the steric effect?

Answer 12

When parts of a molecule block access to another part of a molecule. An example would be the three CH3 groups in a tertiary haloalkane blocking access to the central carbon atom. It’s a good reason why tertiary haloalkanes don’t tend to undergo nucleophilic substitution via SN2 mechanisms.

Question 13

Why do alcohols have high boiling points?

Answer 13

Due to hydrogen bonding between molecules.

Question 14

Why can smaller alcohols mix in water, but larger alcohols can’t?

Answer 14

The polar heads of the alcohol have an effect on its solubility in smaller alcohols while in larger alcohols the long non polar carbon chains mask the polar effect the OH- group has so it does not dissolve.

Question 15

How can you prepare alcohols?

Answer 15

There are three main ways to create alcohols:
- Through nucleophilic substitution with haloalkanes.
Monohaloalkane + KOH/NaOH -> Alcohol

• Reacting water and alkenes with a sulfuric acid catalyst.
  Alkene + Water -> Alcohol
• By reducing aldehydes and ketones with Lithium Aluminium Hydride dissolved in ethers.

Question 16

What reactions can alcohols undergo?

Answer 16

Alcohol + Metal -> Metal Alkoxide
This reaction is a displacement/redox reaction.

Alcohol + Al2O3/ H2SO4 -> Alkene + Water
This is an elimination reaction.

Alcohol + Carboxylic acid/acid chloride -> Ether + Water
These are condensation/esterification reactions.

Question 17

How do you name ethers?

Answer 17

The longest carbon chain is the parent part of the name and the shorter chain acts as a branch. Examples would be:

Ethoxypropane
Methoxyethane
Hexoxyheptane

Question 18

Why do ethers have lower boiling points than alcohols?

Answer 18

They can’t form H bonds with other ether molecules (although they can form them with water).

Question 19

How can you prepare an alkene?

Answer 19

By dehydrating alcohols. Alcohols can be dehydrated by passing them over hot aluminium oxide, or treating them with H2SO4 or H3PO4. Two products can be formed if the alcohol is asymmetric. This is an elimination reaction.

Base-induced elimination of HX from monohaloalkanes.
This is done by refluxing a monohaloalkane with ethanolic KOH or NaOH. This reaction can also end up with two different products.

Question 20

What reactions can alkenes undergo?

Answer 20

The main reactions alkenes undergo is addition reactions with

• Water (called hydration)
• Hydrogen (called hydrogenation)
• HX (called hydrohalogenation)
• Halogens (called halogenation)

Question 21

What is Markovnikov’s Rule?

Answer 21

It states that when H-X or water is added to an asymmetric alkene, the major product produced is the one in which the hydrogen ends up on the carbon which originally had the most hydrogens on it at the start.

Question 22

Explain the mechanism of Halogenation of alkenes.

Answer 22

It is a two step process.
Take the reaction of an alkene and Br2.

Step 1:
As the Br2 molecule approaches the double bond, since the double bond is electron rich it pushes the electrons on the nearest bromine to the farthest away bromine making it become polarised. The closest Br has a slight positive charge and attacks the double bond forming a cyclic ion intermediate.

Step 2:
The remaining Br atom which is negatively charged attacks the cyclic ion intermediate from the side opposite the other bromine due to that bromine blocking it from joining on that side. In this reaction Br is a nucleophile and the CH is the electrophile.

Question 23

Explain the mechanism of hydrohalogenation of alkenes.

Answer 23

It is a two step process.
Take the reaction of an alkene and HBr.

Step 1:
HBr is already polar and is an electrophile. The slightly positive hydrogen atom attacks the double bond forming a carbocation. The Br of the HBr is now a Br- ion.

Step 2:
The remaining Br- ion attacks the carbocation. Depending on which carbon of the double bond the hydrogen bonds to, two different products can be formed.

Question 24

Explain the mechanism of acid-catalysed hydration of alkenes.

Answer 24

This is a three step process.
Take the reaction of water and an alkene.

Step 1:
The H+ ion of the acid catalyst attacks the double bond of the alkene and forms a carbocation.

Step 2:
The carbocation then undergoes fast nucleophilic attack by the water molecule to give a protonated alcohol molecule (an alcohol with an extra hydrogen).

Step 3:
This protonated alcohol is a strong acid and readily loses an H+ ion. Once it loses this proton the final product is there.

Question 25

How can we prepare carboxylic acids?

Answer 25

By oxidising primary alcohols or aldehydes. This is done by refluxing them with acidified potassium dichromate. Although alcohols are first oxidised into aldehydes, the reaction usually can’t be stopped there and goes further into carboxylic acids.

By hydrolysing nitriles. This is done by heating them under reflux with an aqueous acid. H+ ions catalyse this reaction.

By hydrolysing esters. This is done by heating esters under reflux with an aqueous acid or alkali as a catalyst. An aqueous alkali is useful here as it drives the equilibrium to the products side.

By hydrolysing amides. This is done by heating amides under reflux with an aqueous acid or alkali as a catalyst.

Question 26

What reactions can carboxylic acids undergo?

Answer 26

Carboxylic acids do act as normal acids in aqueous solution so they can react with metals, carbonates and alkalis to make salts.

They also react with alcohols to make esters. This is a condensation and esterification reaction.

They can react with amines to form alkylammonium salts. Which when heated can make amides.

They can react with lithium aluminium hydride in ether to form primary alcohols. This reaction is a reduction reaction. LiAlH4 is such a strong reducing agent that it reduces the carboxylic acid straight to the alcohol.

Question 27

What is a _ amine?

a) Primary
b) Secondary
c) Tertiary

Answer 27

A primary amine is an amine in which the nitrogen is attached to one alkyl group.

A secondary amine is an amine in which the nitrogen is attached to two alkyl groups.

A tertiary amine is an amine in which the nitrogen is attached to three alkyl groups.

Question 28

Why do primary and secondary amines have higher boiling points than tertiary amines?

Answer 28

Due to the possibility of hydrogen bonding. Since in tertiary amines there are no N-H bonds, hydrogen bonding cannot occur.

Question 29

What reactions do amines undergo?

Answer 29

They react with mineral acids (HCl, HNO3, H2SO4)to form salts.

They also react with carboxylic acids to form salts.

Question 30

Explain the bonding in benzene.

Answer 30

Every carbon in benzene is sp2 hybridised. These three half filled sp2 orbitals form sigma bonds with a hydrogen atom and the carbon atoms behind them. A p orbital in each carbon is therefore left unhybridised. These p orbitals overlap and form pi orbitals. All of these p orbitals overlap each other forming one big pi orbital also known as a delocalised electron cloud as the electrons in this pi bond are free to move.

Question 31

What reactions can benzene undergo?

Answer 31

Due to benzene’s delocalised electrons it is stable so it resists addition reactions. However, it can undergo electrophilic substitution reactions much more easily.. Benzene undergoes four main electrophilic substitution reactions.

The first is alkylation. This is the electrophilic substitution of a alkyl group. This happens when it reacts with a haloalkane in the presence of AlCl3 as a catalyst.

The next is chlorination/bromination. This is when a chlorine or bromine atom reacts with AlCl3 or FeCl3 to form a positive chlorine or bromine ion. This ion is the electrophile.

The next is nitration. The electrophile here is the nitronium ion (NO2+). This ion is generated through the reaction of concentrated sulfuric and nitric acid.

The final one is sulfonation. The electrophile in this reaction is HOSO2 and is formed by concentrated sulfuric acid.