Organic Chemistry

Question 1

How do molecular orbitals form

Answer 1

When atomic orbitals combine.

Question 2

How do sigma bonds form

Answer 2

Molecular orbitals that form by end on overlap of atomic orbitals along the axis.

Sp3

Question 3

How do pi bonds form

Answer 3

Pi bonds form by side on overlap that lie perpendicular to the axis Sp2

Question 4

Hybridization

Answer 4

The process of mixing atomic orbitals within an atom to generate a set of new atomic orbitals called hybrid orbitals.

Question 5

What happens when the 2s orbital and three 2p orbitals hybridize

Answer 5

They form four degenerate sp3 hybrid orbitals. These adopt a tetrahedral arrangement.the sp3 hybrid orbitals overlap end on with other atomic orbitals to form sigma bonds.

Question 6

Why can bonding in alkenes be described as

Answer 6

The bonding in alkenes can be described in terms of sp2 hybridisation.

The 2s orbital and two of the 2p orbitals hybridise to form three degenerate sp2 hybrid orbitals.

These adopt a trigonal planar arrangement. The hybrid sp2 orbitals overlap end-on to form
Sigma bonds.

The remaining 2p orbital on each carbon atom of the double bond is unhybridised and lies perpendicular to the axis of the
Sigma bond. The unhybridised p orbitals overlap side-on to form pi bonds.

Question 7

How can the bonding in benzene and other aromatic systems be described.

Answer 7

in terms of sp2 hybridisation.

The six carbon atoms in benzene are arranged in a cyclic structure with sigma bonds between the carbon atoms.

The unhybridised p orbitals on each carbon atom overlap side-on to form a pi molecular system, perpendicular to the plane of the sigma bonds.

This pi molecular system extends across all six carbon atoms. The electrons in this system are delocalised.

Question 8

Why are organic molecules colored or colour less.

Answer 8

Electrons fill bonding molecular orbitals, leaving higher energy antibonding orbitals unfilled.

The highest bonding molecular orbital containing electrons is called the highest occupied molecular orbital (HOMO).

The lowest antibonding molecular orbital is called the lowest unoccupied molecular orbital (LUMO).

Absorption of electromagnetic energy can cause electrons to be promoted from HOMO to LUMO.

Question 9

Chromophore

Answer 9

A chromophore is a group of atoms within a molecule that is responsible for absorption of light in the visible region of the spectrum.

Light can be absorbed when electrons in a chromophore are promoted from the HOMO to the LUMO.

Chromophores exist in molecules containing a conjugated system

Question 10

conjugated system

Answer 10

a system of adjacent unhybridised p orbitals that overlap side-on to form a molecular orbital across a number of carbon atoms.

Electrons within this conjugated system are delocalised.

Molecules with alternating single and double bonds, and aromatic molecules have conjugated systems

The more atoms in the conjugated system the smaller the energy gap between HOMO and LUMO.
A lower frequency of light (longer wavelength, lower energy) is absorbed by the compound.
When the wavelength of light absorbed is in the visible region, the compound will exhibit the complementary colour.

Question 11

Homolytic fission

Answer 11

results in the formation of two neutral radicals

 occurs when each atom retains one electron from the sigma covalent
bond and the bond breaks evenly

 normally occurs when non-polar covalent bonds are broken

Question 12

Heterolytic fission

Answer 12

results in the formation of two oppositely charged ions

 occurs when one atom retains both electrons from the sigma covalent bond and the bond breaks unevenly

 normally occurs when polar covalent bonds are broken

Reactions involving heterolytic fission tend to result in far fewer products than reactions involving homolytic fission, and so are better suited for organic synthesis.

Question 13

Why does a single headed arrow represent

Answer 13

The movement of a single electron

Question 14

What does a double headed arrow indicate

Answer 14

The movement of an electron pair

Question 15

What does the tail of an arrow show

Answer 15

The source of the electrons

Question 16

What does the head of the arrow represent

Answer 16

The destination of the electrons

Question 17

Why does two single headed arrows starting at the middle of a covalent bond indicate

Answer 17

Homolytic bond fission

Question 18

What does a double headed arrow starting at the middle of a covalent bond represent

Answer 18

Heterolytic bond fission

Question 19

What does an arrow drawn with the head pointing to the space between two atoms indicate

Answer 19

That a covalent bond will be formed between these two atoms

Question 20

Nucleophiles

Answer 20

 negatively charged ions or neutral molecules that are electron
rich, such as Cl−, Br−, OH−, CN− , NH3 and H2O

 attracted towards atoms bearing a partial (+ ) or full positive charge

 capable of donating an electron pair to form a new covalent bond

Question 21

Electrophiles

Answer 21

positively charged ions or neutral molecules that are electron deficient, such as
H+,NO+ andSO 23

 attracted towards atoms bearing a partial (− ) or full negative charge

 capable of accepting an electron pair to form a new covalent bond

Question 22

Monohaloalkane

Answer 22

contain only one halogen atom

 can be classified as primary, secondary or tertiary according to the number of alkyl groups attached to the carbon atom containing the halogen atom

 take part in elimination reactions to form alkenes using a strong base, such as potassium or sodium hydroxide in ethanol

Question 23

Monohaloalkanes nucleophillic substitution reactions

Answer 23

aqueous alkalis to form alcohols

— alcoholic alkoxides to form ethers

— ethanolic cyanide to form nitriles (chain length increased by one carbon atom) that can be hydrolysed to carboxylic acids

Monohaloalkanes can undergo SN1 and SN2 reactions.

Question 24

Steric Hinderance

Answer 24

describes how a molecule’s physical structure can affect its ability to react.

When a molecule is bulky, meaning it has multiple bonds to compounds or groups other than hydrogen, it can slow down

Question 25

Inductive stabilization of carbocation intermediate

Answer 25

The inductive effect affects the stability as well as acidity or basicity of a chemical species. Electronegative atoms draw electrons toward themselves, which can stabilize a conjugate base

Question 26

How can alcohols be prepared

Answer 26

haloalkanes by substitution

 alkenes by acid-catalysed hydration (addition)

 aldehydes and ketones by reduction using a reducing agent such as lithium aluminium hydride

Question 27

Reactions of alcohols

Answer 27

dehydration to form alkenes using aluminium oxide, concentrated sulfuric acid or concentrated phosphoric acid

 oxidation of primary alcohols to form aldehydes and then carboxylic acids and secondary alcohols to form ketones, using acidified permanganate, acidified dichromate or hot copper(II) oxide

 formation of alcoholic alkoxides by reaction with some reactive metals such as potassium or sodium, which can then be reacted with monohaloalkanes to form ethers

 formation of esters by reaction with carboxylic acids using concentrated sulfuric acid or concentrated phosphoric acid as a catalyst

Question 28

Hydrogen bonding in alcohols

Answer 28

Hydroxyl groups make alcohols polar, which gives rise to hydrogen bonding. Hydrogen bonding can be used to explain the properties of alcohols including boiling points, melting points, viscosity and solubility or miscibility in water.

Question 29

Esters

Answer 29

Ethers can be regarded as substituted alkanes in which a hydrogen atom is replaced with an alkoxy functional group, –OR, and have the general structure R’ – O – R’’, where R’ and R’’ are alkyl groups.

Question 30

Naming esters

Answer 30

Ethers are named as substituted alkanes. The alkoxy group is named by adding the ending ‘oxy’ to the alkyl substituent, and this prefixes the name of the longest carbon chain.

Question 31

Boiling points of Esters

Answer 31

Due to the lack of hydrogen bonding between ether molecules, they have lower boiling points than the corresponding isomeric alcohols.

Question 32

Solubility of Esters

Answer 32

Methoxymethane and methoxyethane are soluble in water. Larger ethers are insoluble in water due to their increased molecular size. Ethers are commonly used as solvents since they are relatively inert chemically and will dissolve many organic compounds.

Question 33

How can alkenes be prepared

Answer 33

dehydration of alcohols using aluminium oxide, concentrated sulfuric acid or concentrated phosphoric acid

 base-induced elimination of hydrogen halides from monohaloalkanes

Question 34

Alkenes electrophillic addition reactions

Answer 34

hydrogen to form alkanes in the presence of a catalyst

 halogens to form dihaloalkanes

 hydrogen halides to form monohaloalkanes

 water using an acid catalyst to form alcohols

Question 35

Markovnikovs rule

Answer 35

Markovnikov’s rule states that when a hydrogen halide or water is added to an unsymmetrical alkene, the hydrogen atom becomes attached to the carbon with the most hydrogen atoms attached to it already.

Question 36

Reaction mechanism for the addition of a hydrogen halide

Answer 36

The reaction mechanisms for the addition of a hydrogen halide and the acid-catalysed addition of water can be represented using curly arrows and showing the intermediate carbocation. The inductive stabilisation of intermediate carbocations formed during these reactions can be used to explain the products formed.
The reaction mechanism for the addition of a halogen can be represented using curly arrows and showing the cyclic ion intermediate.

Question 37

Preparation of carbonyl is acids

Answer 37

oxidising primary alcohols using acidified permanganate, acidified dichromate and hot copper(II) oxide

 oxidising aldehydes using acidified permanganate, acidified dichromate, Fehling’s solution and Tollens’ reagent

 hydrolysing nitriles, esters or amides

Question 38

Reactions of carboxylic acids

Answer 38

formation of salts by reactions with metals or bases

 condensation reactions with alcohols to form esters in the presence of concentrated sulfuric or concentrated phosphoric acid

 reaction with amines to form alkylammonium salts that form amides when heated

 reduction with lithium aluminium hydride to form primary alcohols

Question 39

Amines

Answer 39

Amines are organic derivatives of ammonia in which one or more hydrogen atoms of ammonia has been replaced by an alkyl group.
Amines can be classified as primary, secondary or tertiary according to the number of alkyl groups attached to the nitrogen atom.
Amines react with acids to form salts

Question 40

Hydrogen bonding in amines

Answer 40

Primary and secondary amines, but not tertiary amines, display hydrogen bonding. As a result, primary and secondary amines have higher boiling points than isomeric tertiary amines.

Question 41

Solubility of amines

Answer 41

The shorter the chain length the more soluble in water an amine is

Question 42

Benzene

Answer 42

The benzene ring has a distinctive structural formula. The stability of the benzene ring is due to the delocalisation of electrons in the conjugated system. The presence of delocalised electrons explains why the benzene ring does not take part in addition reactions.

Question 43

Bonding in benzene

Answer 43

Bonding in benzene can be described in terms of sp2 hybridisation, sigma and pi bonds, and electron delocalisation.

Question 44

Benzene ring reactions

Answer 44

 halogenation by reaction of a halogen using aluminium chloride or iron(III) chloride for chlorination and aluminium bromide or iron(III) bromide for bromination

 alkylation by reaction of a haloalkane using aluminium chloride

 nitration using concentrated sulfuric acid and concentrated nitric
acid

 sulfonation using concentrated sulfuric acid

Question 45

Geometric Isomers

Answer 45

can occur when there is restricted rotation around a carbon- carbon double bond or a carbon-carbon single bond in a cyclic compound

 must have two different groups attached to each of the carbon atoms that make up the bond with restricted rotation

 can be labelled cis or trans according to whether the substituent groups are on the same side (cis) or on different sides (trans) of the bond with restricted rotation

 have differences in physical properties, such as melting point and boiling point

 can have differences in chemical properties

Question 46

Optical Isomers

Answer 46

 occur in compounds in which four different groups are arranged tetrahedrally around a central carbon atom (chiral carbon or chiral centre)
 are asymmetric
 are non-superimposable mirror images of each other
 can be described as enantiomers

 have identical physical properties, except for their effect on plane- polarised light

 have identical chemical properties, except when in a chiral environment such as that found in biological systems (only one optical isomer is usually present)

 rotate plane-polarised light by the same amount but in opposite directions and so are optically active

Question 47

Racemic Mixtures

Answer 47

when mixed in equal amounts are optically inactive because the rotational effect of the plane-polarised light cancels out — this is called a racemic mixture

Question 48

Mass Spectrometry

Answer 48

In mass spectrometry, a small sample of an organic compound is bombarded by high-energy electrons. This removes electrons from the organic molecule generating positively charged molecular ions known as parent ions. These molecular ions then break into smaller positively charged ion fragments

Question 49

Infrared spectroscopy

Answer 49

Infrared spectroscopy is used to identify certain functional groups in an organic compound

In infrared spectroscopy, infrared radiation is passed through a sample of the organic compound and then into a detector that measures the intensity of the transmitted radiation at different wavelengths. The absorbance of infrared radiation is measured in wavenumbers, the reciprocal of wavelength, in units of cm-1.

Question 50

Proton NMR

Answer 50

Proton nuclear magnetic resonance spectroscopy (proton NMR or 1H NMR) can give information about the different chemical environments of hydrogen atoms (protons or 1H) in an organic molecule, and about how many hydrogen atoms there are in each of these environments

Question 51

What do peaks mean in proton NMR

Answer 51

The area under the peak is related to the number of 1H atoms in that environment and is often given by an integration curve on a spectrum. The height of an integration curve is proportional to the number of 1H atoms in that environment, and so a ratio of 1H atoms in each environment can be determined.

Question 52

Standard Reference used in proton NMR

Answer 52

tetramethylsilane (TMS)

Question 53

High Resolution NMR

Answer 53

In a high-resolution 1H NMR an interaction with 1H atoms on neighbouring carbon atoms can result in the splitting of peaks into multiplets. The number of 1H atoms on neighbouring carbon atoms will determine the number of peaks within a multiplet and can be determined using the n+1 rule, where n is the number of 1H atoms on the neighbouring carbon atom.

Question 54

Drugs

Answer 54

Drugs are substances that alter the biochemical processes in the body.
Drugs that have beneficial effects are used in medicines.

Question 55

Medicines

Answer 55

A medicine usually contains the drug plus other ingredients such as fillers to add bulk or sweeteners to improve the taste.

Question 56

Agonist

Answer 56

An agonist mimics the natural compound and binds to the receptor molecules to produce a response similar to the natural active compound.

Question 57

Antagonist

Answer 57

An antagonist prevents the natural compound from binding to the receptor, and so blocks the natural response from occurring.

Question 58

Interactions formed by drugs to receptors

Answer 58

van der Waals forces and/or ionic bonds.

Question 59

Structural fragment of a drug

Answer 59

The structural fragment of a drug molecule that allows it to form interactions with a receptor binding site or to an enzyme active site normally consists of different functional groups correctly orientated with respect to each other.