Standard Questions

Question 1

Outline a simple chemical test to detect the presence of a carbonyl and determine its identity

Answer 1

Add 2,4-Dinitrophenylhydrazine (2,4-DNP)

 Orange Precipitate hydrazone will form if a carbonyl is present (Aldehyde or Ketone)

 To determine the identity purify the hydrazone precipitate by recrystallization

 Measure the melting point of the purified precipitate

 Compare the melting point with data table values to identify the aldehyde or ketone

Question 2

Outline a simple chemical test to distinguish between a Ketone and an aldehyde

Answer 2

 Add Tollens Reagent (Ammoniacal Silver Nitrate) Warm in water bath to 60°C

 Aldehyde – A silver mirror will form as Silver reduced Ag+(aq) + e-  Ag(s)

 Ketone – No visible observation

Question 3

Why does Sodium Borohydride have to be aqueous to act as a reducing agent?

Answer 3

 It dissolves in water so the ions dissociate NaBH4 -> Na+ + BH4

 Dative bond breaks releasing Hydride ions (H-)in solution

Question 4

What type of reaction takes place in reduction of a carbonyl and what type of bond fission occurs?

Answer 4

 Nucleophilic addition; Heterolytic fission as both electrons go to one atom

Question 5

Describe in words the mechanism of reduction of a carbonyl

Answer 5

 Nucleophile is attracted to δ+ C

 The lone pair of electrons is donated to form a dative bond

 The π electron pair goes to the oxygen atom

 Causing the π bond to break

Question 6

Why are short chain (

Answer 6

 Highly polar C=O bond and O-H bonds in carboxylic acid molecules for H bonds with polar water molecules

 As the number of carbons increases a greater proportion of the molecule is non polar which is hydrophobic. This does not attract H2O and cannot form H bonds with H2O molecules.

 The more O’s with lone pairs on a compound the more soluble

Question 7

Describe a simple chemical test to differentiate a carboxylic acid from phenol and an alcohol

Answer 7

 Add a carbonate, the carboxylic acid will react the others wont

Question 8

Why are saturated fats solid at room temperature?

Answer 8

 Higher melting point

 More intermolecular forces between molecules  More energy required to overcome intermolecular forces

Question 9

Why do unsaturated fats have lower melting points than saturated fats?

Answer 9

 Lower Boiling point

 Less intermolecular forces between molecules #

 Less energy to overcome

Question 10

Explain why triglycerides are soluble in non-polar solvents and not in water

Answer 10

 There are Van Der Waals forces between triglycerides

 There is Van Der Waals forces between triglycerides and non-polar solvent

 Triglycerides cannot hydrogen bond to water (enough)

 Because there are not enough suitable oxygen atoms/Long Hydrocarbons chains

Question 11

Why do bromine atoms substitute the H atoms on positions 2, 4 and 6 in a reaction with phenol?

Answer 11

 Most reactive hydrogens due to the position of the highly electronegative O atom

Question 12

Uses of Phenols

Answer 12

 Plastics, Antiseptics, disinfectants & Resins for Plants

Question 13

What can be deduced from an a 13C NMR spectrum

Answer 13

 The different types of carbon present, from chemical shift values

 The possible structures of the molecule

Question 14

What can be deduced from a High resolution Proton NMR spectrum

Answer 14

 TMS is a standard for chemical shift measurements and produces a peak at δ=0 (ppm)

Question 15

What information can be deduced from an integration trace

Answer 15

 Relative number of protons in each environment

Question 16

Why are deuterated solvents used? State an example

Answer 16

 So they do not produce a chemical shift/peak an example is CDCl3

Question 17

What is the use of MRI and what is its relation to NMR?

Answer 17

 MRI uses the same technology as NMR to obtain diagnostic information about internal structures in body scanners

Question 18

Describe a proton exchange using D2O

Answer 18

 A proton NMR spectrum is run on the sample of the compound under investigation

 A small amount of D2O is added to another sample; and another scan run

 You can now compare the two spectra produced, to identify the –NH, -OH by identifying which peak/s are not present in the second spectra as these will have been removed

Question 19

Explain how D2O can be used to confirm which protons are responsible for peak ‘x’

Answer 19

 D replaces OH (or NH) proton/Protons are labile

 The Peak for OH (or NH) protons disappears

Question 20

Describe and explain the different ways that a high resolution NMR spectrum can give information of a molecule

Answer 20

 Chemical Shift values give the type of proton environment

 The number of peaks gives the number of different proton environments

 Relative peak areas give the number of protons in each environment

 Splitting gives the number of adjacent protons o Using n+1 rule so e.g. Doublet means one proton on adjacent carbon

 D2O can be used to identify –OH and –NH protons

Question 21

In TLC how can you use the chromatogram to identify the compounds

Answer 21

 Measure how far each spot travels relative to the solvent front

 Calculate the Rf value Using Equation 𝑅𝑓 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑚𝑜𝑣𝑒𝑑 𝑓𝑟𝑜𝑚 𝑏𝑎𝑠𝑒𝑙𝑖𝑛𝑒 𝑏𝑦 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡/ 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑚𝑜𝑣𝑒𝑑 𝑓𝑟𝑜𝑚 𝑏𝑎𝑠𝑒𝑙𝑖𝑛𝑒 𝑏𝑦 𝑆𝑜𝑙𝑣𝑒𝑛𝑡 𝑓𝑟𝑜𝑛𝑡

 Compare the Rf value found with data table values for the type of compound

Question 22

State disadvantages of chromatography

Answer 22

 Similar compounds have similar retention times/Rf values due to similar structutes

 Unknown compounds have no reference retention times/Rf values

 It can be difficult to find a solvent that separates all the components in a mixture (too soluble will not retain on the stationary phase and if not soluble enough then components will hardly move)

Question 23

Discuss the advantages of GC-MS (Gas Chromatography-Mass Spectroscopy)

Answer 23

 Provides a far more analytical tool than chromatography alone

 Gas chromatography separates but cannot conclusively identify components

 Gas Chromatography can give the relative proportions from the relative peak areas

 Mass Spectroscopy identifies structures by a computer comparing fragment ions with a spectral database

Question 24

Describe (Solid) Thin Layer Chromatography (TLC)

Answer 24

 Stationary Phase: Solid, Separation by adsorption

 Mobile Phase: Liquid e.g. Solvent

Question 25

Describe (Liquid or Gas) Gas Chromatography

Answer 25

 Stationary Phase: Liquid or Solid (on a solid support) separation by relative solubility into liquid or adsorption onto solid

Question 26

What are disadvantages of nitration of hagenoalkane; what is a way to minimize this?

Answer 26

 Further substation can occur forming secondary and tertiary amines
 Use an excess if NH3 to ensure this acts as nucleophile

Question 27

Why do Amines act as bases?

Answer 27

 Lone pair of electrons on N accepts a proton/H+

 Lone pair of electrons on N can be donated to H+;H+ accepts lone pair

 Forms Dative coordinate bond

Question 28

Why must a diazonium salt/ion be formed at a temperature

Answer 28

 Diazonium salt/ion is very unstable and will readily hydrolyses to form phenol

Question 29

Why must HNO2 be created in Situ using HCl and NaNO2 when forming a diazonium salt/ion?

Answer 29

 Nitrous acid decomposes very readily and so is always made in situ

Question 30

Suggest how the carboxylic acid group and the primary amine group are able to react with each

Answer 30

 -COOH donates H+ ion to amine group

 N of amine group donate lone pair of electrons to H+

 H+ from acid accepts lone pair

 Dative coordinate bond is formed

Question 31

What are the conditions for there to be e/z isomerism?

Answer 31

 C=C which cannot rotate

 Two different groups attached to each C of the C=C

 Two of the substituent groups on either side of the C=C are the same

Question 32

What are the conditions for there to be a chiral centre?

Answer 32

 Four different groups attached

Question 33

Advantages of a Single Optically pure isomer

Answer 33

 Fewer Side effects

 Increases the Pharmacological activity

 Reduces the need/cost/difficulty in separating stereoisomers

 Drug Doses are reduced

Question 34

Method to prepare a single isomer

Answer 34

 Using enzymes as biological catalysts – As Single isomers are made naturally in living systems, enzymes are being employed to mimic these natural processes

 Chiral Synthesis – Starting with naturally occurring chiral molecules such as amino acids and sugars as starting materials can lead to production of one of a desired optical isomer

 Chiral Catalyst – Such as a transition metal complex resulting in the formation of one optical isomer

 Chiral Synthesis

Question 35

Why are biodegradable polymers increasing in importance?

Answer 35

 Minimize environmental waste and Reduce Landfill Waste

 Polymers produces toxic gases when burnt

Question 36

Why are condensation polymers biodegradable?

Answer 36

 May be hydrolysed at the ester or amide group

 May be photodegradable due to the C=O bond which absorbs radiation

Question 37

What are uses of Kevlar?

Answer 37

 Kevlar is very strong/fire resistant so can be used as protective clothing for firefighters or bullet proof vests