Rings, Acids and Amines

Question 1

Describe the oxidation of primary alcohols

Answer 1

Acidified dichromate ions (Cr(2)O(7)2-/H+)
Colour change from orange to green
First oxidised to an aldehyde (to stop here need to distil as formed)
Then to a carboxilyic acid (to get to here, heat under reflux)
Also get water

Question 2

Describe the oxidation of secondary alcohols

Answer 2

Acidified dichromate ions (Cr(2)O(7)2-/H+)
Colour change from orange to green
Oxidised to a ketone
Also get water

Question 3

Describe the oxidation of aldehydes

Answer 3

Acidified dichromate ions (Cr(2)O(7)2-/H+)
Colour change from orange to green
Under reflux conditions
Get a carboxylic acid
No water formed

Question 4

How can carbonyls be reduced?

Answer 4

With NaBH(4): sodium borohydride (have Na+ and BH(4)-)
Water as solvent
Warm carbonyl with reducing agent
Need 2 reducing agents to reduce aldehydes and ketones (can us (H) as reducing agent)

Question 5

What is the mechanism for the reduction of a carbonyls?

Answer 5

Treat BH(4)- as a source of hydride ions, :H- (semi colon is electron pair)
Carbon is d+ and oxygen is d-
:H- attacks C d+, donating an electron pair; pi bond between C and O breaks, with electrons going to the oxygen
Have an intermediate of a carbon with a negative oxygen and 2 hydrogens
Negative oxygen donates an electron pair to a hydrogen on a water molecule; O-H bond breaks, with an electron pair going to the O
End up with an alcohol and an OH- ion

Question 6

What is the reagent that can be used to detect aldehydes and ketones?

Answer 6

2,4-dinitrophenylhydrazine or 2,4-DNPH

Question 7

How would you test for a carbonyl group in a compound?

Answer 7

Make Brady’s reagent: add 2,4-DNPH to a mixture of methanol and sulfuric acid
Add to compound. If a yellow-orange precipitate (a 2,4-DNPH derivative) forms, a carbonyl group is present

Question 8

Once a compound has tested positive for containing a carbonyl group, how would you identify the compound?

Answer 8

Purify the 2,4-DNPH derivative by filtering and recrystallising it
Measure the melting point of the compound and compare it to a database to identify it.

Question 9

How would you distinguish between an aldehyde and a ketone?

Answer 9

Use Tollen’s reagent
Make Tollen’s reagent: Add sodium hydroxide to silver nitrate, forming a brown precipitate; redissolve the precipitate with ammonia
Tollen’s reagent oxidises aldehydes (to carboxylic acids, reducing silver ions to silver) but not ketones, so a silver mirror forms only if an aldehyde is present.

Question 10

Explain the water solubility of carboxylic acids

Answer 10

Carboxylic acids with between 1 and 4 carbons are very soluble in water. The highly polar C=O and O-H bonds allow carboxylic acid molecules to form hydrogen bonds with water molecules. As the number of carbon atoms in the carboxylic acids increases, the solubility decreases. This is the result of the longer non-polar hydrocarbon chain in the molecule, which doesn’t interact with water molecules.

Question 11

How do carboxylic acids react with metals?

Answer 11

-COOH + M -> -COO(-)M(+) + 0.5 H(2)

Get a carboxylate salt, e.g. M ethanoate

Question 12

How do carboxylic acids react with bases?

Answer 12

-COOH + MOH -> COO(-)M(+) + H(2)O

Get a carboxylate salt, e.g. M ethanoate

Question 13

How do carboxylic acids react with carbonates?

Answer 13

2-COOH + Na(2)CO(3) -> 2-COO(-)Na(+) + CO(2) + H(2)O

Question 14

What happens when carboxylic acids are reacted with alcohols? What conditions are required?

Answer 14

Get an ester. OH of carboxylic acid and H of alcohol become water, and an ester link is formed. Requires an acid catalyst. As is a reversible reaction, use conc. acid catalyst and heat under reflux

Question 15

What are the 2 ways of making an ester?

Answer 15

Carboxylic acid + alcohol

Acid anhydride + alcohol (better yield)

Question 16

What happens when an acid anhydride is reacted with an alcohol? What conditions are required?

Answer 16

Get an ester and a carboxylic acid. Heat gently.

Question 17

What are the 2 ways esters can be hydrolysed?

Answer 17

In hot aqueous acid to form carboxylic acids and alcohols

In hot aqueous alkali to form carboxylate salts and alcohols

Question 18

Describe the acid hydrolysis of esters

Answer 18

Ester heated under reflux with dilute sulfuric/hydrochloric acid as a catalyst. Ester is broken down by water. Reversible reaction.
Ester + water carboxylic acid + alcohol

Question 19

Describe the alkali hydrolysis of esters

Answer 19

Aqueous sodium/potassium hydroxide refluxed with ester. Alkali not a catalyst. Non-reversible reaction. Often known as saponification
Ester + alkali -> carboxylate + alcohol

Question 20

What is saponification?

Answer 20

The alkali hydrolysis of esters

Question 21

What are esters used for?

Answer 21

Perfumes and flavourings

Question 22

What is the difference between saturated and unsaturated fats?

Answer 22

Saturated fats have no carbon-carbon double bonds, unsaturated fats do.

Question 23

What is the shorthand notation for fatty acids?

Answer 23

2 numbers separated by a colon: x:y (a,b….)
x is the number of carbon atoms
y is the number of double bonds
a,b….. are the positions of the double bonds

Question 24

What is the difference between cis and trans fats?

Answer 24

Cis fats are alright for us: the hydrogen groups are together
Trans fats are very bad for us: the hydrogen groups are opposite each other
Cis fats can’t pack close together, and so exist as a liquid. Trans fats can exist close together and can be solid at higher temperatures

Question 25

What are HDLs and LDLs?

Answer 25

HDLs: High density lipoproteins: the ‘good’ cholesterol. Transport cholesterol out of blood and body
LDLs: Low density lipoproteins: deposit lipids onto artery walls building up fatty deposits that restrict blood flow

Question 26

Why are trans fats though to be bad for us?

Answer 26

Trans fatty acids are thought to behave like saturated fats in the body, raising LDL levels and increasing the risk of heart disease. They also lower HDL levels.

Question 27

How are esters used in the biofuel industry?

Answer 27

Biodiesel is made through reacting triglycerides with methanol or ethanol under a sodium/potassium hydroxide catalyst. This gives biodiesel (an ester of fatty acids) and glycerol (which is sold to the cosmetics industry).

Question 28

Describe Kekule’s model of benzene

Answer 28

A 6-carbon ring, with 3 C=C double bonds. He suggested that the molecule oscillated rapidly with the position of the double bonds. There were 3 pairs of overlapping p-orbitals, forming 3 pi bonds in Kekule’s model, whilst in the current model, there is 1 delocalised pi cloud.

Question 29

What is the evidence for the delocalised model of benzene?

Answer 29

Bond lengths; enthalpy change of hydrogenation and resistance to reaction.
Bond lengths: C=C and C-C have different bond lengths (0.134nm and 0.153nm). All bonds in benzene are the same length, halfway in between these two (0.139nm)
Enthalpy change of hydrogenation (EoH): when a hydrogen molecule is added across a C=C, the enthalpy change is known to be -120kJ/mol. This means benzene ought to have EoH of -360kJ/mol, but it actually has -208kJ/mol. Therefore, benzene is much more stable than originally thought
Resistance to reaction: benzene doesn’t react with bromine like other alkenes. However, Kekule modified his structure to account for this (oscillation).

Question 30

Describe the structure and bonding in benzene

Answer 30

6 carbon ring, each with 1 hydrogen (sigma bonds between carbons and hydrogens)
Bond angle of 120deg (trigonal planar around carbons)
p orbital above and below plane of ring
p orbitals overlap to p orbitals on both sides, forming a ring of electron density above and below the plane of the carbon ring
The p electrons are delocalised in a pi-cloud of electron density

Question 31

Describe the electrophilic substitution by nitration of benzene (conditions, equation and mechanism)

Answer 31

Conditions: conc. nitric acid (reactant); conc. sulfuric acid (catalyst); 50degC (or more nitro groups added)
Equation: C(6)H(6) + HNO(3) -> C(6)H(5)NO(2) + H(2)O
Mechanism:
- HNO(3) + H(2)SO(4) -> NO(2)(+) + HSO(4)(-) + H(2)O
- curly arrow from electron ring to NO(2)(+)
- makes benzene with an NO(2) group (draw the H on the carbon with the added group), and a broken electron ring with positive charge
- curly arrow from C-H bond into ring
- makes nitrobenzene and H+
- H(+) + HSO(4)(-) -> H(2)SO(4)

Question 32

Describe the general mechanism for electrophilic substitution with benzene

Answer 32

• electron dense ring attracts electrophile (electron deficient thing)
• intermediate forms with both electrophile and hydrogen attached to a carbon, with a broken ring of electron density in benzene and overall positive charge
• H lost from carbon, forming H(+)

Question 33

Describe the electrophilic substitution by halogenation of benzene (conditions, equation and mechanism)

Answer 33

Conditions: halogen (X) (reagent); FeX(3) or AlX(3) or Fe (as Fe reacts to form FeX(3))
Equation: C(6)H(6) + X(2) -> C(6)H(5)X + HX
Mechanism:
- X(2) + FeX(3) -> X(+) + FeX(4)(-)
- curly arrow from electron ring to X(+)
- makes benzene with an X group (draw the H on the carbon with the X group), and a broken electron ring with positive charge
- curly arrow from C-H into ring
- makes halagenobenzene and H(+)
- H(+) + FeX(4)(-) -> FeX(3) + HX

Question 34

Why doesn’t benzene undergo bromination?

Answer 34

In an alkene, the localised electrons in the pi bond repel the electrons in the Br-Br bond, forming a temporary dipole. The pi-electron pair from the double bond is now attracted to the Br(d+), causing the double bond to break heterolytically. This makes a carbocation and bromine anion which then form a dibromoalkane.
In benzene, the pi-electrons are spread over all 6 carbon atoms in the ring, giving it a lower electron density than alkenes. The low electron density means that the non-polar bromine isn’t sufficiently polarised. So, we need to create a more polar Br(+) for the reaction to proceed

Question 35

Describe the reaction of phenols to form salts

Answer 35

Phenol is weakly acidic: C(6)H(5)OH + aq C(6)H(5)O(-) + H(+)
Can neutralise it by adding NaOH:
C(6)H(5)OH + NaOH -> C(6)H(5)O(-)Na(+) + H(2)O
Can also add a reactive metal like sodium:
2 C(6)H(5)OH + 2Na -> 2 C(6)H(5)O(-)Na(+) + H(2)
Both reactions form sodium phenoxide

Question 36

Describe the reaction of phenol with bromine

Answer 36

C(6)H(5)OH + 3Br(2) -> C(6)H(2)OHBr(3) + 3HBr

Bromine is added to carbon 2,4 and 6 to make 2,4,6-tribromophenol. No halogen carrier or special conditions needed

Question 37

Why can phenol react with bromine easily whilst benzene can’t?

Answer 37

A lone pair of electrons on the oxygen in phenol is drawn into the benzene ring, creating a higher electron density in the ring structure, ‘activating’ the ring. The increased electron density polarises bromine molecules, which are then attracted more strongly to the ring than in benzene

Question 38

What are some of the uses of phenols?

Answer 38

+- chlorophenols used in antiseptics and disinfectants
+- bisphenol in production of epoxy resins for paints
+- used in production of plastics
- alkyl phenols used in surfactants and detergents
- salicylic acid (phenol + COOH on carbon 2) in preparation of aspirin and other drugs

Question 39

Why are amines basic?

Answer 39

A base is defined as a proton acceptor. Amines have a lone pair on the nitrogen atom, which can form a dative covalent bond with a proton. This results in a positively charged thing.

Question 40

Describe the reactions of amines with acids

Answer 40

base + acid -> salt
amine accepts proton of acid first forming (amine + H)+ and then attracts negative group (lets call it A(-)) to make (amine + H)+A(-)

Question 41

How are primary aliphatic amines prepared?

Answer 41

Aliphatic = straight chained
Conditions: halogenoalkane and excess ammonia (reagents); warm gently; ethanol as solvent
Equation:
halogenoalkane + NH(3) -> primary amine + HX
NH(3) + HX -> NH(4)(+)Cl(-)
Mechanism:
Ammonia acts as a nucleophile with lone pair of electrons; attacts d+ carbon atom. Have to add excess ammonia because primary amine formed also has lone electron pair that can attack another halogenoalkane.

Question 42

How are aromatic amines prepared?

Answer 42

Aromatic = cyclic
Conditions: nitrobenzene; concentrated hydrochloric acid and tin (reagents); heated under reflux
Equation:
nitroarene + 6[H] -> arylamine (e.g. phenylamine) + 2H(2)O
This is a reduction of the nitroarene

Question 43

How are azo dyes synthesised?

Answer 43

Diazotisation:
Mix phenylamine and nitrous acid below 10degC, forming a diazonium salt. Nitrous acid is made in situ (in reaction mixture) by reacting sodium nitrate, NaNO(2) with excess HCl:
NaNO(2) + HCl -> HNO(2) + NaCl
phenylamine + HNO(2) + 2HCl -> benzene with N(+)[Cl(-)]=-N + 2H(2)O (the organic product is: 2 Ns, one positive triple bonded, N+ is the one closest to the benzene; Cl(-) next to N(+))
Coupling:
Diazonium ion is reacted with a phenol (or another aromatic compound) under alkaline conditions:
Diazonium ion + phenol + NaOH -> azo dye + NaCl + H(2)O
Azo dye is 2 linked benzene rings through a diazo group (-N=N-)
Azo dyes are brightly coloured

Question 44

What are triglycerides?

Answer 44

Triesters of glycerol and fatty acids