m6 - y13 organic

Question 1

arenes def

Answer 1

aromatic compounds that contain a benzene ring

Question 2

problems with Kekule structure of benzene (bromine water, bond lengths, enthalpy change of hydrogenation)

Answer 2

doesn’t decolourise bromine water, no electrophilic addition reactions 

double bonds are stronger than single bonds so are shorter, meaning K thought benzene would be an irregular hexagon, BUT all carbon-carbon bond lengths were actually equal

ΔH less exothermic than expected, meaning real benzene is more stable than K suggested

Question 3

Kekule structure of benzene vs new idea in terms of p orbitals and electrons

Answer 3

K = p orbitals overlap in pairs (localised)
new = all p orbitals overlap (sideways in both directions), a ring above the sigma bond, a ring below (delocalised π bonding system) there is less repulsion between electrons so benzene is more stable
there are 6 π electrons one from each carbon

Question 4

phenyl group functional group

Answer 4

C6H5

Question 5

nitro group atoms

Answer 5

NO2

Question 6

alkylation def

Answer 6

hydrogen replaced by an alkyl group

Question 7

Friedel-Crafts reactions: reagents and conditions

Answer 7

a haloalkane in the presence of a hydrogen carrier catalyst

Question 8

halogenation of benzene reagents and conditions

Answer 8

halogen and halogen carrier catalyst

Question 9

acyl group

Answer 9

R - C = O

Question 10

conditions for acyl chloride reactions with benzene

Answer 10

acyl chloride in the presence of a halogen carrier catalyst
reflux
ANHYDROUS conditions

Question 11

benzene vs alkene in terms of bromination

Answer 11

higher e- density in alkene localised π bond, so bigger induced dipole in Br2
- polarises electrophile more strongly, attracted more strongly, reacts without a catalyst (by addition)

lower e- density in benzene delocalised π bonding system (spread over more than 2 carbons)
- polarises electrophile less strongly, needs a catalyst to make Br+ (by substitution)
- retains extra stability of delocalised π system, addition would permanently destroy it

Question 12

phenol def in benzene

Answer 12

-OH directly bonded to a benzene ring

Question 13

physical properties of phenol

Answer 13

hydrogen bonding => high BP compared to benzene (london forces only)
slightly soluble in water (because the benzene ring is non-polar)

Question 14

how can phenol decolorise bromine easily

Answer 14

the oxygen has a lone pair of e- in its p-orbital, can overlap with and become part of π bonding system, so has a higher e- density than benzene

Question 15

reagent for nucleophilic addition (turning carbonyls back into alcohols)

Answer 15

NaBH4 (sodium borohydride)
provides H:- nucleophile

Question 16

test to detect C=O carbonyl group, and then whether it’s an aldehyde or ketone

Answer 16

add 2,4-dinitrophenylhydrazine, colour goes orange. recrystallise, find melting point and compare to literature values

add Tollens reagent (ammoniacal silver nitrate), silver mirror appears with aldehyde (been oxidised to carboxylic acid, silver ions reduced to silver),
remain colourless with ketone

Question 17

NMR spectroscopy info

Answer 17

uses radiowaves that are absorbed by the nuclei of certain atoms eg C-13 and H-1 (proton)

need an odd number of nucleons

the frequency (called chemical shift) absorbed depends on the environment of the atom

Question 18

NMR standard used to compare everything to

Answer 18

TMS (tetramethylsilane) is used as standard. the peak is given a value of 0 and all other peaks are compared to it and given a chemical shift δ in ppm
- TMS is inert and non toxic
- volatile so low BP (easy to remove)
- equivalent, all bonds in the same environment so gives a single peak only

Question 19

NMR solvent used

Answer 19

sample needs to be dissolved in a solvent to dilute it

common solvents are water, benzene, trichloromethane (but we can’t use hydrogen bc it will swamp the spectrum so we use deuterium)

• D2O
• C6D6
• CDCl3
  deuterated solvents

Question 20

interpreting carbon NMR

Answer 20

• number of peaks => gives number of different carbon environments

Question 21

chromatography is…

Answer 21

an analytical technique used to separate components of a mixture

Question 22

thin layer chromatography TLC, what is the stationary phase and mobile phase

Answer 22

S = solid eg Al2O3 coating a sheet of glass
M = liquid eg a solvent which moves vertically up the plate

Question 23

gas chromatography, what is the stationary phase and mobile phase

Answer 23

S = solid or liquid on an inert solid support eg long chain alkane coating the inside of a coiled tube
M = carrier gas eg inert gas like nitrogen

Question 24

how does separation occur

Answer 24

the components of the mixture have different attractions for the solid and liquid phases so travel at different speeds

TLC = the stronger the adsorption to the stationary phase, the slower they are carried by the mobile phase

GC = the greater the solubility in the stationary phase the slower they are carried by the mobile phase

Question 25

interpreting gas chromatograms - retention time

Answer 25

time taken for a component to pass from the column inlet to the detector

Question 26

chemical shift gives us

Answer 26

gives the type of carbon environment ie tells us what groups/bonds surround the carbon

Question 27

proton NMR gives you this info

Answer 27

number of peaks (or groups of peaks) => gives us number of different proton environments chemical shift => gives us type of proton environment relative peak areas => area under peak is proportional to number of protons. usually given as a number above the peak telling us *the relative number of protons of each type* (ratio) spin spin coupling => splitting only occurs if protons on neighbouring carbon are different => to check if -NH or -OH add D2O shake vigorously and run spectrum again, peak won’t show up as deuterium exchanges with those H

Question 28

-:CN group called

Answer 28

nitrile group HCN -><- H+ and -:CN

Question 29

carboxylic acid properties

Answer 29

soluble - hydrogen bonds with water boiling points - higher than alcohols, produces a dimer (2 molecules of carboxylic acids form hydrogen bonds between them)

Question 30

how to make an acid anhydride

Answer 30

2 carboxylic acids react water is eliminated

Question 31

using acid anhydride + alcohol to make esters

Answer 31

acid anhydride (splits) and the OH from the alcohol joins one of them to reform a carboxylic acid other half acid anhydride reacts with alcohol to form ester

Question 32

hydrolysis of ester using an alkali

Answer 32

ester + alkali —> CARBOXYLATE SALT + ALCOHOL carboxylate salt = O-

Question 33

formation of acyl chlorides => SOCl2 + carboxylic acid

Answer 33

SOCl2 + carboxylic acid (eg Ch3COOH) —> Ch3COCl + SO2(g) + HCL(g)

Question 34

amines - naming them

Answer 34

add methyl/ethyl/propyl/dimethyl etc to amine also remmeber they only have 3 bonds not 4

Question 35

addition polymers => mechanism is called

Answer 35

free radical substitution

Question 36

reduction of a nitrile to an amine (Ctriple bondN) to NH2 reagents

Answer 36

Ni/H2

Question 37

chirality

Answer 37

carbon with 4 different environments attached => no plane of symmetry so they don’t have a superimposable mirror image so they are optical isomers

Question 38

enantiomers

Answer 38

optical isomers that rotate plane polarised light by equal amounts in opposite directions => a mixture containing 50% of each enantiomer is called racemic mixture

Question 39

which 2 poly groups are formed by condensation polymerisation

Answer 39

polyesters polyamides

Question 40

polyesters

Answer 40

dicarboxylic acid and a diol react and form an ester linkage -COO- produces polyester and water => useful as they can be hydrolysed due to the polarity. => biodegradable in nature by water, so will gradually break down when put into landfill

Question 41

polyamides

Answer 41

formed from dicarboxylic acid and a diamine => amide linkage (-CONH-)

Question 42

hydrolysis of polyesters

Answer 42

ester group / amide group broken down by acid or base hydrolysis because they are polar acidic: (reversible reaction, slow) H+ and H2O forms dicarboxylic acid and diol basic: ester + base —> carboxylate salt + alcohol

Question 43

hydrolysis of polyamides

Answer 43

acidic: H+ and H2O forms dicarboxylic acid and diamine HOWEVER acid means NH2 can become NH3+Cl- basic: NaOH forms diamine and dicarboxylate salt

Question 44

amino acid zwitterions

Answer 44

COOH and NH2 reacting internally to form an internal salt => exists at a specific pH value called the *isoelectric point*

Question 45

amides functional group

Answer 45

R - C =O | N

Question 46

making amides (primary then secondary)

Answer 46

acyl chloride + ammonia = P acyl chloride + primary amine = S