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a chemistry > m6 - y13 organic > Flashcards

m6 - y13 organic Flashcards

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1
Q

arenes def

A

aromatic compounds that contain a benzene ring

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2
Q

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

A

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

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3
Q

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

A

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

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4
Q

phenyl group functional group

A

C6H5

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5
Q

nitro group atoms

A

NO2

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6
Q

alkylation def

A

hydrogen replaced by an alkyl group

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7
Q

Friedel-Crafts reactions: reagents and conditions

A

a haloalkane in the presence of a hydrogen carrier catalyst

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8
Q

halogenation of benzene reagents and conditions

A

halogen and halogen carrier catalyst

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9
Q

acyl group

A

R - C = O

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10
Q

conditions for acyl chloride reactions with benzene

A

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

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11
Q

benzene vs alkene in terms of bromination

A

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

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12
Q

phenol def in benzene

A

-OH directly bonded to a benzene ring

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13
Q

physical properties of phenol

A

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

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14
Q

how can phenol decolorise bromine easily

A

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

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15
Q

reagent for nucleophilic addition (turning carbonyls back into alcohols)

A

NaBH4 (sodium borohydride)
provides H:- nucleophile

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16
Q

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

A

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

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17
Q

NMR spectroscopy info

A

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

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18
Q

NMR standard used to compare everything to

A

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

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19
Q

NMR solvent used

A

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
20
Q

interpreting carbon NMR

A
  • number of peaks => gives number of different carbon environments
21
Q

chromatography is…

A

an analytical technique used to separate components of a mixture

22
Q

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

A

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

23
Q

gas chromatography, what is the stationary phase and mobile phase

A

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

24
Q

how does separation occur

A

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

25
Q

interpreting gas chromatograms - retention time

A

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

26
Q

chemical shift gives us

A

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

27
Q

proton NMR gives you this info

A

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

28
Q

-:CN group called

A

nitrile group

HCN -><- H+ and -:CN

29
Q

carboxylic acid properties

A

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

30
Q

how to make an acid anhydride

A

2 carboxylic acids react
water is eliminated

31
Q

using acid anhydride + alcohol to make esters

A

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

32
Q

hydrolysis of ester using an alkali

A

ester + alkali —> CARBOXYLATE SALT + ALCOHOL

carboxylate salt = O-

33
Q

formation of acyl chlorides => SOCl2 + carboxylic acid

A

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

34
Q

amines - naming them

A

add methyl/ethyl/propyl/dimethyl etc to amine

also remmeber they only have 3 bonds not 4

35
Q

addition polymers => mechanism is called

A

free radical substitution

36
Q

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

A

Ni/H2

37
Q

chirality

A

carbon with 4 different environments attached

=> no plane of symmetry so they don’t have a superimposable mirror image so they are optical isomers

38
Q

enantiomers

A

optical isomers that rotate plane polarised light by equal amounts in opposite directions

=> a mixture containing 50% of each enantiomer is called racemic mixture

39
Q

which 2 poly groups are formed by condensation polymerisation

A

polyesters
polyamides

40
Q

polyesters

A

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

41
Q

polyamides

A

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

42
Q

hydrolysis of polyesters

A

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

43
Q

hydrolysis of polyamides

A

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

basic:
NaOH
forms diamine and dicarboxylate salt

44
Q

amino acid zwitterions

A

COOH and NH2 reacting internally to form an internal salt

=> exists at a specific pH value called the isoelectric point

45
Q

amides functional group

A

R - C =O
|
N

46
Q

making amides (primary then secondary)

A

acyl chloride + ammonia = P

acyl chloride + primary amine = S

a chemistry (21 decks)

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