Mechanisms a2 organic

Question 1

Optical isomerism mechanism (nucleophillic addition) aldehydes to hydroxynitriles

Answer 1

acidified kcn and h+ or acidified hcn. so– :CN- arrow to the delta positive carbon under the c=o bond. then arrow from middle of c=o bond to the oxygen. that makes :o- then join your cn at the bottom, put an arrow to a h+ floating around and then you form a hydroxyalkanenitrile.

Question 2

carboxylic acids are weak acids, what do they disassociate to make

Answer 2

carboxylate ion and a h+. equilibrium lies to the left because it disassociates poorly.

Question 3

what do acyl chlorides react with

Answer 3

water, alcohol, ammonia and primary amines

Question 4

acyl chloride and water and observations

Answer 4

carboxylic acid +hcl. vigorous reaction, white misty fumes of hcl produced

Question 5

acyl chloride and ammonia

Answer 5

produce amides. again vigorous, white misty fumes of hcl produced

Question 6

acyl chloride and alcohol

Answer 6

esters, again hcl produced as well white misty fumes vig rec

Question 7

acyl chloride and primary amines

Answer 7

produce n- substituted amides. and hcl. the h is the one thats substitued from the n. so its n substitued because it donates the hydrogen.
nhch3 +hcl for example when you react ch3nh2 with an acyl chloride.

Question 8

what do acid anhydrides react with

Answer 8

the same as amines, water ammonia alcohol and primary amines. except these reactions are less vigorous and this is why acid anhydrides are used as opposed to acyl chlorides.

Question 9

acid anhydride and water

Answer 9

2 carboxylic acids

Question 10

acid anhydride and alcohol

Answer 10

ester and carboxylic acid

Question 11

acid anhydride and primary amine

Answer 11

n substituted amide and carboxylic acid

Question 12

the actual mechanism for reacting an acyl chloride with an alcohol. you know it already forms an ester and hcl from the simplified way. what mechanism is this

Answer 12

nucleophillic addition elimination
you have your acyl chloride and set out your alochol below it like this. R-O:: - –H on the other side and put it like a upside down L. (the o has 2 lone pairs o it)

arrow from o to c. then from c=o bond from c to o.
then this forms intermediate w c-cl on top o:- on the right and at the bottom the alcohol joined but w o+-r-r

then arrow from the o:- to the c to reform double bond. then arrow from c to cl to eliminate cl:-

then you have your semi ester w alcohol attached. then cl:- attacks hydrogen on alcohol and then arrow from h to o+ this eliminates h as well forming hcl. thats it thats your ester.

Question 13

how is aspirin made

Answer 13

aspirin is an ester made by reacting ethanoic anhydride or ethanoic chloride with salicylic acid. makes aspirin and a carboxylic acid.

Question 14

why use ethanoic anhydride in industry vs ethanoic chloride

Answer 14

safer as less corrosive, does not react as vigorously with water, does not produce the toxic hcl gas, it is cheaper. re

Question 15

reactions of arenes what mechanism is it and why

Answer 15

they undergo electrophillic substitution reactions.
because, benzene has a high electron density as it has a delocalised ring of electrons, this is attractive to electrophiles.
Benzene is also stable so unlike traditional alkenes they do not undergo electrophillic addition reactions as this would disrupt the stable ring of electrons.
instead they undergo electrophillic substitution reactions where the hydrogen or a functional group are substituted for an electrophile

Question 16

the two mechanisms of arenes

Answer 16

friedel crafts acylation and nitration reactions

Question 17

fridel crafts acylation

in order to add onto the benzene what must the electrophile have

Answer 17

benzene is too stable and so is difficult to react, when an acyl group (RCO- ) is added onto a benzene it makes the structure weaker and more easier to modify to make useful products.
it must have a very strong positive charge, the acyl group isnt strong enough therefore we have to carry out a prep step

Question 18

our prep step

Answer 18

we can use a halogen carrier eg alcl3 which will produce a much stronger electrophile with a much stronger positive charge. so in fridel crafts acylation we have to react an acyl chloride with AlCl3 to form a strongly postive electrophile.
alcl3 accepts a pair of electrons away from the acyl group, this forms a positive carbocation r-c (+)=o and Alcl4- (this will be reformed bc it is your catalyst so alcl3 will be back)

Question 19

the actual mechanism

Answer 19

we react the electrophile with a benzene to make a less stable phenylketone under reflux and a dry ether solvent.

from benzene ring a arrow to the positive carbocation. this then disrupts the ring of delocalised electrons and it forms a positive charge in the benzene with a ring not extending beyond the adjacent carbons to the ketone. you also have your hydrogen c-h on that remember.

then you have your alcl4 structurally drawn next to it and arrow from the cl to the hydrogen as your alcl4- is attracted to the postively charged ring and cl breaks away and then from the c-h bond to inside of the benzene ring to reform the benzene ring.

that then forms your less stable phenylketone and hcl + AlCl3

Question 20

nitration of benzene

Answer 20

useful as it allows us to make dyes for clothes and explosives. if we heat benzene with conc h2so4 and conc hno3 then we form nitrobenzene. but we have to make a poweful electrophile first.

Question 21

the prep step

Answer 21

hno3+ h2so4–> h2no3+ +hso4-
the h2no3+ decomposes to form the electrophile
hno3+–>No2+ + h2o
We know use the nitronium ion and react this with benzene to form nitrobenzene

Question 22

the mechanism

Answer 22

pretty much same as friedels acylation. arrow from ring to that, then you have your o2n and h off the same carbon. the ring is halved with a positive charge in the middle. from the c-h bond an arrow moves back into the ring to reform it. that forms your nitrobenzene +h+

The h+ reacts with the hso4- in the prep step and that reforms your h2so4 catalyst.

Question 23

important about nitration and its uses

Answer 23

a temp below 55 degrees will ensure a single No2 subsitution, above this will result in multiple substitutions.

dyes and pharmaceuticals can be made by reducing nitrobenzenes to aromatic amines. (benzene with nh2 attached)
explosives- made from nitrobenzene

Question 24

reduction of nitrobenzene (reducing agents for nitrobenzene to aromatic amines- benzene w nh2 attached)

Answer 24

reacted with sn and conc hcl. so sn or hcl

Question 25

when you reduce nitrobenzene how many hydrogens do you need

Answer 25

3h2 or 6h to make the aromatic amine nh2 and 2h2o

Question 26

how to make aliphatic amines pros and cons

Answer 26

haloalkane and excess ammonia. nucleophillic substitution.
reducing nitriles (nickel or pt catalyst and h2 gas- catalytic hydrogenation)
or use a strong reducing agent eg LiAlH4 and dilute acid dissolved in dry ether

Question 27

pros cons

Answer 27

impure, lone pairs of electrons can form sec tert quat amines.
this forms primary amines. and this reac is also more expensive, nitrogen can act as a nucleophile.

catalytic hydr= cheapest, high temp and high pressure, pure product

using LiAlH4= more expensive as it is more exp.

Question 28

reacting haloalkanes w excess ammonia

Answer 28

nucleophillic sub reaction mechanism. ammonia is a nucleophile that attacks the positive carbon on the c-hx and then arrow to the halogen. this eliminates the cl:- and instead we have the nh3 attached on the end. and then draw it out properly with n+-h-h. then another ammonia molecule’s lone pair attacks the hydrogen on the nh3 and then that moves up to the n+
that forms your primary amine and nh4+cl- (ammonium chloride salt)
the primary amine can still act as a nucleophile though. and can react with the haloalkane to form other types

Question 29

reducing nitriles- catalytic hydrogenation to make primary amines

Answer 29

h2 gas and nickel or platinum catalyst, high temp and pressure pure product made. cheapest way

r-ch2-cn +2h2–>r-ch2-ch2-nh2

Question 30

Reducing nitriles w LiAlH4 to make primary amines

Answer 30

strong reducing agent and dilute acid. more expensive, reducing agent dissolved in a dry ether.

R-ch2-cn + 4(H) —> R-ch2-ch2-nh2

Question 31

making aromatic amines

Answer 31

made by reducing nitro compounds eg nitrobenzene to form phenylamines. used to make dye stuff and pharmaceuticals

reduction and we use 6h or 3h2.

Question 32

how does reducing nitro compounds work

Answer 32

we react it with 6(h) we use tin and conc hcl. —-> so they go above and below the arrow (we heat nitrobenzene under reflux with conc hcl and tin to form a phenyl ammonium salt. eg c6h5nnh3cl

Question 33

then what

Answer 33

when you have your c6h5nh3cl react this with naoh so alkali. this will then give you your phenyl amine c6h5nh2 +nacl+ h2o.

this is done at room temp bc it goes from less stable to more stable.
so in conclusion, tin conc hcl and then naoh to form the phenyl amine from the phenyl ammonium salt

Question 34

reduction of a ketone or aldehyde with nahbh4

Answer 34

add 2(h) so the first h is a h:- and that attacks the c=o carbonyl group and then that forms o:- and then arrow from the lone on the o towards another h, but this time h+ to form a hydroxy. so you form your primary or secondary alcohol depending on your aldehyde or ketone

Question 35

carboxylic acids and sodium carbonate and hydrogencarbonate

Answer 35

ch3cooh+ nahco3–> ch3cooNa +h2o+ co2

ch3cooh + Na2Co3–> 2ch3cooNa +h2o +co2

Question 36

esterification

Answer 36

carboxylic acids and alochols/ acid anhydrides in prescence of strong acid catalyst eg sulfuric acid. this is for cooh oh esterification forming a carboxylic acid and water.
for acid anhydrie it forms an ester and a carboxylic acid

Question 37

ester hydrolysis

Answer 37

using h2o, can be sped up by using an acid or a base

Question 38

acid hydrolysis

Answer 38

dilute acid eg hcl. ester+h2o–> (h+) on top of arrow carboxylic acid and alcohol.
this reaction is reversible so not as high atom
economy
conducted under reflux

Question 39

base hydrolysis

Answer 39

uses a base eg naoh under reflux.
this forms a carboxylate ion and alcohol bc we have oh- reacting not water
coo- + alcohol.
if we use naoh we make coona which is a soap (carboxylate salt) and this is saponification

Question 40

making soap with hydrolysing animal fats and vegetable oils with sodium hydroxide

Answer 40

this uses animal fats to make soap

glycerol+ fat reacts w 3naoh–> glycerol and your sodium salt which is your soap. 3the bits on the end, coo-na+

Question 41

vegetable oils

Answer 41

can be converted into biodiesel by reacting oils with methanol and using koh as the catalyst
oil+ 3methanol–> glycerol+ methyl ester

methyl esters can be used for biodiesel