Download top 5 organic - No Brain Too Small

 No Brain Too Small  CHEMISTRY 
TOP FIVE ORGANIC THINGS TO KNOW
(It’s hard to choose just 5 areas but these are all very important ones)
Esters - know how to make, break and name an ester (and that esters are structural isomers
with c.acids)
•
Naming ___yl ______oate. “-yl part from alcohol (“in-line -O), -oate part from carboxylic
acid (dangly =O”)
These 3 below are all identical – all are ethyl butanoate
•
H
H
C
H
H
C
H
H
C
H
O
H
H
H
C
C
H
C
C
H
H
H
H
H
H
H
O
C
C
C
C
O
•
O
O
H
O
H
C
H
H
H
H
C
H
H
C
H
H
H
H
C
H
HH
C
C
H
C
H
H
Two ways to make
o
Heat alcohol + c.acid + conc H2SO4 – conc. H2SO4 as a catalyst, and as a dehydrating
agent
•
Lose the -OH of c.acid and -H of alcohol as water
o
Mix acid chloride + alcohol – don’t need heat or any conc. H2SO4
•
Lose the -Cl of acid chloride and -H of alcohol as HCl(g)
Preparation method (alc + c.acid); heat under reflux; add carbonate to neutralise acids,
separate ester from impurities by distillation
Breaking – reaction with H2O - hydrolysis (break where you make – i.e between O and C=O)
o
Acid hydrolysis: heat with dilute acid (H+/H2O, heat) – get alcohol + c. acid
o
Alkaline hydrolysis: heat with dilute NaOH, (NaOH(aq), heat) - get alcohol + sodium salt
of the carboxylic acid RCOO-Na+
Something for nothing  If you can make / break an ester you can also do / draw / explain
o
Formation of fats/oils from propan-1,2,3-triol & 3 fatty acids
o
Alkaline hydrolysis of fats/oils for form propan-1,2,3-triol & the sodium salts of fatty
acids
o
Acid hydrolysis of fats/oils for form propan-1,2,3-triol & the fatty acids
o
Formation of polyesters from HO-X-OH & HOOC-Y-COOH or HO-X-OH & ClOC-Y-COCl
o
Hydrolysis of polyesters under acidic or alkaline conditions
•
•
•
•
Isomerism e.g. C4H8O2 could be an ester or a carboxylic acid (although they would be easy
to distinguish by their reactions and physical properties)
H
H
H
C
C
H
H
O
C
H
O
H
C
H
H
H
H
H
C
C
C
H
H
H
O
C
O
H
 No Brain Too Small  CHEMISTRY 
Alcohols, their reactions and how to distinguish between them
Reaction - Oxidation
Use orange coloured H+/Cr2O72- (reduced to green Cr3+) or purple coloured H+/MnO4-(reduced to
colourless Mn2+)
•
•
•
Primary alc  aldehyde (-CHO)  c.acid (-COOH)
Secondary alc ketone  X (not further oxidised)
Tertiary alc  X (not further oxidised)
Because the aldehyde is “on its way” to becoming oxidised to a c.acid….
•
•
If you want the aldehyde distill it off as soon as it is made or it will be fully oxidised to c.acid.
Aldehydes can be oxidised futher (unlike ketones) and so this allows them to be
distinguished from ketones by using mild oxidising agents like:
o
Tollens / Silver mirror test. Ag+/NH3, heat. Result – silver mirror Ag+ + e-  Ag
o
Benedicts / Fehlings solution. Cu2+, heat. Result – orange ppt of Cu2O Cu2+ + e-  Cu+
o
H+/Cr2O72- , heat. Result – orange to green colour change Cr2O72- + 14H+ + 6e-  2Cr3+ +
7H2O
o
In all 3 above the aldehyde  c. acid
•
Lucas reagent – to distinguish 1o, 2o and 3o alcohols
Warm alcohol with Lucas reagent: anhydrous ZnCl2/conc HCl, heat
o
Primary alc  no change (for a very, very long time)
o
Secondary alc cloudy after about 10 mins
o
Tertiary alc  cloudy almost immediately
Cloudiness due to formation of insoluble haloalkane in substitution reaction, -OH off and -Cl
on
Reaction – Esterification
•
•
R-OH + R’-COOH ⇌ R’COOR + H2O ; Conc sulfuric acid acts as catalyst but also removes the
water product, shifting eqm. position to right in favour of ester product
R-OH can be primary, secondary or tertiary
Reaction – formation of haloalkane (a substitution reaction)
•
•
R-OH + SOCl2  R-Cl + SO2 + HCl
SOCl2 will work for primary, secondary & tertiary (and therefore safer bet than PCl5 or PCl3 or
conc HCl or conc HCl/ZnCl2).
Reaction – dehydration to form alkenes C=C & H2O (an elimination reaction)
•
•
Heat, conc H2SO4
If the alcohol is unsymmetrical, get 2 products (the poor get poorer)
 No Brain Too Small  CHEMISTRY 
Haloalkanes
R-X where X is -F, -Cl, -Br, -I (but usually -Cl)
Can be primary, secondary or tertiary
Reaction - Formation
•
•
from (any) alcohol using SOCl2 (substitution)
R-OH  R-Cl
from alkenes using H-X (addition) e.g. HCl
…but remember if alkene is unsymmetrical = 2 products (the rich get richer)
Reactions – Elimination & Substitution with haloalkanes
•
“Eliminate (with) the alcohol!!!!” meaning NaOH(alc) favours
elimination from the haloalkane!
Warm with NaOH(alc) or KOH(alc); NaOH or KOH dissolved in ethanol
R-X + NaOH(alc)  alkene + NaX + H2O
… but remember if haloalkane is unsymmetrical = 2 products (the poor get poorer)
•
“Substitute (with) the water!!!!” meaning NaOH(aq) favours
substitution of the haloalkane!
Uses NaOH(aq) or KOH(aq); NaOH or KOH dissolved in water
R-X + NaOH(aq)  R-OH + NaX
 No Brain Too Small  CHEMISTRY 
Amino acids
•
Are building blocks of peptides & proteins
•
Have a –NH2 and a –COOH group attached to the same C atom
•
Can be chiral if the “middle C” has 2 different groups/atoms attached to it e.g.
H
•
H
H
C
N
C
H
H
H
H
O
C
O
Join together in a condensation reaction, eliminating the small molecule H2O, forming the
“amide / peptide” link of -CONHH
H
H
C
N
C
H
H
H
O
O
H
C
H
C
C
N
H
H
•
O
If you join 2 different amino acids you can make 2 different dipeptides
H
H
H
C
N
C
H
O
O
H
C
H
H
N
H
H
N
C
H
H
O
O
H
C
H
C
C
H
C
C
H
O
O
H
N
C
H
H
H
H
and
NH2 end bonded to -CH(CH3)•
NH2 end bonded to -CH2-
And where you break is where you make…. in the reaction with dilute acids or alkalis
(hydrolysis).
H
H
H
C
N
C
H
H
H
O
O
H
C
C
N
H
o
o
H
C
O
H
Under acidic conditions end up with a -COOH end and a -NH3+ end to the amino acids
Under alkaline conditions end up with a -COO- (or -COO-Na+) end and a -NH2 end to
the amino acids
 No Brain Too Small  CHEMISTRY 
Addition reactions of alkenes (Level 2 stuff but still VERY important…)
Addition of an unsymmetrical reagent, e.g. HCl or H2O to an unsymmetrical alkene; the rich get
richer.
Addition of HCl to
propene
H
major product
H
C
C
H
H
H
C
H
H
H
H
C
C
C
H
H
Cl
H
Cl
H
minor product
H
H
H
C
C
C
H
H
H
H
Addition of HCl to cis and trans isomers of but-2-ene only forms ONE product because but-2-ene is
symmetrical – and the product – a haloalkane has single C-C bonds which can freely rotate
H
H
H
H
H
C
H
C
H
H
C
C
H
H
C
H
H
H
C
C
C
H
H
H
and
both give the same product
H
Cl
H
H
H
C
C
C
C
H
H
H
H
H
For geometrical isomers to exist you need:
o A C=C double bond, around which there is NO free rotation
o The groups/atoms on the C atom at EACH END OF the C=C double bond need to be
different to each other.
different
same
H
C
H
H
no cis / trans
Br
C
C
C
different
H
H
H
H
different
C
H
H
C
C
H
H
H
H
can exist as cis and trans
Addition of water to an alkene – hydration, an addition reaction.
Heat with dilute acid e.g. sulfuric acid. We normally write this as H+/H2O, heat
 No Brain Too Small  CHEMISTRY 
Amides and amines
H
H
H
H
H
C
C
C
N
H
H
H
H
propylamine /1-aminopropane
•
H
C
C
H
H
O
C
H
N
H
propanamide
Be able to distinguish between them – use UI or litmus
o
o
•
H
H
Amines are basic since RNH2 + H2O ⇌ RNH3+ + OH- but amides are essentially neutral
Amines smell bad / fishy / rotten; amides don’t (ethanamide smells of “mouse
droppings” but apparently that’s due to an impurity in the chemical not the
ethanamide itself!)
How to form them
o
Amine: R-OH + NH3  R-NH2 + H2O; use conc NH3 – reaction is substitution
These reactions require catalysts, specialised apparatus, and additional purification
measures.
The same amines can be prepared by treatment of haloalkanes with ammonia. You
get amines formed together with their salts
e.g. R-X + NH3(alc)  R-NH3+ X- and then R-NH3+ X- + NH3  R-NH2 + NH4X.
o
Amide: R-COOH + conc NH3 WILL NOT directly form R-CONH2 + H2O even though it
looks like it will do a simple substitution reaction from looking at the formula.

This is because R-COOH is an acid and NH3 is a base – so their reaction will be an
acid-base reaction and will make the salt of R-COO- NH4+.

To make the amide you need react excess R-COOH with (NH4)2CO3 and heat the
R-COO- NH4+ formed, which will break down to give R-CONH2 + H2O