NaCl Find LEf

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Welcome and thanks for visiting.
Module 3 Practical Observation and
Deduction 2002-7

NaClFind LEf
F Scullion. JustChemy.Com
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Inorganic Observation and Deduction
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The observations and deductions come in pairs of
activities. One involves inorganic chemistry and
the other organic chemistry. The inorganic
exercise typically involves the analysis of a
so-called Double Salt. One of the ions in these
mixed salts is common to both salts in most
cases. Here are some examples - Sodium Sulphate
and Sodium Chloride Calcium Carbonate and
Calcium Chloride
Na SO42- Cl-
Ca2 CO32- Cl-
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Deductions.
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Observations
A is Ammonium Chloride NH4Cl
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Identifying halide ions in solution
Na
The equations for the reactions are represented
thusNaX(aq)    AgNO3(aq)      AgX(s)   
NaNO3(aq) 
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Why Cant You
hite
ream
ellow
NaCl(aq)    AgNO3(aq)      AgCl(s)   
NaNO3(aq)                                        
        white ppt NaBr(aq)    AgNO3(aq)     
AgBr(s)    NaNO3(aq)                            
                    cream ppt NaI(aq)   
AgNO3(aq)      AgI(s)    NaNO3(aq) 
                                           
yellow ppt
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Both AgCl and AgBr are light sensitive. They
have darkened noticeably after 5 minutes
AgBr
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Solubility of Silver Halides in Ammonia solution.
Silver halide Colour Solubility in NH3(aq) Solubility in NH3(aq)
Silver halide Colour Dilute Concentrated
AgCl White v v
AgBr Cream x v
AgI Yellow x x
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Ammonium Compounds
warm Ammonium
Alkali ? Salt NH3 H2O Compound
warm NH4Cl NaOH ? NaCl NH3 H2O
warm (NH4)2SO4
Ca(OH)2 ? CaSO4 2NH3 2H2O
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(ii) Describe how you would carry out the
test for hydrogen chloride gas and state what
you would observe. White 1 fumes/smoke 1
glass rod 1 dipped in conc 1 ammonia 1
(max 4)
AMMONIUM ALKALI ? SALT AMMONIA
WATER COMPD NH4Cl NaOH ? NaCl NH3
H2O
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Observations
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Observations
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It is Potassium Carbonate
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Potassium flame test Emission spectrum Lilac
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Using HCl Limewater to test for CO32-
Effervescence
Limewater
Sample of dilute HCl (aq)
Sign of CO2 gas
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Using Magnesium Nitrate to test for the presence
of CO32-(aq)
CO32-(aq) Mg2(aq) ? MgCO3(s)
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Reactions of the Carbonate Ion. Metal Carbonate
Acid ? Salt CO2 H2O Partial Ionic
Equation CO32- H ? H2O CO2 Balanced
Symbol Equation K2CO3 2HCl ? 2KCl H2O
CO2 ----------------------------------------------
---------- Precipitation Reaction AB(aq)
CD(aq) ? AD(s)
CB(aq) Partial Ionic Equation CO32- Mg2 ?
MgCO3 Balanced Symbol Equation K2CO3 MgCl2 ?
MgCO3 2KCl
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Observation
A is a double salt
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A is a double salt
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Colour is a feature of the compounds of TMs
Cobalt chloride above
Potassium chromate above Nickel chloride opposite
Above is from the Chemguide Website
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Colour of Aqueous Ions
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Colour of Aqueous Ions
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Ammonium Compounds
warm Ammonium
Alkali ? Salt NH3 H2O Compound
warm NH4Cl NaOH ? NaCl NH3 H2O
warm (NH4)2SO4
Ca(OH)2 ? CaSO4 2NH3 2H2O
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(ii) Describe how you would carry out the
test for hydrogen chloride gas and state what
you would observe. White 1 fumes/smoke 1
glass rod 1 dipped in conc 1 ammonia 1
(max 4)
AMMONIUM ALKALI ? SALT AMMONIA
WATER COMPD NH4Cl NaOH ? NaCl NH3
H2O
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NH3(aq) A weak alkali
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Dilute ammonia solution with UI about pH 11
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Lithium Calcium Crimson Red Brick
Red
Strontium Red
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Reactions of the Halide Ions with
AgNO3(aq). Silver Sodium ? Silver
Sodium Nitrate Halide Halide
Nitrate Partial Ionic Equation using X- for
Halide Ions Ag X- ? AgCl Balanced Symbol
Equation AgNO3 NaX ? AgX NaNO3 AgNO3
NaCl ? AgCl NaNO3 AgNO3 NaBr ? AgBr
NaNO3 AgNO3 NaI ? AgI NaNO3
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Mixture of Ammonium Chloride and Li/Sr/Ca Chloride
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Acidified Barium Chloride (or Nitrate) is used to
test for SO 42-(aq)
Ba2(aq) SO42-(aq) BaSO4(s)
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Has Ammonium Sulphate and Sodium Sulphate
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Ammonium Compounds
warm Ammonium
Alkali ? Salt NH3 H2O Compound
warm NH4Cl NaOH ? NaCl NH3 H2O
warm (NH4)2SO4
Ca(OH)2 ? CaSO4 2NH3 2H2O
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(ii) Describe how you would carry out the
test for hydrogen chloride gas and state what
you would observe. White 1 fumes/smoke 1
glass rod 1 dipped in conc 1 ammonia 1
(max 4)
AMMONIUM ALKALI ? SALT AMMONIA
WATER COMPD NH4Cl NaOH ? NaCl NH3
H2O
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NH3(aq) A weak alkali
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Dilute ammonia solution with UI about pH 11
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Flame test for Sodium Orange-yellow
Na NH4 ?
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Ba2 Cl- Cl- Cu2 SO42-
Barium nitrate Copper(II) sulphate
BaCl2 CuSO4
Complete Formula Equation BaCl2(aq)  CuSO4(aq)
? BaSO4(s)   CuCl2(aq) Complete Ionic
Equation Ba2(aq) 2 Cl-(aq) Cu2(aq)
SO42-(aq) ? BaSO4(s) Cu2(aq) 2
Cl-(aq) Net Ionic Equation Ba2 2 Cl- Cu2
SO42-  ? BaSO4(s) Cu2 2 Cl-         
      Ba2(aq) SO42-(aq) BaSO4(s)
Na NH4 SO42-
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Observations
A is a mixture of 2 salts
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Observations A is a
White solid
A is a mixture of 2 salts
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Observations
X is a double salt
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Observations X is a white solid
X is a double salt
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Has sodium chloride and ammonium chloride
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A is a double salt
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A is a double salt
A has NaBr and Na2SO4
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• Oxidising ability of Concentrated Sulphuric
  Acid.Its reactions with halide ions in aqueous
  solution during the formation of the Hydrogen
  Halides.
• CARE CARRY OUT ALL OF THESE IN A FUME CUPBOARD
  WITH GREAT CARE
• NaCl(s)           H2SO4(l)      ?     
  NaHSO4(aq)     HCl(g)
• The concentrated sulphuric acid will not oxidise
  HCl
• Dip a glass rod into concentrated
• ammonia solution and then into the
• test tube where you suspect the
• presence of HCl. The HCl will form
• a white smoke with NH3 if it is
• present.
• NH3 HCl ? NH4Cl

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Oxidising ability of Concentrated Sulphuric
Acid.Its reactions with halide ions in aqueous
solution during the formation of the Hydrogen
Halides. CARE CARRY OUT ALL OF THESE IN A FUME
CUPBOARD WITH GREAT CARE 2a)   KBr(s)    
     H2SO4(l)      ?     KHSO4(aq)        
HBr(g) However, the concentrated sulphuric acid
will oxidise some of the HBr as follows
-1
0 2b)    2HBr         H2SO4       
?    Br2          SO2         2H2O Therefore,
one will observe a Reddish Vapour due to some
bromine being present. The SO2 is and acidic
gas. It is also a reducing agent and will, for
example, decolourise purple potassium
permanganate solution    
An increase in O.N.
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Oxidising ability of Concentrated Sulphuric
Acid.Its reactions with halide ions in aqueous
solution during the formation of the Hydrogen
Halides. CARE CARRY OUT ALL OF THESE IN A FUME
CUPBOARD WITH GREAT CARE 3a)  KI(s)      
H2SO4(l)    ?   KHSO4(aq)       HI(g) The
concentrated sulphuric acid will oxidise some of
the HI as follows  3b)    H2SO4(l)      
2HI(g)     SO2         I2           
2H20  3c)       H2SO4(l)         6HI(g)    
S            3I2          4 H20  3d)       H2SO4
(l)         8HI(g)      H2S         4I2   
      4 H20  During the reaction one will
observe -         Violet Iodine Vapour being
evolved,         The violet vapour cooling and
subliming to form dark solid iodine,         A
smell of rotten eggs (H2S)         Some free
yellow sulphur         Some HI(g) which could be
identified in the way one shows the
presence of HCl. Use concentrated NH3 solution
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SUMMARYOxidising ability of Concentrated
Sulphuric Acid.Its reactions with halide ions in
aqueous solution during the formation of the
Hydrogen Halides. CARE CARRY OUT ALL OF THESE
IN A FUME CUPBOARD WITH GREAT CARE 1)
    NaCl(s)           H2SO4(l)           
NaHSO4(aq)     HCl(g) The concentrated sulphuric
acid will not oxidise HCl 2a)    KBr(s)      
      H2SO4(l)            KHSO4(aq)        
HBr(g) The concentrated sulphuric acid will
oxidise some of the HBr as follows      2b)
   2HBr            H2SO4            Br2   
      SO2         2H2O Therefore, one will
observe a Reddish Vapour  due to some bromine
being present.    3a)    KI(s)             
H2SO4(l)    à     KHSO4             HI(g) The
concentrated sulphuric acid will oxidise some of
the HI as follows  3b)    H2SO4(l)       2HI(g)
    SO2         I2           
2H20  3c)       H2SO4(l)         6HI(g)     
S            3I2          4 H20  3d)       H2SO4
(l)         8HI(g)     H2S         4I2   
      4 H20  During the reaction one will
observe -         Violet Iodine Vapour being
evolved,         The violet vapour cooling and
subliming to form dark solid iodine,         A
smell of rotten eggs (H2S)         Some free
yellow sulphur         Some HI(g) which could be
identified in the way one shows the
presence of HCl. Use concentrated NH3 solution
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Organic Observation and Deduction
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We are informed that B is a mixture of
hydrocarbons
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B has a CC functional group
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Cyclohexene
Only a very weak permanent dipole. Essentially
non-polar Molecules held together by only van der
Waals forces Memo Like dissolves Like.
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Alcohols, carboxylic acids, aldehydes and ketones
are miscible with water. In other words, they are
soluble in water. As the organic molecule
increases in length, masking of the functional
group by the hydrocarbon chain occurs. This
reduces solubility. H-bonding between alcohol
and water molecules is shown to the left.
d
d
d
d
Ethanal H-bonded to water
d
d
d
d
d
d
d
d
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The hydrogen/carbon ratio has an influence on how
cleanly a fuel burns. In general, the higher this
ratio, the cleaner the flame. Methane CH4 (Ratio
4/1 4.00) Ethane C2H6 (Ratio 6/2
3.00) Ethanol C2H5OH (Ratio 6/2
3.00) Propane C3H8 (Ratio 8/3 2.67) Large
alkane C30H62 (Ratio 62/30 2.07) Cyclohexane
C6H12 (Ration 12/6 2.00) Ethene C2H4
(Ration 4/2 2.00) Cyclohexene C6H10 (Ration
10/6 1.67) Methylbenzene C6H5CH3 (Ration
8/6 1.34) Benzene C6H6 (Ration 6/6
1.00)
Decreasing hydrogen/carbon ratio and more smoke
and soot

The black smoke and soot is caused by unburnt
carbon
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Increasing carbon/hydrogen ratio Increasing
smoke and soot
Ethanol Hexane Cyclohexane
Cyclohexene
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Ethene Bromine ? 1,2-dibromoethane
Br Br
Br2 ?
Cyclohexene Bromine ? 1,2-dibromocyclohexa
ne Clear Clear
Clear colourless brown
colourless Liquid liquid
liquid
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George Bush gets the smell of this chemical
Spirit burner
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A primary or secondary alcohol
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Alcohols, carboxylic acids, aldehydes and ketones
are miscible with water. In other words, they are
soluble in water. As the organic molecule
increases in length, masking of the functional
group by the hydrocarbon chain occurs. This
reduces solubility. H-bonding between alcohol
and water molecules is shown to the left.
d
d
d
d
Ethanal H-bonded to water
d
d
d
d
d
d
d
d
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Alcohols undergo combustion in air, burning with
a clean blue flame. The hydrogen/carbon ratio
has an influence on how cleanly a fuel burns. In
general, the higher this ratio, the cleaner the
flame. Methane CH4 (Ratio 4/1
4.00) Ethane C2H6 (Ratio 6/2 3.00) Ethanol
C2H5OH (Ratio 6/2 3.00) Propane C3H8
(Ratio 8/3 2.67) Large alkane C30H62 (Ratio
62/30 2.07) Cyclohexane C6H12 (Ration 12/6
2.00) Ethene C2H4 (Ration 4/2
2.00) Benzene C6H6 (Ration 6/6 1.00)
Decreasing hydrogen/carbon ratio and more smoke
and soot
The alcohol has its on inbuilt oxygen that
helps it to burn. ROH O2 CO2 H2O C2H5OH
3O2 ?2CO2 3H2O
Soot is unburnt carbon
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Potassium dichromate (acidified) is an oxidising
agent. Its formula is K2Cr2O7. The following
equation shows it accepting electrons - Cr2O72-
6 e- 2 Cr3 The role of the acid in mopping
up the oxygens is seen in this next
equation Cr2O72- 6 e- 14H 2 Cr3 7H2O
Cr(VI)
Cr (III)
At this stage (AS) one only need to learn the
following - H H
H I I
I H C
C O-H O ? H C C O
H2O I I
I I H H
H H
ethanal
Warm
ethanol
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Magnesium reacting with 2M Ethanoic Acid
2CH3COOH Mg ? Mg(CH3COO)2 H2
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Deduction
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B is a carboxylic acid RCOOH
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• Vinegar is a dilute
• aqueous solution of
• ethanoic acid.
• It is approximately
• 5 CH3COOH
•  

2CH3COOH Na2CO3 2CH3COO-Na CO2
H2O 2CH3COOH Mg Mg(CH3COO)2 H2
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THE TRIIODOMETHANE (IODOFORM) REACTION Gives a
positive result for two groupings -
H I WITH CH3 C - R
ALCOHOL I
OH where R H, CH3, C2H5,
etc --------------------------------------
O You will see that the alcohol above
is oxidised II to this
carbonyl structure in step 1 of 3 steps! WITH
CH3 C - R Ethanal and Methyl Ketones
therefore
also give ve tests where R H, CH3, C2H5,
etc
Iodoform molecule HCI3
Triiodomethane Yellow crystals with an antiseptic
smell.
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Which of these alcohols would give a ve iodoform
test?
H H H
I
I I H C OH H C C -
OH I I I
H H H H H
H H H H I
I I I I I
H C C C OH H C C C - H
I I I I I
I H H H H HO H
Ethanol is the only primary alcohol to give the
iodoform If "R"is a hydrocarbon group, then you
have a secondary alcohol. Lots of secondary
alcohols give this reaction, but those that do
all have a methyl group attached to the carbon
with the -OH group.
No tertiary alcohols can contain this group
because no tertiary alcohols can have a hydrogen
atom attached to the carbon with the -OH group.
No tertiary alcohols give the triiodomethane
(iodoform) reaction.
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Which of these carbonyls would give a ve
iodoform test?
H

I H C O
H C C O I
I I H
H H H H
H H I I
I I H C C
C O H C C C - H I I
I I II I
H H H H O H
Ethanal is the only aldehyde that gives a ve
iodoform test
Carbonyls not considered
All the 2-ones of the ketones will give a ve
iodoform test
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Version 1 of the Iodoform Test
Sample
Iodine in KI(aq)
Pure iodoform
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Version 1 of the Iodoform Test I2/NaOH To about
5 cm3 of a saturated solution of iodine in
potassium iodide in a test-tube add 5 drops
of alcohol.  
Add alcohol
Iodine in KI(aq)
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Version 1 of the Iodoform Test I2/NaOH To about
5 cm3 of a saturated solution of iodine in
potassium iodide in a test-tube add 5 drops
of ethanol.   Add sodium hydroxide solution
carefully until the colour has almost gone.
The cloudiness is a sign of precipitation.
Iodoform is a pale yellow solid with an
antiseptic smell
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Version 1 of the Iodoform Test I2/NaOH To about
5 cm3 of a saturated solution of iodine in
potassium iodide in a test-tube add 5 drops
of ethanol.   Add sodium hydroxide solution
carefully until the colour has almost gone. 3
REACTIONS or STEPS occur 1. OXIDATION Alcohol
O ? Carbonyl
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Version 1 of the Iodoform Test I2/NaOH To about
5 cm3 of a saturated solution of iodine in
potassium iodide in a test-tube add 5 drops
of ethanol.   Add sodium hydroxide solution
carefully until the colour has almost gone. 3
REACTIONS or STEPS occur 1. OXIDATION Alcohol
O ? Carbonyl 2. All 3 of the H atoms of the
methyl group are substituted by I atoms  
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Version 1 of the Iodoform Test I2/NaOH To about
5 cm3 of a saturated solution of iodine in
potassium iodide in a test-tube add 5 drops
of ethanol.   Add sodium hydroxide solution
carefully until the colour has almost gone. 3
REACTIONS or STEPS occur 1. OXIDATION Alcohol
O ? Carbonyl 2. All 3 of the H atoms of the
methyl group are substituted by I atoms 3. The
CI3COR formed then goes on to form HCI3 and
RCOO-Na Stand the test-tube in water at about
70 oC for two or three minutes, then remove and
allow to cool.  
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Version 1 of the Iodoform Test I2/NaOH To about
5 cm3 of a saturated solution of iodine in
potassium iodide in a test-tube add 5 drops
of ethanol. Add sodium hydroxide solution
carefully until the colour has almost gone. 3
REACTIONS or STEPS occur 1. OXIDATION Alcohol
O ? Carbonyl 2. All 3 of the H atoms of the
methyl group are substituted by I atoms 3. The
CI3COR formed then goes on to form HCI3 and
RCOO-Na Stand the test-tube in water at about
70 oC for two or three minutes, then remove and
allow to cool. Yellow crystals of iodoform
separate out on standing and the smell is like
that of antiseptic  
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Reactions taking place. H I CH3
C - OH I2 2OH- ? CH3 C O 2I-
2H2O I
I R
R CH3 C O
3I2 3OH- ? CI3 C O 3 I-
3H2O I
I R
R CI3 C O
NaOH ? R C O HCI3 I
I
R
O-Na

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Iodoform. Triiodomethane CHI3 A fine yellow
precipitate Antiseptic smell
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Version 2. KI/NaClO Using potassium iodide and
sodium chlorate(I) solutions Sodium chlorate(I)
is also known as sodium hypochlorite. Sodium
chlorate is the active ingredient in most
household bleaches. It is an oxidising agent and
will oxidise Iodide ions to Iodine (I2).
NaClO
Potassium Iodide Iodine
liberated
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Version 2. KI/NaClO Using potassium iodide and
sodium chlorate(I) solutions Sodium chlorate(I)
is also known as sodium hypochlorite. Potassium
iodide solution is added to a small amount of
organic sample,
Sample KI
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Version 2. KI/NaClO Using potassium iodide and
sodium chlorate(I) solutions Sodium chlorate(I)
is also known as sodium hypochlorite. Potassium
iodide solution is added to a small amount of
organic sample, This is followed by sodium
chlorate(I) soln.
Sample KI
NaClO
Iodoform
NaClO is alkaline (source of OH-) and oxidises I-
to I2 As a result KI/NaClO is equivalent to
using I2/NaOH
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Summary of Version 2. KI/NaClO Using potassium
iodide and sodium chlorate(I) solutions Sodium
chlorate(I) is also known as sodium
hypochlorite. Potassium iodide solution is added
to a small amount of organic sample, This is
followed by sodium chlorate(I) soln. NOTE THAT
THE NaClO OXIDISES IODIDE (I-) TO IODINE (I2) So
as well as any possible yellow precipitate, you
will also see the typical reddish-brown colour of
iodine solution being formed during the
reaction. Note also, that sodium chlorate(I)
solution is alkaline and contains a sufficietly
high OH- to carry out the second half of the
reaction. In effect you are making I2 in situ
so the tests are essentially the same. If no
precipitate is formed in the cold, it may be
necessary to warm the mixture very gently. Look
for the formation of a pale yellow precipitate
with antiseptic smell
Sample KI NaClO