HERMIN HARDYANTI UTAMI (111 304 0202)
CHEMISTRY DEPARTMENT
MATHEMATIC AND SCIENCE FACULTY
STATE UNIVERSITY OF MAKASSAR
2012
ORGANIC CHEMISTRY 2
COMPLETE REPORT
RATIFICATION PAGE
The complete report of Organic Chemistry II with the title of ―Synthetic
of Iodoform‖ which made by :
Name : Hermin Hardyanti Utami
ID : 111 304 0202
Group : VII
Class : ICP B
Have been checked by assistant and assistant coordinator. So, this report is
accepted.
Makkassar, December 2012
Assistant Coordinator
Fandi Ahmad, S.Pd.
Assistant
Sabaruddin
Known By,
Responsibility Lecturer
Dra. Hj. Ramdani, M.Si.
A. Tittle of Experiment
Synthetic of Iodoform
B. Purposes
At the last experiment, student can be understand about :
1. Work principles and crystallization technique of solid organic substance.
2. Haloform reaction.
3. Advantage of haloform reaction fo haloform synthetic and carboxylic acid
and to showing there are a methyl group and secondary alcohol.
C. Background
Functional group transformations allow the conversion of a functional group
to transformations an aldehyde or a ketone without affecting the carbon skeleton
of the molecule. Aldehydes can be synthesized by the oxidation of primary
alcohols, or by the reduction of esters, acid chlorides, or nitriles. Since nitriles can
be obtained from alkyl halides, this is a way of adding an aldehyde unit (CHO) to
an alkyl halide. Ketones can be synthesized by the oxidation of secondary
alcohols. Methyl ketones can be synthesized from terminal alkynes (Patrick,
2004:167).
According to Rasyid (2009:136-137) alcohol with one hydrogen that chain at
carbon as bring of carboxyl group can be oxidated become carbonyl compounds.
Primary alcohol that can be axidated with continue become to acid. Secondary
alcohol producing the ketone and tertiary alcohol is can’be oxidated:
H H OH
R C OH R C O R C O
H
Primary alcohol aldehyde acid
R’ R’
R C OH [O]
R C O
H
Secondary alcohol ketone
Most acidic protons are attached to heteroatoms like halogen, oxygen, and
nitrogen. Protons attached to carbon are not normally acidic but there are
exceptions. One such exception occurs with aldehydes or ketones when there is a
CHR2, CH2R or CH3 group next to the carbonyl group. The protons indicated are
acidic and are attached to the a (alpha) carbon. They are therefore called as
protons. A lone pair on the hydroxide oxygen forms a new bond to an proton.
Simultaneously, the C–H bond breaks. Both electrons of that bond end up on the
carbon atom and give it a lone pair of electrons and a negative charge (a
carbanion). However, carbanions are generally very reactive, unstable species that
are not easily formed. Therefore, some form of stabilisation is involved here
(Coulter, 2009:102 ).
According to Fescenden (1986:35-36) chaining of carbon hydrogen bonding
ussualy stable, nonpolar, and non acodic properties. But with there is one group
carbonyl happening alpha hidroden that has acid properties. If one hydrogen has
alpha position to two carbonyl group so its hydrogen is so acid so can be formed a
salt with sythetic of it compound with one alcogxyde. As for the reaction :
O O O
CH3 CCH3 CH3CCH2COCH2CH3
Acetone ethyl aceto acetic
pKa = 20 pKa = 11
According to Tim Doesen (2012:1) compound that contain the CH3CO group
or that produce this group, if occur the oxidation in one condition o one
experiment. Example acetaldehyde CH3CHO from ethanol CH3CH2OH react with
sodium hypochlorous forming iodofom.
O O
R C CH3 + 3NaOH R C CI3 + 3NaOH
O O
R C CH3 + NaOH R C O-Na
+ + CHI3
OH O
R C CH3 + NaOI R C CH3 + NaI + H2O
OH O
R C CH3 + 3NaOI R C O-Na
+ + CHI3 + H2O
According to Miller (1999:300) the radical initially produced by homolytic
decomposition of a dialkyl peroxide can undergo further scission. The rate of
scission depends on the temperature and the stability of the resulting radical. For
example, butoxy radicals decompose on heating to methyl radicals and acetone:
BuO• —> •CH3 +CH3COCH3
D. Chemical and Equipment
1. Chemical
a. Potassium Iodide (KI) solution
b. Acetone (CH3COCH3) solution
c. Sodium hypochlorous (NaOCl) solution
d. 1-propanol (CH3CH2CH2OH) solution
e. Dioxane (C4H8O2) solution
f. Sodium hydroxide (NaOH) 1% solution
g. Potassium Iodine Iodide (KI.I2) solution
h. Ethyl acetoacetic (CH3COH2COOC2H5) solution
i. Acetophenon (C6H5COCH3) solution
j. Aquadest (H2O)
k. Whatmann filter paper
l. Aluminium foil
m. Capillary tube
n. Matches
2. Equipment
a. Dropping pipette (10 pieces)
b. Electronical scale ( 1 piece )
c. Graduated cylinder 100 ml ( 1 piece )
d. Graduated cylinder 10 ml ( 2 pieces)
e. Stir bar ( 2 pieces)
f. Buchner funnel ( 1 piece )
g. Beaker glass 250 ml ( 2 pieces)
h. Oven ( 1 piece )
i. Erlenmeyer 500 ml ( 1 piece )
j. Thiele equipment ( 1 piece )
k. Bunsen burner ( 1 piece )
l. Test tube ( 6 pieces)
m. Test tube rack ( 1 piece )
n. Triangle ( 1 piece )
o. Asbestos gauze ( 1 piece )
p. Spray flask ( 1 piece )
q. Soft cloth and rough cloth ( 2 pieces)
E. Work Procedure
1. Synthetic of Iodoform
a. 6 g of KI was balance by electronical balance.
b. It was added by water 100 ml in beaker glass.
c. It was added by acetone 2 ml and closed by aluminium foil on acidic
cupboard.
d. It was added ba NaOCl 65 ml with slowly and stirred until the mixture
change color to greenish yellow solution.
e. In was let stand until 10th
minutes in acidic cupboard.
f. It was filtered by Buchner funnel.
g. It was rinsed by water until the precipitate in beaker glass is loss.
h. The residue (crystal) was took and it was dried in the oven.
i. It was balanced by electronical scale.
j. It was filled into capillary tube and the melting point was determined.
2. Iodoform Testing
a. 1-propanol test
5 drops of 1-propanol was dropped into test tube.
It was added by dioxane 5 ml.
It was added by 1 ml of NaOH 1% until form 2 layers.
It was added by some drops of KI.I2 until the color change to
brown.
It was added by 20th
drops of NaOH.
It was divided into two observation, namely heated and let stand.
b. Ethyl acetoacetic test
5 drops of ethyl acetoacetic was dropped into test tube.
It was added by dioxane 5 ml.
It was added by 1 ml of NaOH 1% until form 2 layers.
It was added by some drops of KI.I2 until the color change to
brown.
It was added by 20th
drops of NaOH.
It was divided into two observation, namely heated and let stand.
c. Acetophenon test
5 drops of acetophenon was dropped into test tube.
It was added by dioxane 5 ml.
It was added by 1 ml of NaOH 1% until form 2 layers.
It was added by some drops of KI.I2 until the color change to
brown.
It was added by 20th
drops of NaOH.
It was divided into two observation, namely heated and let stand.
F. Observation Result
No. Observation Result
1.
2.
3.
4.
5.
Synthetic of Iodoform :
KI(s) was balanced on electronical
balance.
It was added by water in beaker
glass.
It was added by acetone
It was added by NaOCl 5% drop
by drop and stirred.
It was let stand until 10th
minutes
- 6 g of KI
- 100 ml of water
- 2 ml of cetone
- 6 ml of NaOCl greenish
yellow mixture.
- Separated :
6.
7.
8.
9.
in cupboard.
It was filtrered by Buchner funnel.
The crystal dried in oven.
It was balanced on electronical
balance.
It was filled in capillary tube and
determined it melting point.
Transparent solution
Greenish yellow precipitate.
- Get yellow crystal.
- Pure crystal dry.
- 1.1 g
- 1200-124
0C
1.
2.
Iodoform Test :
1-propanol test
a. 1-propanol
b. It was added by dioxane
c. It was added by NaOH 1%
d. It was added by KI.I2
e. It was added by NaOH
f. It was divided into 2 part:
Heated
Let stand
Ethyl acetoacetic test
a. Ethyl acetoacetic
b. It was added by dioxane
- 5 drops of 1-propanol
- 5 ml of dioxane
- 1 ml of NaOH, form 2 layers:
Transparent solution
Turbidity solution
- Brown solution
- 20 drops of NaOH
- Transparent solution.
- Yellow dark solution.
- 5 drops of ethyl acetoacetic
- 5 ml of dioxane
3.
1.
c. It was added by NaOH 1%
d. It was added by KI.I2
e. It was added by NaOH
f. It was divided into 2 part:
Heated
Let stand
Acetophenon test
a. acetophenon
b. It was added by dioxane
c. It was added by NaOH 1%
d. It was added by KI.I2
e. It was added by NaOH
f. It was divided into 2 part:
Heated
Let stand
- 1 ml of NaOH, form 2 part:
Bubbles
Turbidity solution
- Brown solution
- 20 drops of NaOH
- Yellow solution.
- Yellow solution.
- 5 drops of acetophenon
- 5 ml of dioxane
- 1 ml of NaOH, form 2 layers:
Turbidity solution
Colorless solution
- Brown solution
- 20 drops of NaOH
- Yellow solution and white
precipitate.
- Yellow solution.
G. Data analysis
Known :
ρ acetone = 0.782 g/ml
Mw. Acetone = 58 g/mol
V. acetone = 2 ml
Mw. Iodoform = 373.79 g/mol
M. iodoform(obs) = 1.10 g
Problem :
Rendement……..?
Solution :
Mass acetone = (ρ x v) acetone
= 0,782 g/ml x 2 ml = 1.584 g
= 0.027 mol
Mol acetone = mol iodoform
Mass iodoform(the)= ( n x mw ) iodoform
= 0.027 mol x 393.79 g/mol = 10.63 g
= 10.34%
H. Discussion
a. Synthetic of Iodoform
In this experiment, that had been done of balancing the KI as the main
compound in iodoform synthetic that dissolved with H2O and added by acetone
and NaOCl with slowly. As the unction of KI to react with NaOClo to form KCl
and NaOI. As the function of NaOCl addition with slowly is to react with
perfectly.
KI + NaOCl KCl + NaOI
NaOI will be decompose in solution become to NaO+ and I
-. Addition of acetone
as function to reacting the ion I- with NaO
+ and H
+ at the solution is let stand. So
that formin CH3COONa solution (carboxylic salt) and CHI3 (iodoform) from
acetone compound. As the function of let stand on acidic cupboard is the
precipitate that precipitated in bottom of beaker glass is not evaporating to around
surrounding, as the reaction is :
1st
step
O O
CH3 C CH3 + KI + NaOCl CH3 C CH3 + KCl + NaOI
2nd
step
O O
CH3 C CH3 + NaOI CH3 C CH2I + NaOH
3rd
step
O O
CH3 C CH2I + NaOI CH3 C CHI2 + NaOH
4th
step
O O
CH3 C CHI2 + NaOI CH3 C CI3 + NaOH
5th
step
O O
CH3 C CI3 + NaOH CH3 C O-Na
+ + CHI3
Picture 1: iodoform solution that acidified in acidic cupboard
Solution with it precipitate filtered by Buchner funnel so that the crystal that
obtained more fast dry. Crystal then put into the oven so that the drying process
more fast and evaporating the left of water that contain in crystal. As the color of
iodoform crystal is yellow crystal (iodoform crystal). Crystal that obtained is nor
recrystallization because it has form an iodoform crystal reviewed from it color,
smell, and it melting point and the crystal weight is 1.10 grams and the rendement
is 10.34%, so if recrystallized again can be produce the less weight than 1.10
grams. As the melting point of iodoform crystal is 1200-124
0C. This is same with
theory that the melting point of iodoform is 1230C. least of rendement that
obtained because there is iodoform in filtrate and much of iodoorm that
evaporating at let stand in acidic cupboard and also the stirring way of solution at
the NaOCl addition.
Picture 2 : iodoform crystal
b. Iodorm testing
I. Closing
1. Conclusion
Based on the experiment that hade been done, so it may be conclude that :
a. The principle of isolation organic compound (EPMS) of phenyl
propanoid from galingale are maceration, separation, extraction, and
purification.
b. The isolation technique in galingale can be obtained with maceration
way, separation, and purification.
2. Suggestion
a. Student can be more careful in doing the experiment.
b. Student can me more compact with friends group.
BIBLIOGRAPHY
Barus, Rosbina. 2009. Amidasi Etil Para Metoksi Sinamat yang Diisolasi dari
Kencur. Tesis, Universitas Sumatra Utara.
Ekowati, Juni, dkk. 2010. Pengaruh Katalis pada Sintesis Asam Orto
Metoksisinamat dengan Materi Awal Arto Metoksi Benzaldehida dan Uji
Aktivitas Analgesiknya. Jurnal, Majalah Farmasi Airlangga, Vol.8, No.2,
Oktober 2010.
Fessenden, Ralph J. dan Joan S. Fessenden. 1982. Kimia Organik Edisi Ketiga.
Jakarta : Erlangga.
Kusumawati Idha dan Helmi Yusuf. 2011. Phospolipid Complex As A Carrier Of
Camempferia Galangal Rhizome Extract To Improve Its Analgecic
Activity. International Journal Of Pharmacy And Pharmaceutical Sciences
ISSN- 0975-1491 Vol 3, Suppl 1, 2011.
Nurdiansyah dan Abdi Radha. Efek Lama Maserasi Bubuk Kopra Terhadap
Rendemen, Densitas, dan Bilangan Asam Biodiesel yang Dihasilkan
dengan Metode Transesterifikasi In Situ. Penelitian Poltek Negeri
Pontianak.
Rasyid, Muhaidah. 2009. Kimia Organik I. Makasar : Badan Penerbit UNM.
Suyatno dan Emil Kurniati. Sintesis Senyawa Tabir Surya Eugenil Para Metoksi
Sinamat Melalui Reaksi Esterifikasi. Penelititan Unversitas Negeri
Surabaya.
Tim Dosen. 2012. Penuntun Praktikum Kimia Organik II. Makassar :Jurusan
Kimia FMIPA UNM.
Question Answer
1. EPMS compound classified as based on its structure of benzene group and
alkyl which is the ester group derivate alkaloid to synthesis of amino acid and
binds cinnamic acid pathway.
2. Isolation of EPMS be masked and dried to ethyl para metoxhy cinnamic
compound contained in galingale can extracted well. In addition to the
teaching process can absorb the desired copound and galingale must be dry to
eliminate or reduce the water molecules contain in galingale.
3. Distillation temperature should be maintained arrounf to 800C because who
want to spent on distillation process is solvent ethanol. Where ethanol boils at
700-80
0C. if temperature less than or more than it temperature will be forming
or that out instead of ethanol but others compound.