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UNIVERSITI TEKNOLOGI MARA
FAKULTI KEJURUTERAAN KIMIA
INSTRUMENTAL LABORATORY
(CHE 515)
NAME : IHSAN SABRI BIN ROSLI
(2010562001)
MOHAMAD FADHLI BIN SAMSUDIN
(201077!")
MUHAMMAD JA#VIR B$ SULAIMAN
(20101"0!"") MUFIDAH BINTI MAHFU#
(201052!66!)
ROHA#IERAH BT CHE OMAR
(201012%%")
GROU& : EH221A
E'&ERIMENT :
DATE SUBMITTED : 21 NOVEMBER 2012
SEMESTER :
&ROGRAMMED CODE : EH 221
SUBMIT TO : DR KAMARIAH NOOR ISMAIL
T*+, A++-./*, M/3
(4)
M/3
A3*/.* 5
I*-.*- 5
T8,-9 5E;,
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$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $$$$
$$$$$$$$$$$$$$$$$$$$$$$$$$$
D/*, :
D/*, :
TABLE OF CONTENTS
TITLE PAGE
• Abstract 3
•
Introduction 4-5
• Theory 6-8
• Procedure 9
• Results !-"
• #iscussions 3-""
• $onclusions "3
• Re%erences "4
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ABSTRACT
The experiment was conducted to determine and draw the structure for unknown
compound A to I. The experiment for solid and liquid samples has almost the same
preparation’s way. The difference is only when for solid, it need to be dissolved first before
been transferred into N! tube. Then for both solid and liquid samples, three drops of a
deuterated solvent "usually chloroform# is added into the tube. Then, the tube is wrapped in a
tissue for paddin$ before inserted it into a holder of the centrifu$e. %pectral window is settin$
by modify the appearance and the spectrum is printed. &rom the analyses that have been
done, each unknown compound was determined. The I'(A) name for compound A is *, +
dimethylfuran-"* H #one, whereas compound is *methylcyclopentane/, -dione. Next,
compound ),0, and 1 are cyclohexane/,+dione, ethenyl "* E #but*enoate, and *
methylprop*enoate respectively. In addition, the analysis on compound & shows that it is
cyclobutylideneacetic acid, while compound 2 and compound 3 are "4methylfuran*yl#
methanol and hydroxy-methylcyclopent*en/one respectively. The last compound that
been analy5ed was compound I, that is hex*ynoic acid.
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INTRODUCTION
Nuclear ma$netic resonance spectroscopy "N!# is the technique to determinin$ the
structure of or$anic compounds. It is the only one for which a complete analysis and
interpretation of the entire spectrum is normally expected. 'nderstandin$ the physical
principle on which the methods are based is important to make sure it is success in usin$
N! as an analytical tool. The nuclei of many element isotopes have a characteristics spin.This includes /3 and /-) "but not /*)#. The N! behaviour of /3 and /-) nuclei has been
exploited by or$anic chemist since they provide valuable information that can be used to
deduce the structure of or$anic compounds. a$netic field will develop since a nucleus is a
char$ed particle in motion. /3 and /-) have nuclear spins of /6* and so they behave in a
similar fashion to a simple, tiny bar ma$net. 7hen a field is applied they line up parallel to
the applied field, either spin ali$ned or spin opposed but in the absence of a ma$netic field,
these are randomly oriented. Two schematic representations of these arran$ements are shown
below8
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There are two $eneral types of N! instrument which is continuous wave and
&ourier transform. 1arly experiments were conducted with continuous wave ").7.#
instruments, and in /9:; the first &ourier transform "&.T.# instruments is introduced.
)ontinuous wave N! spectrometer is in principle to optical spectrometer. The
sample is held in a stron$ ma$netic field, and the frequency of the source is slowly scanned.
The ma$nitude of the ener$y chan$es involved in N! spectroscopy is small. This means
that sensitivity is a ma because noise is random, it adds as the square root of the
number of spectra recorded. ecause of this, &ourier transform "&.T.# instruments became
available. In &TN!, all frequencies in a spectrum are irradiated simultaneously with a
radio frequency pulse. &ollowin$ the pulse, the nuclei return to thermal equilibrium. A time
domain emission si$nal is recorded by the instrument as the nuclei relax. A frequency domain
spectrum is obtained by &ourier transformation.
The N! instrumentation is shown as below>
&i$ure /8 basic arran$ement of an N! spectrometer
The sample is positions in the ma$netic field and excited via pulsations in the radio frequency
input circuit. The reali$ned ma$netic field induced a radio si$nal in the output circuit which
is used to $enerate the output si$nal. &ourier analysis of the complex output produces the
actual spectrum. The pulse is repeated as many times as necessary to be identified from the
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back$round noise. The pulse is actually a si$nal of frequency, F , is turned on and then off
a$ain very rapidly, then the result is an output consistin$ of many frequencies centred
about F with a bandwidth of /6t , where t is the duration of the pulse. This means that
radiation is produced of all frequencies in the ran$e F ? /6t . If t is very small, then a lar$e
ran$e of frequencies will be produced simultaneously, and all tar$et nuclei in a sample will be
excited.
THEORY
Basic principles of NMR
Nuclei with an odd mass or odd atomic number have @nuclear spin’. In the presence of
an applied external ma$netic field, /3 /-) nuclei exist in two nuclear spin states of different
ener$y. The difference in ener$y between the two spin states is dependent on the external
ma$netic field stren$th and is always small. . The spin states ener$y difference shown from
the equation below8
∆ E=γ h
2 π B
0
7here8 h (lanck’s constant; stren$th of external ma$netic field
$yroma$netic ratio, is the constant which is a property of the particularᵞ
nucleus
In the absence of a ma$netic field, these are randomly oriented but when a field is
applied they line up parallel to the applied field, either spin ali$ned or spin opposed. The
arran$ements of the spin orientation are shown below8
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&i$ure /8 %pin orientation
Chemical Shif
An N! spectrum is a plot of the radio frequency applied a$ainst absorption. A
si$nal in spectrum is known as resonance and frequency of si$nal is chemical shift. In other
word, chemical shift is defined as the frequency of the resonance expressed with reference to
a standard compound which is defined to be at ; ppm. The scale is in parts per million and it
is independent of the spectrometer frequency.
In an N! spectrum, a peak at a chemical shift of /; ppm is said to be downfield or
deshielded with respect to a peak at 4 ppm and the peak at 4 ppm is upfield or shielded with
respect to the peak at /; ppm.
There are several factors that may affect the chemical shift. This means that different types of proton will occur at different chemical shifts. The various factors include8
a! Elecrone"ai#i$ effec
The resonance position of protons bonded to carbon is shifted downfield by
electrone$ativity elements also bonded to the carbon. The electrone$ativity element
wtithdraws electron density from the carbon and its directly bonded protons which
diminished the ma$nitude of ma$netic field. Thus, the hi$her the electrone$ativity of
the directly bonded atom, the lar$er the downfield shift.
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&i$ure *8 The effect of electrone$ativity element on chemical shift
%! Ma"neic Anisorop$
Anisotropy means nonuniform. a$netic anisotropy means that there is a
nonuniform ma$netic field. 1lectron in systems such as aromatics, alkenes,
alkynes, carbonyl and others interact with the applied field which induced a ma$netic
field that causes the anisotropy. The effects of anisotropy are the nearby proton willexperience three fields which is the applied field, the shieldin$ field of the valence
electron and the field due to the system. Thus, shielded "smaller B# or deshielded
"lar$er B# and also ener$y required as well as the chan$es of frequency absorption are
depend on the position of the proton in the this third field. &i$ure - shows schematic
representation of anisotropy effect.
&i$ure - 8 Anisotropy effect
c! H$&ro"en Bon&in"
(rotons that are involved in hydro$en bondin$ usually C=3 and CN3 have a
lar$e ran$e of chemical shift values. The more hydro$en bondin$ there is, the more
protons are deshielded and the hi$her its chemical shift will be. 3owever, it is difficult
to predict since the amount of hydro$en bondin$ is susceptible to several factor such
as salvation, acidity, concentration and temperature.
'ROCEDURE
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Sample preparaion for li()i&
/. %ample liquid is transfer into the N! tube by usin$ lon$tipped (asteur pipette.
*. About three drop of a deuterated solvent "usually chloroform# is added into the tube.
-. Then, the tube is caped and spins in the hand centrifu$e to collect the residual sample
from the walls and collapse air pockets and bubbles.
+. The tube is wrapped in a tissue for paddin$ before inserted it into a holder of the
centrifu$e.
4. A thin strip of (arafilm around the seam between the tubes and its cap is wrapped to
prevent the sample from dryin$ out.
D. The tube is labeled.
Sample preparaion for soli&
/. A small amount of solid used is measured at the end of spatula. Then, sample solid is
put into a small test tube.
*. About three drop of a deuterated solvent "usually chloroform# is added into the tube.
Noted that the N! will N=T work if deuterated solvent is not usedE
-. The solid sample is dissolved and the solution is transferred into the N! tube.
+. The tube is caped and spins in the hand centrifu$e to collect the residual sample from
the walls and collapse air pockets and bubbles.
4. The tube is wrapped in a tissue for paddin$ before inserted it into a holder of the
centrifu$e.
D. A thin strip of (arafilm around the seam between the tubes and its cap is wrapped to
prevent the sample from dryin$ out.
:. The tube is labeled.
NMR Operain" 'roce&)re
/. Fock on the sample is obtained by ad
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*.* "-, s# • %at. alkanes , !3
/.+ "-,d# • %at. alkanes, !3
/-)
*;4.D • Getone, ! *)=
• Aldehydes, !)3=
/H9.H • 1ster, !)=*!’
/;-.4 • ! *))3*
H*.H • )=!
/D.9 • %at. alkane, !3
/D.- • %at. alkane, !3
/3
*.+ "+,t# • %at. alkane, !3
*.* "/,qrt# • %at. alkane, !3
/.4 "-,d# • %at. alkane, !3
/-)
/9+ • Getone, ! *)=
///.D • Aromatics
-;./ • ))=!
4.: • %at. alkane, !3
)
/3 *.: "H,t# • %at. alkane, !3
/-)
*;H.- • Getone, ! *)=
-:.D • ))=!
0
/3
:.+ "/,t# • Alkene, !)3)3!
:./ "/,qnt# • Alkene, !)3)3!
4.9 "/,d# • Alkene, !)3)3!
+.: "*,d# • 1ster, !)=*)3
/.9 "-,d# • %at. alkane, !3
/-)
/D-.* • 1ster, !)=*)3
/+: • ! *))3*
/+/.- • Alkene, !3))3!
/*/.: • Alkene, !3))3!
9:.+ • )=!
/H.* • %at. alkane
1 /3 :.- "/,t# • Alkene, !3))3!
D.; "*,s# • Alkene, !3))3!
+.: "*,d# • 1ster, !)=*)3
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*.; "-,s# • %at. alkane
/-)
/D+.- • 1ster, !)=*!’
/+/.4 • Alkene, !3))3!
/-4.4 • Alkene, !3))3! /*:.* • Alkene, ! *))3*
9:.H • )=!
/H.* • %at. alkane
&
/3
//.- "/,s# • )arboxylic acid, !)=*3
4.4 "/,s# • Alkene, !3))3!
*.4 "*,t# • %at. alkane
*.+ "*,t# • %at. alkane
*.; "*,qnt# • %at. alkane
/-)
/9- • )arboxylic acid, !)=*3
/;+.- • Alkene, ! *))3*
-*.- • %at. alkane
*/./ • %at. alkane
2
/3
D./ "/,d# • Alkene, !3))3!
4.9 "/,d# • Alkene, !3))3!
+.4 "*,s# • Alkene, ! *))3*
*.- "/,s# • Alcohol, !=3
*.- "-,s# • %at. alkane
/-
)
/4*.D • Alkene, !3))3!
/4*./ • Alkene, !3))3!
/;H.D • Alkene, !3))3!
/;D.- • Alkene, ! *))3*
4:./ • Alcohol, )=3
/-.4 • %at. alkane
3
/3
D.: "/,s# • Alkene, !3))3!
*.+ "*,t# • %at. alkane
*.- "*,t# • %at. alkane
*.; "-,s# • %at. alkane
/-) *;-.H • Getone, ! *)=
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/+9.4 • Alkene, ! *))3*
/+4.9 • Alkene, !3))3!
-*./ • )))
*:.- • %at. alkane/+.+ • %at. alkane
I
/3
9.: "/,s# • )arboxylic acid, !)=*3
*.- "*,t# • %at. alkane
/.D "*,sxt# • %at. alkane
/.; "-,t# • %at. alkane
/-)
/4H.4 • )arboxylic acid, !)=*3
9*.D • Alkyne
:*.9 • Alkyne
*/.; • %at. alkane
*;.: • %at. alkane
/-.+ • %at. alkane
DISCUSSION
7e are required to determine and draw a viable structure for compound A to I. &rom
the elemental composition and the relative molecular mass of the $iven compound, we can
determine the number of carbon, hydro$en and oxy$en present in the compounds. The
calculations involved are shown below8
1lemental composition8 ) C D+.-, 3 C *H.D, = C :.*
!elative molecular mass of the compounds8 //*./ $mol/
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!elative atomic mass8 ) C /*.;/$mol/, 3 C /.;/$mol/, = /D$mol/
Total /;;./
• Number of carbon, ) present is8
64.3
100.1×112.1gmol
−1=72.01 gmol
−1
72.01 gmol−1
12.01 gmol−1=6
• Number of hydro$en, 3 present is8
7.2
100.1×112.1gmol
−1=8.063gmol
−1
8.063gmol−1
1.01gmol−1 =7.98≈8
• Number of oxy$en, = present is8
28.6
100.1×112.1gmol
−1=32.028 gmol
−1
32.028 gmol−1
16 gmol−1
=2
3ence, from the above calculation, the molecular formula for the compounds to be
determined is )D3H=*
Compo)n& A the I'(A) name for this compound is *, +dimethylfuran-"* H #one
*
H NMR
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&
&
$'3
'
'
$'3
*+C NMR
&
&
$'3
'
'
$'3
The possible structure for compound A is *, +dimethylfuran-"* H #one. ased on the
data that had been $iven in experiment, we know that compound A has D peaks for ) and +
peak for 3. &or /-) N!, one of the chemical shifts that $iven in the experiment is *;4.D
ppm. &rom the table of chemical shift "refer above *;; ppm#, the functional $roup that
present is carbonyl $roup which is aldehyde and ketone. &or aldehyde, it must have /3 N!
chemical shift in the ran$e of 9 ppm to/; ppm whereas for ketone, it must have /3 N!
chemical shift in the ran$e of * ppm to - ppm. %o, it is proved that ketone present in the
compound A because the compound shows /3 N! chemical shift in the ran$e of * ppm to-
ppm. esides that, compound A must have double bond because it has chemical shift in the
ran$e of /;; ppm until *;; ppm, that is for /3 N!, the ran$e of chemical shift in the ran$e
of 4 ppm to D.4 ppm. ased on the structure above, it shown that ) that attached to the =
$roup ")=# will have hi$her chemical shift because the presence of stron$ electrone$ative
element that decrease the electron density.
Compo)n& B the I'(A) name for this compound is *methylcyclopentane/, -dione
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*H NMR
$'3
&&
*+C NMR
$'3
&&
&or compound , the possible structure is *methylcyclopentane/, -dione . &rom the
value $iven, we can identify the functional $roup for the compound. &rom the /-) N!
)hemical %hifts, it have δ
/9+.; "ketone 6 aldehyde#, ///.D "alkene#, -;./ "alkene# and 4.:
"alkanes#. Then, for /3 N! )hemical %hift, it have δ *.+ and *.* "ketone 6 alcohol#, and
/.4 "alkanes#. In this compound, it consists of several functional $roup include ketone,
cycloalkane. In /3 N!, for "+,t#, it has three peaks where it has two carbon "contain four
hydro$en# next to carbon that has two hydro$en. &or "/,qrt#, it has four peaks where it has
carbon "contain one hydro$en# next to carbon that has three hydro$en. Fastly "-, d#, it has
two peaks where it has carbon "contain three hydro$en# next to carbon that has one hydro$en.
-3 has lowest chemical shifts because in shielded area. /3 has hi$her chemical shift because
next to pi system and *3 has hi$hest chemical shift because hi$her in hydro$en bondin$ and
next to pi system.
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Compo)n& C the I'(A) name for this compound is cyclohexane/,+dione
*H NMR
&
&
*+C NMR
&
&
&or compound ), the possible structure is cyclohexane/, +dione. &rom the value
$iven, we can identify the functional $roup for the compound. &rom the /-) N! )hemical
%hifts, it have δ *;H.- "ketone 6 aldehyde# and -:.D "ketone 6 alkene 6 alkanes#. Then, for
/3 N! )hemical %hift, it has δ *.: "ketone#. &or this compound, it must have
symmetric compound which it has two peaks, so it has same value of chemical shift. In this
compound, the most accurate compound consists of ketone and cycloalkane functional $roup.
In /3 N!, for "H,t#, it means that each side contains two carbon where each carbon contain
two hydro$en.
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Compo)n& D the I'(A) name for this compound is ethenyl "* E #but*enoate
*H NMR
$'3 & $'"
&
*+C NMR
$'3 & $'"
&
&or compound 0, the possible structure is ethenyl "* E #but*enoate. &rom the value
$iven, we can identify the functional $roup for the compound. &rom the /-) N! )hemical
%hifts, it have δ /D-.* "ester 6 alkene#, /+:.; "alkene 6 aromatic#, /+/.- and /*/.: "alkene#,
9:.+ "aromatic# and /H.* "alkene 6 alkanes#. Then, for /3 N! )hemical %hift, it has δ
:.+ and :./ "aromatic#, 4.9 and +.: "alkene#, and /.9 "alcohol 6 alkene#. &or this compound, it
is a strai$ht chain carbon. It consists of some functional $roup include alkene, alkanes and
ester $roup. In /3 N!, it consist of "/,t#, means that it has three peak where it has carbon
"contain one hydro$en# next to carbon that has two hydro$en. &or "/,qnt#, it has five peaks
where it has carbon "contain one hydro$en# next to carbon that has four hydro$en. &or "/,d#,
it has two peaks where it has carbon "contain one hydro$en# next to carbon that has one
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hydro$en. &or "*,d#, it has two peaks where it has carbon "contain two hydro$en# next to
carbon that has one hydro$en. Then, for "-,d#, it has two peaks where it has carbon "contain
three hydro$en# next to carbon that has one hydro$en. "/3, t# has hi$hest chemical shift value
because located beside the electrone$ative element "=#.
Compo)n& E the I'(A) name for this compound is ethenyl *methylprop*enoate
*H NMR
$'" &$'3
&
$'"
*+C NMR
$'" &$'3
&
$'"
&or compound 1, the possible structure is ethenyl *methylprop*enoate. &rom the
value $iven, we can identify the functional $roup for the compound. &rom the /-) N!
)hemical %hifts, it have δ /D+.-, /+/.4, /-4.4, /*:.*, "alkene#, 9:.H "aromatic# and /H.*
"alkanes#. Then, for /3 N! )hemical %hift, it have δ :.- "aromatic#, D.; and +.:
"alkenes# and *.; "ketone 6 aldehyde 6 alkynes#. In this compound, it consists of some
functional $roup include ester $roup, alkenes, and cycloalkane. In/
3 N!, for "/,t#, it has
three peaks where it has carbon "contain one hydro$en# next to carbon that has two hydro$en.
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&or "*,s#, it has one peaks where it has carbon "contain two hydro$en# next to carbon that has
no hydro$en. Then, for "*,d#, it has two peaks where it has carbon "contain two hydro$en#
next to carbon that has one hydro$en. &or "-,s#, it has one peaks where it has carbon "contain
three hydro$en# next to carbon that has no hydro$en. It can see that "/3,t# has hi$hest
chemical shift value because of electrone$ative element "=# lower the electron density around
the 3 and ).
Compo)n& F the I'(A) name for this compound is cyclobutylideneacetic acid
*H NMR
&'
&
*+C NMR
&'
&
&or compound &, the possible structure is cyclobutylideneacetic acid. &rom the /-)
N! )hemical %hifts, it has δ /9-.; "ketone 6 aldehyde#, /;+.- "aromatic#, :;.; "ether
6alcohol#, -*.- "alkanes# and */./ "alkene 6 alkanes#. Then, for /3 N! )hemical %hift, it
have δ //.- "carboxylic acid#, 4.4 "alkenes#, *.4 and *.+ "ketone 6 alkynes#, and *.;
"alkenes 6 alkynes 6 ketone#. &or this compound, it is a strai$ht chain carbon. It consists of
several functional $roup include carboxyl acid, alkenes and alkanes. In /3 N!, for //.-
"/,s#, it refers to hydroxyl $roup. It has hi$hest chemical shift because located beside
electrone$ative element "=#. Then, for 4.4 "/,s#, it refers to the carbon that has one peak where a carbon "contain one hydro$en# next to carbon that has no hydro$en. &rom the
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compound, it can be seen at the alkenes $roup. &or "*,t#, it has three peaks where it has
carbon "contain two hydro$en# next to carbon that has two hydro$en and for "*,qnt#, it
has five peaks where it has carbon "contain two hydro$en# next to carbon that has four
hydro$en.
Compo)n& , the I'(A) name for this compound is "4methylfuran*yl# methanol
*H NMR
&
&'
$'3
*+C NMR
&
&'
$'3
&or compound 2, the possible structure is "4methylfuran*yl# methanol .&rom the
/3 N! and /-) N! data $iven, we know that this compound has D peaks for carbon, )
and 4 peaks for hydro$en, 3. The chemical shifts data for /-) N! that $iven to us is B
/4*.D, /4*./, /;H.D, /;D.-, 4:./ and /-.4. &rom the value $iven, we can identify the
functional $roup for the compound. &rom the data, B /4*.D and /4*./ must be nei$hborin$
carbon, and by referrin$ to the /-) N! chemical shifts table, the possible functional $roup
at these values is alkene "double bond#. B /;H.D and /;D also came from the nei$hborin$
carbon as their values are very close and the possible functional $roup at these values is also
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alkene. &or B4:./, the possible functional $roup is ether or alcohol. 7hereas, for B/-.4 is
saturated alkanes. The chemical shifts data for /3 N! $iven is B 4.9"/,d#, D./"/,d#, +.4"*,s#,
*.-"/,s# and *.-"-,s#. y referrin$ to the /3 N! chemical shifts table, the possible
functional $roup for B4.9"/,d# and D./"/,d# is alkene, for B+.4"*,s# is alcohol, esters or alkene.
Fastly, for B*.- is alcohol, ketones or alkyne "triple bond#. y comparin$ the information
$ained from both table, the similarities of the functional $roup they have is alkene and
alcohol, which means this compound must have this functional $roup. esides these *
functional $roups, we need one more functional $roup to complete the structure that has *
oxy$en atoms, so the most possible functional $roup is ether.
Compo)n& H the I'(A) name for this compound is *hydroxy-methylcyclopent*
en/one
*H NMR
&
&'
$'3
*+C NMR
&
&'
$'3
&or compound 3, the possible structure is *hydroxy-methylcyclopent*en/one.
&rom the /3 N! and /-) N! data $iven, we know that this compound has D peaks for
carbon, ) and + peaks for hydro$en, 3. The chemical shifts data for /-) N! that $iven to us
is B *;-.H, /+9.4, /+4.9, -*./, *:.- and /+.+. &rom the value $iven, we can identify the
functional $roup for the compound. &rom the data, B /+9.4 and /+4.9 must be nei$hborin$
carbon, and by referrin$ to the /-) N! chemical shifts table, the possible functional $roup
at these values is alkene "double bond system#. B-*./ and *:.- also came from the
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nei$hborin$ carbon as their values are very close and the possible functional $roup at these
values is ketones, saturated alkane or alkene. &or B*;-.H, the possible functional $roup is
ketones or aldehydes. 7hereas, for B/+.+ is saturated alkanes. The chemical shifts data for /3
N! $iven is B *.;"-,s#, *.+"*,t#, D.:"/,s# and *.-"*,t#. y referrin$ to the /3 N! chemical
shifts table, the possible functional $roup for B*.+, *.-, *.; is alcohol, ketones or alkene, for
BD.+ is alkene. y comparin$ the information $ained from both table, the similarities of the
functional $roup they have is alkene, saturated alkanes or ketones which means this
compound must have this functional $roup. esides these functional $roups, we need one
more functional $roup to complete the structure that has * oxy$en atoms, so the most possible
functional $roup is alcohol.
Compo)n& I the I'(A) name for this compound is hex*ynoic acid
*H NMR
&' $'3
&
*+C NMR
&' $'3
&
&rom the /3 N! and /-) N! data $iven, we know that this compound has D peaks
for carbon, ) and + peaks for hydro$en, 3. The chemical shifts data for /-) N! that $iven
to us is B /4H.4, 9*.D, :*.9, */.;, *;.: and /-.+. &rom the value $iven, we can identify the
functional $roup for the compound. &rom the data, B */.; and *;.: must be nei$hborin$
carbon, and by referrin$ to the /-) N! chemical shifts table, the possible functional $roup
at these values is alkene "double bond system# or saturated alkanes. &or B/4H.4, the possible
functional $roup is carboxylic acid. &or B9*.D is ether or alcohol. &or B:*.9, the possible
functional $roup is alcohol, ether, or alkyl "triple bond system#.7hereas, B/-.+ is alkene andsaturated alkane. The chemical shifts data for /3 N! $iven is B 9.:"/,s#, *.-"*,t#, /.D"*,sxt#
22
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and /.;"-,t#. y referrin$ to the /3 N! chemical shifts table, the possible functional $roup
for B*.- and /.D is alcohol, ketones and alkyl, for B9.: is ketones and for B/.; is saturated
alkane. y comparin$ the information $ained from both table, the similarities of the
functional $roup they have is saturated alkanes, alcohol, ketones and alkyl $roup. ut instead
of alcohol and ketones $roup, the carboxylic acid is the most accurate to construct this
structure.
CONCLUSION
The experiment were successfully conducted where the entire unknown compound
were determined. &rom the analysis and investi$ation that have been done, it shows that the
I'(A) name for compound A to I are *, +dimethylfuran-"* H #one, *methylcyclopentane
/,-dione, cyclohexane/,+dione, ethenyl "* E #but*enoate, *methylprop*enoate,
cyclobutylideneacetic acid, "4methylfuran*yl# methanol, *hydroxy-methylcyclopent*
en/one, and hex*ynoic acid respectively.
The compound can be determined by analyses the peak and the chemical shifts shown
in the N! spectrum. The splittin$ number of peak was known by lookin$ at the proton
presents near the sin$le carbon. Fet say there are two nei$hborin$ hydro$en present at the
carbon, the number of peaks will be triplet.
0eterminin$ the chemical shifts of every structure of compound part is very important
as to know the compound. )hemical shifts are the frequency of a resonance si$nal influenced
by if the compound is shielded or deshielded. The shieldin$ effect in 3N! is influenced
by factors such as inductive effects by electrone$ative $roups, ma$net anisotropy, and
hydro$en bondin$. &rom all these knowled$e, it explains why the compounds A to I were
determined to be the compound that have been mentioned above. In conclusion, the
experiment was successfully conducted and the ob
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REFERENCE
Boo-s
/. 2raham %., )rai$ ., "*;;:#.Organic chemistry." 9th edition#8 John 7iley K %on
*. Laboratory Manual CHE 515,&aculty of )hemical 1n$ineerin$
-. Notes NMR spectroscopy, &aculty of )hemical 1n$ineerin$
Inerne
/. Nuclear a$netic !esonance, Instrumentalion 0ecember //, *;/*. !etrieve from
http866teachin$.shu.ac.uk6hwb6chemistry6tutorials6molspec6nmr-.htm
*. Nuclear a$netic !esonance "N!# %pectroscopy. 0ecember /;, *;/*. !etrieve
from http866www.chem.ucal$ary.ca6courses6-4/6)arey4th6)h/-6ch/-nmr/.html
-. Nuclear a$netic !esonance %pectroscopy. 0ecember /;, *;/*. !etrieve from
http866www*.chemistry.msu.edu6faculty6reusch6LirtTxtJml6%pectrpy6nmr6nmr/.htm
http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/nmr3.htmhttp://www.chem.ucalgary.ca/courses/351/Carey5th/Ch13/ch13-nmr-1.htmlhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/nmr/nmr1.htmhttp://www.chem.ucalgary.ca/courses/351/Carey5th/Ch13/ch13-nmr-1.htmlhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/nmr/nmr1.htmhttp://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/nmr3.htm