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SPEKTROMETRI MASSASPEKTROMETRI MASSAmembom molekul-molekul fasa uap dengan membom molekul-molekul fasa uap dengan
energi energi elektron yang sangat tinggi elektron yang sangat tinggi ionion
ion-ion dipisahkan menurut perbandingan ion-ion dipisahkan menurut perbandingan massa massa terhadap muatan (mz atau me)terhadap muatan (mz atau me)
ion-ion yang mempunyai perbandingan mz ion-ion yang mempunyai perbandingan mz khusus khusus dideteksi oleh suatu alat yang dapat dideteksi oleh suatu alat yang dapat menghitung menghitung ΣΣ ion-ion yang menabraknya dibaca ion-ion yang menabraknya dibaca oleh detektor dan oleh detektor dan outputnya diperkuat oampli outputnya diperkuat oampli dan dibaca orekorderdan dibaca orekorder
spektrum massaspektrum massa
grafik yang memplot limpahan ion relatif (ordinat) grafik yang memplot limpahan ion relatif (ordinat) dan massa (absis tepatnya mz dan massa (absis tepatnya mz perbandingan massa dengan perbandingan massa dengan ΣΣ muatan suatu ion) muatan suatu ion) Gambar 1(benzamida ) Gambar 1(benzamida )
Puncak ion + pada mz 121 molekul yg Puncak ion + pada mz 121 molekul yg utuh(M)-1 elektron yang dilepaskan oleh utuh(M)-1 elektron yang dilepaskan oleh tabrakan sinar tabrakan sinar ion molekulion molekul M M++ seri seri fragmen ionfragmen ion ReaksiReaksi
C6H5-C-NH2
O
+ e C6H5-C-NH2
O -NH2
C6H5-C =O
-CO
mz 105
mz 77
M mz 121
-C6H5
CNH2
O
mz 44
SUSUNAN ALAT SPEKTROMETER MASSASUSUNAN ALAT SPEKTROMETER MASSA
inlet system
ion source
mass analyzer
detector
signal processor
readout
vacuum system
sample
10-6 - 10-8 torr
Yang paling tinggi intensitas puncak dasar Yang paling tinggi intensitas puncak dasar (base peak)(base peak)
100 (tinggi x factor sensitivitas) dari 100 (tinggi x factor sensitivitas) dari puncak2 lain termasuk puncak ion puncak2 lain termasuk puncak ion molekul sebagai base peakmolekul sebagai base peak
Ion molekul MIon molekul M++ sinar electron sinar electron mengubah molekul2 menjadi ion positipmengubah molekul2 menjadi ion positip
Nilai mz ion hanya kehilangan satu Nilai mz ion hanya kehilangan satu electron berat molekuler dari molekul electron berat molekuler dari molekul yang asliyang asli
Untuk mengidentifikasi puncak ion Untuk mengidentifikasi puncak ion molekul dalam spektrum molekul dalam spektrum
mampu menggunakan mampu menggunakan spektrumspektrum untuk untuk menentukan berat molekul suatu menentukan berat molekul suatu komponen yang tidak diketahui komponen yang tidak diketahui
isotop-isotop tampak dalam spektrumisotop-isotop tampak dalam spektrum
Molekul-molekul di alam tidak hanya sebagai spesies Molekul-molekul di alam tidak hanya sebagai spesies isotop murni semua atom mempunyai isotop-isotop isotop murni semua atom mempunyai isotop-isotop yang menjadi sifat khas kelimpahan alamyang menjadi sifat khas kelimpahan alam
H H 11H ~ 002 H sebagai isotop H ~ 002 H sebagai isotop 22HH C C 1212C ~ 11 C sebagai isotop C ~ 11 C sebagai isotop 1313CC Isotop 1 atau 2 unit massa gt massa atom Isotop 1 atau 2 unit massa gt massa atom
normalnormal M+ M+1 dan M+2M+ M+1 dan M+2 Pada gambar 1 Puncak M+1 ~ 8 dari intensitas Pada gambar 1 Puncak M+1 ~ 8 dari intensitas
puncak ion molekul puncak ion molekul M+1 M+1 1313C C 22H H 1515N dan N dan 1717OO Tabel limpahan relatip isotop terhadap isotop yang Tabel limpahan relatip isotop terhadap isotop yang
paling melimpahpaling melimpah M+2 M+2 1818O kelimpahannya sangat rendah tidak O kelimpahannya sangat rendah tidak
terdeteksiterdeteksi Jika hanya C H N O F P dan IJika hanya C H N O F P dan I
(M+1) asymp 11 x sum atom C + 036 x sum atom N(M+1) asymp 11 x sum atom C + 036 x sum atom N (11 x sum atom C)(11 x sum atom C)22
(M+2) asymp + 020 x sum (M+2) asymp + 020 x sum atom O atom O 200 200
Jika puncak2 isotop ini cukup intens diukur Jika puncak2 isotop ini cukup intens diukur secara tepat perhitungan di atas dapat secara tepat perhitungan di atas dapat berguna untuk menentukan rumus molekulberguna untuk menentukan rumus molekul
Jika S dan Si ada M+2 lebih intensJika S dan Si ada M+2 lebih intens 1 atom S 1 atom S 3434S ~ 44 pada puncak M+2S ~ 44 pada puncak M+2 1 atom Si 1 atom Si 3030Si ~ 335 pada puncak Si ~ 335 pada puncak
M+2M+2
Mengenal puncak ion molekul (M)Mengenal puncak ion molekul (M) EI (Electron Impact)EI (Electron Impact) Problem Puncak sangat lemah atau tidak Problem Puncak sangat lemah atau tidak
nampaknampak Puncak ion molekul dan bukan Puncak ion molekul dan bukan
puncak fragmentasi atau tidak murnipuncak fragmentasi atau tidak murni CI (Chemical Ionization) puncak yang CI (Chemical Ionization) puncak yang
kuat pada M+1 dan sedikit fragmentasikuat pada M+1 dan sedikit fragmentasi ldquoNitrogen rulerdquo molekul dengan ldquoNitrogen rulerdquo molekul dengan
berat molekul genap tidak mengandung N berat molekul genap tidak mengandung N atau nitrogen genap dan berat molekul atau nitrogen genap dan berat molekul ganjil mempunyai suatu atom nitrogen ganjil mempunyai suatu atom nitrogen ganjilganjil
C H O N S dan X C H O N S dan X PP B Si Ar dan B Si Ar dan alkali tanahalkali tanah
Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul
Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul
IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni
Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens
Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya
AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18
(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul
Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)
(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen
(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin
(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul
PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation
index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains
= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1
Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5
Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1
wherewhereα is H D or halogen (ie any α is H D or halogen (ie any
monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen
atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index
is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter
the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides
molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly
However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass
A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass
THE RULE OF THIRTEENTHE RULE OF THIRTEEN
a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313
The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r
the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2
Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method
if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula
For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
Puncak ion + pada mz 121 molekul yg Puncak ion + pada mz 121 molekul yg utuh(M)-1 elektron yang dilepaskan oleh utuh(M)-1 elektron yang dilepaskan oleh tabrakan sinar tabrakan sinar ion molekulion molekul M M++ seri seri fragmen ionfragmen ion ReaksiReaksi
C6H5-C-NH2
O
+ e C6H5-C-NH2
O -NH2
C6H5-C =O
-CO
mz 105
mz 77
M mz 121
-C6H5
CNH2
O
mz 44
SUSUNAN ALAT SPEKTROMETER MASSASUSUNAN ALAT SPEKTROMETER MASSA
inlet system
ion source
mass analyzer
detector
signal processor
readout
vacuum system
sample
10-6 - 10-8 torr
Yang paling tinggi intensitas puncak dasar Yang paling tinggi intensitas puncak dasar (base peak)(base peak)
100 (tinggi x factor sensitivitas) dari 100 (tinggi x factor sensitivitas) dari puncak2 lain termasuk puncak ion puncak2 lain termasuk puncak ion molekul sebagai base peakmolekul sebagai base peak
Ion molekul MIon molekul M++ sinar electron sinar electron mengubah molekul2 menjadi ion positipmengubah molekul2 menjadi ion positip
Nilai mz ion hanya kehilangan satu Nilai mz ion hanya kehilangan satu electron berat molekuler dari molekul electron berat molekuler dari molekul yang asliyang asli
Untuk mengidentifikasi puncak ion Untuk mengidentifikasi puncak ion molekul dalam spektrum molekul dalam spektrum
mampu menggunakan mampu menggunakan spektrumspektrum untuk untuk menentukan berat molekul suatu menentukan berat molekul suatu komponen yang tidak diketahui komponen yang tidak diketahui
isotop-isotop tampak dalam spektrumisotop-isotop tampak dalam spektrum
Molekul-molekul di alam tidak hanya sebagai spesies Molekul-molekul di alam tidak hanya sebagai spesies isotop murni semua atom mempunyai isotop-isotop isotop murni semua atom mempunyai isotop-isotop yang menjadi sifat khas kelimpahan alamyang menjadi sifat khas kelimpahan alam
H H 11H ~ 002 H sebagai isotop H ~ 002 H sebagai isotop 22HH C C 1212C ~ 11 C sebagai isotop C ~ 11 C sebagai isotop 1313CC Isotop 1 atau 2 unit massa gt massa atom Isotop 1 atau 2 unit massa gt massa atom
normalnormal M+ M+1 dan M+2M+ M+1 dan M+2 Pada gambar 1 Puncak M+1 ~ 8 dari intensitas Pada gambar 1 Puncak M+1 ~ 8 dari intensitas
puncak ion molekul puncak ion molekul M+1 M+1 1313C C 22H H 1515N dan N dan 1717OO Tabel limpahan relatip isotop terhadap isotop yang Tabel limpahan relatip isotop terhadap isotop yang
paling melimpahpaling melimpah M+2 M+2 1818O kelimpahannya sangat rendah tidak O kelimpahannya sangat rendah tidak
terdeteksiterdeteksi Jika hanya C H N O F P dan IJika hanya C H N O F P dan I
(M+1) asymp 11 x sum atom C + 036 x sum atom N(M+1) asymp 11 x sum atom C + 036 x sum atom N (11 x sum atom C)(11 x sum atom C)22
(M+2) asymp + 020 x sum (M+2) asymp + 020 x sum atom O atom O 200 200
Jika puncak2 isotop ini cukup intens diukur Jika puncak2 isotop ini cukup intens diukur secara tepat perhitungan di atas dapat secara tepat perhitungan di atas dapat berguna untuk menentukan rumus molekulberguna untuk menentukan rumus molekul
Jika S dan Si ada M+2 lebih intensJika S dan Si ada M+2 lebih intens 1 atom S 1 atom S 3434S ~ 44 pada puncak M+2S ~ 44 pada puncak M+2 1 atom Si 1 atom Si 3030Si ~ 335 pada puncak Si ~ 335 pada puncak
M+2M+2
Mengenal puncak ion molekul (M)Mengenal puncak ion molekul (M) EI (Electron Impact)EI (Electron Impact) Problem Puncak sangat lemah atau tidak Problem Puncak sangat lemah atau tidak
nampaknampak Puncak ion molekul dan bukan Puncak ion molekul dan bukan
puncak fragmentasi atau tidak murnipuncak fragmentasi atau tidak murni CI (Chemical Ionization) puncak yang CI (Chemical Ionization) puncak yang
kuat pada M+1 dan sedikit fragmentasikuat pada M+1 dan sedikit fragmentasi ldquoNitrogen rulerdquo molekul dengan ldquoNitrogen rulerdquo molekul dengan
berat molekul genap tidak mengandung N berat molekul genap tidak mengandung N atau nitrogen genap dan berat molekul atau nitrogen genap dan berat molekul ganjil mempunyai suatu atom nitrogen ganjil mempunyai suatu atom nitrogen ganjilganjil
C H O N S dan X C H O N S dan X PP B Si Ar dan B Si Ar dan alkali tanahalkali tanah
Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul
Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul
IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni
Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens
Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya
AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18
(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul
Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)
(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen
(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin
(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul
PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation
index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains
= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1
Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5
Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1
wherewhereα is H D or halogen (ie any α is H D or halogen (ie any
monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen
atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index
is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter
the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides
molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly
However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass
A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass
THE RULE OF THIRTEENTHE RULE OF THIRTEEN
a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313
The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r
the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2
Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method
if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula
For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
SUSUNAN ALAT SPEKTROMETER MASSASUSUNAN ALAT SPEKTROMETER MASSA
inlet system
ion source
mass analyzer
detector
signal processor
readout
vacuum system
sample
10-6 - 10-8 torr
Yang paling tinggi intensitas puncak dasar Yang paling tinggi intensitas puncak dasar (base peak)(base peak)
100 (tinggi x factor sensitivitas) dari 100 (tinggi x factor sensitivitas) dari puncak2 lain termasuk puncak ion puncak2 lain termasuk puncak ion molekul sebagai base peakmolekul sebagai base peak
Ion molekul MIon molekul M++ sinar electron sinar electron mengubah molekul2 menjadi ion positipmengubah molekul2 menjadi ion positip
Nilai mz ion hanya kehilangan satu Nilai mz ion hanya kehilangan satu electron berat molekuler dari molekul electron berat molekuler dari molekul yang asliyang asli
Untuk mengidentifikasi puncak ion Untuk mengidentifikasi puncak ion molekul dalam spektrum molekul dalam spektrum
mampu menggunakan mampu menggunakan spektrumspektrum untuk untuk menentukan berat molekul suatu menentukan berat molekul suatu komponen yang tidak diketahui komponen yang tidak diketahui
isotop-isotop tampak dalam spektrumisotop-isotop tampak dalam spektrum
Molekul-molekul di alam tidak hanya sebagai spesies Molekul-molekul di alam tidak hanya sebagai spesies isotop murni semua atom mempunyai isotop-isotop isotop murni semua atom mempunyai isotop-isotop yang menjadi sifat khas kelimpahan alamyang menjadi sifat khas kelimpahan alam
H H 11H ~ 002 H sebagai isotop H ~ 002 H sebagai isotop 22HH C C 1212C ~ 11 C sebagai isotop C ~ 11 C sebagai isotop 1313CC Isotop 1 atau 2 unit massa gt massa atom Isotop 1 atau 2 unit massa gt massa atom
normalnormal M+ M+1 dan M+2M+ M+1 dan M+2 Pada gambar 1 Puncak M+1 ~ 8 dari intensitas Pada gambar 1 Puncak M+1 ~ 8 dari intensitas
puncak ion molekul puncak ion molekul M+1 M+1 1313C C 22H H 1515N dan N dan 1717OO Tabel limpahan relatip isotop terhadap isotop yang Tabel limpahan relatip isotop terhadap isotop yang
paling melimpahpaling melimpah M+2 M+2 1818O kelimpahannya sangat rendah tidak O kelimpahannya sangat rendah tidak
terdeteksiterdeteksi Jika hanya C H N O F P dan IJika hanya C H N O F P dan I
(M+1) asymp 11 x sum atom C + 036 x sum atom N(M+1) asymp 11 x sum atom C + 036 x sum atom N (11 x sum atom C)(11 x sum atom C)22
(M+2) asymp + 020 x sum (M+2) asymp + 020 x sum atom O atom O 200 200
Jika puncak2 isotop ini cukup intens diukur Jika puncak2 isotop ini cukup intens diukur secara tepat perhitungan di atas dapat secara tepat perhitungan di atas dapat berguna untuk menentukan rumus molekulberguna untuk menentukan rumus molekul
Jika S dan Si ada M+2 lebih intensJika S dan Si ada M+2 lebih intens 1 atom S 1 atom S 3434S ~ 44 pada puncak M+2S ~ 44 pada puncak M+2 1 atom Si 1 atom Si 3030Si ~ 335 pada puncak Si ~ 335 pada puncak
M+2M+2
Mengenal puncak ion molekul (M)Mengenal puncak ion molekul (M) EI (Electron Impact)EI (Electron Impact) Problem Puncak sangat lemah atau tidak Problem Puncak sangat lemah atau tidak
nampaknampak Puncak ion molekul dan bukan Puncak ion molekul dan bukan
puncak fragmentasi atau tidak murnipuncak fragmentasi atau tidak murni CI (Chemical Ionization) puncak yang CI (Chemical Ionization) puncak yang
kuat pada M+1 dan sedikit fragmentasikuat pada M+1 dan sedikit fragmentasi ldquoNitrogen rulerdquo molekul dengan ldquoNitrogen rulerdquo molekul dengan
berat molekul genap tidak mengandung N berat molekul genap tidak mengandung N atau nitrogen genap dan berat molekul atau nitrogen genap dan berat molekul ganjil mempunyai suatu atom nitrogen ganjil mempunyai suatu atom nitrogen ganjilganjil
C H O N S dan X C H O N S dan X PP B Si Ar dan B Si Ar dan alkali tanahalkali tanah
Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul
Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul
IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni
Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens
Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya
AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18
(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul
Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)
(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen
(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin
(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul
PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation
index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains
= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1
Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5
Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1
wherewhereα is H D or halogen (ie any α is H D or halogen (ie any
monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen
atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index
is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter
the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides
molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly
However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass
A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass
THE RULE OF THIRTEENTHE RULE OF THIRTEEN
a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313
The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r
the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2
Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method
if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula
For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
Yang paling tinggi intensitas puncak dasar Yang paling tinggi intensitas puncak dasar (base peak)(base peak)
100 (tinggi x factor sensitivitas) dari 100 (tinggi x factor sensitivitas) dari puncak2 lain termasuk puncak ion puncak2 lain termasuk puncak ion molekul sebagai base peakmolekul sebagai base peak
Ion molekul MIon molekul M++ sinar electron sinar electron mengubah molekul2 menjadi ion positipmengubah molekul2 menjadi ion positip
Nilai mz ion hanya kehilangan satu Nilai mz ion hanya kehilangan satu electron berat molekuler dari molekul electron berat molekuler dari molekul yang asliyang asli
Untuk mengidentifikasi puncak ion Untuk mengidentifikasi puncak ion molekul dalam spektrum molekul dalam spektrum
mampu menggunakan mampu menggunakan spektrumspektrum untuk untuk menentukan berat molekul suatu menentukan berat molekul suatu komponen yang tidak diketahui komponen yang tidak diketahui
isotop-isotop tampak dalam spektrumisotop-isotop tampak dalam spektrum
Molekul-molekul di alam tidak hanya sebagai spesies Molekul-molekul di alam tidak hanya sebagai spesies isotop murni semua atom mempunyai isotop-isotop isotop murni semua atom mempunyai isotop-isotop yang menjadi sifat khas kelimpahan alamyang menjadi sifat khas kelimpahan alam
H H 11H ~ 002 H sebagai isotop H ~ 002 H sebagai isotop 22HH C C 1212C ~ 11 C sebagai isotop C ~ 11 C sebagai isotop 1313CC Isotop 1 atau 2 unit massa gt massa atom Isotop 1 atau 2 unit massa gt massa atom
normalnormal M+ M+1 dan M+2M+ M+1 dan M+2 Pada gambar 1 Puncak M+1 ~ 8 dari intensitas Pada gambar 1 Puncak M+1 ~ 8 dari intensitas
puncak ion molekul puncak ion molekul M+1 M+1 1313C C 22H H 1515N dan N dan 1717OO Tabel limpahan relatip isotop terhadap isotop yang Tabel limpahan relatip isotop terhadap isotop yang
paling melimpahpaling melimpah M+2 M+2 1818O kelimpahannya sangat rendah tidak O kelimpahannya sangat rendah tidak
terdeteksiterdeteksi Jika hanya C H N O F P dan IJika hanya C H N O F P dan I
(M+1) asymp 11 x sum atom C + 036 x sum atom N(M+1) asymp 11 x sum atom C + 036 x sum atom N (11 x sum atom C)(11 x sum atom C)22
(M+2) asymp + 020 x sum (M+2) asymp + 020 x sum atom O atom O 200 200
Jika puncak2 isotop ini cukup intens diukur Jika puncak2 isotop ini cukup intens diukur secara tepat perhitungan di atas dapat secara tepat perhitungan di atas dapat berguna untuk menentukan rumus molekulberguna untuk menentukan rumus molekul
Jika S dan Si ada M+2 lebih intensJika S dan Si ada M+2 lebih intens 1 atom S 1 atom S 3434S ~ 44 pada puncak M+2S ~ 44 pada puncak M+2 1 atom Si 1 atom Si 3030Si ~ 335 pada puncak Si ~ 335 pada puncak
M+2M+2
Mengenal puncak ion molekul (M)Mengenal puncak ion molekul (M) EI (Electron Impact)EI (Electron Impact) Problem Puncak sangat lemah atau tidak Problem Puncak sangat lemah atau tidak
nampaknampak Puncak ion molekul dan bukan Puncak ion molekul dan bukan
puncak fragmentasi atau tidak murnipuncak fragmentasi atau tidak murni CI (Chemical Ionization) puncak yang CI (Chemical Ionization) puncak yang
kuat pada M+1 dan sedikit fragmentasikuat pada M+1 dan sedikit fragmentasi ldquoNitrogen rulerdquo molekul dengan ldquoNitrogen rulerdquo molekul dengan
berat molekul genap tidak mengandung N berat molekul genap tidak mengandung N atau nitrogen genap dan berat molekul atau nitrogen genap dan berat molekul ganjil mempunyai suatu atom nitrogen ganjil mempunyai suatu atom nitrogen ganjilganjil
C H O N S dan X C H O N S dan X PP B Si Ar dan B Si Ar dan alkali tanahalkali tanah
Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul
Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul
IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni
Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens
Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya
AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18
(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul
Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)
(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen
(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin
(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul
PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation
index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains
= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1
Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5
Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1
wherewhereα is H D or halogen (ie any α is H D or halogen (ie any
monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen
atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index
is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter
the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides
molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly
However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass
A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass
THE RULE OF THIRTEENTHE RULE OF THIRTEEN
a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313
The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r
the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2
Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method
if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula
For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
Molekul-molekul di alam tidak hanya sebagai spesies Molekul-molekul di alam tidak hanya sebagai spesies isotop murni semua atom mempunyai isotop-isotop isotop murni semua atom mempunyai isotop-isotop yang menjadi sifat khas kelimpahan alamyang menjadi sifat khas kelimpahan alam
H H 11H ~ 002 H sebagai isotop H ~ 002 H sebagai isotop 22HH C C 1212C ~ 11 C sebagai isotop C ~ 11 C sebagai isotop 1313CC Isotop 1 atau 2 unit massa gt massa atom Isotop 1 atau 2 unit massa gt massa atom
normalnormal M+ M+1 dan M+2M+ M+1 dan M+2 Pada gambar 1 Puncak M+1 ~ 8 dari intensitas Pada gambar 1 Puncak M+1 ~ 8 dari intensitas
puncak ion molekul puncak ion molekul M+1 M+1 1313C C 22H H 1515N dan N dan 1717OO Tabel limpahan relatip isotop terhadap isotop yang Tabel limpahan relatip isotop terhadap isotop yang
paling melimpahpaling melimpah M+2 M+2 1818O kelimpahannya sangat rendah tidak O kelimpahannya sangat rendah tidak
terdeteksiterdeteksi Jika hanya C H N O F P dan IJika hanya C H N O F P dan I
(M+1) asymp 11 x sum atom C + 036 x sum atom N(M+1) asymp 11 x sum atom C + 036 x sum atom N (11 x sum atom C)(11 x sum atom C)22
(M+2) asymp + 020 x sum (M+2) asymp + 020 x sum atom O atom O 200 200
Jika puncak2 isotop ini cukup intens diukur Jika puncak2 isotop ini cukup intens diukur secara tepat perhitungan di atas dapat secara tepat perhitungan di atas dapat berguna untuk menentukan rumus molekulberguna untuk menentukan rumus molekul
Jika S dan Si ada M+2 lebih intensJika S dan Si ada M+2 lebih intens 1 atom S 1 atom S 3434S ~ 44 pada puncak M+2S ~ 44 pada puncak M+2 1 atom Si 1 atom Si 3030Si ~ 335 pada puncak Si ~ 335 pada puncak
M+2M+2
Mengenal puncak ion molekul (M)Mengenal puncak ion molekul (M) EI (Electron Impact)EI (Electron Impact) Problem Puncak sangat lemah atau tidak Problem Puncak sangat lemah atau tidak
nampaknampak Puncak ion molekul dan bukan Puncak ion molekul dan bukan
puncak fragmentasi atau tidak murnipuncak fragmentasi atau tidak murni CI (Chemical Ionization) puncak yang CI (Chemical Ionization) puncak yang
kuat pada M+1 dan sedikit fragmentasikuat pada M+1 dan sedikit fragmentasi ldquoNitrogen rulerdquo molekul dengan ldquoNitrogen rulerdquo molekul dengan
berat molekul genap tidak mengandung N berat molekul genap tidak mengandung N atau nitrogen genap dan berat molekul atau nitrogen genap dan berat molekul ganjil mempunyai suatu atom nitrogen ganjil mempunyai suatu atom nitrogen ganjilganjil
C H O N S dan X C H O N S dan X PP B Si Ar dan B Si Ar dan alkali tanahalkali tanah
Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul
Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul
IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni
Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens
Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya
AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18
(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul
Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)
(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen
(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin
(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul
PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation
index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains
= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1
Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5
Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1
wherewhereα is H D or halogen (ie any α is H D or halogen (ie any
monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen
atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index
is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter
the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides
molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly
However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass
A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass
THE RULE OF THIRTEENTHE RULE OF THIRTEEN
a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313
The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r
the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2
Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method
if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula
For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
(M+1) asymp 11 x sum atom C + 036 x sum atom N(M+1) asymp 11 x sum atom C + 036 x sum atom N (11 x sum atom C)(11 x sum atom C)22
(M+2) asymp + 020 x sum (M+2) asymp + 020 x sum atom O atom O 200 200
Jika puncak2 isotop ini cukup intens diukur Jika puncak2 isotop ini cukup intens diukur secara tepat perhitungan di atas dapat secara tepat perhitungan di atas dapat berguna untuk menentukan rumus molekulberguna untuk menentukan rumus molekul
Jika S dan Si ada M+2 lebih intensJika S dan Si ada M+2 lebih intens 1 atom S 1 atom S 3434S ~ 44 pada puncak M+2S ~ 44 pada puncak M+2 1 atom Si 1 atom Si 3030Si ~ 335 pada puncak Si ~ 335 pada puncak
M+2M+2
Mengenal puncak ion molekul (M)Mengenal puncak ion molekul (M) EI (Electron Impact)EI (Electron Impact) Problem Puncak sangat lemah atau tidak Problem Puncak sangat lemah atau tidak
nampaknampak Puncak ion molekul dan bukan Puncak ion molekul dan bukan
puncak fragmentasi atau tidak murnipuncak fragmentasi atau tidak murni CI (Chemical Ionization) puncak yang CI (Chemical Ionization) puncak yang
kuat pada M+1 dan sedikit fragmentasikuat pada M+1 dan sedikit fragmentasi ldquoNitrogen rulerdquo molekul dengan ldquoNitrogen rulerdquo molekul dengan
berat molekul genap tidak mengandung N berat molekul genap tidak mengandung N atau nitrogen genap dan berat molekul atau nitrogen genap dan berat molekul ganjil mempunyai suatu atom nitrogen ganjil mempunyai suatu atom nitrogen ganjilganjil
C H O N S dan X C H O N S dan X PP B Si Ar dan B Si Ar dan alkali tanahalkali tanah
Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul
Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul
IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni
Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens
Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya
AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18
(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul
Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)
(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen
(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin
(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul
PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation
index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains
= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1
Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5
Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1
wherewhereα is H D or halogen (ie any α is H D or halogen (ie any
monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen
atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index
is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter
the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides
molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly
However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass
A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass
THE RULE OF THIRTEENTHE RULE OF THIRTEEN
a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313
The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r
the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2
Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method
if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula
For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
Mengenal puncak ion molekul (M)Mengenal puncak ion molekul (M) EI (Electron Impact)EI (Electron Impact) Problem Puncak sangat lemah atau tidak Problem Puncak sangat lemah atau tidak
nampaknampak Puncak ion molekul dan bukan Puncak ion molekul dan bukan
puncak fragmentasi atau tidak murnipuncak fragmentasi atau tidak murni CI (Chemical Ionization) puncak yang CI (Chemical Ionization) puncak yang
kuat pada M+1 dan sedikit fragmentasikuat pada M+1 dan sedikit fragmentasi ldquoNitrogen rulerdquo molekul dengan ldquoNitrogen rulerdquo molekul dengan
berat molekul genap tidak mengandung N berat molekul genap tidak mengandung N atau nitrogen genap dan berat molekul atau nitrogen genap dan berat molekul ganjil mempunyai suatu atom nitrogen ganjil mempunyai suatu atom nitrogen ganjilganjil
C H O N S dan X C H O N S dan X PP B Si Ar dan B Si Ar dan alkali tanahalkali tanah
Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul
Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul
IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni
Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens
Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya
AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18
(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul
Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)
(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen
(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin
(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul
PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation
index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains
= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1
Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5
Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1
wherewhereα is H D or halogen (ie any α is H D or halogen (ie any
monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen
atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index
is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter
the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides
molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly
However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass
A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass
THE RULE OF THIRTEENTHE RULE OF THIRTEEN
a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313
The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r
the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2
Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method
if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula
For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul
Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul
IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni
Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens
Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya
AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18
(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul
Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)
(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen
(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin
(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul
PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation
index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains
= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1
Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5
Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1
wherewhereα is H D or halogen (ie any α is H D or halogen (ie any
monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen
atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index
is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter
the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides
molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly
However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass
A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass
THE RULE OF THIRTEENTHE RULE OF THIRTEEN
a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313
The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r
the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2
Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method
if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula
For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya
AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18
(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul
Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)
(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen
(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin
(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul
PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation
index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains
= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1
Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5
Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1
wherewhereα is H D or halogen (ie any α is H D or halogen (ie any
monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen
atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index
is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter
the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides
molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly
However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass
A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass
THE RULE OF THIRTEENTHE RULE OF THIRTEEN
a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313
The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r
the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2
Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method
if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula
For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin
(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul
PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation
index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains
= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1
Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5
Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1
wherewhereα is H D or halogen (ie any α is H D or halogen (ie any
monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen
atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index
is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter
the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides
molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly
However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass
A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass
THE RULE OF THIRTEENTHE RULE OF THIRTEEN
a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313
The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r
the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2
Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method
if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula
For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1
wherewhereα is H D or halogen (ie any α is H D or halogen (ie any
monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen
atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index
is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter
the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides
molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly
However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass
A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass
THE RULE OF THIRTEENTHE RULE OF THIRTEEN
a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313
The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r
the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2
Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method
if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula
For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides
molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly
However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass
A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass
THE RULE OF THIRTEENTHE RULE OF THIRTEEN
a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313
The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r
the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2
Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method
if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula
For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313
The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r
the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2
Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method
if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula
For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula
For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound
CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements
TABELTABEL
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides
9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =
(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must
contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
A possible structure might be A possible structure might be
If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme
1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)
CH3
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH
5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =
44A possible substance which fits these A possible substance which fits these
data is data is
The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular
formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu
C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0
OH