tugas penstruk
TRANSCRIPT
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Penentuan Struktur
Senyawa Organik
PRIMA AGUSTI LUKIS
1412 201 011
PROBLEM 225
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PROBLEM 225
Rumus Molekul : C6H7N DBE= ΣC – ΣH/2 – ΣX/2 + ΣN/2 + 1
= 6 – 7/2 – 0 + ½ + 1
= 6 – 3,5 – 0 + 0,5 + 1
= 4
MW = 93
Maksimal 1 siklik4 Ikatan rangkap asiklik
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Spektra 13C NMR
(ppm) Resonance decouple Decouple Prediksi
29 4 sinyal 1 sinyal CH3
122 2 sinyal 1 sinyal CH
145 1 sinyal 1 sinyal C
149 2 sinyal 1 sinyal CH
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Spektra 13C NMR
020406080100120140160PPM
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Spektra 1H NMR
(ppm) Tipe sinyal H Prediksi
2,35 Singlet 3 CH3
7,02 Doublet 2 2-CH
8,51 Doublet 2 2-CH
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Spektra 1H NMR
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Spektra IR
C-H aromatikC-H sp3
C=N aromatik
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Spektra UV
maks = 250 nm, A = 0,70
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Perkiraan Struktur
• RM = C6H7N
• DBE = 4• MW = 93
• 1H NMR
δ ~2,35 ppm → 3H,s
δ ~7,02 ppm → 2H,d
δ ~8,51 ppm → 2H,d
• 13C NMR
minimal ada 4 jenis karbon, yaitu ada
–CH3, ada 2 jenis –CH yang ekuivalen
dan ada C kuarterner.
• IR- ada C-H aromatik
- ada C-H sp3
- ada C=N aromatik 4-metil-piridin
N
CH3
H
H
H
H
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Pola Fragmentasi
N
CH 3
N
H
H
H
N
CH2
+ H
m/z = 93
m/z = 92
N
CH 2
m/z = 92
N
HH
H
N
+
N
m/z = 66
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