2302169 Chem Med Studt
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แบบฝึ กหัด Amine & Spectroscopy
(มีเฉลยท้ ายแบบฝึ กหัด)
1.1
CH3 NH(CH 2) 3CH3
1.7
N C(CH 3) 3
CH 2CH3
NH 2
1.2
1.8
NH 2
NH2
1.3
O
CH2 CH 3
1.9
N
H
CH3
N
CH2 CH 2CH 3
1.4
1.10
1.5
(CH 3CH 2CH2 )3 N
1.11
1.6
(C6 H5 )2 NH
1.12
CH 3CH 2CH2 CH(NH 2)CH(CH3 )2
NH2
N(CH2 CH 3 )2
2) Which compound in each pair is the stronger base?
2.1
(CH 3CH 2) 2NH
2.3
or
(CH 3CH 2) 2NH or (ClCH2CH 2 )2NH
N
2.2
HCON(CH3 )2
or
(CH 3) 3N
or
2.4
N
H
NH
3) How would you prepare 3-phenyl-1-propanamine (C6H5CH2CH2CH2NH2) from each compound?
3.1 C6H5CH2CH2CH2Br
3.2 C6H5CH2CH2Br
3.3 C6H5CH2CH2CH2NO2
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3.4 C6H5CH2CH2CONH2
3.5 C6H5CH2CH2CHO
4) Draw the product of each reductive amination reaction.
CHO
NH2
4.1
O
O
4.2
NH 3
4.3
NaBH 3CN
NaBH 3CN
(CH 3) 2NH
NH 2
4.4
NaBH 3CN
O
NaBH 3CN
5) What products are formed when N-ethylaniline is treated with each reagent?
5.1 HCl
5.5 CH3I (excess)
5.2 acetic acid
5.6 CH3I (excess), followed by Ag2O and heat
5.3 acetone
5.7 CH3CH2COCl
5.4 fomaldehyde, NaBH3CN
5.8 The product in 5.7, then HNO3, H2SO4
6) Devise the synthesis of each compound strating from acetophenone (C6H5COCH3). In some cases, more
than one step is required.
OH
N(CH3 )2
6.1
6.4
C6 H 5
NHCH 3
C6 H 5
NH2
6.2
C6 H 5
6.5
NCH 2CH2 CH3
C6 H 5
O
6.3
6.6
C6 H 5
C6 H 5
NHCH 2CH 2CH2 CH3
7) Draw the organic products formed in each reaction.
7.1
CH 2CH2 Cl
NH3
(C 6H 5CO) 2O
7.6
excess
CH2 CH 2NH 2
O
7.2
Cl +
N
OH-
7.7
NH
H 2O
NaNO 2
HCl
O
7.3 Br
NO2
Sn
HCl
7.8
NH +
CHO
NaBH3 CN
---3--[1] LiAlH4
7.4
CN
7.5
7.9
[2] H2 O
CONHCH 2CH3
[1] LiAlH4
7.10
NH
+
O
H
H 2CH 2CH3 C N CH(CH3 )2
[2] H2 O
[1] CH3 I (excess)
[2] Ag2 O
[3] heat
8) Draw a stepwise mechanism for the following reaction.
OH
OH
f ormaldehyde
HN
N
+
H 2O
mild acid
9) Synthesis each compound from benzene. Use a diazonium salt as one of the synthetic intermediate.
Cl
COOH
9.1
Cl
9.3
9.2
Cl
HO
N N
CH 2
Cl
Cl
10) Which compound (C6H5CH2CH2CH3 or C6H5COCH2CH3 or C6H5OCH2CH3) gives a molecular ion at
m/z = 122
11) Match each structure to its mass spectrum.
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12) How would each of the following pairs of compounds differ in their IR spectra?
O
12.1
and
12.3
O
O
12.2
H 3CH2 C
OH
and
H3 C
OCH 3
12.4
H 3CH2 C
CH3
OCH 3
OCH 3
and CH3 CH=CHCH2 OH
O
and
CH 3 (CH2 )5
OCH 3
13) Match each compound to its IR spectrum.
O
CH 3CH2 CH2 CH 2
OH
13.1
H 3C
H 3 CH 2CH 2CH2 C
CH2
13.2
CH(CH3 )2
13.3
(CH 3) 2CHOCH(CH 3) 2
13.4
O
H 3C
OC(CH 3 )3
13.5
(CH 3CH 2)3 COH
13.6
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14) A chiral compound Y has a strong absorption at 2970-2840 cm-1 in its IR spectrum and gives the
following mass spectrum. Propose the structure for Y.
15) How many different types of protons are present in each compound?
15.1
(CH 3) 3CH
15.4
O
15.2
(CH 3) 3CC(CH3 )3
15.5
15.3
CH 3CH 2OCH 2CH2 CH2CH 2CH 3
H3 C
CH 2CH3
H3 C
Br
16) Which of the indicated protons in each pair absorbs father downfield?
a
16.1
CH 3CH 2CH2 OCH3
a
16.2
a
b
FCH 2CH2 CH 2 CH 2Br
CH 2CH2 CH2 CH 2 CH 3
16.3
a
b
16.4
b
b
CH 3CHBrCH2 CH 2 CH 2CHBr 2
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17) Identify the structures of isomers C and D (molecular formula C4H8O2).
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18) Propose a structure consistent with each set of data.
18.1) C9H10O2, IR absorption at 1718 cm-1
18.2) C9H12, IR absorption at 2850-3150 cm-1
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19) How many 13C NMR signals does each compound exhibit?
19.1
CH(CH3 )3
19.4
19.7
CH 2
O
19.5 CH 3CH2
19.2
H
19.3
CH 3OCH(CH3)2
19.6
CH 2CH3
19.8
O
H
OH
19.9
20) Propose a structure consistent with each set of data.
20.1 A compound X (C6H12O2) gives a strong peak in its IR spectrum at 1740 cm-1. The 1H NMR
spectrum of X shows only two singlets, including one at 3.5 ppm. The 13C NMR spectrum is given below.
Propose a structure for X.
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20.2 A compound Y (C6H10) gives four lines in its 13C NMR spectrum (27, 30, 67 and 93 ppm), and
the IR spectrum given here. Propose a structure for Y.
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เฉลย Amine & Spectroscopy
1.1) N-methylbutanamine; 1.2) octanamine; 1.3) 4,6-dimethylheptanamine; 1.4) N-methyl-Npropylcyclohexanamine; 1.5) N,N-dipropylpropanimine; 1.6) N-phenylaniline; 1.7) N-tert-butyl-Nethylaniline; 1.8) 3S-aminocyclohexanone; 1.9) 2-ethylpyrrolidine; 1.10) 2-methyl-3-hexanamine; 1.11) 3ethyl-2-methylcyclohexanamine; 1.12) N,N-diethylcycloheptanamine
2.1) (CH3CH2) 2NH
2.3) (CH3CH2) 2NH
2.2) (CH3)3N
2.4)
NH
3.1) NH3 (excess); 3.2) step 1 NaCN, step 2 1) LiAlH4, 2) H2O; 3.3) H2/Pd-C; 3.4) step 2 of 3.2); 3.5) NH3,
NaBH3CN
NH2
4.1)
4.2)
NH
N
H2
N
5.1)
4.3)
H2
N
Cl
5.4)
5.3)
N
5.6)
O
N
N
+ H 2 C CH 2
5.5)
O
O
5.8)
5.7)
(CH 3) 2
N
I
N
N
OAc
5.2)
HN
4.4)
NO2
N
+
O2 N
6.1) [1] NaBH4, CH3OH, [2] H2SO4; or step 1 NH3, NaBH3CN, step 2 [1] CH3I (excess), [2] AgO2, [3] heat;
6.4) compound 6.3 [1] mCPBA, [2] CH3NH2; 6.5) NH3, NaBH3CN; 6.6) [1] Br2, acetic acid, [2]
NH2CH2CH2CH2CH3
CO 27.1)
CH 2CH2 NH2
7.2)
7.3) Br
+
NH 2
7.4)
NH 2
CO 2-
7.5)
CH 2NHCH 2CH3
7.6)
7.9)
N
N
7.8)
8)
CH 2 =O
HN
-H2 O
7.10)
H 2C
N
7.7)
H3 CH 2 C CH 2 + (CH 3 )2 NCH(CH 3) 2 +
O-
OH
proton transfer
HN
HO
OH
2
CH 2CH2 NHCOC 6H 5 + benzylate salt
OH
OH
CH2 NH
AH
N
OH
N
OH
N
N N O
CH 3 CH 2CH 2N(CH3 )2
H-A
---11--9.1)
NH2
[1] HNO3, H 2SO 4
[2] H 2, Pd-C
CN
[1] NaNO2, HCl
CO 2H
[1] H 3O+
[2] Cl2, FeCl3
[2] CuCN
Cl
Cl
Cl
[1] HNO3, H 2SO 4
9.2)
[1] H 2, Pd-C
[2] Cl2 , FeCl3 2 times
O 2N
Cl
[2] NaNO2, HCl
HO
Cl
[3] H 2O
Cl
Cl
[1] NaNO2, HCl
N N
NH 2
9.3)
[2] H 2N
Cl
CH 3
CH 3
Cl
(from 9.2)
10) C6H5OCH2CH3
11) mass spectrum [1] and B; mass spectrum [2] and C; mass spectrum [3] and A
12.1) C=C bond (1650 cm-1) vs C-C triple bond (2250 cm-1); 12.2) O-H bond (>3000 cm-1) vs no OH bond; 12.3) C=O bond (1700 cm-1) vs O-H bond (3200-3600 cm-1), C=C bond (1650 cm-1); 12.4) no C=O
bond vs C=O (1700 cm-1); 13.1) spectrum [5]; 13.2) spectrum [1]; 13.3) spectrum [4]; 13.4) spectrum [3];
13.5) spectrum [6]; 13.6) spectrum [2]
14)
Br
Br
two possible enantiomers
and
15.1) 2 types; 15.2) 1 type; 15.3) 7 types; 15.4) 5 types; 15.5) 4 types
16.1) and 16.4) b; 16.2) and 16.3) a
17)
H 3C
O
O
O
OCH 2CH 3
C
H3C
CH 2OCH3
18.1)
CH 3
OCH 2CH3
18.2)
CH 3
D
19.1) 2 signals; 19.2) and 19.7) 5 signals; 19.3), 19.5) and 19.9) 3 signals; 19.4) and 19.6) 7 signals; 19.8) 4
signals
O
20.1)
H3CO
CH 3
CH
CH 3 3
20.2) HC C
CH 3
CH3
CH3