Реакции N-алкил(арил)альд- и кетиминов с эфирами кислот фосфора. Монография
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ISBN 978-5-7882-1787-1
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, 3 ……………………………… |
6 |
+……………………………………………………………… |
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1. 0 , 1 1………………………………………….. |
9 |
1.1 0 , 1 3 1 1 1………….. |
9 |
1.1.1 4, 3 … |
9 |
1.1.2 4, 3 … |
18 |
1.1.2.1 4 ………………... |
18 |
1.1.2.2 4 …………………… |
25 |
1.1.3 ( 1 - …………… |
36 |
1.1.3.1) ………………………. 38
1.1.3.2………………………….. 43
1.1.3.3+ 1 1 1
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1 1…………………………………….. 65 1.2 1 3 1
N-1 1………………………………………………... 68 1.2.1 + 1 '" N- -2- 1 1…….. 71
1.2.1.1 4 3 |
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………………………….... |
71 |
1.2.1.2 4 3 |
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1 ……………………… |
74 |
1.2.2 + 1 |
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N- -2-1 1 1…………………………………….. |
79 |
1.2.3 . |
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…………………………………………... |
86 |
1.2.4 N- - -2,2- 1 |
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1 1…………………………….. |
90 |
1.2.5 + 1 *,*- |
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N- -2-1-2- 1 1……………………..... |
92 |
2. -1 ……………………………………… |
100 |
2.1 + 1 2-1 '"…………… |
101 |
2.1.1 4 3 |
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……………………….. |
101 |
2.1.2 4 3 |
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106 |
2.1.3 . |
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109 |
2.1.3.1 , *,*-1(1- - -3,3-1- |
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109 |
2.1.3.2 + 1 |
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112 |
2.2 *,*- (40) |
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N- -2-1-2- 1 1 (2)………………….. |
115 |
, …………………………….. |
120 |
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– Acyl –
Alk – alkyl –
All –allyl –
Am – amyl – 1 Ar –aryl –
Bn – benzyl –
Boc – tert-butoxycarbonyl – -
i-Bu – iso-butyl –
n-Bu – n-butyl – -
s-Bu – sec. butyl – .-
t-Bu – tert. butyl – .-
Cbz – carbobenzyloxy – benzyloxycarbonyl –
Hex – hexyl –
c-Hex – cyclohexyl –
tPcAlCl - ( - )1
Prot – protecting group – 3
i-Pr – isopropyl –
n-Pr – n-propyl – -
Py – pyridine –
,$ –
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>P(O)H + >C=NR |
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>P(O)C |
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NHR |
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1
1.1.1, !
N- -[9], - [9], -
- [10], 1 (2) 2
1 ('") (1)
1 1 (3).
(RO) PHO + (CF ) C = NX |
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(RO) P(O)C(CF ) NHX |
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2 |
1 |
3 |
2 |
2 |
2 |
3 |
3 |
2 |
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3, X = SO2Ph, ROCO, COCF3, (RO)2P(O)
* 1 (5) -
1 1 '" 1 1 (4) 70°, [11].
700C
R1R2C=NR3 + (R4O)2PHO
(R 4O)2P(O)CR1R2-NHR3
5
5, R1=Me, n-Pr, n-Bu; R2=Me, n-Bu, n-Pent, i-Bu, t-Bu;
R3=n-Bu, n-Hex, Furyl, n -Oct, n-Pentyl, n -Decyl, n-Tetradecyl; R 4=Me, Et, i-Pr, n-Bu
0 , N-1 1 1 -
1 1 -
, 1 1 [12-13].
9
R1 |
R2 |
R1 |
R2 |
CH N |
+ |
R3R4P(O)H |
CH NH |
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O=PR3R4 |
1 1
1 1 1 1 1,
1 0 , 1 -
90-100°, [14]. ' 1
1 1 (6) 60°, [15].
C6F5N = CHC6H4R-4 + (MeO)2PHO |
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C6F6NH |
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CHP(O)(OMe)2 |
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6 |
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C6H4R-4 |
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6, R=H, Me, MeO, Cl |
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4 0 , N-1 1 (7) -
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1 1 [16]. , 1
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R1C |
H |
CH = NBn + |
(R2O) |
PHO |
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R1C |
H |
CHNHBn |
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6 |
4 |
7 |
2 |
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6 |
4 |
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2 |
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O=P(OR |
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$1 -2, -3 -4 (8) in situ -
2 [17].
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RNH2 |
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+ (EtO) PHO |
P(O)(OEt)2 |
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CHO |
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2 |
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N |
PhMe |
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CH = NR |
HC |
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N |
PhMe |
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N |
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8 |
NHR |
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R=Bu, Ph, Bn, CHPh2 |
10
