Stereochemistry of (-)-deoxynupharidine. Synthesis of (R) (-)-.alpha.-methyladipic acid
Stereochemistry of (-)-deoxynupharidine. Synthesis of (R) (-)-.alpha.-methyladipic acid
复制标题
(-)-脱氧nupharine 的立体化学。
DOI：
10.1021/jo00827a055
复制
发表时间：
1970
影响因子：
3.6
通讯作者：
R. T. Lalonde
中科院分区：
文献类型：
--
作者：
C. Wong;E. Auer;R. T. Lalonde
根据已知的方法，通过(+)-(/Z)-3-甲基环己酮的Baeyer-Villiger氧化生成的-和/3-甲基-e-己内酯混合物的重复低温分离结晶得到9(-)-(？)-甲基-e-己内酯(光学纯度，100%)。(-)-(R)-甲基-e-己内酯小心地碱解，然后用过量的高锰酸钾氧化生成(-)-(7Z)-甲基己二酸，它与降解的(-)酸在旋转符号和大小上一致。在合成方法和红外光谱研究的基础上，对(-)-脱氧大豆碱的相对构型和构象进行了令人信服的指认。10*·11早期的核磁共振研究提供了支持存在一个轴向和一个赤道甲基的证据，4a和后来通过纽法拉明转化为脱氧洋地拉定和7-表氧洋地拉定的甲基的特定立体化学指认成为可能。12由于核磁共振显示7个同分异构体有两个赤道甲基，所以脱氧呋喃大豆碱必须有Ci和C7轴向甲基。这种立体化学也通过对脱氧呋喃大豆碱的100 MHz核磁共振中的C6质子的检查而得到证明。H6a(7.30 r)和0(3(8.12 r)信号都是相同的四重奏。H6a四重奏由双星(12.5赫兹)和赤道-赤道(6a-7/3)耦合(Je，e=2.5赫兹)产生，而He(S四重奏)来自双星(12.5赫兹)和
According to a known procedure, 9(—)-(ß)--methyl-e-caprolactone was obtained by repeated lowtemperature fractional crystallization of the mixture of-and/3-methyl-e-caprolactones resulting from the Baeyer-Villiger oxidation of (+)-(/Z)-3-methylcyclo-hexanone (optical purity, 100%). Careful alkaline hydrolysis of the (-)-(R)--methyl-e-caprolactone and subsequent oxidation with an excess of potassium permanganate produced (—)-(7Z)--methyladipic acid which agreed in sign and magnitude of rotationwith the (—) acid of degradation. A convincing assignment of the relative configuration and conformation of (—)-deoxynupharidine has been made on the basis of synthetic methods and infrared studies. 10*· 11Supporting evidence for the presence of one axial and one equatorial methyl group was furnished by an early nmr study, 4a and a specific stereochemical assignment of methyl groups was made possible later by the transformation of nupharamine to deoxynuphari-dine and 7-epideoxynupharidine. 12 Since the 7 epimer was shown by nmr to have two equatorial methyl groups, deoxynupharidine must have Ci equatorial and C7 axial methyl groups. This stereochemistry is also demonstrated through examination of the C6 protons in the 100 MHz nmr of deoxynupharidine. Both H6a (7.30 r) and 0 (3 (8.12 r) signals are identical quartets. The H6a quartet arises from geminal (12.5 Hz) and equatorial-equatorial (6a-7/3) coupling (Je, e= 2.5 Hz), while the He (s quartet arises from geminal (12.5 Hz) and