Compound $\mathbf{E}$ can be synthesized from ethyl 3-chloropropionate as a starting material:

<MultiModal>![Image describes a partially missing organic synthesis route:
ClCCC(=O)OCC>[Na].CCOCC>[X], this step is carried out under "reflux" conditions, the reagent is sodium metal, and the solvent is diethyl ether;
[X]>C[Si](C)(C)Cl>[A], the reagent in this step is trimethylchlorosilane, and the molecular formula of $\mathbf{A}$ is $\mathrm{C_8H_{18}O_2Si}$;
[A]>CCCC[N+](CCCC)(CCCC)CCCC.[F-].CO>[B], the reagent in this step is tetrabutylammonium fluoride, the solvent is methanol, and the molecular formula of $\mathbf{B}$ is $\mathrm{C_5H_{10}O_2}$;
[B]>[Na]N.N>[C], the reagent in this step is sodium amide, the solvent is ammonia, and the molecular formula of $\mathbf{C}$ is $\mathrm{C_3H_7NO}$;
CC(OC([C@@H](NC(OC(C)(C)C)=O)CCC(O)=O)=O)(C)C>C1CCC(CC1)N=C=NC2CCCCC2.C1CC=C2C(=C1)N=NN2O.O1CCCC1>[Y], tetrahydrofuran is used as the solvent in this step;
[Y].[C]>O1CCCC1>[D], tetrahydrofuran is used as the solvent in this step;
[D]>Cl>[E], the condition for this step is "$\mathrm{HCl}$(5% aq), 40°C"](/media/uploads/3c7ba5d3-ce57-4ff4-8e95-09376832db96.png)</MultiModal>

$\mathbf{A}$~$\mathbf{E}$ all contain a three-membered carbon ring. The $^1\mathrm{H ~NMR}$ of $\mathbf{C}$ has peaks at 8.52, 6.42, 0.65, 0.40, with a peak area ratio of 2:1:2:2 (predicted value, the solvent is deuterated dimethyl sulfoxide).

Analyze this route step by step and check the known information, including the molecular formula and proton NMR spectrum, to deduce the unknown substances and select the correct one.
化合物$\mathbf{E}$可由3-氯丙酸乙酯为原料合成: 

<MultiModal>![图片描述了一个部分缺失的有机合成路线:
ClCCC(=O)OCC>[Na].CCOCC>[X], 这一步在"reflux"条件下完成, 试剂为金属钠, 溶剂为乙醚; 
[X]>C[Si](C)(C)Cl>[A], 这一步的试剂为三甲基氯硅烷, $\mathbf{A}$的分子式为$\mathrm{C_8H_{18}O_2Si}$; 
[A]>CCCC[N+](CCCC)(CCCC)CCCC.[F-].CO>[B], 这一步的试剂为四正丁基氟化铵, 溶剂为甲醇, $\mathbf{B}$的分子式为$\mathrm{C_5H_{10}O_2}$; 
[B]>[Na]N.N>[C], 这一步的试剂为氨基钠, 溶剂为氨, $\mathbf{C}$的分子式为$\mathrm{C_3H_7NO}$; 
CC(OC([C@@H](NC(OC(C)(C)C)=O)CCC(O)=O)=O)(C)C>C1CCC(CC1)N=C=NC2CCCCC2.C1CC=C2C(=C1)N=NN2O.O1CCCC1>[Y], 这一步中四氢呋喃作为溶剂; 
[Y].[C]>O1CCCC1>[D], 这一步中四氢呋喃作为溶剂; 
[D]>Cl>[E], 这一步的条件为"$\mathrm{HCl}$(5% aq), 40°C"](/media/uploads/3c7ba5d3-ce57-4ff4-8e95-09376832db96.png)</MultiModal>

$\mathbf{A}$~$\mathbf{E}$均含有一个三元碳环, $\mathbf{C}$的$^1\mathrm{H ~NMR}$ 在8.52, 6.42, 0.65, 0.40处有峰, 峰面积之比为2:1:2:2 (预测值, 溶剂为氘代二甲亚砜). 

逐步分析此路线并核对包括分子式, 核磁共振氢谱等已知信息后, 推断未知的物质, 选出正确的一项.
Theoretically, for every $1 ~\mathrm{mol}~\mathbf{A}$ produced, $4~\mathrm{mol~Na}$ are consumed.
**E** has only one chiral center, with R configuration.
$\mathbf{C}$ is more prone to spontaneous dehydration compared to $\mathrm{C_3H_9NO}$ and its analogs that open the three-membered ring.
In the $^1\mathrm{H}$ NMR of $\mathbf{A}$, the chemical shifts of all peaks are less than 2.0.
In the reaction that generates $\mathbf{D}$ from <MultiModal>![CC(OC([C@@H](NC(OC(C)(C)C)=O)CCC(O)=O)=O)(C)C](/media/uploads/fbb6ad98-c8a8-439d-988a-6ce2c9f27fc8.png)</MultiModal> and $\mathbf{C}$, the balanced chemical equation includes $\mathrm{H_2O}$ in addition to $\mathbf{D}$ in the products.
In the reaction to obtain $\mathbf{E}$ from $\mathbf{D}$, if no hydrocarbons are produced, then the balanced chemical equation involves 4 compounds.
The check digit for the CAS registry number of $\mathbf{E}$ is 2.
The CAS registry number for $\mathbf{E}$ has a check digit of 3.
$\mathbf{B}$ has optical activity and diastereomers.
All of the above options are incorrect.
理论上, 每生成$1 ~\mathrm{mol}~\mathbf{A}$, 需要消耗$4~\mathrm{mol~Na}$
$\mathbf{E}$只有一个手性中心, 为R构型
$\mathbf{C}$和打开三元环的类似物$\mathrm{C_3H_9NO}$相比, $\mathbf{C}$更容易自发脱水
$\mathbf{A}$的$^1\mathrm{H}$ NMR中, 所有峰的化学位移均小于2.0
从<MultiModal>![CC(OC([C@@H](NC(OC(C)(C)C)=O)CCC(O)=O)=O)(C)C](/media/uploads/fbb6ad98-c8a8-439d-988a-6ce2c9f27fc8.png)</MultiModal>和$\mathbf{C}$生成$\mathbf{D}$的反应中, 配平的化学方程式的产物中除$\mathbf{D}$以外还有$\mathrm{H_2O}$
从$\mathbf{D}$得到$\mathbf{E}$的反应中, 没有生成烃类, 则配平的化学方程式涉及4种化合物. 
$\mathbf{E}$的CAS编码的校验码为2
$\mathbf{E}$的CAS编码的校验码为3
$\mathbf{B}$有旋光活性和非对映异构体
以上选项均不正确
Start with the simplest structure, $\mathbf{C}$: The molecular formula of $\mathbf{C}$ is $\mathrm{C_3H_7NO}$, and the structure of $\mathbf{C}$ contains a three-membered carbon ring. The degree of unsaturation is calculated to be 1, with no additional unsaturation. Therefore, the structure of $\mathbf{C}$ can only be 1-aminocyclopropanol (NC1(O)CC1) or 2-aminocyclopropanol (NC1C(O)C1) or N-cyclopropylhydroxylamine (ONC1CC1) or O-cyclopropylhydroxylamine (NOC1CC1).

<Checkpoint>[The structure of $\mathbf{C}$ can only be 1-aminocyclopropanol (NC1(O)CC1) or 2-aminocyclopropanol (NC1C(O)C1) or N-cyclopropylhydroxylamine (ONC1CC1) or O-cyclopropylhydroxylamine (NOC1CC1).](2)</Checkpoint>

Considering the NMR hydrogen spectrum data, the chemical shift of hydrogen on heteroatoms is higher, and the chemical shift of hydrogen on the three-membered ring is lower. Combined with the ratio of peak areas, it can be known that $\mathbf{C}$ is 1-aminocyclopropanol (NC1(O)CC1). <MultiModal>![NC1(O)CC1](/media/uploads/277deeb6-0e73-4b07-9157-7dedda0d63a9.png)</MultiModal>

<Checkpoint>[$\mathbf{C}$ is 1-aminocyclopropanol (NC1(O)CC1)](3)</Checkpoint>

Reversing the route, it can be found that only functional group transformations occurred from $\mathbf{A}$ to $\mathbf{C}$. $\mathbf{B}$ undergoes a substitution reaction with sodium amide to obtain $\mathbf{C}$. Therefore, $\mathbf{B}$ is 1-ethoxycyclopropanol (CCOC1(O)CC1). <MultiModal>![CCOC1(O)CC1](/media/uploads/3339d953-8e76-47e1-87e8-935c41e28889.png)</MultiModal>

<Checkpoint>[$\mathbf{B}$ is 1-ethoxycyclopropanol (CCOC1(O)CC1)](2)</Checkpoint>

$\mathbf{A}$ undergoes deprotection of the trimethylsilyl group by tetrabutylammonium fluoride to obtain $\mathbf{A}$, which is (1-ethoxycyclopropoxy)trimethylsilane (C1C(O[Si](C)(C)C)(OCC)C1). <MultiModal>![C1C(O[Si](C)(C)C)(OCC)C1](/media/uploads/52036e26-6937-4e3e-855b-3b8a9b4c4d3d.png)</MultiModal>

<Checkpoint>[$\mathbf{A}$ is (1-ethoxycyclopropoxy)trimethylsilane (C1C(O[Si](C)(C)C)(OCC)C1)](2)</Checkpoint>

Continue the route from $\mathbf{C}$: First, the amino group and a carboxyl group protected L-glutamic acid are activated by DCC. After adding $\mathbf{C}$, the more nucleophilic amino group forms an amide $\mathbf{D}$ with the protected glutamic acid. The structure of $\mathbf{D}$ is CC(OC([C@@H](NC(OC(C)(C)C)=O)CCC(NC1(O)CC1)=O)=O)(C)C. <MultiModal>![CC(OC([C@@H](NC(OC(C)(C)C)=O)CCC(NC1(O)CC1)=O)=O)(C)C](/media/uploads/5f6b82ef-4436-4853-b603-923dee198216.png)</MultiModal>

<Checkpoint>[$\mathbf{C}$ condenses with protected L-glutamic acid under the action of DCC to obtain amide $\mathbf{D}$, with the structure CC(OC([C@@H](NC(OC(C)(C)C)=O)CCC(NC1(O)CC1)=O)=O)(C)C](2)</Checkpoint>

$\mathbf{D}$ is deprotected with hydrochloric acid to remove the Boc and t-Bu protecting groups, yielding $\mathbf{E}$, with the structure OC(=O)[C@H](CCC(=O)NC1(CC1)O)N. <MultiModal>![OC(=O)[C@H](CCC(=O)NC1(CC1)O)N](/media/uploads/047ce2f2-b8b1-4d1b-92a8-88bf83c1cb05.png)</MultiModal>

<Checkpoint>[$\mathbf{D}$ is deprotected by hydrochloric acid to obtain $\mathbf{E}$, with the structure OC(=O)[C@H](CCC(=O)NC1(CC1)O)N](2)</Checkpoint>

$\mathbf{E}$ is coprine, also known as ink cap amino acid, CAS code is 58919-61-2. The last digit of the CAS code is a checksum, which is 2 here. Option G is correct, option H is incorrect.

<Checkpoint>[The CAS code of $\mathbf{E}$ is 58919-61-2, and the checksum is 2](2)</Checkpoint>

Option A: Judging from the molecular formula, only the $\mathrm{C-Cl}$ bond of the starting compound is reduced, corresponding to 2 equivalents of $\mathrm{Na}$.

<Checkpoint>[Theoretically, 2 equivalents of $\mathrm{Na}$ are needed to obtain $\mathbf{A}$](1)</Checkpoint>

Option B: Starting from glutamic acid, the only chiral center in the molecule has not changed, and the final configuration is S.

<Checkpoint>[The only chiral center in $\mathbf{E}$ has S configuration](0.5)</Checkpoint>

Option C: $\mathbf{C}$ is limited by the ring strain of the three-membered ring, and the lone pair electrons do not overlap well with the $\mathrm{C-O}$ antibonding orbital, making it more difficult to eliminate dehydration compared to non-cyclic analogs.

<Checkpoint>[$\mathbf{C}$ is more difficult to spontaneously dehydrate than non-cyclic analogs](1)</Checkpoint>

Option D: The chemical shift of the $\alpha-\mathrm{H}$ of the oxygen on the ethoxy group in $\mathbf{A}$ is greater than 2.0.

<Checkpoint>[The chemical shift of the $\alpha-\mathrm{H}$ of the oxygen in $\mathbf{A}$ is greater than 2.0](0.5)</Checkpoint>

Option E: The 1 molecule of water generated by the condensation is transferred to the dicyclohexylurea, which is the byproduct corresponding to DCC. There will be no $\mathrm{H_2O}$ in the products in the balanced chemical equation.

<Checkpoint>[In the amidation reaction, $\mathrm{H_2O}$ will not appear in the products in the balanced chemical equation](1)</Checkpoint>

Option F: In the deprotection step, the possible forms of the removed tert-butyl group include tert-butanol and 2-chloro-2-methylpropane. The balanced chemical equation corresponding to tert-butanol involves the following substances: $\mathbf{D}$, water, $\mathbf{E}$, tert-butanol, carbon dioxide. The balanced chemical equation corresponding to 2-chloro-2-methylpropane involves the following substances: $\mathbf{D}$, hydrogen chloride, $\mathbf{E}$, 2-chloro-2-methylpropane, carbon dioxide. Neither can be four kinds.

<Checkpoint>[The balanced chemical equation in deprotection cannot involve only 4 substances](1)</Checkpoint>

Option I: The structure of $\mathbf{B}$ has a plane of symmetry and no chiral center, so it has no optical activity and no diastereomers.

<Checkpoint>[$\mathbf{B}$ has no optical activity and no diastereomers. ](0.5)</Checkpoint>
从结构最简单的$\mathbf{C}$入手: $\mathbf{C}$的分子式为$\mathrm{C_3H_7NO}$, 而$\mathbf{C}$的结构中含有一个三元碳环, 计算不饱和度为1, 没有额外的不饱和度. 因此, $\mathbf{C}$的结构只可能为1-氨基环丙醇(NC1(O)CC1)或2-氨基环丙醇(NC1C(O)C1)或N-环丙基羟胺(ONC1CC1)或O-环丙基羟胺(NOC1CC1). 

<Checkpoint>[$\mathbf{C}$的结构只可能为1-氨基环丙醇(NC1(O)CC1)或2-氨基环丙醇(NC1C(O)C1)或N-环丙基羟胺(ONC1CC1)或O-环丙基羟胺(NOC1CC1).](2)</Checkpoint>

考虑核磁共振氢谱数据, 杂原子氢化学位移较高, 三元环上氢化学位移较低. 结合峰面积之比,  可知$\mathbf{C}$是1-氨基环丙醇(NC1(O)CC1). <MultiModal>![NC1(O)CC1](/media/uploads/277deeb6-0e73-4b07-9157-7dedda0d63a9.png)</MultiModal>

<Checkpoint>[$\mathbf{C}$是1-氨基环丙醇(NC1(O)CC1)](3)</Checkpoint>

倒推路线, 可以发现$\mathbf{A}$至$\mathbf{C}$仅发生了官能团转化, $\mathbf{B}$与氨基钠发生取代反应得到$\mathbf{C}$. 因此, $\mathbf{B}$是1-乙氧基环丙醇(CCOC1(O)CC1). <MultiModal>![CCOC1(O)CC1](/media/uploads/3339d953-8e76-47e1-87e8-935c41e28889.png)</MultiModal>

<Checkpoint>[$\mathbf{B}$是1-乙氧基环丙醇(CCOC1(O)CC1)](2)</Checkpoint>

$\mathbf{A}$经四正丁基氟化铵脱去三甲基硅基保护, 得到$\mathbf{A}$是(1-乙氧基环丙氧基)三甲基硅烷(C1C(O[Si](C)(C)C)(OCC)C1). <MultiModal>![C1C(O[Si](C)(C)C)(OCC)C1](/media/uploads/52036e26-6937-4e3e-855b-3b8a9b4c4d3d.png)</MultiModal>

<Checkpoint>[$\mathbf{A}$是(1-乙氧基环丙氧基)三甲基硅烷(C1C(O[Si](C)(C)C)(OCC)C1)](2)</Checkpoint>

从$\mathbf{C}$继续路线: 首先, 氨基和一个羧基被保护的L-谷氨酸被DCC活化, 加入$\mathbf{C}$后, 亲核性更强的氨基与被保护的谷氨酸形成酰胺$\mathbf{D}$, $\mathbf{D}$的结构是CC(OC([C@@H](NC(OC(C)(C)C)=O)CCC(NC1(O)CC1)=O)=O)(C)C. <MultiModal>![CC(OC([C@@H](NC(OC(C)(C)C)=O)CCC(NC1(O)CC1)=O)=O)(C)C](/media/uploads/5f6b82ef-4436-4853-b603-923dee198216.png)</MultiModal>

<Checkpoint>[$\mathbf{C}$与被保护的L-谷氨酸在DCC作用下缩合得到酰胺$\mathbf{D}$, 结构为CC(OC([C@@H](NC(OC(C)(C)C)=O)CCC(NC1(O)CC1)=O)=O)(C)C](2)</Checkpoint>

$\mathbf{D}$在盐酸作用下脱去Boc和t-Bu保护基, 得到$\mathbf{E}$, 结构为OC(=O)[C@H](CCC(=O)NC1(CC1)O)N. <MultiModal>![OC(=O)[C@H](CCC(=O)NC1(CC1)O)N](/media/uploads/047ce2f2-b8b1-4d1b-92a8-88bf83c1cb05.png)</MultiModal>

<Checkpoint>[$\mathbf{D}$在盐酸作用下脱去保护基, 得到$\mathbf{E}$, 结构为OC(=O)[C@H](CCC(=O)NC1(CC1)O)N](2)</Checkpoint>

$\mathbf{E}$即鬼伞毒素, 又名墨盖蘑菇氨酸, CAS编码为58919-61-2. CAS编码的最后一位是校验码, 在这里是2. 选项G正确, 选项H错误. 

<Checkpoint>[$\mathbf{E}$的CAS编码为58919-61-2, 校验码是2](2)</Checkpoint>

选项A: 从分子式判断, 只有起始化合物的$\mathrm{C-Cl}$键被还原, 对应于2倍量的$\mathrm{Na}$. 

<Checkpoint>[得到$\mathbf{A}$理论上需要2倍量的$\mathrm{Na}$](1)</Checkpoint>

选项B: 从谷氨酸开始, 分子中唯一的手性中心都未改变, 最终是S构型. 

<Checkpoint>[$\mathbf{E}$中唯一的手性中心为S构型](0.5)</Checkpoint>

选项C: $\mathbf{C}$受限于三元环的环张力, 孤对电子同$\mathrm{C-O}$反键轨道重叠不好, 相比与非环状类似物更难消除脱水. 

<Checkpoint>[$\mathbf{C}$和非环状类似物相比,更难自发脱水](1)</Checkpoint>

选项D: $\mathbf{A}$中乙氧基上氧的$\alpha-\mathrm{H}$的化学位移大于2.0. 

<Checkpoint>[$\mathbf{A}$中氧的$\alpha-\mathrm{H}$的化学位移大于2.0](0.5)</Checkpoint>

选项E: 缩合产生的1分子水转移至DCC对应的副产物二环己基脲中, 在配平的化学方程式中产物不会有$\mathrm{H_2O}$出现. 

<Checkpoint>[酰胺化反应中, 在配平的化学方程式中产物不会出现$\mathrm{H_2O}$](1)</Checkpoint>

选项F: 去保护一步中, 脱去的叔丁基可能的存在形式包括叔丁醇, 2-氯-2-甲基丙烷. 叔丁醇对应的配平的化学方程式中涉及物质有: $\mathbf{D}$, 水, $\mathbf{E}$, 叔丁醇, 二氧化碳,  2-氯-2-甲基丙烷对应的配平的化学方程式中涉及物质有: $\mathbf{D}$, 氯化氢, $\mathbf{E}$, 2-氯-2-甲基丙烷, 二氧化碳, 都不可能是四种. 

<Checkpoint>[脱保护中配平的化学方程式不可能只涉及4种物质](1)</Checkpoint>

选项I: $\mathbf{B}$的结构中有对称面, 没有手性中心, 因此没有旋光活性和非对映异构体. 

<Checkpoint>[$\mathbf{B}$的没有旋光活性和非对映异构体. ](0.5)</Checkpoint>