The following is a total synthesis route:

<MultiModal>![Compound $\bf K$ (SMILES O=C(N1)NC2C1CSC2CCCCC(O)=O) was synthesized through a multi-step reaction: C1=CCCCCC1>>[A], [A]>>[B], [B] >>[C], [C]>>[D], [D]>>[E], [E]>>[F], [F]>>[G], [G]>>[H], [H]>>[I], [I]>>[J], [J]>>[K]. The reaction conditions are: (1) $\rm NBS, AIBN, CCl_4$, reflux for 3 hours (2) $\rm AcSH, Et_3N, CH_3CN, $, react at 0 degrees Celsius for 4 hours (3) $\rm EtONa, EtOH$, reflux for 15 minutes (4) $\rm AcOCH_2CH_2NO_2, EtOH$, react at 0 degrees Celsius for 3 hours (5) $\rm PhNCO, Et_3N$(catalytic amount), $\rm PhH$, react at 25 degrees Celsius for 24 hours (6) $\rm LiAlH_4, EtOH$, reflux for 4 hours (7) $\rm MeOCOCl, NaHCO_3(aq), MeOH$, react at 25 degrees Celsius for half an hour (8) $\rm Ac_2O, DMSO$, react at 25 degrees Celsius for 18 hours (9) $\rm NH_2OH·HCl, EtOH, Py$, reflux for half an hour (10) $\rm PPA$, react at 100 degrees Celsius for 15 minutes (11) Divided into two steps, first $\rm Ba(OH)_2, H_2O$, reflux for 20 hours, and then add $\rm COCl_2$ and react at 0 degrees Celsius ](/media/uploads/70ccefca-fe3c-43df-b3f7-e6bdee25072e.png)</MultiModal>

Given that the molecular formula of $\bf D$ is $\rm C_9H_{15}NO_2S$, and $\bf C$ is a sodium salt. Which of the following statements is correct:

1.  The conversion of $\bf D$ to $\bf E$ involves an intermediate containing a carbon-nitrogen triple bond.
2.  $\bf E$ contains 2 rings.
3.  The molecular formula of $\bf H$ is $\rm C_{11}H_{17}NO_3S$.
4.  $\bf J$ contains a seven-membered ring.
5.  The transformation from $\bf I$ to $\bf J$ may produce a compound with high ring strain, with a molecular formula of $\rm C_{11}H_{16}N_2O_2S$.
以下为一条全合成路线：

<MultiModal>![通过多步反应合成了化合物$\bf K$（SMILES为O=C(N1)NC2C1CSC2CCCCC(O)=O）：C1=CCCCCC1>>[A], [A]>>[B], [B] >>[C], [C]>>[D], [D]>>[E], [E]>>[F], [F]>>[G], [G]>>[H], [H]>>[I], [I]>>[J], [J]>>[K]。反应条件为：(1)$\rm NBS, AIBN, CCl_4$, 回流3小时 (2)$\rm AcSH, Et_3N, CH_3CN, $, 0摄氏度反应4小时 (3)$\rm EtONa, EtOH$, 回流15分钟 (4)$\rm AcOCH_2CH_2NO_2, EtOH$, 0摄氏度反应3小时 (5)$\rm PhNCO, Et_3N$(催化量), $\rm PhH$, 25摄氏度反应24小时 (6)$\rm LiAlH_4, EtOH$, 回流4小时 (7)$\rm MeOCOCl, NaHCO_3(aq), MeOH$, 25摄氏度反应半小时 (8)$\rm Ac_2O, DMSO$, 25摄氏度反应18小时 (9)$\rm NH_2OH·HCl, EtOH, Py$, 回流半小时 (10)$\rm PPA$, 100摄氏度反应15分钟 (11)分为两步，先是$\rm Ba(OH)_2, H_2O$, 回流20小时，再是加入$\rm COCl_2$在0摄氏度下反应 ](/media/uploads/70ccefca-fe3c-43df-b3f7-e6bdee25072e.png)</MultiModal>

已知$\bf D$的分子式为$\rm C_9H_{15}NO_2S$，$\bf C$为钠盐。下面说法正确的是：

1. $\bf D$转化为$\bf E$经历一个含有碳氮三键的中间体
2. $\bf E$中含有2个环
3. $\bf H$的分子式为$\rm C_{11}H_{17}NO_3S$
4. $\bf J$含七元环
5. $\bf I$到$\bf J$时可能产生一个环张力较大的化合物，分子式为$\rm C_{11}H_{16}N_2O_2S$
All other options are incorrect
2, 4
1, 3, 5
1, 2
2, 4, 5
1, 4, 5
1, 2, 4, 5
1, 2, 3, 4
2, 3, 4
3, 4, 5
4
3, 5
1, 2, 3, 4, 5
2, 5
其他选项均不正确
2, 4
1, 3, 5
1, 2
2, 4, 5
1, 4, 5
1, 2, 4, 5
1, 2, 3, 4
2, 3, 4
3, 4, 5
4
3, 5
1, 2, 3, 4, 5
2, 5
(1) The reaction is allylic radical halogenation, $\bf A$ is BrC1C=CCCCC1

<Checkpoint>[$\bf A$ is BrC1C=CCCCC1](1)</Checkpoint>

(2) The reaction is the substitution of $\rm Br^-$ by $\rm AcS^-$, $\bf B$ is CC(SC1C=CCCCC1)=O

<Checkpoint>[$\bf B$ is CC(SC1C=CCCCC1)=O](1)</Checkpoint>

(3) The reaction is the removal of the acetyl group by $\rm EtO^-$, $\bf C$ is [Na]SC1C=CCCCC1

<Checkpoint>[$\bf C$ is [Na]SC1C=CCCCC1](1)</Checkpoint>

(4) Acetylation can occur here, but it is meaningless since the acetyl group was just removed in the previous step, so it can only be the substitution of $\rm AcO^-$, $\bf D$ is O=[N+](CCSC1C=CCCCC1)[O-]

<Checkpoint>[$\bf D$ is O=[N+](CCSC1C=CCCCC1)[O-]](1)</Checkpoint>

(5) This condition is very classic, it can convert nitro compounds to nitrile oxide [O-][N+]#CCSC1C=CCCCC1, which is relatively reactive, and subsequent intramolecular [3+2] reaction easily occurs to give $\bf E$: C12C3CCCCC1ON=C2CS3

<MultiModal>![SMILES is [O-][N+]#CCSC1C=CCCCC1](/media/uploads/39bd3e26-a8ed-447a-aaf6-ff2401590c63.png)</MultiModal>

<Checkpoint>[$\bf E$ is C12C3CCCCC1ON=C2CS3](1)</Checkpoint>

Therefore, the intermediate contains a carbon-nitrogen triple bond, and $\bf E$ contains three rings, statement 1 is correct, statement 2 is wrong

(6) $\rm LiAlH_4$ reductively cleaves the N-O bond, and then reduces the carbon-nitrogen double bond to give $\bf F$: NC(CS1)C2C1CCCCC2O

<Checkpoint>[$\bf F$ is NC(CS1)C2C1CCCCC2O](1)</Checkpoint>

(7) $\rm MeOCOCl$ reacts with the most nucleophilic amino group in $\bf F$ to give $\bf G$: OC1CCCCC2C1C(NC(OC)=O)CS2

<Checkpoint>[$\bf G$ is OC1CCCCC2C1C(NC(OC)=O)CS2](1)</Checkpoint>

(8) If the hydroxyl group in $\bf F$ is acetylated, it is meaningless to add hydroxylamine later; therefore, DMSO has oxidizing properties, here it is similar to Swern oxidation, but the dehydrating agent is replaced by $\rm Ac_2O$. $\bf H$: O=C1CCCCC2C1C(NC(OC)=O)CS2

<Checkpoint>[$\bf H$ is O=C1CCCCC2C1C(NC(OC)=O)CS2](1)</Checkpoint>

The molecular formula of $\bf H$ is $\rm C_{11}H_{17}NO_3S$, statement 3 is correct

(9) Hydroxylamine condenses with the carbonyl group to give $\bf I$: O/N=C1CCCCC2C\1C(NC(OC)=O)CS2

<Checkpoint>[$\bf I$ is O/N=C1CCCCC2C\1C(NC(OC)=O)CS2](1)</Checkpoint>

(10) Beckmann rearrangement occurs to give $\bf J$: O=C(N1)CCCCC2C1C(NC(OC)=O)CS2

<Checkpoint>[$\bf J$ is O=C(N1)CCCCC2C1C(NC(OC)=O)CS2](1)</Checkpoint>

$\bf J$ contains a five-membered ring and an eight-membered ring, statement 4 is wrong

Beckmann rearrangement may also not involve alkyl migration, but rather cleavage of the C-C bond to form a carbocation (which can be stabilized by the neighboring sulfur atom), followed by closing the three-membered ring and dehydration to form O=C(N1C2C1CSC2CCCCC#N)OC, the molecular formula is $\rm C_{11}H_{16}N_2O_2S$, statement 5 is correct

<MultiModal>![SMILES is O=C(N1C2C1CSC2CCCCC#N)OC](/media/uploads/763aa6f8-82a8-4b2b-9130-82198f116a88.png)</MultiModal>

<Checkpoint>[Beckmann rearrangement can form O=C(N1C2C1CSC2CCCCC#N)OC byproduct](1)</Checkpoint>

(11) Hydrolyze the amide, then react the two amino groups with $\rm COCl_2$ to close the five-membered ring

<MultiModal>![The figures are $\bf A$~$\bf J$ respectively, see the above process for SMILES](/media/uploads/bfa00329-9d29-4037-877e-c3432929d9b1.png)</MultiModal>
(1)该反应为烯丙位的自由基卤化，$\bf A$为BrC1C=CCCCC1

<Checkpoint>[$\bf A$为BrC1C=CCCCC1](1)</Checkpoint>

(2)该反应为$\rm AcS^-$对于$\rm Br^-$的取代，$\bf B$为CC(SC1C=CCCCC1)=O

<Checkpoint>[$\bf B$为CC(SC1C=CCCCC1)=O](1)</Checkpoint>

(3)该反应为$\rm EtO^-$反应脱去乙酰基，$\bf C$为[Na]SC1C=CCCCC1

<Checkpoint>[$\bf C$为[Na]SC1C=CCCCC1](1)</Checkpoint>

(4)这里可以乙酰化，但上一步刚脱掉乙酰基，没有意义，只能是取代$\rm AcO^-$，$\bf D$为O=[N+](CCSC1C=CCCCC1)[O-]

<Checkpoint>[$\bf D$为O=[N+](CCSC1C=CCCCC1)[O-]](1)</Checkpoint>

(5)这个条件十分经典，可以将硝基化合物转化为氧化腈[O-][N+]#CCSC1C=CCCCC1，它比较活泼，后续容易发生分子内[3+2]反应得到$\bf E$：C12C3CCCCC1ON=C2CS3

<MultiModal>![SMILES为[O-][N+]#CCSC1C=CCCCC1](/media/uploads/39bd3e26-a8ed-447a-aaf6-ff2401590c63.png)</MultiModal>

<Checkpoint>[$\bf E$为C12C3CCCCC1ON=C2CS3](1)</Checkpoint>

因此中间体含碳氮三键，$\bf E$含三个环，说法1正确，说法2错误

(6)$\rm LiAlH_4$将N-O键还原裂解，接着还原碳氮双键得到$\bf F$：NC(CS1)C2C1CCCCC2O

<Checkpoint>[$\bf F$为NC(CS1)C2C1CCCCC2O](1)</Checkpoint>

(7)$\rm MeOCOCl$与$\bf F$中亲核性最强的氨基反应得到$\bf G$：OC1CCCCC2C1C(NC(OC)=O)CS2

<Checkpoint>[$\bf G$为OC1CCCCC2C1C(NC(OC)=O)CS2](1)</Checkpoint>

(8)若$\bf F$中羟基发生乙酰化，后续加入羟胺没有意义；因此而DMSO有氧化性，这里类似Swern氧化，将脱水剂换成了$\rm Ac_2O$。$\bf H$：O=C1CCCCC2C1C(NC(OC)=O)CS2

<Checkpoint>[$\bf H$为O=C1CCCCC2C1C(NC(OC)=O)CS2](1)</Checkpoint>

$\bf H$的分子式为$\rm C_{11}H_{17}NO_3S$，说法3正确

(9)羟胺与羰基缩合得到$\bf I$：O/N=C1CCCCC2C\1C(NC(OC)=O)CS2

<Checkpoint>[$\bf I$为O/N=C1CCCCC2C\1C(NC(OC)=O)CS2](1)</Checkpoint>

(10)发生Beckmann重排得到$\bf J$：O=C(N1)CCCCC2C1C(NC(OC)=O)CS2

<Checkpoint>[$\bf J$为O=C(N1)CCCCC2C1C(NC(OC)=O)CS2](1)</Checkpoint>

$\bf J$含五元环和八元环，说法4错误

Beckmann重排也可能不发生烷基迁移，而是断裂C-C键形成碳正离子（这里可以被硫原子邻基参与稳定），随后关三元环脱水，形成O=C(N1C2C1CSC2CCCCC#N)OC，分子式为$\rm C_{11}H_{16}N_2O_2S$，说法5正确

<MultiModal>![SMILES为O=C(N1C2C1CSC2CCCCC#N)OC](/media/uploads/763aa6f8-82a8-4b2b-9130-82198f116a88.png)</MultiModal>

<Checkpoint>[Beckmann重排可以形成O=C(N1C2C1CSC2CCCCC#N)OC副产物](1)</Checkpoint>

(11)将酰胺水解，随后两个氨基与$\rm COCl_2$反应关五元环

<MultiModal>![图中分别为$\bf A$~$\bf J$，SMILES见上面的过程](/media/uploads/bfa00329-9d29-4037-877e-c3432929d9b1.png)</MultiModal>