A synthetic route for $\bf{G}$ is as follows:

<MultiModal>![Synthetic route: Compound $\bf{1}$, SMILES is ClCC1CO1, generates compound $\bf{2}$ in the presence of $\bf{A}$, the SMILES of compound $\bf{2}$ is OC1CN(C(C2=CC=CC=C2)C3=CC=CC=C3)C1, compound $\bf{2}$ generates $\bf{B}$ under the action of $MsCl$, its chemical formula is $C_{17}H_{20}ClNO_3S$, $\bf{B}$ undergoes a one-step hydrogen reaction, a one-step reaction with C=CCCC(Cl)=O to generate $\bf{C}$, $\bf{C}$ reacts with potassium tert-butoxide in tert-butanol to generate $\bf{D}$ with the chemical formula $C_8H_{11}NO$, $\bf{D}$ generates $\bf{E}$ under FVP conditions at $500℃$, $\bf{E}$ generates $\bf{F}$ upon heating, $\bf{F}$ reacts with hydrogen and lithium aluminum hydride to generate $\bf{G}$ with the chemical formula $C_8H_{15}N$](/media/uploads/bfa849dc-2f3d-4d31-8431-e26e12890ad2.jpg)</MultiModal>

The following statements are made:

1. $\bf{C}$ has one chiral carbon atom

2. The degree of unsaturation of $\bf{C}$ is 3

3. $\bf{E}$ contains a four-membered ring

4. $\bf{G}$ has two rings

5. $\bf{A}$ has 13 carbon atoms

6. $\bf{A}$ can be replaced by $NH_3$

Which of the following options is completely correct
$\bf{G}$的一种合成路线如下：

<MultiModal>![合成路线：化合物$\bf{1}$，SMILES为ClCC1CO1，在$\bf{A}$的存在下生成化合物$\bf{2}$，化合物$\bf{2}$的SMILES为OC1CN(C(C2=CC=CC=C2)C3=CC=CC=C3)C1，化合物$\bf{2}$在$MsCl$作用下生成$\bf{B}$，其化学式为$C_{17}H_{20}ClNO_3S，$\bf{B}$在经历一步氢气反应，一步与C=CCCC(Cl)=O反应后生成$\bf{C}$，$\bf{C}$在叔丁醇中与叔丁醇钾反应生成化学式为$C_8H_{11}NO$的$\bf{D}$，$\bf{D}$在$500℃$的FVP条件下生成$\bf{E}$，$\bf{E}$加热生成$\bf{F}$，$\bf{F}$与氢气、氢化铝锂反应生成化学式为$C_8H_{15}N$的$\bf{G}$](/media/uploads/bfa849dc-2f3d-4d31-8431-e26e12890ad2.jpg)</MultiModal>

有以下说法：

1.$\bf{C}$中有一个手性碳原子

2.$\bf{C}$的不饱和度为3

3.$\bf{E}$含有四元环

4.$\bf{G}$中有两个环

5.$\bf{A}$中有13个碳原子

6.$\bf{A}$可以用$NH_3$替代

下列选项中完全正确的是
1,3,6
1,4
2,3,5
2,4,5
2,4,5,6
1,5,6
3,4,6
3,5,6
4,5,6
2,3,5
2,3,6
All of the above options are incorrect or incomplete.
1,3,6
1,4
2,3,5
2,4,5
2,4,5,6
1,5,6
3,4,6
3,5,6
4,5,6
2,3,5
2,3,6
以上选项均不正确或不完整
Compound $\bf{1}$, SMILES is ClCC1CO1, generates compound $\bf{2}$ under condition $\bf{A}$. In $\bf{2}$, the epoxide ring opens, forming a new nitrogen-containing four-membered ring. Combined with the position of the hydroxyl group, it can be known that the nitrogen atom attacks the 1st and 3rd positions of $\bf{1}$. Then the structure of $\bf{A}$ is NC(C1=CC=CC=C1)C2=CC=CC=C2, containing 13 carbon atoms.

<Checkpoint>[The structure of $\bf{A}$ is NC(C1=CC=CC=C1)C2=CC=CC=C2, containing 13 carbon atoms, statement 5 is correct](1)</Checkpoint>

$\bf{2}$ generates $\bf{B}$ under the action of $MsCl$. There is only 1 $S$ in $\bf{B}$, which should be one molecule of $MsCl$ reacting with the hydroxyl group. Therefore, it is easy to obtain its structure as CS(OC1C[N+](C1)([H])C(C2=CC=CC=C2)C3=CC=CC=C3)(=O)=O.[Cl-], which conforms to the chemical formula $C_{17}H_{20}ClNO_3S$.

<Checkpoint>[The structure of $\bf{B}$ is CS(OC1C[N+](C1)([H])C(C2=CC=CC=C2)C3=CC=CC=C3)(=O)=O.[Cl-]](1)</Checkpoint>

The purpose of introducing the sterically hindered diphenylmethyl group on the amino group here is to prevent it from reacting with $MsCl$. If ammonia is used in the previous step, side reactions may occur in this step, so $\bf{A}$ cannot be replaced with ammonia.

<Checkpoint>[The purpose of introducing the sterically hindered diphenylmethyl group on the amino group is to prevent it from reacting with $MsCl$. If ammonia is used in the previous step, side reactions may occur in this step, so $\bf{A}$ cannot be replaced with ammonia, statement 6 is incorrect](1)</Checkpoint>

$\bf{B}$ is hydrogenolyzed to remove the diphenylmethyl group, and then reacted with acyl chloride C=CCCC(Cl)=O to obtain O=C(CCC=C)N1CC(OS(C)(=O)=O)C1, which is $\bf{C}$. There is a four-membered ring and two double bonds, with a degree of unsaturation of 3, and it does not have a chiral center.

<Checkpoint>[The structure of $\bf{C}$ is O=C(CCC=C)N1CC(OS(C)(=O)=O)C1, which has a four-membered ring and two double bonds, with a degree of unsaturation of 3, and does not have a chiral center, statement 1 is incorrect, statement 2 is correct](2)</Checkpoint>

After $\bf{C}$ is treated with potassium tert-butoxide, $S$ disappears, then elimination occurs to obtain $\bf{D}$: O=C(CCC=C)N1C=CC1, which conforms to the chemical formula $C_8H_{11}NO$.

<Checkpoint>[The structure of $\bf{D}$ is O=C(CCC=C)N1C=CC1](1)</Checkpoint>

$\bf{D}$ undergoes four-membered ring opening under flash vacuum pyrolysis conditions at $500℃$ to obtain $\bf{E}$: O=C(CCC=C)/N=C/C=C

<Checkpoint>[The structure of $\bf{E}$ is O=C(CCC=C)/N=C/C=C, does not contain a four-membered ring, statement 3 is incorrect](1)</Checkpoint>

$\bf{E}$ undergoes a Diels-Alder reaction upon heating to obtain $\bf{F}$: O=C1N2C=CCCC2CC1

<Checkpoint>[The structure of $\bf{F}$ is O=C1N2C=CCCC2CC1](1)</Checkpoint>

$\bf{F}$ hydrogenates to reduce the double bond, and lithium aluminum hydride reduces the amide to amine, to obtain $\bf{G}$: N12CCCCC1CCC2, which conforms to the chemical formula $C_8H_{15}N$, and has two rings.

<Checkpoint>[The structure of $\bf{G}$ is N12CCCCC1CCC2, the chemical formula has two rings, statement 4 is correct](1)</Checkpoint>

Select option D
化合物$\bf{1}$，SMILES为ClCC1CO1，在条件$\bf{A}$的作用下生成化合物$\bf{2}$，$\bf{2}$中环氧打开，形成了新的含氮四元环，结合羟基位置，可以知道是氮原子进攻$\bf{1}$的1号位和3号位，则$\bf{A}$的结构为NC(C1=CC=CC=C1)C2=CC=CC=C2，含13个碳原子

<MultiModal>![NC(C1=CC=CC=C1)C2=CC=CC=C2](/media/uploads/002a023c-04a1-409d-96c8-383040bd58b3.png)</MultiModal>

<Checkpoint>[$\bf{A}$的结构为NC(C1=CC=CC=C1)C2=CC=CC=C2，含13个碳原子，说法5正确](1)</Checkpoint>

$\bf{2}$在$MsCl$作用下生成$\bf{B}$，$\bf{B}$中只有1个$S$，应该为一分子$MsCl$与羟基反应，因此容易得到其结构为CS(OC1C[N+](C1)([H])C(C2=CC=CC=C2)C3=CC=CC=C3)(=O)=O.[Cl-]，符合化学式$C_{17}H_{20}ClNO_3S$

<MultiModal>![CS(OC1C[N+](C1)([H])C(C2=CC=CC=C2)C3=CC=CC=C3)(=O)=O.[Cl-]](/media/uploads/f5e0562a-0ad2-486b-8fd3-96118b5948b3.png)</MultiModal>

<Checkpoint>[$\bf{B}$的结构为CS(OC1C[N+](C1)([H])C(C2=CC=CC=C2)C3=CC=CC=C3)(=O)=O.[Cl-]](1)</Checkpoint>

此处引入氨基上大位阻二苯甲基的目的是阻止其与$MsCl$反应，若上一步换用氨则这一步可能出现副反应，故不能将$\bf{A}$替换为氨

<Checkpoint>[引入氨基上大位阻二苯甲基的目的是阻止其与$MsCl$反应，若上一步换用氨则这一步可能出现副反应，故不能将$\bf{A}$替换为氨，说法6错误](1)</Checkpoint>

$\bf{B}$氢解，脱去二苯甲基，随后与酰氯C=CCCC(Cl)=O反应，得到O=C(CCC=C)N1CC(OS(C)(=O)=O)C1，即$\bf{C}$，其中有一个四元环和两个双键，不饱和度为3，且不具有手性中心。

<MultiModal>![O=C(CCC=C)N1CC(OS(C)(=O)=O)C1](/media/uploads/1daae6cc-f7de-4521-af3c-edd180c622a8.png)</MultiModal>

<Checkpoint>[$\bf{C}$的结构为O=C(CCC=C)N1CC(OS(C)(=O)=O)C1，其中有一个四元环和两个双键，不饱和度为3，不具有手性中心，说法1错误，说法2正确](2)</Checkpoint>

$\bf{C}$在经叔丁醇钾处理后$S$消失，则发生消除得到$\bf{D}$：O=C(CCC=C)N1C=CC1，符合化学式$C_8H_{11}NO$

<MultiModal>![O=C(CCC=C)N1C=CC1](/media/uploads/a7139ba3-8a27-45a2-a4a4-88a3b6b05aaf.png)</MultiModal>

<Checkpoint>[$\bf{D}$的结构为O=C(CCC=C)N1C=CC1](1)</Checkpoint>

$\bf{D}$在$500℃$的闪式真空热解条件下四元环打开，得到$\bf{E}$：O=C(CCC=C)/N=C/C=C

<MultiModal>![O=C(CCC=C)/N=C/C=C](/media/uploads/4eac81ab-f93c-4dfa-9d3a-f33c93c22e96.png)</MultiModal>

<Checkpoint>[$\bf{E}$的结构为O=C(CCC=C)/N=C/C=C，不含有四元环，说法3错误](1)</Checkpoint>

$\bf{E}$加热发生Diels-Alder反应，得到$\bf{F}$：O=C1N2C=CCCC2CC1

<MultiModal>![O=C1N2C=CCCC2CC1](/media/uploads/f4e231da-1fce-444f-8901-23657e8c6fb7.png)</MultiModal>

<Checkpoint>[$\bf{F}$的结构为O=C1N2C=CCCC2CC1](1)</Checkpoint>

$\bf{F}$加氢还原双键，氢化铝锂还原酰胺为胺，得到$\bf{G}$：N12CCCCC1CCC2，符合化学式$C_8H_{15}N$，有两个环

<MultiModal>![N12CCCCC1CCC2](/media/uploads/ed99ab04-f2ec-4bbb-a263-c2539aeca566.png)</MultiModal>

<Checkpoint>[$\bf{G}$的结构为N12CCCCC1CCC2，化学式有两个环，说法4正确](1)</Checkpoint>

选择选项D