People utilized the following reaction to construct the azepine ring system:

<MultiModal>![In a mixed solvent of toluene and THF, at 70 degrees Celsius, `N#CC1=CC=NC=C1` and $\bf{A}$ react with $Cs_2CO_3$, $H_2O$, and 4 equivalents of iodine to yield `O=CC1=C(C#N)C=CN=C(C2=CC=CC=C2)C1`](/media/uploads/ff61b12d-f29f-4770-a405-0d06ee8948e6.png)</MultiModal>

Mechanism studies indicate that the reaction proceeds through the key intermediate ${\bf{X^+}}(C_{14}H_{12}IN_2)$; the salt of $\bf{X^+}$ can be isolated even if $CsCO_3$ and $H_2O$ are not added to the reaction. $\bf{X^+}$ further reacts with $I_2$ under the above reaction conditions to generate the final product.

Choose the correct option from the following:
人们利用以下反应构建了氮杂七元环系：

<MultiModal>![在甲苯、THF混合溶剂，70摄氏度下，`N#CC1=CC=NC=C1`和$\bf{A}$与$Cs_2CO_3$、 $H_2O$和4当量碘反应得到`O=CC1=C(C#N)C=CN=C(C2=CC=CC=C2)C1`](/media/uploads/ff61b12d-f29f-4770-a405-0d06ee8948e6.png)</MultiModal>

机理研究表明，反应经历关键中间体${\bf{X^+}}(C_{14}H_{12}IN_2)$；反应即使不加入$CsCO_3$和 $H_2O$ 也可以分离得到$\bf{X^+}$的盐。$\bf{X^+}$在以上反应条件下继续与$I_2$反应生成最终产物。

选择以下选项中正确的一项：
All other options are incorrect
The molecular weight of $\bf{A}$ is approximately 92.
The degree of unsaturation of $\bf{A}$ is 4.
$\bf{X^+}$ contains only 1 hexagon
$\bf{X^+}$ to obtain the product also requires consuming 1 molecule of iodine and 4 molecules of cesium carbonate (considered as a monobasic base)
To obtain `COC(=O)C1=CN=C(CC(=C1)C=O)C2=CC=C3C=CC=CC3=C2` under the same conditions, the two required reactants are `O=C(OC)C1=CC=NC=C1` and `C=CC1=CC2=CC=CC=C2C=C1`
其他选项均不正确
$\bf{A}$的分子量约为92
$\bf{A}$的不饱和度为4
$\bf{X^+}$中仅有1个六元环
$\bf{X^+}$得到产物还需消耗1分子碘和4分子碳酸铯（视为一元碱）
若想要在相同条件下得到`COC(=O)C1=CN=C(CC(=C1)C=O)C2=CC=C3C=CC=CC3=C2`，需要的两种反应物为`O=C(OC)C1=CC=NC=C1`和`C=CC1=CC2=CC=CC=C2C=C1`
$\bf{X^+}$ has $C_8H_8$ and one iodine more than `N#CC1=CC=NC=C1`, so the molecular formula of $\bf{A}$ should be $C_8H_8$. The most reasonable $C_8H_8$ alkene is styrene. Therefore, the molecular weight of $\bf{A}$ is 104 and the degree of unsaturation is 5.

<Checkpoint>[The molecular formula of $\bf{A}$ should be $C_8H_8$, the molecular weight is 104 and the degree of unsaturation is 5, options BC are incorrect](2)</Checkpoint>

After the double bond in $\bf{A}$ reacts with iodine, it is nucleophilically attacked by `N#CC1=CC=NC=C1` to obtain $\bf{X^+}$: `ICC(C1=CC=CC=C1)[N+]2=CC=C(C#N)C=C2`, which has two six-membered rings.

<MultiModal>![`ICC(C1=CC=CC=C1)[N+]2=CC=C(C#N)C=C2`](/media/uploads/ff18cdd5-7ba4-4d7c-be21-8a9c80ec983d.png)</MultiModal>

<Checkpoint>[The structure of $\bf{X^+}$ is `ICC(C1=CC=CC=C1)[N+]2=CC=C(C#N)C=C2`, which has two six-membered rings, option D is incorrect](1)</Checkpoint>

The mechanism for obtaining the product from $\bf{X^+}$ is:

<MultiModal>![First, cesium carbonate, as a base, eliminates $HI$ in $\bf{X^+}$ to generate a double bond: `C=C(C1=CC=CC=C1)[N+]2=CC=C(C#N)C=C2`, then a water molecule attacks and departs a proton to generate a hydroxyl group: `C=C(N1C=CC(C#N)=CC1O)C2=CC=CC=C2`, the newly generated double bond is attacked by a molecule of new iodine: `N#CC1=CC(O)N(C2(C3=CC=CC=C3)C[I+]2)C=C1`, the hydrogen of the newly generated hydroxyl group is abstracted by a base: `N#CC1=CC([O-])N(C2(C3=CC=CC=C3)C[I+]2)C=C1`, then the six-membered ring opens: `N#CC(/C=C\[N-]C1(C2=CC=CC=C2)C[I+]1)=C/C=O`, the negative charge on the nitrogen attacks along the conjugated double bond to open the iodine three-membered ring to generate a seven-membered ring: `N#CC1=CC=NC(I)(C2=CC=CC=C2)CC1C=O`, finally the base abstracts the α-hydrogen of the aldehyde group with the strongest acidity, and the iodine atom leaves to obtain the product](/media/uploads/835bcdb2-a1db-44be-b997-bb55473bb089.png)</MultiModal>

According to the mechanism, one molecule of iodine is consumed and 4 protons need to be consumed. Each molecule of cesium carbonate can neutralize 1 proton (cesium bicarbonate is not basic enough), so 1 molecule of iodine and 4 molecules of base need to be consumed.

<Checkpoint>[Obtaining the product from $\bf{X^+}$ requires consuming 1 molecule of iodine and 4 molecules of base, option E is correct](1)</Checkpoint>

According to the mechanism, the aryl group on the alkene is located at the 2-position, the newly generated aldehyde group is located at the 4-position, the two 2-positions of the original pyridine are the 7-position and the aldehyde carbon of the seven-membered ring, respectively, the two 3-positions of the pyridine are the 6-position and the 4-position of the seven-membered ring, respectively, and the 4-position of the pyridine is the 5-position of the seven-membered ring. The aryl group of `COC(=O)C1=CN=C(CC(=C1)C=O)C2=CC=C3C=CC=CC3=C2` is a naphthyl group, so the corresponding alkene in the reaction is `C=CC1=CC2=CC=CC=C2C=C1`; the substituent methyl ester group is located at the 6-position of the seven-membered ring, corresponding to the 3-position of the original pyridine, and the pyridine derivative is `O=C(OC)C1=CC=CN=C1`.

<Checkpoint>[The two reactants are `C=CC1=CC2=CC=CC=C2C=C1` and `O=C(OC)C1=CC=CN=C1` respectively, option F is incorrect](1)</Checkpoint>

The structure of `COC(=O)C1=CN=C(CC(=C1)C=O)C2=CC=C3C=CC=CC3=C2` is:

<MultiModal>![`COC(=O)C1=CN=C(CC(=C1)C=O)C2=CC=C3C=CC=CC3=C2`](/media/uploads/c65aa423-3ef4-41ed-8092-b6b1f25eb6cd.png)</MultiModal>
$\bf{X^+}$比`N#CC1=CC=NC=C1`多出 $C_8H_8$并含一个碘，故$\bf{A}$的分子式应为$C_8H_8$。最合理的$C_8H_8$烯烃就是苯乙烯。因此$\bf{A}$的分子量为104且不饱和度为5。

<Checkpoint>[$\bf{A}$的分子式应为$C_8H_8$，分子量为104且不饱和度为5，选项BC错误](2)</Checkpoint>

$\bf{A}$中的双键与碘反应后，被`N#CC1=CC=NC=C1`亲核进攻，得到$\bf{X^+}$：`ICC(C1=CC=CC=C1)[N+]2=CC=C(C#N)C=C2`，有两个六元环。

<MultiModal>![`ICC(C1=CC=CC=C1)[N+]2=CC=C(C#N)C=C2`](/media/uploads/ff18cdd5-7ba4-4d7c-be21-8a9c80ec983d.png)</MultiModal>

<Checkpoint>[$\bf{X^+}$的结构为`ICC(C1=CC=CC=C1)[N+]2=CC=C(C#N)C=C2`，有两个六元环，选项D错误](1)</Checkpoint>

从$\bf{X^+}$得到产物的机理为：

<MultiModal>![首先碳酸铯作为碱，消除$\bf{X^+}$中的$HI$，产生双键：`C=C(C1=CC=CC=C1)[N+]2=CC=C(C#N)C=C2`，随后水分子进攻，离去质子产生羟基：`C=C(N1C=CC(C#N)=CC1O)C2=CC=CC=C2`，新产生的双键被一分子新的碘进攻：`N#CC1=CC(O)N(C2(C3=CC=CC=C3)C[I+]2)C=C1`，新产生羟基的氢被碱拔去：`N#CC1=CC([O-])N(C2(C3=CC=CC=C3)C[I+]2)C=C1`，随后六元环开环：`N#CC(/C=C\[N-]C1(C2=CC=CC=C2)C[I+]1)=C/C=O`，氮上的负电荷沿共轭双键进攻打开碘三元环产生七元环：`N#CC1=CC=NC(I)(C2=CC=CC=C2)CC1C=O`，最后碱拔去酸性最强的醛基α氢，离去碘原子得到产物](/media/uploads/835bcdb2-a1db-44be-b997-bb55473bb089.png)</MultiModal>

根据机理可知，共消耗一分子碘且需消耗4个质子，每分子碳酸铯可中和1个质子（碳酸氢铯碱性不足），则还需消耗1分子碘和4分子碱。

<Checkpoint>[$\bf{X^+}$得到产物还需消耗1分子碘和4分子碱，选项E正确](1)</Checkpoint>

结合机理可知，烯烃所带的芳基位于2号位，新产生的醛基位于4号位，原先吡啶的两个2号位分别为七元环的7号位和醛基碳，吡啶的两个3号位分别为七元环的6号位和4号位，吡啶的4号位为七元环的5号位。`COC(=O)C1=CN=C(CC(=C1)C=O)C2=CC=C3C=CC=CC3=C2`的芳基为萘基，则对应反应中的烯烃为`C=CC1=CC2=CC=CC=C2C=C1`；其中的取代基甲酯基位于七元环6号位，对应原先吡啶的3号位，吡啶衍生物为`O=C(OC)C1=CC=CN=C1`。

<Checkpoint>[两种反应物分别为`C=CC1=CC2=CC=CC=C2C=C1`和`O=C(OC)C1=CC=CN=C1`，选项F错误](1)</Checkpoint>

`COC(=O)C1=CN=C(CC(=C1)C=O)C2=CC=C3C=CC=CC3=C2`的结构为：

<MultiModal>![`COC(=O)C1=CN=C(CC(=C1)C=O)C2=CC=C3C=CC=CC3=C2`](/media/uploads/c65aa423-3ef4-41ed-8092-b6b1f25eb6cd.png)</MultiModal>