The final product of the illustrated synthetic route is a hydrocarbon with a relative molecular mass between 150 and 160, whose molecule contains hydrogen atoms in two chemical environments and belongs to the $D_\mathrm{2h}$ point group. Identify the correct statement about it from the following.
<MultiModal>![The image shows a multi-step reaction: CC1=CC=CC=C1>F>[G],[G]>H>[I],[I]>J>[K],[K]>O=NO[Na].Cl>[L],[L]>>[Final Product], the reaction from K to L is carried out at 0~5℃, and the last step reaction is carried out under heating and produces carbon dioxide and nitrogen](/media/uploads/9d36bc4c-b2fa-4fea-afca-6efe28bcdcde.png)</MultiModal>
图示合成路线的最终产物为某种相对分子质量介于150至160之间的烃，其分子中有两种化学环境的氢，属于$D_\mathrm{2h}$点群。关于其的以下说法中，找出正确的一项。
<MultiModal>![图片中为多步反应：CC1=CC=CC=C1>F>[G],[G]>H>[I],[I]>J>[K],[K]>O=NO[Na].Cl>[L],[L]>>[最终产物]，K到L的反应在0~5℃下进行，最后一步反应在加热下进行且产生了二氧化碳和氮气](/media/uploads/9d36bc4c-b2fa-4fea-afca-6efe28bcdcde.png)</MultiModal>
The condition $\mathbf{F}$ may include only one substance (a certain acid).
The condition $\mathbf{H}$ contains a reducing agent.
The condition $\mathbf{J}$ contains an oxidant.
The two substituents in $\mathbf{G}$ are in the para position.
The two substituents of $\mathbf{I}$ exhibit different types of electronic effects on the benzene ring.
$\mathbf{K}$ can form an inner salt, but the conjugated system of its inner salt is smaller compared to the form without formal charges.
The formal charge within the $\mathbf{L}$ molecule can be eliminated through the formation of a quinoid structure.
The final product of this synthetic route does not contain structures with ring strain.
条件$\mathbf{F}$可以仅包含一种物质（某种酸）。
条件$\mathbf{H}$中含有还原剂。
条件$\mathbf{J}$中含有氧化剂。
$\mathbf{G}$中两个取代基处于对位。
$\mathbf{I}$的两个取代基对苯环的电子效应类型不同。
$\mathbf{K}$可形成内盐，但其内盐的共轭体系相较于不带形式电荷的形态更小。
$\mathbf{L}$分子内的形式电荷可通过形成醌式结构消除。
该合成路线的最终产物不含具有环张力的结构。
The product is a hydrocarbon, and its relative molecular mass is between 150 and 160. Considering that the relative molecular mass of benzene is $78.108$, and the last two steps are the formation and elimination of carboxy diazonium salt, it is considered that there may be coupling of two benzene rings in the product.

<Checkpoint>[The product has two benzene rings](1)</Checkpoint>

Since the hydrogen of the product has two kinds of chemical environments, it cannot be biphenyl, and a reasonable result is biphenylene.

<MultiModal>![C12=CC=CC=C1C3=C2C=CC=C3](/media/uploads/7a105200-ccc4-4ccb-9bd3-577a9408bf96.png)</MultiModal>

<Checkpoint>[The final product is C12=CC=CC=C1C3=C2C=CC=C3](1)</Checkpoint>

Since the product molecule contains a four-membered ring, option $\mathrm{H}$ is incorrect.

<Checkpoint>[The product contains a four-membered ring](1)</Checkpoint>

Retrosynthetic analysis starts from the product. From $\mathbf{L}$ to the product, the condition is heating, removing $\mathrm{N_2,CO_2}$, so $\mathbf{L}$ is carboxy diazonium salt, and inferred from the four-membered ring structure of the product, the carboxy and diazonium groups in $\mathbf{L}$ are in the ortho position.

<Checkpoint>[$\mathbf{L}$ is ortho-carboxy diazonium salt](1)</Checkpoint>

Its structure is

<MultiModal>![O=C(C1=C([N+]#N)C=CC=C1)[O-]](/media/uploads/4a09d73e-7604-4918-8bec-15ae9c844ef1.png)</MultiModal>

<Checkpoint>[$\mathbf{L}$ is O=C(C1=C([N+]#N)C=CC=C1)[O-]](1)</Checkpoint>

The formal charge of diazonium salt cannot be eliminated by the quinone structure, option $\mathrm{G}$ is incorrect.

<Checkpoint>[The formal charge of diazonium salt cannot be eliminated by the quinone structure](1)</Checkpoint>

From $\mathbf{K}$ to $\mathbf{L}$, acidic nitrous acid conditions generate diazonium salt. A reasonable reaction is to convert the amino group to diazonium salt, so $\mathbf{K}$ is amino acid.

<MultiModal>![NC1=C(C(O)=O)C=CC=C1](/media/uploads/dea84e39-c1ec-4e45-840e-5632aa8f5079.png)</MultiModal>

<Checkpoint>[$\mathbf{K}$ molecule contains an amino group](1)</Checkpoint>

<Checkpoint>[$\mathbf{K}$ is NC1=C(C(O)=O)C=CC=C1](1)</Checkpoint>

After $\mathbf{K}$ forms an inner salt, the lone pair electrons originally involved in conjugation on the nitrogen atom are used to form an $\mathrm{N-H}$ bond, and the conjugated system has one less nitrogen atom than the form without a formal charge. Therefore, option $\mathrm{F}$ is correct.

<Checkpoint>[The conjugated system of the inner salt of $\mathbf{K}$ has one less nitrogen atom than the structure without a formal charge](1)</Checkpoint>

From the initial substrate toluene to $\mathbf{K}$, after three steps of reaction, the carboxyl group in $\mathbf{K}$ is oxidized from the methyl group, and the common method for introducing amino groups on the benzene ring is to first introduce a nitro group and then reduce it. On the other hand, since the methyl group is an electron-donating group and the carboxyl group is an electron-withdrawing group, in order to place the nitro group in the ortho position, the nitro group needs to be introduced first and then oxidized. Since the amino group has strong reducing properties, the step of oxidizing the methyl group to a carboxyl group needs to be before reducing the nitro group to an amino group. From this, the conditions of $\mathbf{F,H,J}$ and the structures of intermediates $\mathbf{G,I}$ can be deduced.

<Checkpoint>[Nitro group introduction step is before methyl oxidation](1)</Checkpoint>

<Checkpoint>[Nitro reduction step is after methyl oxidation](1)</Checkpoint>

Therefore, $\mathbf{F}$ is $\mathrm{H_2SO_4,HNO_3}$. $\mathbf{F}$ involves two acids, and $\mathrm{A}$ is incorrect.

<Checkpoint>[$\mathbf{F}$ is $\mathrm{H_2SO_4,HNO_3}$](1)</Checkpoint>

$\mathbf{G}$ is CC1=C([N+]([O-])=O)C=CC=C1

<MultiModal>![CC1=C([N+]([O-])=O)C=CC=C1](/media/uploads/35dac384-3619-4db5-9fda-3ec4908398d8.png)</MultiModal>

<Checkpoint>[$\mathbf{G}$ is CC1=C([N+]([O-])=O)C=CC=C1](1)</Checkpoint>

Since $\mathbf{L}$ is ortho, it is inferred that $\mathbf{G}$ is also ortho, and $\mathrm{D}$ is incorrect.

In order to protect the amino group from oxidation, it should be oxidized first and then reduced. Therefore, $\mathbf{H}$ is an oxidizing agent and $\mathbf{J}$ is a reducing agent, and both $\mathrm{B}$ and $\mathrm{C}$ are incorrect.

<Checkpoint>[$\mathbf{H}$ is an oxidizing agent, $\mathbf{J}$ is a reducing agent,](1)</Checkpoint>

The structure of $\mathbf{I}$ is O=C(C1=C([N+]([O-])=O)C=CC=C1)O

<MultiModal>![O=C(C1=C([N+]([O-])=O)C=CC=C1)O](/media/uploads/64a31f92-d2b5-4f96-98d2-ab8c417b4331.png)</MultiModal>

<Checkpoint>[$\mathbf{I}$ is O=C(C1=C([N+]([O-])=O)C=CC=C1)O](1)</Checkpoint>

Among them, both carboxyl and nitro groups have strong electron-withdrawing effects, so option $\mathrm{E}$ is incorrect.

<Checkpoint>[Carboxyl and nitro groups are both electron-withdrawing groups](1)</Checkpoint>
产物为烃，且其相对分子质量介于150至160之间。考虑到苯的相对分子质量是$78.108$，且最后两步分别是羧基重氮盐的形成和消除，故考虑产物中可能有两个苯环的偶联。

<Checkpoint>[产物有两个苯环](1)</Checkpoint>

由于产物的氢有两种化学环境，故不能是联苯，合理的结果为双苯并环丁烯。

<MultiModal>![C12=CC=CC=C1C3=C2C=CC=C3](/media/uploads/7a105200-ccc4-4ccb-9bd3-577a9408bf96.png)</MultiModal>

<Checkpoint>[最终产物为C12=CC=CC=C1C3=C2C=CC=C3](1)</Checkpoint>

由于产物分子中含有四元环，选项$\mathrm{H}$错误。

<Checkpoint>[产物含有四元环](1)</Checkpoint>

从产物开始进行逆合成分析，从$\mathbf{L}$到产物，条件为加热，脱去$\mathrm{N_2,CO_2}$，故$\mathbf{L}$为羧基重氮盐，且由产物的四元环结构推断，$\mathbf{L}$中的羧基和重氮基团处于邻位。

<Checkpoint>[$\mathbf{L}$为邻位羧基重氮盐](1)</Checkpoint>

其结构为

<MultiModal>![O=C(C1=C([N+]#N)C=CC=C1)[O-]](/media/uploads/4a09d73e-7604-4918-8bec-15ae9c844ef1.png)</MultiModal>

<Checkpoint>[$\mathbf{L}$为O=C(C1=C([N+]#N)C=CC=C1)[O-]](1)</Checkpoint>

重氮盐的形式电荷不可通过醌式结构消除，选项$\mathrm{G}$错误。

<Checkpoint>[重氮盐的形式电荷不可通过醌式结构消除](1)</Checkpoint>

从$\mathbf{K}$到$\mathbf{L}$，酸性亚硝酸条件生成重氮盐，合理的反应是将氨基转化为重氮盐，故$\mathbf{K}$为氨基酸

<MultiModal>![NC1=C(C(O)=O)C=CC=C1](/media/uploads/dea84e39-c1ec-4e45-840e-5632aa8f5079.png)</MultiModal>

<Checkpoint>[$\mathbf{K}$分子中有氨基](1)</Checkpoint>

<Checkpoint>[$\mathbf{K}$为NC1=C(C(O)=O)C=CC=C1](1)</Checkpoint>

$\mathbf{K}$形成内盐后，氮原子上原本参与共轭的孤对电子用于形成$\mathrm{N-H}$键，共轭体系比不带形式电荷的形式少一个氮原子。故选项$\mathrm{F}$正确。

<Checkpoint>[$\mathbf{K}$的内盐的共轭体系比无形式电荷结构少一个氮原子](1)</Checkpoint>

从最初的底物甲苯到$\mathbf{K}$，经过三步反应，$\mathbf{K}$中的羧基由甲基氧化而来，而苯环上氨基的常见引入方法为先引入硝基再还原。另一方面，由于甲基为给电子基团，而羧基为吸电子基团，为了将硝基上在邻位，需要先引入硝基后氧化。由于氨基具有较强还原型，将甲基氧化为羧基的步骤需位于将硝基还原为氨基之前。由此可推导出$\mathbf{F,H,J}$的条件和中间体$\mathbf{G,I}$的结构。

<Checkpoint>[硝基引入步骤在甲基氧化前](1)</Checkpoint>

<Checkpoint>[硝基还原步骤在甲基氧化后](1)</Checkpoint>

因此$\mathbf{F}$为$\mathrm{H_2SO_4,HNO_3}$，$\mathbf{F}$涉及两种酸，$\mathrm{A}$错误。

<Checkpoint>[$\mathbf{F}$为$\mathrm{H_2SO_4,HNO_3}$](1)</Checkpoint>

$\mathbf{G}$为CC1=C([N+]([O-])=O)C=CC=C1

<MultiModal>![CC1=C([N+]([O-])=O)C=CC=C1](/media/uploads/35dac384-3619-4db5-9fda-3ec4908398d8.png)</MultiModal>

<Checkpoint>[$\mathbf{G}$为CC1=C([N+]([O-])=O)C=CC=C1](1)</Checkpoint>

由于$\mathbf{L}$是邻位，反推出$\mathbf{G}$也是邻位，$\mathrm{D}$错误。

为了保护氨基不受氧化，应当先氧化后还原，因此$\mathbf{H}$为氧化剂，$\mathbf{J}$为还原剂，$\mathrm{B}$和$\mathrm{C}$均错误。

<Checkpoint>[$\mathbf{H}$为氧化剂，$\mathbf{J}$为还原剂，](1)</Checkpoint>

$\mathbf{I}$的结构为O=C(C1=C([N+]([O-])=O)C=CC=C1)O

<MultiModal>![O=C(C1=C([N+]([O-])=O)C=CC=C1)O](/media/uploads/64a31f92-d2b5-4f96-98d2-ab8c417b4331.png)</MultiModal>

<Checkpoint>[$\mathbf{I}$为O=C(C1=C([N+]([O-])=O)C=CC=C1)O](1)</Checkpoint>

其中，羧基和硝基都具有强的吸电子效应，故选项$\mathrm{E}$错误。

<Checkpoint>[羧基和硝基都是吸电子基团](1)</Checkpoint>