$\mathbf{A}$ and $\mathbf{B}$ undergo [4+2] cycloaddition reaction to obtain compound $\mathbf{C}$ ($\mathrm{C_8H_{10}O_2}$). $\mathbf{C}$ is hydrolyzed in acidic environment to $\mathbf{D}$ and methanol, and hydrolyzed in alkaline environment to $\mathbf{E}$ and methanol. $\mathbf{E}$ can spontaneously convert to $\mathbf{D}$ under acidic conditions. $\mathbf{C}$ reacts with excess methylmagnesium chloride in tetrahydrofuran to obtain $\mathbf{F}$. $\mathbf{F}$ is treated with aqueous ammonium chloride solution to obtain the main product $\mathbf{G}$ $(\mathrm{C_9H_{14}O})$ and a small amount of $\mathbf{H}$. $\mathbf{G}$ can be converted relatively completely to $\mathbf{H}$ by treatment with concentrated acid, with the release of gas.

The following options are correct:
$\mathbf{A}$和$\mathbf{B}$发生[4+2]环加成反应得到化合物$\mathbf{C}$($\mathrm{C_8H_{10}O_2)}$。$\mathbf{C}$在酸性环境中水解为$\mathbf{D}$和甲醇, 碱性环境中水解成$\mathbf{E}$和甲醇, $\mathbf{E}$可在酸性环境下自发转化为$\mathbf{D}$。$\mathbf{C}$ 和足量甲基氯化镁在四氢呋喃中反应, 得到 $\mathbf{F}$, $\mathbf{F}$ 用氯化铵水溶液处理, 获得主要产物 $\mathbf{G}$ $(\mathrm{C_9H_{14}O})$ 和少量$\mathbf{H}$。$\mathbf{G}$用浓酸处理可以较完全地转化为$\mathbf{H}$,同时有气体放出。

下列选项正确的是：
All other options are incorrect
$\mathbf{A，B，D}$ can all be deprotonated by a strong base (such as potassium tert-butoxide)
$\mathbf{H}$ does not possess strong toxicity.
Generation of $\mathbf{H}$ from $\mathbf{G}$ is accompanied by the formation of one molecule of a small molecule with a relative molecular mass of 44.
$\mathbf{F}$ possesses a chiral carbon atom.
$\mathbf{G}$'s product in a strong alkaline solution (such as an ethanol solution of potassium hydroxide) has a tetrasubstituted double bond.
More than 3 correct items are in the options.
Exactly two of the other options are correct.
其他选项均不正确
$\mathbf{A，B，D}$均可被强碱（如叔丁醇钾）夺取质子
$\mathbf{H}$不具有很强的毒性
由$\mathbf{G}$生成$\mathbf{H}$的同时伴随着一分子相对分子质量为44的小分子生成
$\mathbf{F}$具有手性碳原子
$\mathbf{G}$在强碱溶液中（如氢氧化钾的乙醇溶液）生成的产物具有四取代双键
选项中有3个以上的正确项
其他选项中恰有两个正确项
$\mathbf{A}$ and $\mathbf{B}$ undergo [4+2] cycloaddition reaction to obtain compound $\mathbf{C}$ $\mathrm{(C_8H_{10}O_2)}$, with an unsaturation degree of 4, and at least one six-membered ring, indicating that $\mathbf{C}$ has three unsaturated bonds;

<Checkpoint>[$\mathbf{C}$ has three unsaturated bonds and a six-membered ring](1)</Checkpoint>

$\mathbf{C}$ is hydrolyzed in an acidic environment to $\mathbf{D}$ and methanol, and hydrolyzed in a basic environment to $\mathbf{E}$ and methanol. $\mathbf{E}$ can spontaneously convert to $\mathbf{D}$ in an acidic environment, which can be inferred that $\mathbf{C}$ contains a methyl ester substituent on the six-membered ring; then $\mathbf{D}$ is the corresponding carboxylic acid, and $\mathbf{E}$ is the sodium salt of the carboxylic acid.

<Checkpoint>[$\mathbf{C}$ contains a methyl ester substituent on the six-membered ring](1)</Checkpoint>

<Checkpoint>[$\mathbf{D}$ is the corresponding carboxylic acid](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{E}$ is the sodium salt of the carboxylic acid](0.5)</Checkpoint>

$\mathbf{C}$ reacts with excess methyl magnesium chloride in tetrahydrofuran to obtain $\mathbf{F}$; it can be inferred that it is the reaction of the methyl ester group with the Grignard reagent, then the product $\mathbf{F}$ is the magnesium salt of a tertiary alcohol;

<Checkpoint>[$\mathbf{F}$ is the magnesium salt of a tertiary alcohol](1)</Checkpoint>

$\mathbf{F}$ is treated with an aqueous solution of ammonium chloride to obtain the main product $\mathbf{G}$, at which point the magnesium salt of the tertiary alcohol is hydrolyzed to obtain the tertiary alcohol. The molecular formula of $\mathbf{G}$ is $(\mathrm{C_9H_{14}O})$, with an unsaturation degree of 3, thus containing a six-membered ring and two double bonds, and containing a tertiary alcohol structure transformed from a methyl ester substituent on a six-membered ring. Due to the characteristics of the cycloaddition reaction, the two double bonds generally constitute a 1,4-cyclohexadiene structure; thus, it is easy to draw the structure of $\mathbf{G}$ as CC(O)(C)C1=CCC=CC1.

<Checkpoint>[The two double bonds generally constitute a 1,4-cyclohexadiene structure due to the characteristics of the cycloaddition reaction](1)</Checkpoint>

<Checkpoint>[The structure of $\mathbf{G}$ is CC(O)(C)C1=CCC=CC1](1)</Checkpoint>

Deducing backwards, $\mathbf{F}$ is CC(C)(O[Mg]Cl)C1=CCC=CC1; $\mathbf{E}$ is O=C(O[Na])C1=CCC=CC1, $\mathbf{D}$ is O=C(O)C1=CCC=CC1,

<Checkpoint>[$\mathbf{F}$ is CC(C)(O[Mg]Cl)C1=CCC=CC1](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{E}$ is O=C(O[Na])C1=CCC=CC1](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{D}$ is O=C(O)C1=CCC=CC1](0.5)</Checkpoint>

$\mathbf{C}$ is O=C(OC)C1=CCC=CC1, which exactly matches its molecular formula.

$\mathbf{A}$ and $\mathbf{B}$ undergo [4+2] to generate $\mathbf{C}$, it is easy to obtain $\mathbf{A}$ and $\mathbf{B}$ as C=CC=C, C#CC(OC)=O, respectively.

<Checkpoint>[$\mathbf{A}$ is C=CC=C](1)</Checkpoint>

<Checkpoint>[$\mathbf{B}$ is C#CC(OC)=O](1)</Checkpoint>

$\mathbf{A}$ is butadiene, because the basicity required to break the carbon-hydrogen bond is much stronger than the basicity of potassium tert-butoxide, so it cannot be deprotonated by potassium tert-butoxide, so option B is incorrect.

<Checkpoint>[$\mathbf{A}$ is butadiene and cannot be deprotonated](1)</Checkpoint>

$\mathbf{G}$ is treated with a strong acid, and the tertiary alcohol becomes a carbocation; at this time, there is no eliminable hydrogen on the ring, so only one molecule of propene can be eliminated to obtain a cyclohexadiene cation, which immediately aromatizes to generate benzene C1=CC=CC=C1; therefore, $\mathbf{H}$ is benzene C1=CC=CC=C1.

<Checkpoint>[$\mathbf{H}$ is benzene C1=CC=CC=C1](1)</Checkpoint>

Benzene is toxic, option C is incorrect; the small molecule released from $\mathbf{G}$ to generate $\mathbf{H}$ is propene, with a relative molecular mass of 42, option D is incorrect.

<Checkpoint>[The small molecule released from $\mathbf{G}$ to generate $\mathbf{H}$ is propene](1)</Checkpoint>

$\mathbf{G}$ is treated with a strong base, and the tertiary alcohol can only be eliminated to a terminal alkene, which is a disubstituted double bond, with the structure C=C(C)C1=CCC=CC1; this structure does not have a tetrasubstituted double bond, so option F is incorrect.

<Checkpoint>[The product of treating $\mathbf{G}$ with a strong base is C=C(C)C1=CCC=CC1](1)</Checkpoint>

In summary, options B-F are all incorrect, and option A is correct.

<MultiModal>![This figure gives the structural formulas of the unknown structures $\mathbf{A-H}$ in this question, and their SMILES are all given in the analysis: $\mathbf{A}$ and $\mathbf{B}$ are C=CC=C and C#CC(OC)=O respectively, $\mathbf{C}$ is O=C(OC)C1=CCC=CC1, $\mathbf{F}$ is CC(C)(O[Mg]Cl)C1=CCC=CC1; $\mathbf{E}$ is O=C(O[Na])C1=CCC=CC1, $\mathbf{D}$ is O=C(O)C1=CCC=CC1, $\mathbf{G}$ is CC(O)(C)C1=CCC=CC1, $\mathbf{H}$ is C1=CC=CC=C1](/media/uploads/a3699f66-11c8-438f-99cc-2724845fcbe4.png)</MultiModal>
$\mathbf{A}$和$\mathbf{B}$发生[4+2]环加成反应得到化合物$\mathbf{C}$ $\mathrm{(C_8H_{10}O_2)}$，不饱和度为4，且至少有一个六元环，说明$\mathbf{C}$存在三个不饱和键；

<Checkpoint>[$\mathbf{C}$存在三个不饱和键和六元环](1)</Checkpoint>

$\mathbf{C}$在酸性环境中水解为$\mathbf{D}$和甲醇, 碱性环境中水解成$\mathbf{E}$和甲醇，$\mathbf{E}$可在酸性环境下自发转化为$\mathbf{D}$，可以推断$\mathbf{C}$含有六元环上的甲酯基取代基；那么$\mathbf{D}$为对应的羧酸，$\mathbf{E}$为羧酸的钠盐。

<Checkpoint>[$\mathbf{C}$含有六元环上的甲酯基取代基](1)</Checkpoint>

<Checkpoint>[$\mathbf{D}$为对应的羧酸](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{E}$为羧酸的钠盐](0.5)</Checkpoint>

$\mathbf{C}$ 和足量甲基氯化镁在四氢呋喃中反应, 得到 $\mathbf{F}$；可推断是甲酯基与格氏试剂的反应，那么产物$\mathbf{F}$为三级醇的镁盐；

<Checkpoint>[$\mathbf{F}$为三级醇的镁盐](1)</Checkpoint>

$\mathbf{F}$ 用氯化铵水溶液处理, 获得主要产物 $\mathbf{G}$，此时三级醇的镁盐水解得到三级醇。$\mathbf{G}$ 分子式为$(\mathrm{C_9H_{14}O})$，不饱和度为3，从而含有一个六元环和两个双键，且含有一个六元环上的甲酯基取代基转化为的三级醇结构。两个双键由于环加成反应特点，一般构成1，4-环己二烯的结构；从而很容易绘制得到$\mathbf{G}$结构为CC(O)(C)C1=CCC=CC1。

<Checkpoint>[两个双键由于环加成反应特点，一般构成1，4-环己二烯的结构](1)</Checkpoint>

<Checkpoint>[$\mathbf{G}$结构为CC(O)(C)C1=CCC=CC1](1)</Checkpoint>

倒推得到$\mathbf{F}$为CC(C)(O[Mg]Cl)C1=CCC=CC1；$\mathbf{E}$为O=C(O[Na])C1=CCC=CC1，$\mathbf{D}$为O=C(O)C1=CCC=CC1，

<Checkpoint>[$\mathbf{F}$为CC(C)(O[Mg]Cl)C1=CCC=CC1](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{E}$为O=C(O[Na])C1=CCC=CC1](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{D}$为O=C(O)C1=CCC=CC1](0.5)</Checkpoint>

$\mathbf{C}$为O=C(OC)C1=CCC=CC1，刚好符合其分子式。

$\mathbf{A}$与$\mathbf{B}$发生[4+2]生成$\mathbf{C}$，易得$\mathbf{A}$，$\mathbf{B}$分别为C=CC=C，C#CC(OC)=O。

<Checkpoint>[$\mathbf{A}$为C=CC=C](1)</Checkpoint>

<Checkpoint>[$\mathbf{B}$为C#CC(OC)=O](1)</Checkpoint>

$\mathbf{A}$为丁二烯，因为断开碳氢键需要的碱性远强于叔丁醇钾的碱性，无法被叔丁醇钾拔除质子，选项B错误。

<Checkpoint>[$\mathbf{A}$为丁二烯，无法被拔除质子](1)</Checkpoint>

$\mathbf{G}$加入强酸处理，三级醇变为碳正离子；此时环上没有可消除的氢，因此只能消去一分子丙烯得到环己二烯正离子，其立即芳构化生成苯C1=CC=CC=C1；因此$\mathbf{H}$为苯C1=CC=CC=C1。

<Checkpoint>[$\mathbf{H}$为苯C1=CC=CC=C1](1)</Checkpoint>

苯具有毒性，选项C错误；$\mathbf{G}$生成$\mathbf{H}$放出的小分子为丙烯，相对分子质量为42，选项D错误。

<Checkpoint>[$\mathbf{G}$生成$\mathbf{H}$放出的小分子为丙烯](1)</Checkpoint>

$\mathbf{G}$加入强碱处理，三级醇只能消除为末端烯烃，为二取代双键，结构为C=C(C)C1=CCC=CC1；该结构不存在四取代的双键，选项F错误。

<Checkpoint>[$\mathbf{G}$加入强碱处理产物为C=C(C)C1=CCC=CC1](1)</Checkpoint>

综上，选项B-F均错误，选项A正确。

<MultiModal>![本图给出了本题未知结构$\mathbf{A-H}$的结构式，其SMILES在解析中已经全都给出：$\mathbf{A}$，$\mathbf{B}$分别为C=CC=C，C#CC(OC)=O，$\mathbf{C}$为O=C(OC)C1=CCC=CC1，$\mathbf{F}$为CC(C)(O[Mg]Cl)C1=CCC=CC1；$\mathbf{E}$为O=C(O[Na])C1=CCC=CC1，$\mathbf{D}$为O=C(O)C1=CCC=CC1，$\mathbf{G}$为CC(O)(C)C1=CCC=CC1，$\mathbf{H}$为C1=CC=CC=C1](/media/uploads/a3699f66-11c8-438f-99cc-2724845fcbe4.png)</MultiModal>