A compound $\mathbf{A}$ with the molecular formula $\mathrm{C_8H_{14}}$ contains a quaternary carbon. After heating and refluxing it with $\mathrm{KMnO_4}$, only a product with the molecular formula $\mathrm{C_7H_{12}O}$ is detected in the reaction solution. $\mathbf{A}$ reacts with hydrobromic acid to give $\mathbf{B}$, while reaction with hydrobromic acid in the presence of peroxide gives isomer $\mathbf{C}$. $\mathbf{B}$ in sodium ethoxide-ethanol gives $\mathbf{D}$. $\mathbf{D}$ reacts with $\mathrm{BH_3}$ and is further treated with acetic acid to give compound $\mathbf{E}$ containing a five-membered ring structure. $\mathbf{D}$ can decolorize $\mathrm{Br_2}$ and produce $\mathbf{F}$. Treatment of $\mathbf{F}$ with sodium ethoxide-ethanol solution yields two isomeric products $\mathbf{G}$ and $\mathbf{H}$. Deduce the structures of all species based on the above information.

Based on your deduced results, choose the correct option. If all options are incorrect, select option A: All other options are incorrect.
分子式为 $\mathrm{C_8H_{14}}$ 的化合物 $\mathbf{A}$ 含有一个四级碳，其与 $\mathrm{KMnO_4}$ 加热回流后，在反应溶液中仅检测到分子式为 $\mathrm{C_7H_{12}O}$ 的产物。$\mathbf{A}$ 与氢溴酸作用得到 $\mathbf{B}$，而与氢溴酸在过氧化物的存在下作用得到异构体 $\mathbf{C}$。$\mathbf{B}$ 在乙醇钠-乙醇中得到 $\mathbf{D}$，$\mathbf{D}$ 与 $\mathrm{BH_3}$ 反应并进一步用乙酸处理后得到含有五元环结构的化合物 $\mathbf{E}$。$\mathbf{D}$ 可以使 $\mathrm{Br_2}$ 褪色并生成 $\mathbf{F}$，$\mathbf{F}$ 用乙醇钠-乙醇溶液处理生成两个异构体产物 $\mathbf{G}$ 和 $\mathbf{H}$。根据以上的信息推导所有物种的结构。

请根据你推断的结果，选择正确的选项。如果所有选项均错误，选择选项 A：其他选项均不正确。
All other options are incorrect.
Compound $\mathbf{A}$ has 5 different chemical shift hydrogens in the nuclear magnetic resonance hydrogen spectrum.
Among compounds $\mathbf{A\sim H}$, there are 4 types of hydrocarbons.
There are 2 pairs of isomers among compounds $\mathbf{A\sim H}$
The substituents on the ring in compound $\mathbf{G/H}$ are completely identical to those in compound $\mathbf{A}$.
The different kinds of substituents on the ring in compound $\mathbf{A}$ are located on adjacent $\rm C$ atoms.
其他选项均不正确
化合物 $\mathbf{A}$ 在核磁共振氢谱中有 5 种不同化学位移的氢
化合物 $\mathbf{A\sim H}$ 中有 4 种烃类物质
化合物 $\mathbf{A\sim H}$ 中有 2 对同分异构体
化合物 $\mathbf{G/H}$ 中与化合物 $\mathbf{A}$ 中环上的取代基均完全相同
化合物 $\mathbf{A}$ 中环上的不同种取代基位于相邻的 $\rm C$ 上
To solve this problem, we first need to deduce the structure of $\mathbf{A}$ based on the known information and the reaction properties of $\mathbf{A}$.

The molecular formula of $\mathbf{A}$ is $\mathrm{C_8H_{14}}$, and its degree of unsaturation $\Omega=2$.

<Checkpoint>[The degree of unsaturation of $\mathbf{A}$ is $\Omega=2$.](0.5)</Checkpoint>

This indicates that it contains an unsaturated $\rm C=C$ double bond or a cyclic structure. Since $\mathbf{A}$ reacts with $\mathrm{KMnO_4}$ under reflux with heating, only a product with the molecular formula $\mathrm{C_7H_{12}O}$ is detected in the reaction solution. The ratio of the species after the reaction shows that $\mathbf{A}$ has one less carbon, which indicates that it contains only one unsaturated $\rm C=C$ double bond and it is a terminal alkene double bond (`[*][CX4][CX3](=[CH2])[CX4][*]`, hereinafter referred to as `[*]CC(=C)C[*]` for short), and the other degree of unsaturation is contributed by a cyclic structure.

<Checkpoint>[$\mathbf{A}$ contains a terminal alkene double bond and a cyclic structure](0.5)</Checkpoint>

$\mathbf{A}$ reacts with hydrobromic acid to give $\mathbf{B}$, then the product after the latter tertiary carbocation combines with a bromide ion is `[*]CC(Br)(C)C[*]` .

<Checkpoint>[$\mathbf{B}$ contains the fragment `[*]CC(Br)(C)C[*]`](0.5)</Checkpoint>

The reaction with hydrobromic acid in the presence of peroxide gives the isomer $\mathbf{C}$. This is a free radical reaction, and the direction of hydrobromic acid addition to the double bond is opposite to that of the ionic reaction. The product obtained should be `[*]CC(CBr)C[*]` .

<Checkpoint>[$\mathbf{C}$ contains the fragment `[*]CC(CBr)C[*]`](0.5)</Checkpoint>

$\mathbf{B}$ gives $\mathbf{D}$ in sodium ethoxide-ethanol. This is a one-step elimination reaction. It should be noted that $\mathbf{A}$ cannot be generated here, but an isomer $\mathbf{D}$ with a different double bond position from $\mathbf{A}$ should be generated. Therefore, the structure of $\mathbf{D}$ can be written as `[*]C=C(C)C[*]` .

<Checkpoint>[$\mathbf{D}$ contains the fragment `[*]C=C(C)C[*]`](0.5)</Checkpoint>

$\mathbf{D}$ reacts with $\mathrm{BH_3}$ and is further treated with acetic acid to obtain a compound $\mathbf{E}$ containing a five-membered ring structure. This step is a hydroboration reduction reaction, which reduces the unsaturated double bond to a saturated bond. Obviously, no ring formation has occurred. Therefore, the "five-membered ring" described here was originally present in the substrate. This also confirms the assumption that "$\mathbf{A}$ has a cyclic structure". Therefore, the cyclic structure here is a five-membered ring. The structure of $\mathbf{E}$ contains a cyclic structure like `C1C(C)CCC1`, while $\mathbf{A}$ has an additional terminal alkene double bond outside the ring `C1C(=C)CCC1`.

<Checkpoint>[$\mathbf{E}$ contains the `C1C(C)CCC1` structure, and $\mathbf{A}$ contains the `C1C(=C)CCC1` structure](1)</Checkpoint>

$\mathbf{D}$ can decolorize $\mathrm{Br_2}$ and generate $\mathbf{F}$. This step is a dibromination of an alkene. According to the previous inference, $\mathbf{F}$ contains the `C1(Br)C(Br)(C)CCC1` structure.

<Checkpoint>[$\mathbf{F}$ contains the `C1(Br)C(Br)(C)CCC1` structure](0.5)</Checkpoint>

$\mathbf{F}$ is treated with sodium ethoxide-ethanol solution to generate two isomeric products $\mathbf{G}$ and $\mathbf{H}$. This is a one-step elimination reaction to obtain a diene. Rewriting `C1(Br)C(Br)(C)CCC1` as the equivalent `[C:1]1[C:2]([Br:a])[C:3]([Br:b])([C:4])[C:5][C:6]1` indicates that the hydrogen on `[C:1]` can undergo elimination with `[Br:a]` and there is only one way, which indicates that `[C:1]` is not the quaternary carbon in $\mathbf{A}$ as indicated in the question. Similarly, the hydrogen on `[C:4]/[C:5]` can undergo two different elimination modes with `[Br:b]` to obtain two isomeric dienes, which also indicates that `[C:4]/[C:5]` is not a quaternary carbon either. Therefore, only `[C:6]` can be a quaternary carbon. Combining with the molecular formula $\mathrm{C_8H_{14}}$ of $\mathbf{A}$ in the question, it is not difficult to find that the quaternary carbon is connected to two methyl groups. Therefore, the structure of $\mathbf{A}$ is `C1C(=C)CC(C)(C)C1`.

<Checkpoint>[The structure of $\mathbf{A}$ is `C1C(=C)CC(C)(C)C1`](3)</Checkpoint>

Based on this, we can get:

- The structure of $\mathbf{B}$ is `C1C(CBr)CC(C)(C)C1`
- The structure of $\mathbf{C}$ is `C1C(Br)(C)CC(C)(C)C1`

<Checkpoint>[The structure of $\mathbf{B}$ is `C1C(CBr)CC(C)(C)C1`, and the structure of $\mathbf{C}$ is `C1C(Br)(C)CC(C)(C)C1`](1)</Checkpoint>

- There are two possibilities for the structure of $\mathbf{D}$: `C1=C(C)CC(C)(C)C1` or `C1C(C)=CC(C)(C)C1`. If it is the latter, then its generated $\mathbf{F}$ cannot undergo an elimination reaction under the condition of sodium ethoxide to obtain two different elimination products (only one bromine can be eliminated, and there are three possible products), so it should be `C1=C(C)CC(C)(C)C1`

<Checkpoint>[The structure of $\mathbf{D}$ is `C1=C(C)CC(C)(C)C1`](3)</Checkpoint>

- The structure of $\mathbf{E}$ is `C1C(C)CC(C)(C)C1`
- The structure of $\mathbf{F}$ is `C1(Br)C(Br)(C)CC(C)(C)C1`
- The structure of $\mathbf{G/H}$ is `C1=CC(C)=CC1(C)C`/`C1=CC(=C)CC1(C)C`

Finally, judge each option one by one:

- Compound $\mathbf{A}$ has 6 different chemical shift hydrogens in the nuclear magnetic resonance hydrogen spectrum (2 types of terminal alkene + 3 types of -CH2- in the ring + 1 type of gem-dimethyl), so option B is incorrect.

<Checkpoint>[Compound $\mathbf{A}$ has 6 different chemical shift hydrogens in the nuclear magnetic resonance hydrogen spectrum](1)</Checkpoint>

- There are 5 types of hydrocarbons among compounds $\mathbf{A\sim H}$ ($\mathbf{A/D/E/G/H}$), so option C is incorrect.

<Checkpoint>[There are 5 types of hydrocarbons among compounds $\mathbf{A\sim H}$](1)</Checkpoint>

- There are 3 pairs of isomers among compounds $\mathbf{A\sim H}$ ($\mathbf{A-D/B-C/G-H}$), so option D is incorrect.

<Checkpoint>[There are 3 pairs of isomers among compounds $\mathbf{A\sim H}$](1)</Checkpoint>

- Compound $\mathbf{A}$ has 2 methyl groups and 1 methylene group on the ring, and compound $\mathbf{G/H}$ has only one that is the same as $\mathbf{A}$, and all the substituents on the other ring are methyl groups. Option E is incorrect.

<Checkpoint>[The substituents on the ring in compound $\mathbf{G/H}$ are not all the same as those on the ring in compound $\mathbf{A}$](1)</Checkpoint>

- The 2 methyl groups and the methylene group on the ring in compound $\mathbf{A}$ are located on spaced $\rm C$ atoms. Option F is incorrect.

<Checkpoint>[The 2 methyl groups and the methylene group on the ring in compound $\mathbf{A}$ are located on spaced $\rm C$ atoms](1)</Checkpoint>

All options are incorrect, choose option A.
解决这道题目，首先需要根据已知信息与 $\mathbf{A}$ 的反应性质，推断 $\mathbf{A}$ 的结构。

$\mathbf{A}$ 的分子式为 $\mathrm{C_8H_{14}}$ ，其不饱和度 $\Omega=2$。

<Checkpoint>[$\mathbf{A}$ 的不饱和度 $\Omega=2$。](0.5)</Checkpoint>

这说明其中含有不饱和 $\rm C=C$ 双键或环状结构。由于 $\mathbf{A}$ 与 $\mathrm{KMnO_4}$ 加热回流后，在反应溶液中仅检测到分子式为 $\mathrm{C_7H_{12}O}$ 的产物。反应后的物种之比 $\mathbf{A}$ 少了一个碳，这说明其中只含有一个不饱和 $\rm C=C$ 双键且为末端烯烃双键（`[*][CX4][CX3](=[CH2])[CX4][*]`，之后简记为`[*]CC(=C)C[*]`），另一个不饱和度为环状结构贡献。

<Checkpoint>[$\mathbf{A}$ 中含有一个末端烯烃双键、一个环状结构](0.5)</Checkpoint>

$\mathbf{A}$ 与氢溴酸作用得到 $\mathbf{B}$，则后者三级碳正离子结合溴离子后的产物 `[*]CC(Br)(C)C[*]`。

<Checkpoint>[$\mathbf{B}$ 中含有`[*]CC(Br)(C)C[*]` 片段](0.5)</Checkpoint>

而与氢溴酸在过氧化物的存在下作用得到异构体 $\mathbf{C}$，这是一个自由基反应，溴化氢加成双键的方向与离子反应相反，得到的产物应当为 `[*]CC(CBr)C[*]`。

<Checkpoint>[$\mathbf{C}$ 中含有`[*]CC(CBr)C[*]` 片段](0.5)</Checkpoint>

$\mathbf{B}$ 在乙醇钠-乙醇中得到 $\mathbf{D}$，这是一步消除反应，需要注意的是此处并无法生成 $\mathbf{A}$，而应该生成双键位置与 $\mathbf{A}$ 不同的同分异构体 $\mathbf{D}$，故 $\mathbf{D}$ 的结构可以记为 `[*]C=C(C)C[*]`。

<Checkpoint>[$\mathbf{D}$ 中含有`[*]C=C(C)C[*]` 片段](0.5)</Checkpoint>

$\mathbf{D}$ 与 $\mathrm{BH_3}$ 反应并进一步用乙酸处理后得到含有五元环结构的化合物 $\mathbf{E}$。这一步反应是硼氢化还原反应，将不饱和的双键还原成饱和键，这显然没有发生成环，因此，此处所描述的“五元环”是底物一开始就有的，这也印证了“$\mathbf{A}$ 中一个环状结构”的设想，因此，此处的环状结构是五元环，$\mathbf{E}$ 的结构包含 `C1C(C)CCC1` 这样的环状结构，而 $\mathbf{A}$ 则增加了环外的末端烯烃双键 `C1C(=C)CCC1`。

<Checkpoint>[$\mathbf{E}$ 中含有 `C1C(C)CCC1` 结构、$\mathbf{A}$ 中含有`C1C(=C)CCC1` 结构](1)</Checkpoint>

$\mathbf{D}$ 可以使 $\mathrm{Br_2}$ 褪色并生成 $\mathbf{F}$，这一步反应发生了烯烃的二溴化，根据之前的推论，$\mathbf{F}$ 中包含 `C1(Br)C(Br)(C)CCC1` 结构。

<Checkpoint>[$\mathbf{F}$ 中含有 `C1(Br)C(Br)(C)CCC1` 结构](0.5)</Checkpoint>

$\mathbf{F}$ 用乙醇钠-乙醇溶液处理生成两个异构体产物 $\mathbf{G}$ 和 $\mathbf{H}$，这是一步消除反应，得到二烯烃，将 `C1(Br)C(Br)(C)CCC1` 改写为等价的 `[C:1]1[C:2]([Br:a])[C:3]([Br:b])([C:4])[C:5][C:6]1` 表示，其中 `[C:1]` 上的氢能与 `[Br:a]` 发生消除且只有一种方式，这说明 `[C:1]` 并非题目中所指的 $\mathbf{A}$ 中的四级碳，同样的，`[C:4]/[C:5]` 的氢能与 `[Br:b]` 发生两种不同的消除方式得到两种异构体二烯烃，同时也说明 `[C:4]/[C:5]` 也不是四级碳，因此，只有 `[C:6]` 可以是四级碳，结合题目中 $\mathbf{A}$ 的分子式 $\mathrm{C_8H_{14}}$，不难发现四级碳上相连了两个甲基，因此，$\mathbf{A}$ 的结构为 `C1C(=C)CC(C)(C)C1`。

<Checkpoint>[$\mathbf{A}$ 的结构为 `C1C(=C)CC(C)(C)C1`](3)</Checkpoint>

据此，可以得到：

- $\mathbf{B}$ 的结构为 `C1C(CBr)CC(C)(C)C1`
- $\mathbf{C}$ 的结构为 `C1C(Br)(C)CC(C)(C)C1`

<Checkpoint>[$\mathbf{B}$ 的结构为 `C1C(CBr)CC(C)(C)C1`、$\mathbf{C}$ 的结构为 `C1C(Br)(C)CC(C)(C)C1`](1)</Checkpoint>

- $\mathbf{D}$ 的结构有两种可能性 `C1=C(C)CC(C)(C)C1` 或 `C1C(C)=CC(C)(C)C1`，若为后者，则其生成的 $\mathbf{F}$ 无法在乙醇钠的条件下发生消除反应得到两种不同的消除产物（只能消除一个溴、且有三种可能的产物），故其应该为 `C1=C(C)CC(C)(C)C1`

<Checkpoint>[$\mathbf{D}$ 的结构为 `C1=C(C)CC(C)(C)C1`](3)</Checkpoint>

- $\mathbf{E}$ 的结构为 `C1C(C)CC(C)(C)C1`
- $\mathbf{F}$ 的结构为 `C1(Br)C(Br)(C)CC(C)(C)C1`
- $\mathbf{G/H}$ 的结构为 `C1=CC(C)=CC1(C)C`/`C1=CC(=C)CC1(C)C`

最终，对于每个选项进行逐个判断：

- 化合物 $\mathbf{A}$ 在核磁共振氢谱中有 6 种不同化学位移的氢（烯烃末端2种+环内三个-CH2-3种+偕二甲基1种），选项 B 错误。

<Checkpoint>[化合物 $\mathbf{A}$ 在核磁共振氢谱中有 6 种不同化学位移的氢](1)</Checkpoint>

- 化合物 $\mathbf{A\sim H}$ 中有 5 种烃类物质（$\mathbf{A/D/E/G/H}$），选项 C 错误。

<Checkpoint>[化合物 $\mathbf{A\sim H}$ 中有 5 种烃类物质](1)</Checkpoint>

- 化合物 $\mathbf{A\sim H}$ 中有 3 对同分异构体（$\mathbf{A-D/B-C/G-H}$），选项 D 错误。

<Checkpoint>[化合物 $\mathbf{A\sim H}$ 中有 3 对同分异构体](1)</Checkpoint>

- 化合物 $\mathbf{A}$ 中环上有2个甲基与1个亚甲基，化合物 $\mathbf{G/H}$ 中只有一个与 $\mathbf{A}$ 相同，另一个环上取代基全部为甲基。选项 E 错误。

<Checkpoint>[化合物 $\mathbf{G/H}$ 中与化合物 $\mathbf{A}$ 中环上的取代基并非都完全相同](1)</Checkpoint>

- 化合物 $\mathbf{A}$ 中环上的2个甲基与亚甲基位于间隔的 $\rm C$ 上。选项 F 错误。

<Checkpoint>[化合物 $\mathbf{A}$ 中环上的2个甲基与亚甲基位于间隔的 $\rm C$ 上](1)</Checkpoint>

所有选项均错误，选择选项 A。