The construction of macrocycles can be achieved through fragmentation reactions:

<MultiModal>![Reaction of `C1CCCCCC(=O)CCCCC1` with `C=C(C)C[Mg]Cl` yields $\bf{A}$, followed by reaction first with borane and then with hydrogen peroxide to yield $\bf{B}$, reaction with tri-n-butylphosphine and `CC1=CC=C(SSC2=CC=C(C)C=C2)C=C1` to yield $\bf{C}$, reaction with peracetic acid to yield $\bf{D}$, reaction with p-toluenesulfonic acid to yield $\bf{E}$, followed by reaction with peracetic acid to yield $\bf{F}$, and finally reaction first with sodium amide and then with potassium tert-butoxide to yield `CC1/C=C/CCCCCCCCCCC(=O)C1`](/media/uploads/bffe40fc-6c31-449e-8d60-25ccf5c46b31.png)</MultiModal>

Select the correct item from the following options:
利用碎裂化反应可实现大环的构建：

<MultiModal>![`C1CCCCCC(=O)CCCCC1`与`C=C(C)C[Mg]Cl`反应得到$\bf{A}$，再先与乙硼烷、随后与过氧化氢反应得到$\bf{B}$，与三正丁基膦和`CC1=CC=C(SSC2=CC=C(C)C=C2)C=C1`反应得到$\bf{C}$，与过乙酸反应得到$\bf{D}$，与对甲苯磺酸反应得到$\bf{E}$，再与过乙酸反应得到$\bf{F}$，最后先与氨基钠反应，随后与叔丁醇钾反应得到`CC1/C=C/CCCCCCCCCCC(=O)C1`](/media/uploads/bffe40fc-6c31-449e-8d60-25ccf5c46b31.png)</MultiModal>

选出以下选项中正确的一项：
All other options are incorrect.
$\bf{B}$ contains two alcohol hydroxyl groups, both of which are tertiary alcohols.
$\bf{C}$ contains two oxygen atoms.
Both the generation of $\bf{D}$ and the generation of $\bf{F}$ reactions utilize peracetic acid. In both reactions, each molecule of reactant consumes one molecule of peracetic acid.
$\bf{E}$ contains 3 cycles
The generation of the final product 1 involves 3 cyclic intermediates.
其他选项均不正确
$\bf{B}$中具有两个醇羟基且均为三级醇
$\bf{C}$中有两个氧原子
生成$\bf{D}$和生成$\bf{F}$这两步反应均用到了过乙酸，这两个反应每分子反应物均消耗一分子过乙酸
$\bf{E}$中具有3个环
生成最终1产物的过程中历经了3个环的中间体
First, the Grignard reagent `C=C(C)C[Mg]Cl` undergoes carbonyl addition to `C1CCCCCC(=O)CCCCC1` to yield $\bf{A}$: `OC1(CC(C)=C)CCCCCCCCCCC1`.

<Checkpoint>[The structure of $\bf{A}$ is `OC1(CC(C)=C)CCCCCCCCCCC1`](1)</Checkpoint>

The generation of $\bf{B}$ from $\bf{A}$ is a typical hydroboration-oxidation condition, used to convert alkenes into alcohols. First, borane adds to the less sterically hindered carbon of the alkene, and is then converted to a hydroxyl group under the action of hydrogen peroxide, yielding $\bf{B}$: `OC1(CC(CO)C)CCCCCCCCCCC1`, which has two alcohol hydroxyl groups, but one of them is a primary alcohol, so option B is incorrect.

<Checkpoint>[The structure of $\bf{B}$ is `OC1(CC(CO)C)CCCCCCCCCCC1`, which has two alcohol hydroxyl groups, but one of them is a primary alcohol, so option B is incorrect](1)</Checkpoint>

When $\bf{B}$ is converted to $\bf{C}$, first, tri-n-butylphosphine attacks the disulfide bond, producing `CCCC[P+](CCCC)(CCCC)SC1=CC=C(C)C=C1`, which then reacts with the least sterically hindered alcohol hydroxyl group, leaving thiophenol, to give `OC1(CC(CO[P+](CCCC)(CCCC)CCCC)C)CCCCCCCCCCC1`. Finally, thiophenol attacks, leaving `O=P(CCCC)(CCCC)CCCC`, yielding $\bf{C}$: `OC1(CC(CSC2=CC=C(C)C=C2)C)CCCCCCCCCCC1`, which has only one oxygen atom, so option C is incorrect.

<Checkpoint>[The structure of $\bf{C}$ is `OC1(CC(CSC2=CC=C(C)C=C2)C)CCCCCCCCCCC1`, which has only one oxygen atom, so option C is incorrect](1)</Checkpoint>

Under the action of peracetic acid, the sulfur atom is oxidized, and the thioether is oxidized to a sulfone, giving $\bf{D}$: `OC1(CC(CS(C2=CC=C(C)C=C2)(=O)=O)C)CCCCCCCCCCC1`

<Checkpoint>[The structure of $\bf{D}$ is `OC1(CC(CS(C2=CC=C(C)C=C2)(=O)=O)C)CCCCCCCCCCC1`](1)</Checkpoint>

Under the action of a strong acid, the tertiary alcohol undergoes elimination, which can produce two alkenes $\bf{E1}$: `CC(CS(C1=CC=C(C)C=C1)(=O)=O)C/C2=C/CCCCCCCCCC2` and $\bf{E2}$: `CC(CS(C1=CC=C(C)C=C1)(=O)=O)/C=C2CCCCCCCCCCC\2`. In the next step, the double bond is oxidized to an epoxide by peracetic acid, corresponding to $\bf{F1}$: `CC(CS(C1=CC=C(C)C=C1)(=O)=O)CC2(O3)C3CCCCCCCCCC2` and $\bf{F2}$: `CC(CS(C1=CC=C(C)C=C1)(=O)=O)C(O2)C32CCCCCCCCCCC3`.
The generation of $\bf{D}$ consumes two molecules of peracetic acid, and the generation of $\bf{F}$ consumes one molecule of peracetic acid, so option D is incorrect.

<Checkpoint>[The generation of $\bf{D}$ consumes two molecules of peracetic acid, and the generation of $\bf{F}$ consumes one molecule of peracetic acid, so option D is incorrect](1)</Checkpoint>

Under the action of a strong base, $\bf{F}$ loses a proton from the α-position of the sulfone, attacking the epoxide. At this time, $\bf{F2}$ cannot obtain a suitable fragmentation intermediate, only $\bf{F1}$ can produce `CC(C1S(C2=CC=C(C)C=C2)(=O)=O)CC3(O)C1CCCCCCCCCC3`. $\bf F1$ has three rings, so option F is correct.

<Checkpoint>[The intermediate of the resulting product is `CC(C1S(C2=CC=C(C)C=C2)(=O)=O)CC3(O)C1CCCCCCCCCC3`, which has three rings, so option F is correct](1)</Checkpoint>

Therefore, the structure of $\bf{F}$ is `CC(CS(C1=CC=C(C)C=C1)(=O)=O)CC2(O3)C3CCCCCCCCCC2`, and the structure of $\bf{E}$ is `CC(CS(C1=CC=C(C)C=C1)(=O)=O)C/C2=C/CCCCCCCCCC2`, which has two rings, so option E is incorrect.

<Checkpoint>[The structure of $\bf{F}$ is `CC(CS(C1=CC=C(C)C=C1)(=O)=O)CC2(O3)C3CCCCCCCCCC2`, and the structure of $\bf{E}$ is `CC(CS(C1=CC=C(C)C=C1)(=O)=O)C/C2=C/CCCCCCCCCC2`, which has two rings, so option E is incorrect](1)</Checkpoint>

Subsequently, the alcohol anion causes the newly formed five-membered ring to fragment, leaving `CC(C=C1)=CC=C1S([O-])=O`, to obtain the product.

<MultiModal>![Shows the fragmentation mechanism, where the negative charge of the alcohol anion is transferred to the connected carbon-carbon single bond, then a trans double bond is generated, leaving `CC(C=C1)=CC=C1S([O-])=O`](/media/uploads/2f0078dc-5b79-4ca1-880c-097be2628370.png)</MultiModal>
首先，格氏试剂`C=C(C)C[Mg]Cl`与`C1CCCCCC(=O)CCCCC1`的羰基加成得到$\bf{A}$：`OC1(CC(C)=C)CCCCCCCCCCC1`。

<Checkpoint>[$\bf{A}$的结构为`OC1(CC(C)=C)CCCCCCCCCCC1`](1)</Checkpoint>

$\bf{A}$生成$\bf{B}$是一个典型的硼氢化氧化条件，用于将烯烃转化为醇。首先硼烷加成在烯烃位阻较小的碳上，随后在过氧化氢作用下被转化为羟基，得到$\bf{B}$：`OC1(CC(CO)C)CCCCCCCCCCC1`，具有两个醇羟基，但其中一个是一级醇，选项B错误

<Checkpoint>[$\bf{B}$的结构为`OC1(CC(CO)C)CCCCCCCCCCC1`，具有两个醇羟基，但其中一个是一级醇，选项B错误](1)</Checkpoint>

$\bf{B}$生成$\bf{C}$时，首先三正丁基膦进攻二硫键，产生`CCCC[P+](CCCC)(CCCC)SC1=CC=C(C)C=C1`，随后与位阻最小的醇羟基反应，离去硫酚，得到`OC1(CC(CO[P+](CCCC)(CCCC)CCCC)C)CCCCCCCCCCC1`，最后硫酚进攻，离去`O=P(CCCC)(CCCC)CCCC`，得到$\bf{C}$：`OC1(CC(CSC2=CC=C(C)C=C2)C)CCCCCCCCCCC1`，仅有一个氧原子，选项C错误

<Checkpoint>[$\bf{C}$的结构为`OC1(CC(CSC2=CC=C(C)C=C2)C)CCCCCCCCCCC1`，仅有一个氧原子，选项C错误](1)</Checkpoint>

在过乙酸作用下，硫原子被氧化，硫醚被氧化为砜，得到$\bf{D}$：`OC1(CC(CS(C2=CC=C(C)C=C2)(=O)=O)C)CCCCCCCCCCC1`

<Checkpoint>[$\bf{D}$的结构为`OC1(CC(CS(C2=CC=C(C)C=C2)(=O)=O)C)CCCCCCCCCCC1`](1)</Checkpoint>

在强酸作用下，三级醇发生消除，可产生两种烯烃$\bf{E1}$：`CC(CS(C1=CC=C(C)C=C1)(=O)=O)C/C2=C/CCCCCCCCCC2`和$\bf{E2}$：`CC(CS(C1=CC=C(C)C=C1)(=O)=O)/C=C2CCCCCCCCCCC\2`，在下一步双键被过乙酸氧化为环氧，对应$\bf{F1}$：`CC(CS(C1=CC=C(C)C=C1)(=O)=O)CC2(O3)C3CCCCCCCCCC2`和$\bf{F2}$：`CC(CS(C1=CC=C(C)C=C1)(=O)=O)C(O2)C32CCCCCCCCCCC3`。
生成$\bf{D}$消耗两分子过乙酸，生成$\bf{F}$消耗一分子过乙酸，选项D错误

<Checkpoint>[生成$\bf{D}$消耗两分子过乙酸，生成$\bf{F}$消耗一分子过乙酸，选项D错误](1)</Checkpoint>

$\bf{F}$在强碱作用下拔除砜α位的氢，进攻环氧，此时$\bf{F2}$无法得到合适的碎裂化中间体，只有$\bf{F1}$可以产生`CC(C1S(C2=CC=C(C)C=C2)(=O)=O)CC3(O)C1CCCCCCCCCC3`。$\bf F1$有三个环，选项F正确

<Checkpoint>[得到产物的中间体为`CC(C1S(C2=CC=C(C)C=C2)(=O)=O)CC3(O)C1CCCCCCCCCC3`，有三个环，选项F正确](1)</Checkpoint>

因此$\bf{F}$的结构为`CC(CS(C1=CC=C(C)C=C1)(=O)=O)CC2(O3)C3CCCCCCCCCC2`，$\bf{E}$的结构为`CC(CS(C1=CC=C(C)C=C1)(=O)=O)C/C2=C/CCCCCCCCCC2`,有两个环，选项E错误。

<Checkpoint>[$\bf{F}$的结构为`CC(CS(C1=CC=C(C)C=C1)(=O)=O)CC2(O3)C3CCCCCCCCCC2`，$\bf{E}$的结构为`CC(CS(C1=CC=C(C)C=C1)(=O)=O)C/C2=C/CCCCCCCCCC2`，有两个环，选项E错误](1)</Checkpoint>

随后醇负离子使得新产生的五元环碎裂，离去`CC(C=C1)=CC=C1S([O-])=O`，得到产物。

<MultiModal>![展示出碎裂化的机理，醇负离子的负电荷转移到所连的碳碳单键上，接着产生反式双键，离去`CC(C=C1)=CC=C1S([O-])=O`](/media/uploads/2f0078dc-5b79-4ca1-880c-097be2628370.png)</MultiModal>