<MultiModal>![The image describes a two-step organic tandem reaction. The substrate is [H][C@@]1([C@H](O)C)CCC2=C1CC[C@@H](CO)C=C2; it reacts with $TBSCl$ followed by $mCPBA$ to obtain $\bf{A}$; $\bf{A}$ reacts with $Et_2AlCN$ to obtain $\bf{B}$.](/media/uploads/13bdf507-ccff-47dd-9908-f5e4e6ba6cce.png)</MultiModal>
Scientists attempted to synthesize a product with the same molecular skeleton but with a cyano substituent from the synthetic route shown above, but surprisingly, compound $\bf{B}$ without a cyano substituent was produced.

Known:
1. $\bf{B}$ contains a six-membered ring.
2. In the first step, 2.1 equivalents of $\rm{TBSCl}$ and 1.0 equivalents of $\rm{mCPBA}$ were added.

Which of the following statements about the structure of $\bf{B}$ is correct:
<MultiModal>![该图片描述了两步有机串联反应。底物为[H][C@@]1([C@H](O)C)CCC2=C1CC[C@@H](CO)C=C2；其与$TBSCl$反应后加入$mCPBA$，得到$\bf{A}$；$\bf{A}$与$Et_2AlCN$反应得到$\bf{B}$。](/media/uploads/13bdf507-ccff-47dd-9908-f5e4e6ba6cce.png)</MultiModal>
科学家们尝试从上图所示的合成路线合成分子骨架不变且带有氰基取代的产物，但令人意外的是生成了没有氰基取代的化合物$\bf{B}$。

已知：
1. $\bf{B}$含有六元环，
2. 第一步反应中加入$\rm{TBSCl}$为2.1当量，加入的$\rm{mCPBA}$为1.0当量。

下列关于$\bf{B}$的结构，说法正确的是：
All other options are incorrect
$\bf{B}$ has 5 chiral carbons
$\bf{B}$ contains 4 oxygen atoms
$\bf{B}$ Contains fused ring systems
$\bf{B}$ contains three unsaturated bonds
$\bf{B}$ first reacts with tetrabutylammonium fluoride, and then undergoes dehydration under acidic conditions. The unsaturation degree of the obtained product is 6.
其他选项均不正确
$\bf{B}$有5个手性碳
$\bf{B}$含有4个氧原子
$\bf{B}$含有并环体系
$\bf{B}$含有三个不饱和键
$\bf{B}$先与四正丁基氟化铵反应，再在酸性条件下加热失水，得到的产物不饱和度为6。
The substrate contains two hydroxyl groups, which react with 2.1 equivalents of dimethyl tert-butylsilyl chloride, which is an alcohol protection reaction, generating $\rm{Si-O}$ bonds.

Then it reacts with one equivalent of $\rm{mCPBA}$, which is an epoxidation reaction of the double bond; $\rm{mCPBA}$ preferentially oxidizes the electron-rich tetrasubstituted fused-ring double bond, so the structure of product $\bf{A}$ is [H][C@@]1([C@H](O[Si](C)(C)C(C)(C)C)C)CCC2(O3)C13CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2.

<Checkpoint>[$\rm{mCPBA}$ preferentially oxidizes the electron-rich tetrasubstituted fused-ring double bond](1)</Checkpoint>

<Checkpoint>[The structure of $\bf{A}$ is [H][C@@]1([C@H](O[Si](C)(C)C(C)(C)C)C)CCC2(O3)C13CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2](1)</Checkpoint>

(This question does not require judging the stereochemistry of the epoxide, which is irrelevant to the problem-solving)

Then adding the Lewis acid $\rm{Et_2AlCN}$ can activate the epoxide; the researcher's original intention was to use cyanide to attack the epoxide to perform ring-opening, but the product does not contain cyanide. At this time, consider that the epoxide is easily ring-opened after being activated, forming a carbocation intermediate; since the stability of the allylic tertiary carbocation is greater than that of the simple tertiary carbocation, the structure of the carbocation intermediate formed is [H][C@@]1([C@H](O[Si](C)(C)C(C)(C)C)C)CC[C+]2C1(O[Al](CC)CC)CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2.

<Checkpoint>[Lewis acid $\rm{Et_2AlCN}$ can activate the epoxide; the epoxide is easily ring-opened after being activated, forming a carbocation intermediate](1)</Checkpoint>

<Checkpoint>[The stability of the allylic tertiary carbocation is greater than that of the simple tertiary carbocation](1)</Checkpoint>

<Checkpoint>[The structure of the carbocation intermediate is [H][C@@]1([C@H](O[Si](C)(C)C(C)(C)C)C)CC[C+]2C1(O[Al](CC)CC)CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2](1)</Checkpoint>

At this time, the substrate is a β-carbocation of the hydroxyl group, which is a typical semi-pinacol rearrangement structure. The electron back donation of the hydroxyl group promotes the migration of the carbon atom; because the formation of a six-membered ring is more stable than the formation of a four-membered ring, the structure of the product $\bf{B}$ after migration is O=C1[C@@]([C@H](O[Si](C)(C)C(C)(C)C)C)([H])CCC12CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2.

<Checkpoint>[Semi-pinacol rearrangement occurs; the formation of a six-membered ring is more stable than the formation of a four-membered ring](1)</Checkpoint>

<Checkpoint>[$\bf{B}$ is O=C1[C@@]([C@H](O[Si](C)(C)C(C)(C)C)C)([H])CCC12CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2](2)</Checkpoint>

Determine the options based on the structure of $\bf{B}$: $\bf{B}$ contains four chiral carbons, two unsaturated bonds, three oxygen atoms, and a spiro system of a five-membered ring and a six-membered ring. Therefore, options B, C, D, and E are all incorrect.

<Checkpoint>[$\bf{B}$ contains four chiral carbons, two unsaturated bonds, three oxygen atoms, and a spiro system of a five-membered ring and a six-membered ring](1)</Checkpoint>

$\bf{B}$ reacts with tetrabutylammonium fluoride to remove all hydroxyl protecting groups, and the alcohol turns into an alkene after heating and dehydration, and the product is O=C1/C(CCC12CC[C@@H](C=C)C=C2)=C\C, which has 6 degrees of unsaturation, so option F is correct.

<Checkpoint>[The reaction product of option F is O=C1/C(CCC12CC[C@@H](C=C)C=C2)=C\C, which has 6 degrees of unsaturation](1)</Checkpoint>

<MultiModal>![This figure shows the organic structural formulas involved in this question. The structure of $\bf{A}$ is [H][C@@]1([C@H](O[Si](C)(C)C(C)(C)C)C)CCC2(O3)C13CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C; the structure of the carbocation intermediate is [H][C@@]1([C@H](O[Si](C)(C)C(C)(C)C)C)CC[C+]2C1(O[Al](CC)CC)CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2; $\bf{B}$ is $O=C1[C@@]([C@H](O[Si](C)(C)C(C)(C)C)C)([H])CCC12CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2; the reaction product of option F is O=C1/C(CCC12CC[C@@H](C=C)C=C2)=C\C.](/media/uploads/db4d1131-1905-4925-8145-b86d8d8b953b.png)</MultiModal>
底物中含有两个羟基，与2.1当量的二甲基叔丁基硅基氯反应，为醇的保护反应，生成$\rm{Si-O}$键。

之后与一当量的$\rm{mCPBA}$反应，为双键的环氧化反应；$\rm{mCPBA}$优先氧化富电子的四取代并环双键，因此产物$\bf{A}$的结构为[H][C@@]1([C@H](O[Si](C)(C)C(C)(C)C)C)CCC2(O3)C13CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2。

<Checkpoint>[$\rm{mCPBA}$优先氧化富电子的四取代并环双键](1)</Checkpoint>

<Checkpoint>[$\bf{A}$的结构为[H][C@@]1([C@H](O[Si](C)(C)C(C)(C)C)C)CCC2(O3)C13CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2](1)</Checkpoint>

（本题不要求判断环氧的立体化学，跟解题无关）

之后加入lewis酸$\rm{Et_2AlCN}$，可以活化环氧；研究人员本意是利用氰根对环氧进攻进行开环，但产物没有氰根，此时考虑环氧被活化后容易开环，形成碳正离子中间体；由于烯丙基三级碳正离子的稳定性大于单纯的三级碳正离子，故形成的碳正离子中间体结构为[H][C@@]1([C@H](O[Si](C)(C)C(C)(C)C)C)CC[C+]2C1(O[Al](CC)CC)CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2。

<Checkpoint>[lewis酸$\rm{Et_2AlCN}$，可以活化环氧；环氧被活化后容易开环，形成碳正离子中间体](1)</Checkpoint>

<Checkpoint>[烯丙基三级碳正离子的稳定性大于单纯的三级碳正离子](1)</Checkpoint>

<Checkpoint>[碳正离子中间体结构为[H][C@@]1([C@H](O[Si](C)(C)C(C)(C)C)C)CC[C+]2C1(O[Al](CC)CC)CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2](1)</Checkpoint>

此时底物为羟基的β-碳正离子，为典型的semi-pinacol重排结构，羟基电子回推促使碳原子进行迁移；因为产生六元环比产生四元环更加稳定，故迁移后产物结构为$\bf{B}$为O=C1[C@@]([C@H](O[Si](C)(C)C(C)(C)C)C)([H])CCC12CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2。

<Checkpoint>[发生semi-pinacol重排；产生六元环比产生四元环更加稳定](1)</Checkpoint>

<Checkpoint>[$\bf{B}$为O=C1[C@@]([C@H](O[Si](C)(C)C(C)(C)C)C)([H])CCC12CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2](2)</Checkpoint>

根据$\bf{B}$的结构判断选项：$\bf{B}$含有四个手性碳，两个不饱和键，三个氧原子和五元环螺六元环的体系。因此选项B,C,D,E均错误。

<Checkpoint>[$\bf{B}$含有四个手性碳，两个不饱和键，三个氧原子和五元环螺六元环的体系](1)</Checkpoint>

$\bf{B}$与四正丁基氟化铵反应脱除所有羟基保护基，加热失水后醇变为烯烃，产物为O=C1/C(CCC12CC[C@@H](C=C)C=C2)=C\C，具有6个不饱和度，选项F正确。

<Checkpoint>[选项F的反应产物为O=C1/C(CCC12CC[C@@H](C=C)C=C2)=C\C，具有6个不饱和度](1)</Checkpoint>

<MultiModal>![本图表示了本题涉及到的有机物结构式。$\bf{A}$的结构为[H][C@@]1([C@H](O[Si](C)(C)C(C)(C)C)C)CCC2(O3)C13CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C；碳正离子中间体结构为[H][C@@]1([C@H](O[Si](C)(C)C(C)(C)C)C)CC[C+]2C1(O[Al](CC)CC)CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2；$\bf{B}$为$O=C1[C@@]([C@H](O[Si](C)(C)C(C)(C)C)C)([H])CCC12CC[C@@H](CO[Si](C)(C)C(C)(C)C)C=C2；选项F的反应产物为O=C1/C(CCC12CC[C@@H](C=C)C=C2)=C\C。](/media/uploads/db4d1131-1905-4925-8145-b86d8d8b953b.png)</MultiModal>