$(S)-\bf{A}$ can utilize the $\left[ 3 { + } 2 \right]$ reaction to construct the basic skeleton of the compound. Pay attention to the stereochemistry of each substance.

<MultiModal>![$(S)-\bf{A}$ reacts with triphenylphosphine and C=C=CC(OC)=O to obtain $\bf{C}$, $\bf{C}$ is hydrogenated under palladium on carbon catalysis to obtain $\bf{D}$, $\bf{D}$ reacts with zinc, diiodomethane and titanium tetraiodide to obtain $\bf{E}$, $\bf{E}$ is first reacted with $p-TsOH$, and then with methyl Grignard reagent and diethyl ether to obtain C[C@H]1CCC=C(C)[C@@]12C[C@H](C(O)(C)C)CC2](/media/uploads/415a534b-1e74-479e-bba4-dd062079cea8.jpg)</MultiModal>

$(S)-\bf{A}$ can be prepared by the following reaction and has the following reactions:

<MultiModal>![O=C1C(CN2CCC[C@H]2COC)=CCCC1 is first reacted with zinc bromide in $THF$, then reacted with lithium dimethylcuprate, the system temperature is raised from minus 78 degrees Celsius to zero degrees to obtain $(S)-\bf{A}$, then $(S)-\bf{A}$ reacts with lithium diphenylcuprate to obtain $\bf{B}$](/media/uploads/035b9ef6-b18b-4446-9971-583dfed15c49.png)</MultiModal>

Choose the correct option from the following
$(S)-\bf{A}$可以借助 $\left[ 3 { + } 2 \right]$ 反应，构建化合物的基本骨架。注意各物质立体化学。

<MultiModal>![$(S)-\bf{A}$在三苯基膦和C=C=CC(OC)=O作用下得到$\bf{C}$，$\bf{C}$在钯碳催化下加氢得到$\bf{D}$，$\bf{D}$在锌、二碘甲烷和四碘化钛作用下得到$\bf{E}$，$\bf{E}$先经过$p-TsOH$作用，再在甲基格氏试剂、乙醚作用下得到C[C@H]1CCC=C(C)[C@@]12C[C@H](C(O)(C)C)CC2](/media/uploads/415a534b-1e74-479e-bba4-dd062079cea8.jpg)</MultiModal>

$(S)-\bf{A}$可通过如下反应制备并具有如下反应：

<MultiModal>![O=C1C(CN2CCC[C@H]2COC)=CCCC1先在$THF$中与溴化锌作用，随后与二甲基铜锂反应，体系温度从零下78摄氏度升至零度得到$(S)-\bf{A}$，随后$(S)-\bf{A}$与二苯基铜锂反应得到$\bf{B}$](/media/uploads/035b9ef6-b18b-4446-9971-583dfed15c49.png)</MultiModal>

选择下列选项中正确的一项
All other options are incorrect
The newly generated chiral center in $\bf{B}$ has an absolute configuration of S.
The most stable conformation of $\bf{B}$ has an axial bond on the six-membered ring.
Generating $(S)-\bf{A}$ can achieve the goal of controlling chirality in one step without adding zinc bromide.
$(S)-\bf{A}$ generating $\bf{C}$ involves an intermediate with the structure

<MultiModal>![O=C1[C@@]2(CC(C(OC)=O)C(C2)=P(C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)[C@@H](C)CCC1](/media/uploads/90854cd8-70c7-4b9e-a66b-a634a2cf67ac.png)</MultiModal>
The chemical formula of $\bf{E}$ is $C_{14}H_{20}O_3$
$\bf{E}$ The two-step reaction sequence to obtain the product can be interchanged.
其他选项均不正确
$\bf{B}$中新产生的手性中心绝对构型为S
$\bf{B}$的最稳定构象中六元环上存在直立键
生成$(S)-\bf{A}$一步不添加溴化锌也可以实现控制手性的目的
$(S)-\bf{A}$生成$\bf{C}$的过程中存在结构为

<MultiModal>![O=C1[C@@]2(CC(C(OC)=O)C(C2)=P(C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)[C@@H](C)CCC1](/media/uploads/90854cd8-70c7-4b9e-a66b-a634a2cf67ac.png)</MultiModal>

的中间体
$\bf{E}$的化学式为$C_{14}H_{20}O_3$
$\bf{E}$得到产物的两步反应先后顺序可以调换
O=C1C(CN2CCC[C@H]2COC)=CCCC1 can form a complex with zinc bromide CO1C[C@@H]2CCC[N@@]23CC4=CCCCC4=O[Zn@]13[R] (R represents other possible ligands), fixing the chiral environment near the nucleophilic site. Subsequently, lithium dimethylcuprate nucleophilically attacks from the less hindered side and the chiral auxiliary leaves to obtain $(S)-\bf{A}$, with the structure C=C1[C@@H](C)CCCC1=O.

<Checkpoint>[The formation of $(S)-\bf{A}$ requires the participation of zinc bromide to form a complex that fixes the chiral environment](1)</Checkpoint>

After the addition of lithium diphenylcuprate, an enolate anion is generated, followed by workup to produce a new carbonyl group. This process minimizes the repulsion of all substituents on the six-membered ring, so they are all in equatorial positions. The structure of $\bf{B}$ is O=C1[C@H](CC2=CC=CC=C2)[C@@H](C)CCC1

<Checkpoint>[The structure of $\bf{B}$ is O=C1[C@H](CC2=CC=CC=C2)[C@@H](C)CCC1, all in equatorial positions](1)</Checkpoint>

From the structure of $\bf{B}$, it can be seen that the absolute configuration of the newly generated chiral center is R

<Checkpoint>[The absolute configuration of the newly generated chiral center is R](1)</Checkpoint>

$(S)-\bf{A}$ reacts with triphenylphosphine and C=C=CC(OC)=O to obtain $\bf{C}$. First, triphenylphosphine adds to C=C=CC(OC)=O to form a zwitterion C=C([CH-]C(OC)=O)[P+](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3, followed by a $\left[ 3 { + } 2 \right]$ reaction, reacting from the less hindered side to produce another intermediate O=C1[C@@]2(CC(C(OC)=O)C(C2)=P(C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)[C@@H](C)CCC1. Finally, the negative charge is transferred to the α-position of the ester group, eliminating triphenylphosphine to obtain $\bf{C}$, with the structure C[C@H]([C@]12CC(C(OC)=O)=CC2)CCCC1=O

<Checkpoint>[The formation of $\bf{C}$ involves the intermediate O=C1[C@@]2(CC(C(OC)=O)C(C2)=P(C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)[C@@H](C)CCC1](1)</Checkpoint>

$\bf{C}$ is hydrogenated over palladium on carbon from the less hindered side to obtain $\bf{D}$, with the structure C[C@H]([C@]12C[C@H](C(OC)=O)CC2)CCCC1=O

The reaction of $\bf{D}$ to obtain $\bf{E}$ is the Lombardo reaction. First, zinc reduces diiodomethane to obtain methylene carbene. The methylene carbene reacts with titanium tetraiodide to obtain C=[Ti](I)I, which reacts with the carbonyl group to produce C[C@H]1CCCC2(O[Ti](I)(I)C2)[C@@]13C[C@H](C(OC)=O)CC3, eliminating O=[Ti](I)I to obtain $\bf{E}$, with the structure C[C@H]([C@]12C[C@H](C(OC)=O)CC2)CCCC1=C, and the chemical formula $C_{14}H_{22}O_2$

<Checkpoint>[The chemical formula of $\bf{E}$ is $C_{14}H_{22}O_2$](1)</Checkpoint>

$\bf{E}$ reacts with $p-TsOH$, and the double bond undergoes isomerization, yielding C[C@H]1CCC=C(C)[C@@]12C[C@H](C(OC)=O)CC2. Subsequently, methyl Grignard reagent reacts with the ester group, adding twice to obtain the product. If the Grignard reagent is used to react with $\bf{E}$ first, the ester group can be smoothly converted to a tertiary alcohol, but the hydroxyl group may be difficult to retain after treatment with $p-TsOH$, so the order of these two reaction steps cannot be reversed.

<Checkpoint>[Reversing the order of the two reaction steps to obtain the product from $\bf{E}$ may lead to the tertiary alcohol being difficult to retain](1)</Checkpoint>

Therefore, choose option E

<MultiModal>![本题中结构](/media/uploads/61470733-a609-404b-a6da-105f4e5367fe.png)</MultiModal>
O=C1C(CN2CCC[C@H]2COC)=CCCC1会与溴化锌形成配合物CO1C[C@@H]2CCC[N@@]23CC4=CCCCC4=O[Zn@]13[R]（R代表其他可能的配位基团），固定亲核位点附近的手性环境，随后二甲基铜锂从位阻小的一侧亲核并离去手性助剂得到$(S)-\bf{A}$，结构为C=C1[C@@H](C)CCCC1=O。

<Checkpoint>[生成$(S)-\bf{A}$需要溴化锌参与形成配合物固定手性环境](1)</Checkpoint>

二苯基铜锂加成后产生烯醇负离子，随后后处理产生新的羰基，这一过程会使六元环上所有取代基排斥最小，因此均处于平伏键，$\bf{B}$的结构为O=C1[C@H](CC2=CC=CC=C2)[C@@H](C)CCC1

<Checkpoint>[$\bf{B}$的结构为O=C1[C@H](CC2=CC=CC=C2)[C@@H](C)CCC1，均处于平伏键](1)</Checkpoint>

由$\bf{B}$的结构可知，新产生的手性中心绝对构型为R

<Checkpoint>[新产生的手性中心绝对构型为R](1)</Checkpoint>

$(S)-\bf{A}$在三苯基膦和C=C=CC(OC)=O作用下得到$\bf{C}$，首先三苯基膦与C=C=CC(OC)=O加成形成偶极体C=C([CH-]C(OC)=O)[P+](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3，随后发生$\left[ 3 { + } 2 \right]$ 反应，从位阻小的一侧反应产生另一中间体O=C1[C@@]2(CC(C(OC)=O)C(C2)=P(C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)[C@@H](C)CCC1，最后负电荷转移到酯基α位，消去三苯基膦得到$\bf{C}$，结构为C[C@H]([C@]12CC(C(OC)=O)=CC2)CCCC1=O

<Checkpoint>[生成$\bf{C}$的过程中经历了O=C1[C@@]2(CC(C(OC)=O)C(C2)=P(C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)[C@@H](C)CCC1中间体](1)</Checkpoint>

$\bf{C}$在钯碳催化下从位阻小的一侧加氢得到$\bf{D}$，结构为C[C@H]([C@]12C[C@H](C(OC)=O)CC2)CCCC1=O

$\bf{D}$得到$\bf{E}$的反应为Lombardo反应，首先锌还原二碘甲烷得到亚甲基卡宾，亚甲基卡宾与四碘化钛反应得到C=[Ti](I)I，其与羰基反应产生C[C@H]1CCCC2(O[Ti](I)(I)C2)[C@@]13C[C@H](C(OC)=O)CC3，离去O=[Ti](I)I得到$\bf{E}$，结构为C[C@H]([C@]12C[C@H](C(OC)=O)CC2)CCCC1=C，化学式为$C_{14}H_{22}O_2$

<Checkpoint>[$\bf{E}$的化学式为$C_{14}H_{22}O_2$](1)</Checkpoint>

$\bf{E}$与$p-TsOH$反应，双键发生异构化，得到C[C@H]1CCC=C(C)[C@@]12C[C@H](C(OC)=O)CC2，随后甲基格氏试剂与酯基反应，加成两次，得到产物。若先用格氏试剂与$\bf{E}$反应，酯基可顺利转化为三级醇，但再经$p-TsOH$处理羟基可能难以保留，因此不能调换这两步的反应顺序。

<Checkpoint>[调换$\bf{E}$得到产物的两步反应先后顺序会导致三级醇难以保留](1)</Checkpoint>

故选选项E

<MultiModal>![本题中结构](/media/uploads/61470733-a609-404b-a6da-105f4e5367fe.png)</MultiModal>