<MultiModal>![O=C(C=C1OC)C(C)=CC1=O>C=C/C=C/C.Cl[Sn](Cl)(Cl)Cl>[A], the reaction was carried out at -20°C](/media/uploads/07d887bb-5ab9-4611-ad86-b3f8419472ae.png)</MultiModal>

<MultiModal>![O=C(C=C1OC)C(C)=CC1=O>C=C/C=C/C.FB(F)(O(CC)CC)F>[B], the reaction was carried out at -20°C](/media/uploads/75820bb1-242a-4861-a1f3-4929dbbe0651.png)</MultiModal>

Please provide the most likely products A and B for the above two reactions, respectively (without considering stereochemistry)
<MultiModal>![O=C(C=C1OC)C(C)=CC1=O>C=C/C=C/C.Cl[Sn](Cl)(Cl)Cl>[A]，反应于-20 ℃进行](/media/uploads/07d887bb-5ab9-4611-ad86-b3f8419472ae.png)</MultiModal>

<MultiModal>![O=C(C=C1OC)C(C)=CC1=O>C=C/C=C/C.FB(F)(O(CC)CC)F>[B]，反应于-20 ℃进行](/media/uploads/75820bb1-242a-4861-a1f3-4929dbbe0651.png)</MultiModal>

请分别给出以上两个反应最有可能得到的产物A和B（不考虑立体化学）
<MultiModal>![O=C(C=C1OC)C(C(C)C=CC2)(C)C2C1=O, Product A](/media/uploads/d06adf26-cf31-497c-971a-bfe5dc2dce55.png)</MultiModal>

Product A

<MultiModal>![O=C(C=C1OC)C(CC=CC2C)(C)C2C1=O, Product B](/media/uploads/91395bca-f973-4001-bd11-dc8b27dcaba9.png)</MultiModal>

Product B
<MultiModal>![O=C(C=C1OC)C(C(C)C=CC2)(C)C2C1=O,Product A](/media/uploads/d06adf26-cf31-497c-971a-bfe5dc2dce55.png)</MultiModal>

Product A

<MultiModal>![O=C(C=C1OC)C(C(C)C=CC2)(C)C2C1=O,Product B](/media/uploads/d06adf26-cf31-497c-971a-bfe5dc2dce55.png)</MultiModal>

Product B
<MultiModal>![O=C(C=C1OC)C(CC=CC2C)(C)C2C1=O,Product A](/media/uploads/91395bca-f973-4001-bd11-dc8b27dcaba9.png)</MultiModal>

Product A

<MultiModal>![O=C(C=C1OC)C(CC=CC2C)(C)C2C1=O,Product B](/media/uploads/91395bca-f973-4001-bd11-dc8b27dcaba9.png)</MultiModal>

Product B
<MultiModal>![O=C(C=C1OC)C(CC=CC2C)(C)C2C1=O, Product A](/media/uploads/91395bca-f973-4001-bd11-dc8b27dcaba9.png)</MultiModal>  

Product A  

<MultiModal>![O=C(C=C1OC)C(C(C)C=CC2)(C)C2C1=O, Product B](/media/uploads/d06adf26-cf31-497c-971a-bfe5dc2dce55.png)</MultiModal>  

Product B
<MultiModal>![O=C(C=C1OC)C(C(C)C=CC2)(C)C2C1=O,产物A](/media/uploads/d06adf26-cf31-497c-971a-bfe5dc2dce55.png)</MultiModal>

产物A

<MultiModal>![O=C(C=C1OC)C(CC=CC2C)(C)C2C1=O,产物B](/media/uploads/91395bca-f973-4001-bd11-dc8b27dcaba9.png)</MultiModal>

产物B
<MultiModal>![O=C(C=C1OC)C(C(C)C=CC2)(C)C2C1=O,产物A](/media/uploads/d06adf26-cf31-497c-971a-bfe5dc2dce55.png)</MultiModal>

产物A

<MultiModal>![O=C(C=C1OC)C(C(C)C=CC2)(C)C2C1=O,产物B](/media/uploads/d06adf26-cf31-497c-971a-bfe5dc2dce55.png)</MultiModal>

产物B
<MultiModal>![O=C(C=C1OC)C(CC=CC2C)(C)C2C1=O,产物A](/media/uploads/91395bca-f973-4001-bd11-dc8b27dcaba9.png)</MultiModal>

产物A

<MultiModal>![O=C(C=C1OC)C(CC=CC2C)(C)C2C1=O,产物B](/media/uploads/91395bca-f973-4001-bd11-dc8b27dcaba9.png)</MultiModal>

产物B
<MultiModal>![O=C(C=C1OC)C(CC=CC2C)(C)C2C1=O,产物A](/media/uploads/91395bca-f973-4001-bd11-dc8b27dcaba9.png)</MultiModal>

产物A

<MultiModal>![O=C(C=C1OC)C(C(C)C=CC2)(C)C2C1=O,产物B](/media/uploads/d06adf26-cf31-497c-971a-bfe5dc2dce55.png)</MultiModal>

产物B
$\mathrm{SnCl}_{4}$ can simultaneously coordinate with both the carbonyl and methoxy groups, thereby activating the lower carbonyl.  

<Checkpoint>[$\mathrm{SnCl}_{4}$ can simultaneously coordinate with both the carbonyl and methoxy groups, thereby activating the lower carbonyl](1)</Checkpoint>  

In contrast, $\mathrm{BF}_{3}$ can only form tetracoordination and preferentially coordinates with the electron-rich upper carbonyl, activating the upper carbonyl.  

<Checkpoint>[In contrast, $\mathrm{BF}_{3}$ can only form tetracoordination and preferentially coordinates with the electron-rich upper carbonyl, activating the upper carbonyl](1)</Checkpoint>  

For the diene component, the alkyl group is electron-donating, so the side connected to the alkyl group becomes positively charged, while the other end becomes negatively charged. Reacting with the corresponding sites of the dienophile yields products A and B:  

A. <MultiModal>![O=C(C=C1OC)C(CC=CC2C)(C)C2C1=O,Product A](/media/uploads/91395bca-f973-4001-bd11-dc8b27dcaba9.png)</MultiModal>  

B. <MultiModal>![O=C(C=C1OC)C(C(C)C=CC2)(C)C2C1=O,Product B](/media/uploads/d06adf26-cf31-497c-971a-bfe5dc2dce55.png)</MultiModal>  

Therefore, option D is correct.  

<Checkpoint>[Product A is O=C(C=C1OC)C(CC=CC2C)(C)C2C1=O](1)</Checkpoint>  

<Checkpoint>[Product B is O=C(C=C1OC)C(C(C)C=CC2)(C)C2C1=O](1)</Checkpoint>
$\mathrm{SnCl}_{4}$ 可以同时与羰基和甲氧基络合, 从而活化下面的羰基



<Checkpoint>[$\mathrm{SnCl}_{4}$ 可以同时与羰基和甲氧基络合, 从而活化下面的羰基](1)</Checkpoint>

而 $\mathrm{BF}_{3}$ 只能四配位, 优先与上方富电子羰基络合, 活化上方的羰基

<Checkpoint>[而 $\mathrm{BF}_{3}$ 只能四配位, 优先与上方富电子羰基络合, 活化上方的羰基](1)</Checkpoint>

对双烯体而言，烷基给电子，因此与烷基连接一侧为正电性，另一端为负电性，与亲双烯体的相应位点反应可以得到产物A和B：

A. <MultiModal>![O=C(C=C1OC)C(CC=CC2C)(C)C2C1=O,产物A](/media/uploads/91395bca-f973-4001-bd11-dc8b27dcaba9.png)</MultiModal>

B.<MultiModal>![O=C(C=C1OC)C(C(C)C=CC2)(C)C2C1=O,产物B](/media/uploads/d06adf26-cf31-497c-971a-bfe5dc2dce55.png)</MultiModal>

因此选项D正确。

<Checkpoint>[产物A为O=C(C=C1OC)C(CC=CC2C)(C)C2C1=O](1)</Checkpoint>

<Checkpoint>[产物B为O=C(C=C1OC)C(C(C)C=CC2)(C)C2C1=O](1)</Checkpoint>