<MultiModal>![O=C1C=C(C)CCC1>BrCCCBr.[Na]N>[$\mathbf{B}$], $\mathbf{B}$ is the product](/media/uploads/f93683e2-8012-4378-9e29-59ed38462e73.png)</MultiModal>

<MultiModal>![O=C1C=C(C)CCC1>BrCCCCBr.[Na]N>[$\mathbf{C}$], $\mathbf{C}$ is the product](/media/uploads/4c046178-b1ac-4c22-a78a-1e02ceaab847.png)</MultiModal>

Given that the products $\mathbf{B}$ and $\mathbf{C}$ obtained in the liquid ammonia solution of sodium amide are slightly different, and without considering enantiomerism, try to give the structural formulas of products $\mathbf{B}$ and $\mathbf{C}$ respectively.
<MultiModal>![O=C1C=C(C)CCC1>BrCCCBr.[Na]N>[$\mathbf{B}$],$\mathbf{B}$是产物](/media/uploads/f93683e2-8012-4378-9e29-59ed38462e73.png)</MultiModal>

<MultiModal>![O=C1C=C(C)CCC1>BrCCCCBr.[Na]N>[$\mathbf{C}$],$\mathbf{C}$是产物](/media/uploads/4c046178-b1ac-4c22-a78a-1e02ceaab847.png)</MultiModal>

已知在氨基钠的液氨溶液中，反应得到的产物$\mathbf{B}$和$\mathbf{C}$略有不同，在不考虑对映异构的条件下，试分别给出产物$\mathbf{B}$和$\mathbf{C}$的结构式
All other options are incorrect
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<MultiModal>![CC1=CCCC2=C1CCCO2](/media/uploads/56e1e57d-066e-436e-ac79-4b7e12dc678c.png)</MultiModal>

<MultiModal>![CC1=CCCC2=C1CCCCO2](/media/uploads/cc4e23a4-0924-4a9c-8007-0c2b4306b55a.png)</MultiModal>
<MultiModal>![CC1=CCCC2=C1CCCO2](/media/uploads/56e1e57d-066e-436e-ac79-4b7e12dc678c.png)</MultiModal>

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<MultiModal>![O=C1C2(CCC2)C(C)=CCC1](/media/uploads/af861ebc-cfae-472e-9425-36fcf6ae2bdd.png)</MultiModal>

<MultiModal>![O=C1C2(CCCC2)C(C)=CCC1](/media/uploads/8d75636d-6abe-46ff-adc5-74446538ded3.png)</MultiModal>
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<MultiModal>![CC1=CCCC2=C1CCCCO2](/media/uploads/cc4e23a4-0924-4a9c-8007-0c2b4306b55a.png)</MultiModal>
<MultiModal>![O=C1C2(CCC2)C(C)=CCC1](/media/uploads/af861ebc-cfae-472e-9425-36fcf6ae2bdd.png)</MultiModal>

<MultiModal>![O=C1C=C(C)CCC12CCCC2](/media/uploads/c6e64d42-4968-4224-8276-439091c1a286.png)</MultiModal>
<MultiModal>![O=C1C2(CCC2)C(C)=CCC1](/media/uploads/af861ebc-cfae-472e-9425-36fcf6ae2bdd.png)</MultiModal>

<MultiModal>![O=C1C=C(C)C2(CCCC2)CC1](/media/uploads/0152b252-d22f-4988-b538-0e68a531bd69.png)</MultiModal>
<MultiModal>![O=C1C2(CCC2)C(C)=CCC1](/media/uploads/af861ebc-cfae-472e-9425-36fcf6ae2bdd.png)</MultiModal>

<MultiModal>![CC1=CC(OCCCC2)=C2CC1](/media/uploads/d66c2d1b-a1a2-4c24-a08e-3398e65c2b4d.png)</MultiModal>
<MultiModal>![CC1=CC(OCCC2)=C2CC1](/media/uploads/90e7889d-a38f-4c5b-ac1e-336ada808dbe.png)</MultiModal>

<MultiModal>![O=C1C2(CCCC2)C(C)=CCC1](/media/uploads/8d75636d-6abe-46ff-adc5-74446538ded3.png)</MultiModal>
<MultiModal>![CC1=CC(OCCC2)=C2CC1](/media/uploads/90e7889d-a38f-4c5b-ac1e-336ada808dbe.png)</MultiModal>

<MultiModal>![CC1=CCCC2=C1CCCCO2](/media/uploads/cc4e23a4-0924-4a9c-8007-0c2b4306b55a.png)</MultiModal>
<MultiModal>![CC1=CC(OCCC2)=C2CC1](/media/uploads/90e7889d-a38f-4c5b-ac1e-336ada808dbe.png)</MultiModal>

<MultiModal>![O=C1C=C(C)CCC12CCCC2](/media/uploads/c6e64d42-4968-4224-8276-439091c1a286.png)</MultiModal>
<MultiModal>![CC1=CC(OCCC2)=C2CC1](/media/uploads/90e7889d-a38f-4c5b-ac1e-336ada808dbe.png)</MultiModal>

<MultiModal>![O=C1C=C(C)C2(CCCC2)CC1](/media/uploads/0152b252-d22f-4988-b538-0e68a531bd69.png)</MultiModal>
<MultiModal>![CC1=CC(OCCC2)=C2CC1](/media/uploads/90e7889d-a38f-4c5b-ac1e-336ada808dbe.png)</MultiModal>

<MultiModal>![CC1=CC(OCCCC2)=C2CC1](/media/uploads/d66c2d1b-a1a2-4c24-a08e-3398e65c2b4d.png)</MultiModal>
其他选项均不正确
<MultiModal>![CC1=CCCC2=C1CCCO2](/media/uploads/56e1e57d-066e-436e-ac79-4b7e12dc678c.png)</MultiModal>

<MultiModal>![O=C1C2(CCCC2)C(C)=CCC1](/media/uploads/8d75636d-6abe-46ff-adc5-74446538ded3.png)</MultiModal>

<MultiModal>![CC1=CCCC2=C1CCCO2](/media/uploads/56e1e57d-066e-436e-ac79-4b7e12dc678c.png)</MultiModal>

<MultiModal>![CC1=CCCC2=C1CCCCO2](/media/uploads/cc4e23a4-0924-4a9c-8007-0c2b4306b55a.png)</MultiModal>

<MultiModal>![CC1=CCCC2=C1CCCO2](/media/uploads/56e1e57d-066e-436e-ac79-4b7e12dc678c.png)</MultiModal>

<MultiModal>![O=C1C=C(C)CCC12CCCC2](/media/uploads/c6e64d42-4968-4224-8276-439091c1a286.png)</MultiModal>

<MultiModal>![CC1=CCCC2=C1CCCO2](/media/uploads/56e1e57d-066e-436e-ac79-4b7e12dc678c.png)</MultiModal>

<MultiModal>![O=C1C=C(C)C2(CCCC2)CC1](/media/uploads/0152b252-d22f-4988-b538-0e68a531bd69.png)</MultiModal>

<MultiModal>![CC1=CCCC2=C1CCCO2](/media/uploads/56e1e57d-066e-436e-ac79-4b7e12dc678c.png)</MultiModal>

<MultiModal>![CC1=CC(OCCCC2)=C2CC1](/media/uploads/d66c2d1b-a1a2-4c24-a08e-3398e65c2b4d.png)</MultiModal>
<MultiModal>![O=C1C2(CCC2)C(C)=CCC1](/media/uploads/af861ebc-cfae-472e-9425-36fcf6ae2bdd.png)</MultiModal>

<MultiModal>![O=C1C2(CCCC2)C(C)=CCC1](/media/uploads/8d75636d-6abe-46ff-adc5-74446538ded3.png)</MultiModal>
<MultiModal>![O=C1C2(CCC2)C(C)=CCC1](/media/uploads/af861ebc-cfae-472e-9425-36fcf6ae2bdd.png)</MultiModal>

<MultiModal>![CC1=CCCC2=C1CCCCO2](/media/uploads/cc4e23a4-0924-4a9c-8007-0c2b4306b55a.png)</MultiModal>

<MultiModal>![O=C1C2(CCC2)C(C)=CCC1](/media/uploads/af861ebc-cfae-472e-9425-36fcf6ae2bdd.png)</MultiModal>

<MultiModal>![O=C1C=C(C)CCC12CCCC2](/media/uploads/c6e64d42-4968-4224-8276-439091c1a286.png)</MultiModal>

<MultiModal>![O=C1C2(CCC2)C(C)=CCC1](/media/uploads/af861ebc-cfae-472e-9425-36fcf6ae2bdd.png)</MultiModal>

<MultiModal>![O=C1C=C(C)C2(CCCC2)CC1](/media/uploads/0152b252-d22f-4988-b538-0e68a531bd69.png)</MultiModal>

<MultiModal>![O=C1C2(CCC2)C(C)=CCC1](/media/uploads/af861ebc-cfae-472e-9425-36fcf6ae2bdd.png)</MultiModal>

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<MultiModal>![CC1=CC(OCCC2)=C2CC1](/media/uploads/90e7889d-a38f-4c5b-ac1e-336ada808dbe.png)</MultiModal>

<MultiModal>![O=C1C=C(C)CCC12CCCC2](/media/uploads/c6e64d42-4968-4224-8276-439091c1a286.png)</MultiModal>
<MultiModal>![CC1=CC(OCCC2)=C2CC1](/media/uploads/90e7889d-a38f-4c5b-ac1e-336ada808dbe.png)</MultiModal>

<MultiModal>![O=C1C=C(C)C2(CCCC2)CC1](/media/uploads/0152b252-d22f-4988-b538-0e68a531bd69.png)</MultiModal>

<MultiModal>![CC1=CC(OCCC2)=C2CC1](/media/uploads/90e7889d-a38f-4c5b-ac1e-336ada808dbe.png)</MultiModal>

<MultiModal>![CC1=CC(OCCCC2)=C2CC1](/media/uploads/d66c2d1b-a1a2-4c24-a08e-3398e65c2b4d.png)</MultiModal>
First, in the liquid ammonia system, the thermodynamically stable linear conjugated anion intermediate $\mathbf{1}$ tends to form preferentially.

<MultiModal>![[O-]C1=CC(C)=CCC1](/media/uploads/18026980-0926-4150-a733-ad8df3da75c3.png)</MultiModal>

<Checkpoint>[Anion intermediate $\mathbf{1}$:[O-]C1=CC(C)=CCC1](1)</Checkpoint>

For the reaction to generate products $\mathbf{B}$ and $\mathbf{C}$, since the carbanion is more stable at the $\alpha$ position of the carbonyl group, it first reacts with the corresponding haloalkanes to obtain intermediate $\mathbf{2}$ and intermediate $\mathbf{3}$ respectively.

<Checkpoint>[Since the carbanion is more stable at the $\alpha$ position of the carbonyl group, it first reacts with the corresponding haloalkanes to obtain intermediate $\mathbf{2}$ and intermediate $\mathbf{3}$](1)</Checkpoint>

<MultiModal>![Intermediate $\mathbf{2}$: O=C1C(CCCBr)C(C)=CCC1](/media/uploads/f1946f24-640e-4f84-9dc7-c0150ae87080.png)</MultiModal>

<Checkpoint>[Intermediate $\mathbf{2}$: O=C1C(CCCBr)C(C)=CCC1](1)</Checkpoint>

<MultiModal>![Intermediate $\mathbf{3}$: O=C1C(CCCCBr)C(C)=CCC1](/media/uploads/65b64c64-788b-409e-af11-85dfbedb592b.png)</MultiModal>

<Checkpoint>[Intermediate $\mathbf{3}$: O=C1C(CCCCBr)C(C)=CCC1](1)</Checkpoint>

Then, a proton is abstracted by a base to obtain intermediate $\mathbf{4}$ and intermediate $\mathbf{5}$ respectively.

<MultiModal>![Intermediate $\mathbf{4}$: [O-]C1=C(CCCBr)C(C)=CCC1](/media/uploads/5abf3c1f-88e3-47c2-bf40-c012a8b6e574.png)</MultiModal>

<Checkpoint>[Intermediate $\mathbf{4}$: [O-]C1=C(CCCBr)C(C)=CCC1](1)</Checkpoint>

<MultiModal>![Intermediate $\mathbf{5}$: [O-]C1=C(CCCCBr)C(C)=CCC1](/media/uploads/96281ac7-43c5-4bc2-b293-42a465be1606.png)</MultiModal>

<Checkpoint>[Intermediate $\mathbf{5}$: [O-]C1=C(CCCCBr)C(C)=CCC1](1)</Checkpoint>

For intermediate $\mathbf{5}$, since the carbon-end nucleophilicity is stronger than the oxygen-end, the product $\mathbf{C}$ containing a five-membered ring is preferentially obtained.

<Checkpoint>[For intermediate $\mathbf{5}$, since the carbon-end nucleophilicity is stronger than the oxygen-end, the product $\mathbf{C}$ containing a five-membered ring is preferentially obtained](1)</Checkpoint>

<MultiModal>![Product $\mathbf{C}$: O=C1C2(CCCC2)C(C)=CCC1](/media/uploads/8d75636d-6abe-46ff-adc5-74446538ded3.png)</MultiModal>

<Checkpoint>[Product $\mathbf{C}$: O=C1C2(CCCC2)C(C)=CCC1](1)</Checkpoint>

For intermediate $\mathbf{4}$, since the four-membered ring formed by carbon-end nucleophilic attack is unstable, it tends to use oxygen-end nucleophilic attack to form product $\mathbf{B}$ containing two six-membered rings.

<Checkpoint>[For intermediate $\mathbf{4}$, since the four-membered ring formed by carbon-end nucleophilic attack is unstable, it tends to use oxygen-end nucleophilic attack to form product $\mathbf{B}$ containing two six-membered rings](1)</Checkpoint>

<MultiModal>![Product $\mathbf{B}$: CC1=CCCC2=C1CCCO2](/media/uploads/56e1e57d-066e-436e-ac79-4b7e12dc678c.png)</MultiModal>

<Checkpoint>[Product $\mathbf{B}$: CC1=CCCC2=C1CCCO2](1)</Checkpoint>
首先，在液氨体系中，倾向于优先形成热力学稳定的线性共轭阴离子中间体$\mathbf{1}$

<MultiModal>![[O-]C1=CC(C)=CCC1](/media/uploads/18026980-0926-4150-a733-ad8df3da75c3.png)</MultiModal>

<Checkpoint>[阴离子中间体$\mathbf{1}$:[O-]C1=CC(C)=CCC1](1)</Checkpoint>

对于生成产物$\mathbf{B}$和$\mathbf{C}$的反应，由于碳负离子在羰基$\alpha$位更加稳定，因此首先都与相应的卤代烷反应分别得到中间体$\mathbf{2}$和中间体$\mathbf{3}$

<Checkpoint>[由于碳负离子在羰基$\alpha$位更加稳定，因此首先都与相应的卤代烷反应分别得到中间体$\mathbf{2}$和中间体$\mathbf{3}$](1)</Checkpoint>

<MultiModal>![中间体$\mathbf{2}$：O=C1C(CCCBr)C(C)=CCC1](/media/uploads/f1946f24-640e-4f84-9dc7-c0150ae87080.png)</MultiModal>

<Checkpoint>[中间体$\mathbf{2}$：O=C1C(CCCBr)C(C)=CCC1](1)</Checkpoint>

<MultiModal>![中间体$\mathbf{3}$：O=C1C(CCCCBr)C(C)=CCC1](/media/uploads/65b64c64-788b-409e-af11-85dfbedb592b.png)</MultiModal>

<Checkpoint>[中间体$\mathbf{3}$：O=C1C(CCCCBr)C(C)=CCC1](1)</Checkpoint>

接着再被碱攫取一个质子分别得到中间体$\mathbf{4}$和中间体$\mathbf{5}$

<MultiModal>![中间体$\mathbf{4}$：[O-]C1=C(CCCBr)C(C)=CCC1](/media/uploads/5abf3c1f-88e3-47c2-bf40-c012a8b6e574.png)</MultiModal>

<Checkpoint>[中间体$\mathbf{4}$：[O-]C1=C(CCCBr)C(C)=CCC1](1)</Checkpoint>

<MultiModal>![中间体$\mathbf{5}$：[O-]C1=C(CCCCBr)C(C)=CCC1](/media/uploads/96281ac7-43c5-4bc2-b293-42a465be1606.png)</MultiModal>

<Checkpoint>[中间体$\mathbf{5}$：[O-]C1=C(CCCCBr)C(C)=CCC1](1)</Checkpoint>

对于中间体$\mathbf{5}$，由于碳端亲核能力强于氧端，因此优先得到含五元环的产物$\mathbf{C}$

<Checkpoint>[对于中间体$\mathbf{5}$，由于碳端亲核能力强于氧端，因此优先得到含五元环的产物$\mathbf{C}$](1)</Checkpoint>

<MultiModal>![产物$\mathbf{C}$：O=C1C2(CCCC2)C(C)=CCC1](/media/uploads/8d75636d-6abe-46ff-adc5-74446538ded3.png)</MultiModal>

<Checkpoint>[产物$\mathbf{C}$：O=C1C2(CCCC2)C(C)=CCC1](1)</Checkpoint>

而对于中间体$\mathbf{4}$，由于碳端亲核形成的四元环不稳定，因此倾向于用氧端亲核，形成含两个六元环的产物$\mathbf{B}$

<Checkpoint>[而对于中间体$\mathbf{4}$，由于碳端亲核形成的四元环不稳定，因此倾向于用氧端亲核，形成含两个六元环的产物$\mathbf{B}$](1)</Checkpoint>

<MultiModal>![产物$\mathbf{B}$：CC1=CCCC2=C1CCCO2](/media/uploads/56e1e57d-066e-436e-ac79-4b7e12dc678c.png)</MultiModal>

<Checkpoint>[产物$\mathbf{B}$：CC1=CCCC2=C1CCCO2](1)</Checkpoint>