The construction of medium-sized rings is different from the construction of five- or six-membered rings. Due to their larger ring strain, some relatively more special reaction conditions are required to achieve this. Radical chemistry can play a good role in the construction of this type of ring system. By using simple reduction conditions, medium-sized rings can also be constructed efficiently. Select the final products of the following 3 reactions, without requiring stereochemistry.
<MultiModal>![Reaction 1: CC1(C)CCC(C/C=C/C2=CC=CC=C2)OC1=O reacts with $SmI_2$ in $H_2O$
Reaction 2: C=C(CCOC(C)(C)CC(CCCC)=O)CCl reacts with $SmI_2$ in THF
Reaction 3: BrC1=CC=CC=C1CCCN(C(C)=O)C#C[Si](C)(C)C reacts with $Bu_3SnH, AIBN, NaOH$ in benzene under reflux](/media/uploads/ecf8fca1-03c9-4382-a93e-7f113ee4bb85.png)</MultiModal>

<MultiModal>![a) CC(C1O)(C)CCCCC1CC2=CC=CC=C2
b) OC(CC1CC2=CC=CC=C2)CCC(C)(C)C1O
c) CC1(C)CCC(CC2CC3=CC=CC=C3)OC12O
d) CC1(C)CCCC(O)CC(C2=CC=CC=C2)C1O
e) C=C1CCOC(C)(C)CC(CCCC)(O)C1
f) C=C1CCOC(C)(C)CC(CCCC)C1
g) CC(N1/C=C([Si](C)(C)C)\C2=C(C=CC=CC=C2)CCC1)=O
h) C[Si](/C1=C/NCCCC2=C1C=CC=C2)(C)C
i) O=C(C)N1/C(C2=CC=CC=C2CCC1)=C/[Si](C)(C)C](/media/uploads/33f03f58-c4b1-4d39-9f89-3134250ab748.png)</MultiModal>
中环的构建不同于五六元环的构建，由于其较大的环张力需要一些相对更特殊的反应条件实现。自由基化学可以在该类环系的构建中起到好的效果。利用简单的还原条件，也可以高效地构筑中环。选出下面3个反应的最终产物，不要求立体化学。

<MultiModal>![反应1：CC1(C)CCC(C/C=C/C2=CC=CC=C2)OC1=O与$SmI_2$在$H_2O$中反应
反应2：C=C(CCOC(C)(C)CC(CCCC)=O)CCl在THF中与$SmI_2$反应
反应3：BrC1=CC=CC=C1CCCN(C(C)=O)C#C[Si](C)(C)C先在benzene中回流下与$Bu_3SnH, AIBN$反应，之后在室温与$1M NaOH$反应。](/media/uploads/ecf8fca1-03c9-4382-a93e-7f113ee4bb85.png)</MultiModal>

<MultiModal>![a) CC(C1O)(C)CCCCC1CC2=CC=CC=C2
b) OC(CC1CC2=CC=CC=C2)CCC(C)(C)C1O
c) CC1(C)CCC(CC2CC3=CC=CC=C3)OC12O
d) CC1(C)CCCC(O)CC(C2=CC=CC=C2)C1O
e) C=C1CCOC(C)(C)CC(CCCC)(O)C1
f) C=C1CCOC(C)(C)CC(CCCC)C1
g) CC(N1/C=C([Si](C)(C)C)\C2=C(C=CC=C2)CCC1)=O
h) C[Si](/C1=C/NCCCC2=C1C=CC=C2)(C)C
i) O=C(C)N1/C(C2=CC=CC=C2CCC1)=C/[Si](C)(C)C](/media/uploads/33f03f58-c4b1-4d39-9f89-3134250ab748.png)</MultiModal>
All other options are incorrect
a.e.g.
c.e.g.
b.e.g.
d.e.h.
b.f.g.
c.e.i.
d.e.i.
b.e.h.
其他选项均不正确
a.e.g.
c.e.g.
b.e.g.
d.e.h.
b.f.g.
c.e.i.
d.e.i.
b.e.h.
Reaction 1:

CC1(C)CCC(C/C=C/C2=CC=CC=C2)OC1=O reacts with $SmI_2$ in $H_2O$, the carbonyl group is reduced to a hydroxyl group and a carbon radical, undergoing a 5-exo ring closure, to obtain OC(CC1C[X, X is a single electron]C2=CC=CC=C2)CCC(C)(C)C1O

<Checkpoint>[OC(CC1C[X, X is a single electron]C2=CC=CC=C2)CCC(C)(C)C1O](1)</Checkpoint>

OC(CC1C[X, X is a single electron]C2=CC=CC=C2)CCC(C)(C)C1O is further reduced and quenched with water to obtain c.

<Checkpoint>[OC(CC1C[X, X is a single electron]C2=CC=CC=C2)CCC(C)(C)C1O is further reduced and quenched with water to obtain c.](1)</Checkpoint>

c can undergo ring-opening in water and is further reduced to obtain b.

<Checkpoint>[c can undergo ring-opening in water and is further reduced to obtain b.](1)</Checkpoint>

Reaction 2:

Allyl chloride is reduced to an allyl radical, and the carbonyl group is also reduced to a carbon radical and an oxygen anion. The two radicals couple and quench to obtain e.

<Checkpoint>[Reaction 2, allyl chloride is reduced to an allyl radical, and the carbonyl group is also reduced to a carbon radical and an oxygen anion. The two radicals couple and quench to obtain e.](1)</Checkpoint>

Reaction 3:

Under radical initiation conditions, $Sn$ radicals are generated, abstracting $Br$ to obtain phenyl radicals, undergoing 8-endo ring closure on the alkyne to obtain CC(N1/C[X, X is a single electron]=C([Si](C)(C)C)\C2=C(C=CC=C2)CCC1)=O

CC(N1/C[X, X is a single electron]=C([Si](C)(C)C)\C2=C(C=CC=C2)CCC1)=O is quenched by Sn-H to obtain g.

Because Si can stabilize β-position radicals, 7-endo ring closure does not occur.

The amide bond is difficult to hydrolyze at room temperature, and the acetyl group will not leave, so the product is not h.

<Checkpoint>[Because Si can stabilize β-position radicals, 7-endo ring closure does not occur.](1)</Checkpoint>

<Checkpoint>[Undergoing 8-endo ring closure to obtain CC(N1/C[X, X is a single electron]=C([Si](C)(C)C)\C2=C(C=CC=C2)CCC1)=O](1)</Checkpoint>

<Checkpoint>[CC(N1/C[X, X is a single electron]=C([Si](C)(C)C)\C2=C(C=CC=C2)CCC1)=O is quenched by Sn-H to obtain g.](1)</Checkpoint>

<Checkpoint>[The amide bond is difficult to hydrolyze at room temperature, the product is not h](1)</Checkpoint>

In summary, option D is reasonable.
反应1：

CC1(C)CCC(C/C=C/C2=CC=CC=C2)OC1=O与$SmI_2$在$H_2O$中反应，羰基被还原成羟基和碳自由基，发生一次5 exo关环，得到OC(CC1C[X,X是单电子]C2=CC=CC=C2)CCC(C)(C)C1O

<Checkpoint>[OC(CC1C[X,X是单电子]C2=CC=CC=C2)CCC(C)(C)C1O](1)</Checkpoint>

OC(CC1C[X,X是单电子]C2=CC=CC=C2)CCC(C)(C)C1O被进一步还原，并被水淬灭得到c.

<Checkpoint>[OC(CC1C[X,X是单电子]C2=CC=CC=C2)CCC(C)(C)C1O被进一步还原，并被水淬灭得到c.](1)</Checkpoint>

c在水中可以开环，并且被进一步还原得到b。

<Checkpoint>[c在水中可以开环，并且被进一步还原得到b。](1)</Checkpoint>

反应2：

烯丙基氯被还原得到烯丙基自由基，羰基也被还原得到碳自由基和氧负离子，两个自由基偶联并淬灭，得到e.

<Checkpoint>[反应2，烯丙基氯被还原得到烯丙基自由基，羰基也被还原得到碳自由基和氧负离子，两个自由基偶联并淬灭，得到e.](1)</Checkpoint>

反应3：

自由基引发条件下产生$Sn$自由基，攫取$Br$得到苯基自由基，对炔基发生8-endo关环得到CC(N1/C[X，X是单电子]=C([Si](C)(C)C)\C2=C(C=CC=C2)CCC1)=O

CC(N1/C[X，X是单电子]=C([Si](C)(C)C)\C2=C(C=CC=C2)CCC1)=O被Sn-H淬灭得到g。

因为Si能稳定β位自由基，因此不发生7-endo关环。

室温下酰胺键难以水解，乙酰基不会离去，因此产物不为h。

<Checkpoint>[因为Si能稳定β位自由基，因此不发生7-endo关环。](1)</Checkpoint>

<Checkpoint>[发生8-endo关环得到CC(N1/C[X，X是单电子]=C([Si](C)(C)C)\C2=C(C=CC=C2)CCC1)=O](1)</Checkpoint>

<Checkpoint>[CC(N1/C[X，X是单电子]=C([Si](C)(C)C)\C2=C(C=CC=C2)CCC1)=O被Sn-H淬灭得到g。](1)</Checkpoint>

<Checkpoint>[室温下酰胺键难以水解，产物不为h](1)</Checkpoint>

综上，D选项是合理的。