For the following organic reaction, select the incorrect description regarding its organic product.

<MultiModal>![The image shows a rightward horizontal arrow. On the left side of the arrow is an organic molecule with the structure c1ccccc1CCCO. Above the arrow is the text [Et_2N=SF_2][BF_4], and below the arrow is the text Et_3N.](/media/uploads/02167c2d-5a6c-4a97-82dc-6f6a63be0503.png)</MultiModal>
对于下列有机反应，关于其有机产物的描述，选出不正确的一项。

<MultiModal>![图中有一个向右的横向箭头，箭头左边为一个有机分子，其结构为c1ccccc1CCCO，箭头上方文字为[Et_2N=SF_2][BF_4]，箭头下方文字为Et_3N。](/media/uploads/02167c2d-5a6c-4a97-82dc-6f6a63be0503.png)</MultiModal>
Under this reaction condition, the mass fraction of the halogen in the generated halogenated hydrocarbon molecule is $13.75\%$.
Under this reaction condition, the two atoms with the longest shortest path in the generated ether pass through 16 chemical bonds.
Under this reaction condition, an organic molecule with a 4-coordinate sulfur atom was generated, and the total number of atoms in the molecule is 45.
Under this reaction condition, an organic compound without a benzene ring was generated, which contains atoms of six different elements.
Under these reaction conditions, among the benzene-ring-containing products generated, the product with the highest relative molecular mass exceeds 300.
Under these reaction conditions, among the generated products, there are organic compounds devoid of benzene rings, with relative molecular masses ranging between 130 and 140.
Under these reaction conditions, there are no two organic products with a relative molecular mass difference of less than 10 between them.
If the reaction is carried out in acetonitrile solvent, a byproduct containing nitrogen, oxygen, and benzene rings but no fluorine or sulfur will be obtained.
该反应条件下生成的一种卤代烃分子中，卤素的质量分数为$13.75\%$。
该反应条件下生成的一种醚中，最短路径最长的两个原子之间经过了16根化学键。
该反应条件下生成了一种具有4配位硫原子的有机分子，其分子中的总原子数为45。
该反应条件下，生成了一种不含苯环的有机物，其包含6种不同元素的原子。
该反应条件下，生成的含苯环的产物中，相对分子质量最大的产物，其相对分子质量超过300。
该反应条件下，生成的产物中，存在不含苯环的有机物，其相对分子质量介于130至140范围。
在该反应条件下，不存在两种彼此之间相对分子质量之差小于10的有机产物。
若反应在乙腈溶剂中进行，会得到同时含有氮、氧、苯环，且不含氟或硫的副产物。
In this reaction system, triethylamine acts as a base to deprotonate the hydroxyl group of the alcohol, yielding an alkoxide intermediate:

<MultiModal>![c1ccccc1CCC[O-]](/media/uploads/9d5c96f6-0d0a-4852-9349-937defe65bab.png)</MultiModal>

In the cation of $\mathrm{\left[Et_2N=SF_2\right]\left[BF_4\right]}$, the sulfur atom is electrophilic. Upon attack by the alkoxide of the substrate, a neutral intermediate containing tetracoordinated sulfur—with both nitrogen and oxygen ligands—is formed:

<MultiModal>![CCN(CC)S(F)(F)OCCCc1ccccc1](/media/uploads/cfd80569-3af1-401e-882f-cbbffe5b9957.png)</MultiModal>

<Checkpoint>[Intermediate CCN(CC)S(F)(F)OCCCc1ccccc1 generated](1)</Checkpoint>

This intermediate subsequently undergoes an intramolecular substitution reaction. The possible mechanism involves the lone pair on $\mathrm{O}$ attacking the antibonding orbital of the $\mathrm{S-F}$ bond, cleaving the $\mathrm{S-F}$ bond to form a tight ion pair. The fluoride ion then attacks the carbon bonded to oxygen, replacing the oxygen of the alkoxide to yield a monofluorinated product:

<MultiModal>![c1ccccc1CCCF](/media/uploads/1636b9e2-8869-48ce-9c54-597bf6bf829d.png)</MultiModal>

<Checkpoint>[Monofluorinated product c1ccccc1CCCF](1)</Checkpoint>

Simultaneously, a product containing a sulfur-oxygen bond is formed:

<MultiModal>![CCN(CC)S(=O)F](/media/uploads/d70ebdb5-6b26-44e3-87fd-2476591d8649.png)</MultiModal>

<Checkpoint>[Byproduct CCN(CC)S(=O)F departs](1)</Checkpoint>

These two are the main products under these conditions. The relative molecular mass of the monofluorinated product is $19.00+12.01\times9+1.008\times11=138.178$, and the mass fraction of fluorine is $\frac{19.00}{138.178}\times100\%=13.75\%$. Thus, option $\mathbf{A}$ is correct.

<Checkpoint>[Relative molecular mass of fluorinated product is $138.178$](1)</Checkpoint>

<Checkpoint>[Mass fraction of fluorine is $13.75\%$](1)</Checkpoint>

The other product does not contain a benzene ring, and its relative molecular mass is $139.2$, falling between $130-140$. Thus, option $\mathbf{F}$ is correct. This molecule contains six elements: $\mathrm{C,H,O,N,F,S}$, so option $\mathbf{D}$ is also correct.

<Checkpoint>[Product lacks a benzene ring; relative molecular mass is $139.2$](1)</Checkpoint>

<Checkpoint>[Product contains $\mathrm{C,H,O,N,F,S}$ six elements](1)</Checkpoint>

Note that the difference in relative molecular mass between the monofluorinated product and the departing oxygen-containing molecule is $1.022<10$, making option $\mathbf{G}$ incorrect and thus a candidate for exclusion.

The intermediate
<MultiModal>![CCN(CC)S(F)(F)OCCCc1ccccc1](/media/uploads/cfd80569-3af1-401e-882f-cbbffe5b9957.png)</MultiModal>
can also be nucleophilically attacked by the alkoxide. If the alkoxide attacks the carbon bonded to oxygen, a stabilized anion leaves by expelling $F^{-}$, yielding an ether:

<MultiModal>![c1ccccc1CCCOCCCc2ccccc2](/media/uploads/0005e100-a574-4e5b-87cd-b28284302fea.png)</MultiModal>

<Checkpoint>[Further nucleophilic attack yields an ether](1)</Checkpoint>

Its relative molecular mass is $254.356$.

<Checkpoint>[Relative molecular mass of ether is $254.356$](1)</Checkpoint>

In this molecule, the longest shortest path along chemical bonds occurs between a hydrogen at the *para*-position of one benzene ring and a hydrogen at the *para*-position of the other benzene ring: $\mathrm{H-C-C-C-C-C-C-C-O-C-C-C-C-C-C-C-H}$, totaling 17 atoms and 16 bonds. Thus, option $\mathbf{B}$ is correct.

<Checkpoint>[Path spans from *p*-hydrogen of one benzene ring to *p*-hydrogen of the other, with 16 bonds](1)</Checkpoint>

If the alkoxide attacks the sulfur atom of the sulfur-containing intermediate, a pentacoordinated intermediate forms, expelling a diethylamino anion to yield a byproduct with two oxygens bonded to sulfur:

<MultiModal>![c1ccccc1CCCOS(F)(F)OCCCc2ccccc2](/media/uploads/25981701-92d1-4c24-a69e-569275b61826.png)</MultiModal>

Its relative molecular mass is $340.426$, so option $\mathbf{E}$ is correct.

<Checkpoint>[Product relative molecular mass is $340.426$](1)</Checkpoint>

In this molecule, the sulfur atom is tetracoordinated, and the molecule contains 45 atoms, making option $\mathbf{C}$ correct.

<Checkpoint>[Contains 45 atoms](1)</Checkpoint>

Finally, option $\mathbf{H}$ remains. If the solvent is acetonitrile, it may participate in the reaction. Acetonitrile is an electrophile, with its electrophilic site at the carbon of the cyano group, and can be attacked by the alkoxide to yield an intermediate:

<Checkpoint>[Cyanocarbon of acetonitrile attacked by alkoxide](1)</Checkpoint>

<MultiModal>![CC(=[N-])OCCCc1ccccc1](/media/uploads/e57113bc-4640-4882-8ca2-ae365d82a8fa.png)</MultiModal>

This intermediate then undergoes a $\left[1,3\right]-\sigma$ migration rearrangement. The carbon with partial positive charge migrates from the oxygen end to the nitrogen end via a suprafacial-antara $\left[1,3\right]-\sigma$ migration. The corresponding molecular orbitals are:

<MultiModal>![The figure shows three dumbbell shapes: two vertical dumbbells on the left and right sides, and one horizontal dumbbell above them. The left vertical dumbbell has a white upper lobe and a gradient gray lower lobe; the right vertical dumbbell has a gradient gray upper lobe and a white lower lobe. The horizontal dumbbell has a white left lobe and a gradient gray right lobe. From a central black point, two single lines and a pair of parallel double lines extend: the double lines point left-down to the middle of the left dumbbell; one single line points right-down to the middle of the right dumbbell; the other single line points right-up toward the gap between the right dumbbell and the upper horizontal dumbbell. A three-segment zigzag line extends upward from the horizontal dumbbell, connecting to a vertex of a hexagon. Inside the hexagon, lines parallel to its top, bottom-left, and bottom-right edges are present.](/media/uploads/c0df38c8-0128-4a49-b6dd-81156abb6aa5.png)</MultiModal>

Subsequent protonation yields a stable amide product.

<Checkpoint>[$\left[1,3\right]-\sigma$ migration rearrangement occurs](1)</Checkpoint>

<MultiModal>![CC(=O)NCCCc1ccccc1](/media/uploads/9016425e-c609-48bd-a22e-f6e630e0afea.png)</MultiModal>

This product contains nitrogen, oxygen, and a benzene ring but lacks fluorine and sulfur, so option $\mathbf{H}$ is correct.

In summary, the only incorrect option is $\mathbf{G}$.
在该反应体系中，三乙胺作为碱，将醇羟基去质子化得到醇负离子中间体：

<MultiModal>![c1ccccc1CCC[O-]](/media/uploads/9d5c96f6-0d0a-4852-9349-937defe65bab.png)</MultiModal>

$\mathrm{\left[Et_2N=SF_2\right]\left[BF_4\right]}$的阳离子中，硫原子具有亲电性，其被底物的醇负离子进攻后，得到含有四配位硫，硫原子上即有氮也有氧配位的的电中性中间体：

<MultiModal>![CCN(CC)S(F)(F)OCCCc1ccccc1](/media/uploads/cfd80569-3af1-401e-882f-cbbffe5b9957.png)</MultiModal>

<Checkpoint>[生成中间体CCN(CC)S(F)(F)OCCCc1ccccc1](1)</Checkpoint>

该中间体紧接着发生分子内取代反应，可能的机理为：$\mathrm{O}$上的孤对电子进攻$\mathrm{S-F}$键的反键轨道，将$\mathrm{S-F}$键断开形成紧密离子对，接着氟离子进攻与氧相连的碳原子，氟原子取代醇负离子中的氧，得到一元氟取代产物

<MultiModal>![c1ccccc1CCCF](/media/uploads/1636b9e2-8869-48ce-9c54-597bf6bf829d.png)</MultiModal>

<Checkpoint>[一元氟取代产物c1ccccc1CCCF](1)</Checkpoint>

同时得到一个含有硫氧键的产物：

<MultiModal>![CCN(CC)S(=O)F](/media/uploads/d70ebdb5-6b26-44e3-87fd-2476591d8649.png)</MultiModal>

<Checkpoint>[离去的产物为CCN(CC)S(=O)F](1)</Checkpoint>

两者为该条件下的主产物。其中，一元氟取代物的相对分子质量为$19.00+12.01\times9+1.008\times11=138.178$，氟的质量分数为$\frac{19.00}{138.178}\times100\%=13.75\%$，故选项$\mathbf{A}$说法正确。

<Checkpoint>[氟代物的相对分子质量为$138.178$](1)</Checkpoint>

<Checkpoint>[氟的质量分数为$13.75\%$](1)</Checkpoint>

生成的另一个产物不含苯环，其相对分子质量为$139.2$，介于$130-140$之间，故选项$\mathbf{F}$说法正确。该分子中包含$\mathrm{C,H,O,N,F,S}$六种元素，选项$\mathbf{D}$说法正确。

<Checkpoint>[产物不含苯环，其相对分子质量为$139.2$](1)</Checkpoint>

<Checkpoint>[产物中存在$\mathrm{C,H,O,N,F,S}$六种元素](1)</Checkpoint>

注意到反应中生成的一元氟取代物和离去的含氧分子的相对分子质量分别为$138.178$和$139.2$，其差值为$1.022<10$，故选项$\mathbf{G}$说法不正确，成为候选选项。

中间体
<MultiModal>![CCN(CC)S(F)(F)OCCCc1ccccc1](/media/uploads/cfd80569-3af1-401e-882f-cbbffe5b9957.png)</MultiModal>
同样可被醇负离子亲核进攻，若醇负离子进攻与氧原子相连的碳，则离去一个可通过离去$F^{-}$达到稳定的负离子，剩下一分子的醚：

<MultiModal>![c1ccccc1CCCOCCCc2ccccc2](/media/uploads/0005e100-a574-4e5b-87cd-b28284302fea.png)</MultiModal>

<Checkpoint>[进一步亲核进攻生成醚](1)</Checkpoint>

其相对分子质量为$254.356$。

<Checkpoint>[醚的相对分子质量为$254.356$](1)</Checkpoint>

在该分子中，沿化学键的最短路径最长的情形为从一个苯环*p*-位的氢原子到另一个苯环*p*-位的氢原子，即$\mathrm{H-C-C-C-C-C-C-C-O-C-C-C-C-C-C-C-H}$共计17个原子和16根化学键，选项$\mathbf{B}$说法正确。

<Checkpoint>[路径为从一个苯环的*p*位氢到另一个苯环的*p*位氢，有16根化学键](1)</Checkpoint>

若醇负离子进攻含硫中间体的硫原子，则经过五配位中间体后，离去二乙基胺基负离子，生成在硫上连接两个氧的副产物：

<MultiModal>![c1ccccc1CCCOS(F)(F)OCCCc2ccccc2](/media/uploads/25981701-92d1-4c24-a69e-569275b61826.png)</MultiModal>

其相对分子质量为$340.426$，故选项$\mathbf{E}$说法正确。

<Checkpoint>[产物相对分子质量为$340.426$](1)</Checkpoint>

分子中，硫原子为4配位，共有45个原子，选项$\mathbf{C}$说法正确。

<Checkpoint>[含有45个原子](1)</Checkpoint>

最后还剩下$\mathbf{H}$选项，若溶剂为乙腈，则乙腈可能参与反应。乙腈是亲电试剂，其亲电位点在氰基的碳原子上，故可被醇负离子进攻，得到中间体

<Checkpoint>[乙腈的氰基碳被醇负离子进攻](1)</Checkpoint>

<MultiModal>![CC(=[N-])OCCCc1ccccc1](/media/uploads/e57113bc-4640-4882-8ca2-ae365d82a8fa.png)</MultiModal>

该中间体接着发生$\left[1,3\right]-\sigma$迁移重排，带有部分正电性的碳原子从氧端迁移至氮端，发生同面-异面$\left[1,3\right]-\sigma$迁移，相应的分子轨道为：

<MultiModal>![图中有三个哑铃形，其中两个竖向的哑铃形排列于左右两侧，左侧的哑铃形上端为白色，下端为渐变的灰色，右侧的哑铃形上端为渐变的灰色，下端为白色。一个横向的哑铃形位于两个竖向哑铃形中间上方的位置，其左端为白色，有右端为渐变的灰色。在三个哑铃形中间有两条单线段和一对平行双线段，三者从中间的一个黑点出发，平行双线段指向左侧偏下，连接到左侧竖向哑铃形的中部，一条单线段指向右侧偏下，连接到右侧竖向哑铃形的中部，另一条单线段指向右侧偏上，指向右侧竖向哑铃形与上方横向哑铃形之间的空隙处。从上方的哑铃形出发，有一条折为三段的折线段按左上-右上-左上的顺序大致向上延伸，折线段的上端连接到一个正六边形的顶点。正六边形的上下两边水平，在正六边形内部，紧邻其上边、左下边和右下边的位置存在与相应边平行且长度相等的线段](/media/uploads/c0df38c8-0128-4a49-b6dd-81156abb6aa5.png)</MultiModal>

之后质子化得到稳定的酰胺产物。

<Checkpoint>[发生$\left[1,3\right]-\sigma$迁移重排](1)</Checkpoint>

<MultiModal>![CC(=O)NCCCc1ccccc1](/media/uploads/9016425e-c609-48bd-a22e-f6e630e0afea.png)</MultiModal>

其含有氮、氧、苯环，不含氟、硫，故$\mathbf{H}$选项正确。

综上，唯一可选的选项为$\mathbf{G}$。