Compound $\mathbf{A}$ can be prepared by reacting 2,4-disulfophenol with concentrated nitric acid. In 1873, Hermann Sprengel discovered that $\mathbf{A}$ could be used as a primary explosive, and it was widely employed in subsequent conflicts including the First Sino-Japanese War, the Russo-Japanese War, and World War I. In 1848, Scottish chemist John Stenhouse mixed calcium hypochlorite with a small amount of $\mathbf{A}$ to produce $\mathbf{B}$, which was used by the German military as a tear gas agent during World War I and is now utilized as an insecticide. $\mathbf{B}$ contains no $H$ atoms or $C-C$ bonds, with $\omega_{C}=7.31\%$. Reducing $\mathbf{B}$ with $Sn$ in hydrochloric acid yields another blister agent $\mathbf{C}$, which was prepared but never deployed by the German military; its relative molecular mass is less than 150. $\mathbf{C}$ can also be obtained by reacting the active component of detonators, $\mathbf{D}$, with chlorine gas in hydrochloric acid. All atoms in $\mathbf{C}$ are coplanar, and no charge separation exists.

Electrolyzing a methanol-sulfuric acid solution of $\mathbf{B}$ yields compound $\mathbf{E}$ at the cathode. $\mathbf{E}$ can also be synthesized by reacting nitrosyl chloride with sodium trichloromethanesulfinate. When heated near its melting point, $\mathbf{E}$ decomposes into equimolar amounts of $\mathbf{B}$, $\mathbf{F}$, and nitrosyl chloride. Heating at $120^{\circ} \mathrm{C}$ produces a common solvent $\mathbf{G}$, compound $\mathbf{H}$, and gas $\mathbf{I}$ in a molar ratio of $1 : 1 : 2$. $\mathbf{F}$ and $\mathbf{H}$ have similar structures, with a molecular weight difference of 16.

Deduce the chemical formulas of $\mathbf{A}$ to $\mathbf{I}$. Among the following statements, the incorrect ones are:
1. In the reaction equation for the formation of $\mathbf{B}$ from $\mathbf{A}$ and calcium hypochlorite, there are two products containing $Ca$.
2. The conversion of 1 equivalent of $\mathbf{B}$ to $\mathbf{C}$ requires the consumption of 2 equivalents of $Sn$.
3. 1 equivalent of $\mathbf{D}$ produces 2 equivalents of $\mathbf{C}$.
4. The formation of 1 equivalent of $\mathbf{E}$ from $\mathbf{B}$ consumes 2 equivalents of electrons.
5. No charge separation exists in $\mathbf{H}$.
6. The structure of $\mathbf{F}$ can be regarded as $\mathbf{C}$ with one substituent replaced.
化合物$\mathbf{A}$可由2,4-二磺酸基苯酚与浓硝酸反应制得。1873年，Hermann Sprengel发现$\mathbf{A}$可作起爆药，此后的甲午战争，日俄战争及第一次世界大战中，$\mathbf{A}$都被广泛地使用。1848年，苏格兰化学家John Stermhouse混合次氯酸钙与少量$\mathbf{A}$制得了$\mathbf{B}$，$\mathbf{B}$在一战中被德军用作催泪剂，现在其被用作杀虫剂。$\mathbf{B}$不含$H$原子或$C-C$键，其中 $\omega_{C}=7.31\%$。用$Sn$在盐酸中还原$\mathbf{B}$，得到另一种曾被德军制备但未应用的糜烂性毒剂$\mathbf{C}$，相对分子质量小于150；$\mathbf{C}$亦可由雷管的有效物质$\mathbf{D}$与氯气在盐酸中反应得到；$\mathbf{C}$中所有原子共面，不存在电荷分离。

电解$\mathbf{B}$的甲醇-硫酸溶液，在阴极得到化合物$\mathbf{E}$；$\mathbf{E}$也可用亚硝酰氯与三氯甲基亚磺酸钠反应制备。在$\mathbf{E}$的熔点附近加热，$\mathbf{E}$分解为等摩尔的$\mathbf{B}$、$\mathbf{F}$和亚硝酰氯；而在$120^{\circ} \mathrm{C}$加热时，则生成摩尔比 $1 : 1 : 2$ 的常见溶剂$\mathbf{G}$、化合物$\mathbf{H}$和气体$\mathbf{I}$。$\mathbf{F}$与$\mathbf{H}$的结构相似，分子量相差16。

推断$\mathbf{A}$至$\mathbf{I}$的化学式，以下说法中错误的有：
1. 由$\mathbf{A}$和次氯酸钙生成$\mathbf{B}$的反应方程式中，含$Ca$的产物有2种
2. 1当量$\mathbf{B}$生成$\mathbf{C}$需要消耗2当量$Sn$
3. 1当量$\mathbf{D}$生成2当量$\mathbf{C}$
4. 1当量$\mathbf{B}$生成$\mathbf{E}$消耗2当量电子
5. $\mathbf{H}$中不存在电荷分离
6. $\mathbf{F}$的结构可以视为$\mathbf{C}$中一个取代基被替换后的结构
1
2
3
4
5
6
2,6
4,5
1,2
1,5
3,5
1,5,6
2,4,5
2,3,6
None of the above options are correct
1
2
3
4
5
6
2,6
4,5
1,2
1,5
3,5
1,5,6
2,4,5
2,3,6
以上选项均不正确
Clearly, $\mathbf{A}$ is trinitrophenol ($C_6H_2(OH)(NO_3)_3$). Since $\mathbf{B}$ contains no $H$ atoms or $C-C$ bonds, $\mathbf{B}$ likely contains only one $C$ atom. The relative molecular mass of $\mathbf{B}$ is $12.01/0.0731=164.3$. After subtracting the mass of $C$, the remaining mass corresponds to 3 $Cl$ atoms and 1 $NO_2$ group, so $\mathbf{B}=CCl_3NO_2$.

<Checkpoint>[$\mathbf{A}$ is trinitrophenol ($C_6H_2(OH)(NO_3)_3$)](1)</Checkpoint>

<Checkpoint>[$\mathbf{B}=CCl_3NO_2$](1)</Checkpoint>

The balanced equation for the reaction forming $\mathbf{B}$ from $\mathbf{A}$ is as follows:

$2C_6H_2(OH)(NO_2)_3+11Ca(OCl)_2=6CCl_3NO_2+6CaCO_3+3Ca(OH)_2+2CaCl_2$

<Checkpoint>[$2C_6H_2(OH)(NO_2)_3+11Ca(OCl)_2=6CCl_3NO_2+6CaCO_3+3Ca(OH)_2+2CaCl_2$](1)</Checkpoint>

There are 3 $Ca$-containing products, so statement 1 is incorrect.

$\mathbf{C}$ can be obtained by reducing $\mathbf{B}$ with $Sn$ in hydrochloric acid. The fully reduced product of $\mathbf{B}$ might be $CCl_3NH_2$, but this is not a planar compound. Therefore, $\mathbf{C}$ is likely a partially reduced product of $\mathbf{B}$. Since all atoms in $\mathbf{C}$ are coplanar, a $C=N$ double bond is probable, suggesting $\mathbf{C}$ could be $Cl_2CN(OH)$ or $Cl_2CNH$, which is consistent with the given information.

<Checkpoint>[$\mathbf{C} = Cl_2CN(OH)$ or $Cl_2CNH$](1)</Checkpoint>

The active substance in detonators, $\mathbf{D}$, could be hydrazoic acid or mercury fulminate. Based on the structure of $\mathbf{C}$, the latter is more plausible, i.e., $\mathbf{D}=Hg(CNO)_2$.

<Checkpoint>[$\mathbf{D}=Hg(CNO)_2$](1)</Checkpoint>

When $\mathbf{D}$ reacts with $Cl_2$ in hydrochloric acid to form $\mathbf{C}$, if $\mathbf{C}$ were $Cl_2CNH$, it would be difficult to account for the oxygen atoms from mercury fulminate. Therefore, $\mathbf{C}$ is $\mathbf{C} = Cl_2CN(OH)$.

<Checkpoint>[Excluding $Cl_2CNH$, $\mathbf{C}$ is $\mathbf{C} = Cl_2CN(OH)$.](1)</Checkpoint>

The equation for the reduction of $\mathbf{B}$ to $\mathbf{C}$ with $Sn$ in hydrochloric acid is as follows:

$CCl_3NO_2+2Sn+3HCl=Cl_2CN(OH)+2SnCl_2+H_2O$

<Checkpoint>[$CCl_3NO_2+2Sn+3HCl=Cl_2CN(OH)+2SnCl_2+H_2O$](1)</Checkpoint>

1 equivalent of $\mathbf{B}$ requires 2 equivalents of $Sn$ to form $\mathbf{C}$, so statement 2 is correct.

The equation for the reaction of $\mathbf{D}$ with $Cl_2$ in hydrochloric acid to form $\mathbf{C}$ is as follows:

$Hg(CNO)_2+2Cl_2+2HCl=2Cl_2CN(OH)+HgCl_2$

<Checkpoint>[$Hg(CNO)_2+2Cl_2+2HCl=2Cl_2CN(OH)+HgCl_2$](1)</Checkpoint>

1 equivalent of $\mathbf{D}$ produces 2 equivalents of $\mathbf{C}$, so statement 3 is correct.

Sodium trichloromethanesulfinate can serve as a trichloromethylation reagent. Since $\mathbf{E}$ can be prepared from nitrosyl chloride and sodium trichloromethanesulfinate, the chemical formula of $\mathbf{E}$ is $Cl_3CNO$.

<Checkpoint>[$\mathbf{E}=Cl_3CNO$](1)</Checkpoint>

Reducing 1 equivalent of $\mathbf{B}$ to $\mathbf{E}$ consumes 2 equivalents of electrons, so statement 4 is correct.

The equation for the thermal decomposition of $\mathbf{E}$ near its melting point is:

$nCl_3CNO=CCl_3NO_2+\mathbf{F}+NOCl$

From the number of $O$ atoms in the products, $n≥3$. When $n=3$, $\mathbf{F}=C_2Cl_5N$. Based on the structure of $\mathbf{E}$, the structure of $\mathbf{F}$ is $Cl_3CNCCl_2$.

<Checkpoint>[$\mathbf{F}=Cl_3CNCCl_2$](1)</Checkpoint>

The structure of $\mathbf{F}$ resembles that of $\mathbf{C}$, with the hydroxyl group replaced by a trichloromethyl group, so statement 6 is correct.

$\mathbf{H}$ has a relative molecular mass 16 units higher than $\mathbf{F}$, so $\mathbf{H}=Cl_3CN(O)CCl_2$.

<Checkpoint>[$\mathbf{H}=Cl_3CN(O)CCl_2$](0.5)</Checkpoint>

The reaction equation for $\mathbf{E}$ at $120℃$ is $Cl_3CNO=CCl_4+Cl_3CN(O)CCl_2+NO$, with a product ratio of $1:1:2$.

The $N-O$ bond in $\mathbf{H}$ exhibits charge separation, so statement 5 is incorrect.

<Checkpoint>[Charge separation exists in $\mathbf{H}$, so statement 5 is incorrect](0.5)</Checkpoint>

In summary, statements 1 and 5 are incorrect, so the answer is J.
显然$\mathbf{A}$是三硝基苯酚（$C_6H_2(OH)(NO_3)_3$）。根据$\mathbf{B}$不含$H$原子或$C-C$键，判断$\mathbf{B}$可能只含一个$C$，$\mathbf{B}$的相对分子质量为$12.01/0.0731=164.3$，除去$C$的质量后，对应3个$Cl$和1个$NO_2$，因此$\mathbf{B}=CCl_3NO_2$。

<Checkpoint>[$\mathbf{A}$是三硝基苯酚（$C_6H_2(OH)(NO_3)_3$）](1)</Checkpoint>

<Checkpoint>[$\mathbf{B}=CCl_3NO_2$](1)</Checkpoint>

配平由$\mathbf{A}$生成$\mathbf{B}$的方程式如下：

$2C_6H_2(OH)(NO_2)_3+11Ca(OCl)_2=6CCl_3NO_2+6CaCO_3+3Ca(OH)_2+2CaCl_2$

<Checkpoint>[$2C_6H_2(OH)(NO_2)_3+11Ca(OCl)_2=6CCl_3NO_2+6CaCO_3+3Ca(OH)_2+2CaCl_2$](1)</Checkpoint>

有3种含$Ca$产物，说法1错误。

$\mathbf{C}$可由$\mathbf{B}$在盐酸中被$Sn$还原得到，$\mathbf{B}$被完全还原的产物可能是$CCl_3NH_2$，但是不是平面化合物，因此$\mathbf{C}$可能是$\mathbf{B}$部分还原的产物。$\mathbf{C}$所有原子共平面，因此很可能形成了$C=N$双键，想到$\mathbf{C}$的结构可能为$Cl_2CN(OH)$或者为$Cl_2CNH$，与已知信息没有矛盾之处。

<Checkpoint>[$\mathbf{C} = Cl_2CN(OH)$或$Cl_2CNH$](1)</Checkpoint>

雷管的有效物质$\mathbf{D}$可能是叠氮化氢或者雷酸汞，结合$\mathbf{C}$的结构，可以判断后者更符合，即$\mathbf{D}=Hg(CNO)_2$

<Checkpoint>[$\mathbf{D}=Hg(CNO)_2$](1)</Checkpoint>

$\mathbf{D}$在盐酸中与$Cl_2$反应生成$\mathbf{C}$，若$\mathbf{C}$为$Cl_2CNH$，很难解释雷酸汞中的氧原子去向，因此$\mathbf{C}$为$\mathbf{C} = Cl_2CN(OH)$。

<Checkpoint>[排除$Cl_2CNH$，$\mathbf{C}$为$\mathbf{C} = Cl_2CN(OH)$。](1)</Checkpoint>


用$Sn$在盐酸中还原$\mathbf{B}$得到$\mathbf{C}$的方程式如下：

$CCl_3NO_2+2Sn+3HCl=Cl_2CN(OH)+2SnCl_2+H_2O$

<Checkpoint>[$CCl_3NO_2+2Sn+3HCl=Cl_2CN(OH)+2SnCl_2+H_2O$](1)</Checkpoint>

1当量$\mathbf{B}$生成$\mathbf{C}$需要消耗2当量$Sn$，说法2正确。

$\mathbf{D}$在盐酸中与$Cl_2$反应生成$\mathbf{C}$的方程式如下：

$Hg(CNO)_2+2Cl_2+2HCl=2Cl_2CN(OH)+HgCl_2$

<Checkpoint>[$Hg(CNO)_2+2Cl_2+2HCl=2Cl_2CN(OH)+HgCl_2$](1)</Checkpoint>

1当量$\mathbf{D}$生成2当量$\mathbf{C}$，说法3正确。

三氯甲基亚磺酸钠可以做三氯甲基化试剂，由$\mathbf{E}$可用亚硝酰氯与三氯甲基亚磺酸钠反应制备，得$\mathbf{E}$的化学式为$Cl_3CNO$

<Checkpoint>[$\mathbf{E}=Cl_3CNO$](1)</Checkpoint>

1当量$\mathbf{B}$还原为$\mathbf{E}$消耗2当量电子，说法4正确。

$\mathbf{E}$在熔点附近加热的反应方程式为：

$nCl_3CNO=CCl_3NO_2+\mathbf{F}+NOCl$

由产物中$O$的个数可知$n≥3$。当$n=3$时，$\mathbf{F}=C_2Cl_5N$，结合$\mathbf{E}$的结构，得$\mathbf{F}$的结构为$Cl_3CNCCl_2$。

<Checkpoint>[$\mathbf{F}=Cl_3CNCCl_2$](1)</Checkpoint>

$\mathbf{F}$的结构与$\mathbf{C}$的结构相似，可以视为$\mathbf{C}$的羟基替换为三氯甲基，因此说法6正确。

$\mathbf{H}$与$\mathbf{F}$相对分子质量相差16，则$\mathbf{H}=Cl_3CN(O)CCl_2$

<Checkpoint>[$\mathbf{H}=Cl_3CN(O)CCl_2$](0.5)</Checkpoint>

$\mathbf{E}$在$120℃$下反应方程式为$Cl_3CNO=CCl_4+Cl_3CN(O)CCl_2+NO$，符合产物比$1:1:2$

$\mathbf{H}$中的$N-O$键存在电荷分离，说法5错误。

<Checkpoint>[$\mathbf{H}$中存在电荷分离，说法5错](0.5)</Checkpoint>

综上，1和5错误，选J。