Iminoboranes $( \mathrm { R B } { \equiv } \mathrm { N R } ^ { \prime } )$ are isoelectronic with alkynes and also readily undergo addition, polymerization, and other reactions. However, the methods for synthesizing iminoboranes are very limited, which to some extent restricts their applications. In 2021, research teams from Southern University of Science and Technology and Nankai University jointly developed an efficient iminoborane transfer reagent $\mathrm{X}$.

Heating $4.84 ~ \mathrm { g } ~ \mathrm { D m p B B r } _ { 2 }$ and $0.25 ~ \mathrm { g } ~ \mathrm { L i N H _ { 2 } }$ to $100 ^ { \circ } \mathrm { C }$, followed by solvent evaporation, yields $\mathrm{A}$. A peak in its mass spectrum (corresponding to $[ \pmb { \mathrm { A } } ] ^ { + } ) \ m / z = 420.1$. Reacting $0.420 \textrm { g } \mathrm{A}$ with $0.335 ~ \mathrm { g } ~ \mathrm { L i H M D S }$ in benzene at room temperature for half an hour allows the isolation of $0.236 \ \mathrm { g }$ of white powder $\mathrm{X}$, with a yield of approximately $45 \%$. The theoretical mass fraction of Br is $30.80 \%$.  

[Dmp: 2,6-bis(2,4,6-trimethylphenyl)phenyl; LiHMDS: lithium bis(trimethylsilyl)amide]  

$\mathrm{X}$ can exhibit the properties of iminoboranes and be used to synthesize heterocycles. Reacting $0.048 ~ g$ of pivaloyl chloride with $0.210 ~ g ~ \mathrm{X}$ yields $0.092 ~ g$ of an eight-membered ring compound $\mathrm{B}$; when $\mathrm{X}$ reacts with an excess of trimethylsilyl azide, $\mathrm{C}$ is obtained.  

Which of the following statements are correct?  

1. $\mathrm{A}$ contains 2 $\mathrm{Br}$ atoms;  
2. The sum of atoms in the chemical formula of $\mathrm{X}$ is 56;  
3. The yield of $\mathrm{B}$ is $55 \%$;  
4. $\mathrm{C}$ contains a six-membered ring;  
5. $\mathrm{C}$ has a $C_2$ axis.
亚氨基硼烷 $( \mathrm { R B } { \equiv } \mathrm { N R } ^ { \prime } )$ 和炔是等电子体，它们也很容易发生加成、聚合等反应。然而合成亚氨基硼烷的方法非常有限，这在一定程度上限制了它的应用。2021年，南方科技大学和南开大学的研究小组共同发展了一种高效的转亚氨基硼烷试剂$\mathrm{X}$。

将 $4 . 8 4 ~ \mathrm { g } ~ \mathrm { D m p B B r } _ { 2 }$ 和 $0 . 2 5 ~ \mathrm { g } ~ \mathrm { L i N H _ { 2 } }$ 加热至 $1 0 0 ^ { \circ } \mathrm { C }$ ，蒸发溶剂后得到$\mathrm{A}$，其质谱中的一个峰（对应 $[ \pmb { \mathrm { A } } ] ^ { + } ) \ m / z = 4 2 0 . 1$ 。取 $0 . 4 2 0 \textrm { g } \mathrm{A}$ 和 $0 . 3 3 5 ~ \mathrm { g } ~ \mathrm { L i H M D S }$ 在苯中室温反应半小时，可以分离出 $0 . 2 3 6 \ \mathrm { g }$ 白色粉末$\mathrm{X}$，产率约为 $45 \%$ ，Br的理论质量分数为 $3 0 . 8 0 \%$ 。

[Dmp：2,6-双（2,4,6-三甲基苯基）苯基；LiHMDS：双（三甲基硅基）氨基锂]

$\mathrm{X}$可以体现亚氨基硼烷的性质，用来合成杂环。取 $0 . 0 4 8 ~ g$ 特戊酰氯与 $0 . 2 1 0 ~ g ~ \mathrm{X}$反应，可以分离得到 $0 . 0 9 2 ~ g$ 含八元环的$\mathrm{B}$；$\mathrm{X}$与过量的叠氮三甲基硅基反应，则得到$\mathrm{C}$。

以下说法正确的有：

1. $\mathrm{A}$中含有2个$Br$；
2. $\mathrm{X}$的化学式中原子数之和为56；
3. $\mathrm{B}$的产率为$55 \%$；
4. $\mathrm{C}$含有六元环；
5. $\mathrm{C}$有$C_2$轴。
1,2,3
1,2,4
1,2,5
2,3,4
2,3,5
1,3,5
1,3,4
3,4,5
1,4,5
2,4,5
1,2,3,4
2,3,4,5
1,3,4,5
1,2,4,5
1,2,3,5
None of the above options are correct
1,2,3
1,2,4
1,2,5
2,3,4
2,3,5
1,3,5
1,3,4
3,4,5
1,4,5
2,4,5
1,2,3,4
2,3,4,5
1,3,4,5
1,2,4,5
1,2,3,5
以上选项均不正确
The amount of substance of $DmpBBr_2$ is: $4.48g/484.1 g \cdot mol^{-1}=1.00 \times 10^{-2}mol$  

The amount of substance of $LiNH_2$ is: $0.25g/23.0 g \cdot mol^{-1}=1.09 \times 10^{-2}mol$  

Thus, $DmpBBr_2$ and $LiNH_2$ react in a 1:1 ratio. Combined with mass spectrometry data, it can be inferred that $\mathbf{A}$ is $DmpB(Br)NH_2$. Since $\mathbf{A}$ contains only one $Br$, statement 1 is incorrect.  

<Checkpoint>[$\mathbf{A}=DmpB(Br)NH_2$, 1 incorrect](1)</Checkpoint>  

The amount of substance of $LiHMDS$ is: $0.335g/167.3 g \cdot mol^{-1}=2.00\times 10^{-3}mol$  

The amount of substance of $\mathbf{A}$ is: $0.420g/420.2 g \cdot mol^{-1}=1.00\times 10^{-3}mol$  

Thus, $\mathbf{A}$ and $LiHMDS$ react in a 1:2 ratio.  

If fully converted, the mass of $\mathbf{X}$ would be $0.236g/45\%=0.524g$  

Assuming the amount of substance of $\mathbf{X}$ is $1.00\times 10^{-3}mol$, its molar mass would be $524 g \cdot mol^{-1}$.  

From the theoretical mass fraction of $Br$, it can be deduced that $\mathbf{X}$ contains $524 \times 0.308/79.9=2$ $Br$ atoms. Substituting back with 2 $Br$ atoms, the precise molar mass of $\mathbf{X}$ is $518.8 g \cdot mol^{-1}$. After subtracting the mass of 2 $Br$ atoms and one $DmpBN$ structure, the remaining mass is 20.7, corresponding to 3 $Li$ atoms. Therefore, the chemical formula of $\mathbf{X}$ is $DmpBNLi_3Br_2$, i.e., $C_{24}H_{25}Li_3BNBr_2$. $\mathbf{X}$ has 56 atoms, so statement 2 is correct.  

<Checkpoint>[$\mathbf{X}=C_{24}H_{25}Li_3BNBr_2$, 2 correct](2)</Checkpoint>  

Considering the reactivity of $\mathbf{X}$ as an iminoborane, and given that $\mathbf{B}$ contains an eight-membered ring, it can be concluded that $\mathbf{B}$ is the dimer of the addition product of $\mathbf{X}$ with pivaloyl chloride. The chemical formula of $\mathbf{B}$ is $(DmpBNCO^tBu)_2$, with the following structure:  

<MultiModal>![[Dmp]B1O/C(C(C)(C)C)=N\B([Dmp])O/C(C(C)(C)C)=N\1  
](/media/uploads/5e3a3660-4114-43cc-84d7-10873c6bd3a9.png)</MultiModal>  

<Checkpoint>[$\mathbf{B}=(DmpBNCO^tBu)_2$](1)</Checkpoint>  

The amount of substance of pivaloyl chloride is: $0.048g/120.6 g \cdot mol^{-1}=3.98 \times 10^{-4}mol$  

The amount of substance of $\mathbf{X}$ is: $0.210g/1037.8 g \cdot mol^{-1}=2.02 \times 10^{-4}mol$  

The amount of substance of $\mathbf{B}$ obtained from the reaction is: $0.092g/846.8 g \cdot mol^{-1}=1.09 \times 10^{-4}mol$  

The yield is $2 \times 1.09 \times 10^{-4}/3.98 \times 10^{-4}=55\%$, so statement 3 is correct.  

<Checkpoint>[Yield of $\mathbf{B}$ is $55\%$, 3 correct](1)</Checkpoint>  

The structure of $\mathbf{C}$, obtained from the reaction of $\mathbf{X}$ with excess trimethylsilyl azide, is as follows:  

<MultiModal>![[Dmp]B1N([Si](C)(C)C)N=NN1[Si](C)(C)C  
](/media/uploads/ebbd3ba1-7e58-45a4-a797-670d17472731.png)</MultiModal>  

The five-membered ring structure of $\mathbf{C}$ is symmetric and possesses a $C_2$ axis, so statement 4 is incorrect, and statement 5 is correct.  

<Checkpoint>[$\mathbf{C}$ has a symmetric five-membered ring structure, 4 incorrect, 5 correct](2)</Checkpoint>
$DmpBBr_2$的物质的量为：$4.84g/484.1 g \cdot mol^{-1}=1.00 \times 10^{-2}mol$

$LiNH_2$的物质的量为：$0.25g/23.0 g \cdot mol^{-1}=1.09 \times 10^{-2}mol$

因此$DmpBBr_2$和$LiNH_2$按照1:1反应，结合质谱数据，可以推测$\mathbf{A}$为$DmpB(Br)NH_2$。$\mathbf{A}$只含有一个$Br$，说法1错误。

<Checkpoint>[$\mathbf{A}=DmpB(Br)NH_2$，1错误](1)</Checkpoint>

$LiHMDS$的物质的量为：$0.335g/167.3 g \cdot mol^{-1}=2.00\times 10^{-3}mol$

$\mathbf{A}$的物质的量为：$0.420g/420.2 g \cdot mol^{-1}=1.00\times 10^{-3}mol$

因此$\mathbf{A}$和$LiHMDS$按照1:2反应。

若完全转化，则$\mathbf{X}$的质量为$0.236g/45\%=0.524g$

假设$\mathbf{X}$的物质的量为$1.00\times 10^{-3}mol$，那么其摩尔质量为$524 g \cdot mol^{-1}$。

由$Br$的理论质量分数，可以推断出$\mathbf{X}$含有$524 \times 0.308/79.9=2$个$Br$，反代入2个$Br$，可得$\mathbf{X}$的精确摩尔质量为$518.8 g \cdot mol^{-1}$。$\mathbf{X}$的摩尔质量除去2个$Br$、一个$DmpBN$的结构，还剩20.7，对应3个$Li$，因此$\mathbf{X}$的化学式为$DmpBNLi_3Br_2$，即$C_{24}H_{25}Li_3BNBr_2$。$\mathbf{X}$有56个原子，因此说法2正确。

<Checkpoint>[$\mathbf{X}=C_{24}H_{25}Li_3BNBr_2$，2正确](2)</Checkpoint>

考虑$\mathbf{X}$作为亚氨基硼烷的反应性，结合$\mathbf{B}$含有八元环，可知$\mathbf{B}$是$\mathbf{X}$与特戊酰氯加成产物的二聚体，$\mathbf{B}$的化学式为$(DmpBNCO^tBu)_2$，结构如下：

<MultiModal>![[Dmp]B1O/C(C(C)(C)C)=N\B([Dmp])O/C(C(C)(C)C)=N\1
](/media/uploads/5e3a3660-4114-43cc-84d7-10873c6bd3a9.png)</MultiModal>

<Checkpoint>[$\mathbf{B}=(DmpBNCO^tBu)_2$](1)</Checkpoint>

特戊酰氯的物质的量为：$0.048g/120.6 g \cdot mol^{-1}=3.98 \times 10^{-4}mol$

$\mathbf{X}$的物质的量为：$0.210g/1037.8 g \cdot mol^{-1}=2.02 \times 10^{-4}mol$

反应得到的$\mathbf{B}$的物质的量为：$0.092g/846.8 g \cdot mol^{-1}=1.09 \times 10^{-4}mol$

产率为$2 \times 1.09 \times 10^{-4}/3.98 \times 10^{-4}=55\%$，说法3正确

<Checkpoint>[$\mathbf{B}$的产率为$55\%$，3正确](1)</Checkpoint>

$\mathbf{X}$与过量叠氮三甲基硅反应得到的$\mathbf{C}$结构如下：

<MultiModal>![[Dmp]B1N([Si](C)(C)C)N=NN1[Si](C)(C)C
](/media/uploads/ebbd3ba1-7e58-45a4-a797-670d17472731.png)</MultiModal>

$\mathbf{C}$的五元环结构对称，具有$C_2$轴，因此说法4错误，说法5正确。

<Checkpoint>[$\mathbf{C}$有对称的五元环结构，4错误，5正确](2)</Checkpoint>