Hexafluoroacetone can undergo significant hydration. In a concentrated solution of $\mathrm{NaOH}$, it reacts with p-methoxybenzaldehyde at 100 ℃ to obtain alcohol $\bf{O}$ and salt $\bf{P}$.

Similarly, benzoyl chloride reacts with trifluoromethyl iodide and $\mathrm{TDAE}$ in a theoretical 1:1:1 ratio to obtain ester $\bf{S}$. It is believed that a benzene ring-containing intermediate $\bf{T}$ is first generated, and $\bf{T}$ then reacts with benzoyl chloride to obtain $\bf{S}$. At the same time, salt $\bf{U}$ is also produced. It is known that $\mathrm{TDAE}$ is tetra(dimethylamino)ethylene.

<MultiModal>![C1C=CC=C(C(Cl)=O)C=1.C(F)(I)(F)F.C(N(C)C)(N(C)C)=C(N(C)C)N(C)C>>[S]](/media/uploads/552af2e3-d2aa-4ba8-897d-dab8790af55c.png)</MultiModal>

Which of the following statements are correct:
1. The chemical formula of $\bf{O}$ is $\mathrm{C_3F_6OH_2}$
2. $\bf{P}$ does not contain hydrogen element
3. In the reaction, TDAE is oxidized to the cation of $\bf{U}$
4. The mass fraction of oxygen in $\bf{S}$ is 9.19%
5. $\bf{S}$ has 3 kinds of chemically different hydrogens
六氟丙酮可以发生明显的水化。在$\mathrm{NaOH}$的浓溶液中，其与对甲氧基苯甲醛于100℃下反应得到醇$\bf{O}$以及盐$\bf{P}$。

类似地，苯甲酰氯与三氟甲基碘以及$\mathrm{TDAE}$理论上1：1：1反应得到酯$\bf{S}$。该过程可认为先生成了含苯环的中间体$\bf{T}$，$\bf{T}$再与苯甲酰氯反应得到$\bf{S}$。同时反应还产生了盐$\bf{U}$。已知$\mathrm{TDAE}$是四（二甲氨基）乙烯。

<MultiModal>![C1C=CC=C(C(Cl)=O)C=1.C(F)(I)(F)F.C(N(C)C)(N(C)C)=C(N(C)C)N(C)C>>[S]](/media/uploads/552af2e3-d2aa-4ba8-897d-dab8790af55c.png)</MultiModal>

下面的说法中正确的有：
1. $\bf{O}$的化学式为$\mathrm{C_3F_6OH_2}$
2. $\bf{P}$中不包含氢元素
3. 在反应中TDAE被氧化为$\bf{U}$的阳离子
4. $\bf{S}$中含氧的质量分数为9.19%
5. $\bf{S}$有3种化学环境不同的氢
All other options are incorrect
2.3.4.
1.2.4.
1.2.3.4.5
2.3.4.5.
1.3.5.
2.3.
其他选项均不正确
2.3.4.
1.2.4.
1.2.3.4.5
2.3.4.5.
1.3.5.
2.3.
The problem states that hexafluoroacetone can undergo significant hydration, and in a concentrated solution of $\mathrm{NaOH}$, it can undergo trifluoromethyl anion transfer reaction to obtain the compound $\bf O$ and sodium trifluoroacetate (salt $\bf P$) as shown in the figure.

<MultiModal>![OC(C(F)(F)F)C1=CC=C(OC)C=C1](/media/uploads/893aad09-3a0a-4270-95a6-397a03036252.png)</MultiModal>

<Checkpoint>[$\bf{O}$ is $\mathrm{OC(C(F)(F)F)C1=CC=C(OC)C=C1}$](1)</Checkpoint>

<Checkpoint>[$\bf{P}$ is $\mathrm{CF_3COONa}$](1)</Checkpoint>

Therefore, 1 is incorrect, and 2 is correct.

In the second reaction, $\mathrm{TDAE}$ can reduce $\mathrm{CF_3I}$ to obtain $\mathrm{CF_3^-}$ equivalent to attack acyl chloride twice to obtain the tertiary alcohol intermediate $\bf T$: OC(C(F)(F)F)(C(F)(F)F)C1=CC=CC=C1. At the same time, the cation $\mathrm{TADF^{2+}}$ of $\bf{U}$ is obtained. Afterwards, the tertiary alcohol undergoes acylation again to obtain the product $\bf S$: O=C(C1=CC=CC=C1)OC(C(F)(F)F)(C(F)(F)F)C2=CC=CC=C2.

The structure of $\bf S$ is shown in the figure:
<MultiModal>![O=C(C1=CC=CC=C1)OC(C(F)(F)F)(C(F)(F)F)C2=CC=CC=C2](/media/uploads/56ff248c-9616-4b17-af16-604602731ad5.png)</MultiModal>

<Checkpoint>[$\bf{T}$ is OC(C(F)(F)F)(C(F)(F)F)C1=CC=CC=C1](1)</Checkpoint>

<Checkpoint>[$\bf{S}$ is O=C(C1=CC=CC=C1)OC(C(F)(F)F)(C(F)(F)F)C2=CC=CC=C2](1)</Checkpoint>

Therefore, 3 and 4 are correct (based on the correct mass fraction of oxygen in the chemical formula), 5 should be 6 types, which is incorrect. Option B is correct.
题目中说六氟丙酮可以发生明显的水化，在$\mathrm{NaOH}$的浓溶液中可以发生三氟甲基负离子转移反应，得到如图所示的化合物$\bf O$和三氟乙酸钠(盐$\bf P$)。

<MultiModal>![OC(C(F)(F)F)C1=CC=C(OC)C=C1](/media/uploads/893aad09-3a0a-4270-95a6-397a03036252.png)</MultiModal>

<Checkpoint>[$\bf{O}$为$\mathrm{OC(C(F)(F)F)C1=CC=C(OC)C=C1}$](1)</Checkpoint>

<Checkpoint>[$\bf{P}$为$\mathrm{CF_3COONa}$](1)</Checkpoint>

因此1错误，2正确。

第二个反应中，$\mathrm{TDAE}$可以还原$\mathrm{CF_3I}$得到$\mathrm{CF_3^-}$等价物进攻两次酰氯，得到三级醇中间体$\bf T$: OC(C(F)(F)F)(C(F)(F)F)C1=CC=CC=C1。与此同时得到$\bf{U}$的阳离子$\mathrm{TADF^{2+}}$。此后三级醇再次发生酰基化得到产物$\bf S$: O=C(C1=CC=CC=C1)OC(C(F)(F)F)(C(F)(F)F)C2=CC=CC=C2。

$\bf S$的结构如图所示：
<MultiModal>![O=C(C1=CC=CC=C1)OC(C(F)(F)F)(C(F)(F)F)C2=CC=CC=C2](/media/uploads/56ff248c-9616-4b17-af16-604602731ad5.png)</MultiModal>

<Checkpoint>[$\bf{T}$是OC(C(F)(F)F)(C(F)(F)F)C1=CC=CC=C1](1)</Checkpoint>

<Checkpoint>[$\bf{S}$是O=C(C1=CC=CC=C1)OC(C(F)(F)F)(C(F)(F)F)C2=CC=CC=C2](1)</Checkpoint>

因此3.4.正确（根据化学式含氧质量分数正确），5应为6种，错误. 选项B正确