Polyhydroxyalkanoates (PHAs) are a class of degradable polyester materials that play an important role in achieving a sustainable circular plastics economy. One class of polymer can be obtained by the ring-opening polymerization reaction of monomer $\bf{A}$ with $^t\rm{Bu-P_4}$ and $\rm{BnOH}$ at $\rm{70~\degree C}$. $^t\rm{Bu-P_4 = 
 ^tBuNP[NP(NMe_2)_3]_3})$

<MultiModal>![As shown in the figure, a polyester compound with a repeating unit structure of $O=C([*])C([R])([R])C(C)O[*]$](/media/uploads/63461b76-c0e3-468e-8793-b1b16968af08.png)</MultiModal>

The following statements are made:

1. $\bf{A}$ contains a four-membered ring, and if $\rm{R = Me}$, then its chemical formula is $\rm{C_6H_{10}O_2}$

2. $^t\rm{Bu-P_4}$ reacts directly with $\bf{A}$ to initiate polymerization

3. $^t\rm{Bu-P_4}$ only acts as a base

4. If $\rm{R = Me}$ and considering the end-capping of the polymer and a number average molecular weight of $5244$, then the average degree of polymerization of the polymer is $45$

5. It is known that when $\rm{R = H}$, the polymer will rapidly decompose at $\rm{180~\degree C}$; then when $\rm{R = Me}$, the polymer will also rapidly decompose at $\rm{180~\degree C}$

6. When $\rm{R = H}$, the structure of the polymer pyrolysis product is OC(/C=C\C)=O

Then the sum of the corresponding numbers of all correct statements is:
聚羟基脂肪酸酯(PHA)是一类可降解的聚酯类材料，对实现可持续发展循环塑料经济有重要作用。其中一类聚合物可由单体$\bf{A}$与 $^t\rm{Bu-P_4}$ 、$\rm{BnOH}$在 $\rm{70~\degree C}$ 下发生开环聚合反应得到。$^t\rm{Bu-P_4 = 
 ^tBuNP[NP(NMe_2)_3]_3})$

<MultiModal>![如图所示为一聚酯类化合物，其重复单元的结构为$O=C([*])C([R])([R])C(C)O[*]$](/media/uploads/63461b76-c0e3-468e-8793-b1b16968af08.png)</MultiModal>

有以下说法：

1.$\bf{A}$中含有四元环，且若$\rm{R = Me}$，则其化学式为$\rm{C_6H_{10}O_2}$

2.$^t\rm{Bu-P_4}$直接与$\bf{A}$反应引发聚合

3.$^t\rm{Bu-P_4}$仅起到碱的作用

4.若$\rm{R = Me}$且考虑聚合物的封端且数均分子量为$5244$，则该聚合物的平均聚合度为$45$

5.已知$\rm{R = H}$时，聚合物在$\rm{180~\degree C}$下会迅速分解；则当$\rm{R = Me}$时，聚合物在$\rm{180~\degree C}$下也会迅速分解

6.当$\rm{R = H}$时，聚合物热解产物的结构为OC(/C=C\C)=O

则所有正确说法对应编号之和为：
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The repeating unit of the polymer is $O=C([*])C([R])([R])C(C)O[*]$, which is an ester structure. Ring-opening polymerization yields a polyester, which means that $\bf{A}$ is a cyclic lactone. It is not difficult to deduce from the repeating unit structure that the structure of $\bf{A}$ is CC1OC(C1([R])[R])=O. When $\rm{R = Me}$, the chemical formula of $\bf{A}$ is $\rm{C_6H_{10}O_2}$.

<Checkpoint>[The structure of $\bf{A}$ is CC1OC(C1([R])[R])=O, containing a four-membered ring. When $\rm{R = Me}$, the chemical formula of $\bf{A}$ is $\rm{C_6H_{10}O_2}$. Statement 1 is correct.](1)</Checkpoint>

$^t\rm{Bu-P_4}$ is a sterically hindered base with weak nucleophilic ability and cannot directly react with $\bf{A}$ to initiate polymerization. Instead, it initiates polymerization by abstracting the hydroxyl hydrogen of benzyl alcohol to generate a negative ion.

<Checkpoint>[$^t\rm{Bu-P_4}$ has large steric hindrance and cannot initiate polymerization nucleophilically. Statement 2 is incorrect.](1)</Checkpoint>

<Checkpoint>[$^t\rm{Bu-P_4}$ initiates polymerization by abstracting the hydroxyl hydrogen of benzyl alcohol, acting as a base. Statement 3 is correct.](1)</Checkpoint>

The benzyl alcohol anion initiates polymerization via nucleophilic attack, so one end of the polymer is capped with benzyl ester. Subsequent polymerization relies on the generated alcohol anion to continue nucleophilic attack until the polymerization stops after protonation. When $\rm{R = Me}$, the repeating unit of the polymer is $\rm{C_6H_{10}O_2}$ and the molecular weight is $114.14$. The capping at both ends can be considered to add additional benzyl alcohol, with a molecular weight of $108.14$. Then the average degree of polymerization $n$ is: $(5244 - 108.14) ÷ 114.14 = 45$

<Checkpoint>[The polymer is capped with benzyl ester and hydrogen. Calculation shows that the average degree of polymerization is $(5244 - 108.14) ÷ 114.14 = 45$. Statement 4 is correct.](1)</Checkpoint>

When $\rm{R = H}$, the polymer will undergo thermal cracking of the ester under heating conditions, producing olefins and carboxylic acids. When there is no hydrogen at the $\alpha$ position of the ester group, thermal cracking cannot occur. Therefore, when $\rm{R = Me}$, the polymer can remain stable.

<Checkpoint>[When there is no hydrogen at the $\alpha$ position of the ester group, the polymer cannot undergo thermal cracking. Therefore, when $\rm{R = Me}$, the polymer can remain stable at $\rm{180~\degree C}$. Statement 5 is incorrect.](1)</Checkpoint>

During thermal cracking, the hydrogen and the ester oxygen are in the para position. At this time, the conformation with the methyl group in the para position to the carbonyl group is the most stable. The methyl group and the carboxyl group are in the trans position in the resulting double bond, and the structure is OC(/C=C/C)=O

<MultiModal>![Shows the two possible Newman projections of the thermal cracking conformation (considering the chirality of the carbon atom attached to the ester oxygen), both adopting staggered conformations, where the ester oxygen is in the para position to the carbonyl α-hydrogen, the methyl group remains in the para position to the carbonyl group, sandwiched between two hydrogen atoms to minimize the potential energy, and then cracking produces a trans double bond. R in the figure represents other parts of the polymer and has no special meaning](/media/uploads/62601f92-9c4a-4822-9073-04c0e4169c22.png)</MultiModal>

<Checkpoint>[In the most stable conformation during thermal cracking, the methyl group is located in the para position to the carbonyl group, and the trans double bond is obtained after pyrolysis, with the structure OC(/C=C/C)=O. Statement 6 is incorrect.](1)</Checkpoint>

The correct statements are $1$, $3$, and $4$, the sum is $8$, choose $\rm{H}$
聚合物的重复单元是$O=C([*])C([R])([R])C(C)O[*]$，这是一个酯的结构。开环聚合得到聚酯，意味着$\bf{A}$是一个环状的内酯，由重复单元结构不难推断出$\bf{A}$的结构为CC1OC(C1([R])[R])=O，当$\rm{R = Me}$时，$\bf{A}$的化学式为$\rm{C_6H_{10}O_2}$。

<Checkpoint>[$\bf{A}$的结构为CC1OC(C1([R])[R])=O，含有四元环，当$\rm{R = Me}$时，$\bf{A}$的化学式为$\rm{C_6H_{10}O_2}$，说法1正确](1)</Checkpoint>

$^t\rm{Bu-P_4}$为大位阻碱，自身亲核能力弱，无法直接与$\bf{A}$反应引发聚合，而是通过拔除苯甲醇的羟基氢产生负离子引发聚合。

<Checkpoint>[$^t\rm{Bu-P_4}$位阻大，无法亲核引发聚合，说法2错误](1)</Checkpoint>

<Checkpoint>[$^t\rm{Bu-P_4}$通过拔除苯甲醇羟基氢引发聚合，起到碱的作用，说法3正确](1)</Checkpoint>

苯甲醇负离子进攻引发聚合，则聚合物一端封端为苯甲酯，后续聚合依靠产生的醇负离子继续亲核，直到质子化后聚合停止。当$\rm{R = Me}$时，聚合物重复单元为$\rm{C_6H_{10}O_2}$，分子量为$114.14$，两端封端可以认为是增加了额外的苯甲醇，分子量为$108.14$。则平均聚合度$n$为：$(5244 - 108.14) ÷ 114.14 = 45$

<Checkpoint>[聚合物封端为苯甲酯和氢，计算得平均聚合度为$(5244 - 108.14) ÷ 114.14 = 45$，说法4正确](1)</Checkpoint>

当$\rm{R = H}$时，加热条件下聚合物会发生酯的热裂解，产生烯烃和羧酸。当酯基$\alpha$位没有氢时，热裂解无法发生，故当$\rm{R = Me}$时聚合物可以保持稳定。

<Checkpoint>[当酯基$\alpha$位不存在氢时，聚合物无法发生热裂解，因此$\rm{R = Me}$时聚合物可在$\rm{180~\degree C}$下保持稳定，说法5错误](1)</Checkpoint>

发生热裂解时，氢与酯基氧处于对位，此时甲基与羰基处于对位的构型最稳定，产生的双键中甲基与羧基处于反式，结构为OC(/C=C/C)=O

<MultiModal>![展示出两种（考虑到酯基氧所连碳原子手性）可能的热裂解构象的纽曼投影式，均采取对交叉构象，其中酯基氧处于羰基α位氢的对位，甲基保持在羰基对位，夹在两个氢原子之间使势能最低，随后裂解产生反式双键，图中的R表示聚合物的其他部分，无特殊意义](/media/uploads/62601f92-9c4a-4822-9073-04c0e4169c22.png)</MultiModal>

<Checkpoint>[热裂解时最稳定的构象中甲基位于羰基对位，热解后得到反式双键，结构为OC(/C=C/C)=O，说法6错误](1)</Checkpoint>

正确的说法为$1$，$3$，$4$，和为$8$，选$\rm{H}$