A student performed the following experiment. $\mathrm{3.00 ~g}$ of phthalic anhydride was taken, added to a flask, and the reaction atmosphere was replaced with nitrogen. Under positive nitrogen pressure, $\mathrm{20 ~mL}$ of acetic anhydride and $\mathrm{10 ~mL}$ of triethylamine mixture dried with $\mathrm{KOH}$ solid were added. After stirring until dissolved, ethyl acetoacetate was added. Shortly after, a pale yellow solid precipitated in the system, and the mixture was stirred at room temperature overnight. The next day, the solution was added to a mixture of $\mathrm{24 ~mL}$ concentrated hydrochloric acid and $\mathrm{24 ~mL}$ water, stirred at room temperature for 30 minutes, and filtered to obtain a pale yellow solid. After washing with water and air-drying, $\mathbf{M}$ was obtained with a yield of 63%.

The NMR data of $\mathbf{M}$ are as follows: $\mathrm{^1H-NMR~δ~ 8.19 (dt, J = 7.9, 0.9 Hz, 1H), 8.02 (dt, J = 7.5, 1.0 Hz, 1H), 7.80 (td, J = 7.7, 1.2 Hz, 1H), 7.73 (td, J = 7.5, 1.0 Hz, 1H), 4.42 (q, J = 7.1 Hz, 2H), 2.64 (s, 3H), 1.38 (t, J = 7.2 Hz, 3H)}$

$\mathrm{^{13}C-NMR ~δ~ 195.5, 164.3, 164.2, 152.8, 136.4, 135.5, 133.0, 126.2, 126.0, 125.4, 116.4, 62.3, 31.8, 14.0}$

Two peaks at m/z = 283.05789 and 261.07593 were observed in the Orbitrap mass spectrum, corresponding to the $\mathrm{Na^+}$ and $\mathrm{H^+}$ adducts of $\mathbf{M}$ in the mass spectrum, respectively.

Based on the above information, analyze the structure of $\mathbf{M}$ and assign the chemical shifts of protons and carbon atoms to answer the following questions: If the two types of protons with chemical shifts of 2.64 and 1.38 in $\mathbf{M}$ are connected along chemical bonds, let the number of chemical bonds passed be $a$; the number of chemical bonds passed when the carbon atom with a chemical shift of 195.5 and the proton with a chemical shift of 4.42 are connected along chemical bonds is $b$; if there are multiple possible connection methods, select the one with the fewest chemical bonds passed to calculate $a$ and $b$; the maximum number of heavy atoms (i.e., non-hydrogen atoms) in the same plane of each molecule is $c$;

Please give the values of $a$, $b$, and $c$.
某同学进行了如下的实验。取$\mathrm{3.00 ~g}$邻苯二甲酸酐，加入烧瓶并将反应氛围置换为氮气后，氮气正压下加入$\mathrm{20 ~mL}$乙酸酐和$\mathrm{10 ~mL}$由$\mathrm{KOH}$固体干燥的三乙胺混合物，搅拌至溶解后加入乙酰乙酸乙酯，不久后体系中就析出淡黄色固体，室温搅拌放置过夜。次日，将溶液加入$\mathrm{24 ~mL}$浓盐酸和$\mathrm{24 ~mL}$水的混合物中，室温搅拌30分钟，抽滤可得淡黄色固体，用水洗涤后自然晾干，得到$\mathbf{M}$，产率63%。

$\mathbf{M}$的核磁信息如下：$\mathrm{^1H-NMR~δ~ 8.19 (dt, J = 7.9, 0.9 Hz, 1H), 8.02 (dt, J = 7.5, 1.0 Hz, 1H), 7.80 (td, J = 7.7, 1.2 Hz, 1H), 7.73 (td, J = 7.5, 1.0 Hz, 1H), 4.42 (q, J = 7.1 Hz, 2H), 2.64 (s, 3H), 1.38 (t, J = 7.2 Hz, 3H)}$

$\mathrm{^{13}C-NMR ~δ~ 195.5, 164.3, 164.2, 152.8, 136.4, 135.5, 133.0, 126.2, 126.0, 125.4, 116.4, 62.3, 31.8, 14.0}$

轨道阱质谱中观察到m/z = 283.05789和261.07593的两个峰，分别对应于$\mathbf{M}$在质谱中加合$\mathrm{Na^+}$和$\mathrm{H^+}$的峰。

请根据以上信息解析$\mathbf{M}$的结构并归属质子和碳原子的化学位移以回答以下问题：若沿着化学键将$\mathbf{M}$中化学位移为2.64和1.38的两种质子连接起来，记其中经过的化学键数量为$a$；化学位移为195.5的碳原子和化学位移为4.42的质子沿化学键连接起来经过的化学键数为$b$；若有多条可能连接方法，选取经过的化学键数最少的那一种以计算$a$和$b$；每个分子最多处于同一平面的重原子（即非氢原子）数量为$c$；

请给出$a$、$b$和$c$的值。
$a = 10, b = 5, c = 19$
$a = 10, b = 8, c= 19$
$a = 8, b = 5, c = 11$
$a = 9, b = 8, c = 11$
$a = 9, b = 8, c = 17$
$a = 12, b = 5, c = 9$
$a = 12, b = 6, c = 19$
All other options are not completely correct.
$a = 8, b = 5, c = 19$
$a = 7, b = 5, c = 19$
$a = 7, b = 8, c = 19$
$a = 7, b = 8, c = 18$
$a = 7, b = 5, c = 18$
$a = 10, b = 5, c = 18$
$a = 10, b = 8, c = 18$
$a = 10, b = 6, c = 18$
$a = 8, b = 5, c = 18$
$a = 7, b = 5, c = 17$
$a = 10, b = 5, c = 19$
$a = 10, b = 8, c= 19$
$a = 8, b = 5, c = 11$
$a = 9, b = 8, c = 11$
$a = 9, b = 8, c = 17$
$a = 12, b = 5, c = 9$
$a = 12, b = 6, c = 19$
其他选项都不完全正确。
$a = 8, b = 5, c = 19$
$a = 7, b = 5, c = 19$
$a = 7, b = 8, c = 19$
$a = 7, b = 8, c = 18$
$a = 7, b = 5, c = 18$
$a = 10, b = 5, c = 18$
$a = 10, b = 8, c = 18$
$a = 10, b = 6, c = 18$
$a = 8, b = 5, c = 18$
$a = 7, b = 5, c = 17$
Through mass spectrometry, it is known that $\mathbf{M}$ has 14 carbons, at least 12 hydrogens, and a relative molecular mass of approximately 195; thus, it can be deduced that the chemical formula of $\mathbf{M}$ is $\mathrm{C_{14}H_{12}O_5}$.

<Checkpoint>[$\mathbf{M}$化学式为$\mathrm{C_{14}H_{12}O_5}$](1)</Checkpoint>

Subsequently, the NMR spectrum was analyzed. From the coupling constant of the proton spectrum, it can be seen that the 3 hydrogens with a chemical shift of 1.38 and the 2 hydrogens with a chemical shift of 4.42 are adjacent, producing an ethyl structure;

<Checkpoint>[化学位移为1.38的3个氢和4.42的两个氢相邻，产生乙基的结构](1)</Checkpoint>

and the relatively low-field chemical shift of 4.42 indicates that the methylene group is connected to an oxygen atom.

<Checkpoint>[4.42的这一较为低场的化学位移暗示该亚甲基连接了氧原子](1)</Checkpoint>

The singlet at 2.64 should correspond to the methyl part of the acetyl group.

<Checkpoint>[2.64的单峰应当对应乙酰基的甲基部分](1)</Checkpoint>

The remaining four protons at 8.19, 8.02, 7.80, and 7.73 should correspond to the four hydrogens on the benzene ring, indicating that the benzene ring of phthalic anhydride did not participate in the reaction;

<Checkpoint>[8.19、8.02、7.80、7.73的四个质子应当对应于苯环上的四个氢](1)</Checkpoint>

<Checkpoint>[邻苯二甲酸酐的苯环没有参与反应](1)</Checkpoint>

At the same time, its splitting into four peaks (instead of two peaks) indicates that the symmetry of the benzene ring has been broken, and the originally symmetrical two carbonyl carbon atoms have become asymmetrical.

<Checkpoint>[苯环的对称性被破坏了，原本对称的两个羰基碳原子变得不对称了](1)</Checkpoint>

Observing the carbon spectrum, there are three low-field peaks at 195.5, 164.3, and 164.2, corresponding to one ketone carbonyl and two ester carbonyls, respectively.

<Checkpoint>[195.5、164.3、164.2三个低场峰分别对应于一个酮羰基和两个酯羰基](1)</Checkpoint>

Observing the substrate structure, it can be seen that only ethyl acetoacetate contains the ethyl moiety, so the ethyl ester group is likely retained in the product, and its molecular formula is $\mathrm{C_3H_5O_2}$; the molecular formula of the acetyl group is $\mathrm{C_2H_3O}$; the molecular formula of the 1,2-phenylene group is $\mathrm{C_6H_4}$; after removing these parts, $\mathrm{C_3O_2}$ remains in the compound, which should be connected to the benzene ring in some way, and make the four protons of the benzene ring inequivalent.

<Checkpoint>[除去酯基、乙氧基和苯环部分，化合物还剩$C_3O_2$碎片](1)</Checkpoint>

<Checkpoint>[$\mathrm{C_3O_2}$碎片应当以某种方式连接到苯环上，且使得苯环的四个质子不等价](1)</Checkpoint>

Since the carbon-carbon bond between the benzene ring and the carbonyl group of the original phthalic anhydride is not easily broken under the above reaction conditions, occupying two carbon atoms, the last carbon atom must form a ring with the two original carbonyl carbon atoms.

<Checkpoint>[最后一个碳原子必然与两个原先的羰基碳原子成环](1)</Checkpoint>

Since the original two carbonyl carbons are electrophilic, the carbon atom used for this ring formation must be nucleophilic, and is very likely derived from the methylene group of ethyl acetoacetate, so this carbon atom is likely to be connected to the ethoxycarbonyl group.

<Checkpoint>[由于原先的两个羰基碳都是亲电性的，因此中间这一成环所用的碳原子必然是亲核性的](1)</Checkpoint>

<Checkpoint>[该亲核性碳原子极有可能来源于乙酰乙酸乙酯的亚甲基](1)</Checkpoint>

<Checkpoint>[这个碳原子很可能和乙氧羰基连在一起](1)</Checkpoint>

The acetyl group can then be attached to one of the oxygen atoms, so that one oxygen atom of the C3O2 fragment is a ketone carbonyl, and the other oxygen atom is an enol ester oxygen atom, making the benzene ring asymmetrical, thus inferring that the structure of $\mathbf{M}$ is O=C1C(C(OCC)=O)=C(OC(C)=O)C2=CC=CC=C21.

<MultiModal>![本图为**M**的结构式，O=C1C(C(OCC)=O)=C(OC(C)=O)C2=CC=CC=C21](/media/uploads/febc44fb-dc92-4870-a664-6f83d393e14e.png)</MultiModal>

<Checkpoint>[乙酰基则可以结合在$C_3O_2$片段中一个氧原子上](1)</Checkpoint>

<Checkpoint>[$C_3O_2$碎片的一个氧原子为酮羰基，另一个氧原子为烯醇酯的氧原子，使苯环变得不对称](1)</Checkpoint>

<Checkpoint>[$\mathbf{M}$的结构为O=C1C(C(OCC)=O)=C(OC(C)=O)C2=CC=CC=C21](5)</Checkpoint>

To judge whether this structure is reasonable, we can briefly judge from the perspective of the reaction mechanism. First, triethylamine, as a base, removes the active proton of ethyl acetoacetate, and the generated anion acts as a nucleophile to react with the electrophilic anhydride carbonyl to form the intermediate O=C(C1=CC=CC=C1C(O)=O)C(C(C)=O)C(OCC)=O;

<Checkpoint>[中间体O=C(C1=CC=CC=C1C(O)=O)C(C(C)=O)C(OCC)=O](2)</Checkpoint>

Subsequently, since acetic anhydride is the reaction medium, the carboxyl group is further activated into a new anhydride O=C(C1=CC=CC=C1C(OC(C)=O)=O)C(C(C)=O)C(OCC)=O through anhydride exchange.

<Checkpoint>[通过酸酐交换的方式进一步活化为一个新的酸酐O=C(C1=CC=CC=C1C(OC(C)=O)=O)C(C(C)=O)C(OCC)=O](2)</Checkpoint>

Since the methine carbon atom after the condensation of ethyl acetoacetate still has an active proton, it is removed under the action of triethylamine, and the generated anion undergoes intramolecular cyclization again to obtain the intermediate O=C(C1=CC=CC=C1C2=O)C2(C(C)=O)C(OCC)=O with a quaternary carbon atom.

<Checkpoint>[负离子再一次进行分子内关环](1)</Checkpoint>

<Checkpoint>[中间体O=C(C1=CC=CC=C1C2=O)C2(C(C)=O)C(OCC)=O](2)</Checkpoint>

This intermediate's quaternary carbon atom is connected to four carbonyl groups, and has a strong tendency to remove one carbonyl group and form a larger conjugated system.

<Checkpoint>[季碳原子连接了四个羰基，有很强的脱除一个羰基并形成更大共轭体系的倾向](1)</Checkpoint>

Therefore, a molecule of acetic acid in the system attacks the acetyl group portion of the original ethyl acetoacetate of the intermediate, removing a molecule of acetic anhydride to obtain 2-ethoxycarbonyl-1,3-indanedione O=C(C1=CC=CC=C1C2=O)C2C(OCC)=O, which has an enol tautomer O=C1C(C(OCC)=O)=C(O)C2=CC=CC=C21.

<Checkpoint>[生成2-乙氧羰基1,3-茚二酮O=C(C1=CC=CC=C1C2=O)C2C(OCC)=O](1)</Checkpoint>

<Checkpoint>[2-乙氧羰基1,3-茚二酮存在烯醇式的互变异构体O=C1C(C(OCC)=O)=C(O)C2=CC=CC=C21](1)</Checkpoint>

This enol intermediate can be captured by excess acetic anhydride in the system to obtain the product O=C1C(C(OCC)=O)=C(OC(C)=O)C2=CC=CC=C21, which is $\mathbf{M}$. This shows that our inference is reasonable.

<Checkpoint>[烯醇式中间体被体系中过量的乙酸酐捕获就能得到产物$\mathbf{M}$](1)</Checkpoint>

After deducing the target product, since the methyl group at 2.64 comes from the acetyl group, and the proton at 1.38 comes from the terminal methyl group of the ethyl group, the connection method between them is H-C-C-O-C-C-C-O-C-C-H, so $a = 10$.

<Checkpoint>[连接方式为H-C-C-O-C-C-C-O-C-C-H，于是$a = 10$](2)</Checkpoint>

The carbon at 195.5 corresponds to the ketone carbonyl carbon on the five-membered ring, and the proton at 4.42 comes from the methylene portion of the ethyl group, so the connection method between them is C-C-C-O-C-H, so $b = 5$.

<Checkpoint>[连接方式为C-C-C-O-C-H，于是$b = 5$](2)</Checkpoint>

Since $\mathbf{M}$ can be regarded as an acetyl indandione derivative, the benzene ring and the $\mathrm{C_3O_2}$ fragment should be coplanar;

<Checkpoint>[苯环和$\mathrm{C_3O_2}$片段应当共平面](1)</Checkpoint>

Since the oxygen atom of the alcohol part of the ester group is still conjugated with the ester carbonyl group, the alcohol part of the ester can also be conjugated with the carbonyl group. Thus, the heavy atoms of the enol ester portion can be coplanar;

<Checkpoint>[烯醇酯部分的重原子可以共平面](1)</Checkpoint>

All heavy atoms of the ethyl ester group can also be coplanar.

<Checkpoint>[乙酯基团的所有重原子也可以共平面](1)</Checkpoint>

Thus, all heavy atoms in the system can be coplanar, making $c = 19$. Therefore, the only correct option is A.

<Checkpoint>[体系中所有重原子都可以共平面](1)</Checkpoint>

<Checkpoint>[$c = 19$](1)</Checkpoint>

<MultiModal>![本图为本题合成**M**的详细机理涉及到的中间体，供审核所看，不写图片描述](/media/uploads/ff819642-03f4-4a21-bdd1-d083a5ef02b9.png)</MultiModal>
通过质谱信息可知，$\mathbf{M}$有14个碳，至少有12个氢，相对分子质量大约在195左右；从而可推出$\mathbf{M}$化学式为$\mathrm{C_{14}H_{12}O_5}$。

<Checkpoint>[$\mathbf{M}$化学式为$\mathrm{C_{14}H_{12}O_5}$](1)</Checkpoint>

随后分析核磁谱图。由氢谱的耦合常数，可知化学位移为1.38的3个氢和4.42的两个氢相邻，产生乙基的结构；

<Checkpoint>[化学位移为1.38的3个氢和4.42的两个氢相邻，产生乙基的结构](1)</Checkpoint>

且4.42的这一较为低场的化学位移暗示该亚甲基连接了氧原子

<Checkpoint>[4.42的这一较为低场的化学位移暗示该亚甲基连接了氧原子](1)</Checkpoint>

2.64的单峰应当对应乙酰基的甲基部分。

<Checkpoint>[2.64的单峰应当对应乙酰基的甲基部分](1)</Checkpoint>

剩余的8.19、8.02、7.80、7.73的四个质子应当对应于苯环上的四个氢，这说明邻苯二甲酸酐的苯环没有参与反应；

<Checkpoint>[8.19、8.02、7.80、7.73的四个质子应当对应于苯环上的四个氢](1)</Checkpoint>

<Checkpoint>[邻苯二甲酸酐的苯环没有参与反应](1)</Checkpoint>

同时其分裂为四个峰（而非两个峰）说明苯环的对称性被破坏了，原本对称的两个羰基碳原子变得不对称了。

<Checkpoint>[苯环的对称性被破坏了，原本对称的两个羰基碳原子变得不对称了](1)</Checkpoint>

观察碳谱，发现有195.5、164.3、164.2三个低场峰，分别对应于一个酮羰基和两个酯羰基。

<Checkpoint>[195.5、164.3、164.2三个低场峰分别对应于一个酮羰基和两个酯羰基](1)</Checkpoint>

观察底物结构，可见只有乙酰乙酸乙酯包含乙基部分，故产物中很可能保留乙酯基团，其分子式为$\mathrm{C_3H_5O_2}$；乙酰基分子式为$\mathrm{C_2H_3O}$；1,2-苯二亚基分子式为$\mathrm{C_6H_4}$；除去这些部分，化合物还剩$\mathrm{C_3O_2}$，应当以某种方式连接到苯环上，且使得苯环的四个质子不等价。

<Checkpoint>[除去酯基、乙氧基和苯环部分，化合物还剩$C_3O_2$碎片](1)</Checkpoint>

<Checkpoint>[$\mathrm{C_3O_2}$碎片应当以某种方式连接到苯环上，且使得苯环的四个质子不等价](1)</Checkpoint>

由于苯环和原先邻苯二甲酸酐羰基的碳碳键不易在上述反应条件下断裂，占据了两个碳原子，因此最后一个碳原子必然与两个原先的羰基碳原子成环。

<Checkpoint>[最后一个碳原子必然与两个原先的羰基碳原子成环](1)</Checkpoint>

由于原先的两个羰基碳都是亲电性的，因此中间这一成环所用的碳原子必然是亲核性的，极有可能来源于乙酰乙酸乙酯的亚甲基，因此这个碳原子很可能和乙氧羰基连在一起。

<Checkpoint>[由于原先的两个羰基碳都是亲电性的，因此中间这一成环所用的碳原子必然是亲核性的](1)</Checkpoint>

<Checkpoint>[该亲核性碳原子极有可能来源于乙酰乙酸乙酯的亚甲基](1)</Checkpoint>

<Checkpoint>[这个碳原子很可能和乙氧羰基连在一起](1)</Checkpoint>

而乙酰基则可以结合在其中一个氧原子上，使得C3O2碎片的一个氧原子为酮羰基，另一个氧原子为烯醇酯的氧原子，使苯环变得不对称，因此推断$\mathbf{M}$的结构为O=C1C(C(OCC)=O)=C(OC(C)=O)C2=CC=CC=C21。

<MultiModal>![本图为**M**的结构式，O=C1C(C(OCC)=O)=C(OC(C)=O)C2=CC=CC=C21](/media/uploads/febc44fb-dc92-4870-a664-6f83d393e14e.png)</MultiModal>

<Checkpoint>[乙酰基则可以结合在$C_3O_2$片段中一个氧原子上](1)</Checkpoint>

<Checkpoint>[$C_3O_2$碎片的一个氧原子为酮羰基，另一个氧原子为烯醇酯的氧原子，使苯环变得不对称](1)</Checkpoint>

<Checkpoint>[$\mathbf{M}$的结构为O=C1C(C(OCC)=O)=C(OC(C)=O)C2=CC=CC=C21](5)</Checkpoint>

为判断这一结构是否合理，我们可以简要地从反应机理的角度判断。首先三乙胺作碱脱除乙酰乙酸乙酯的活泼质子，生成的阴离子作亲核试剂与亲电的酸酐羰基反应，形成中间体O=C(C1=CC=CC=C1C(O)=O)C(C(C)=O)C(OCC)=O；

<Checkpoint>[中间体O=C(C1=CC=CC=C1C(O)=O)C(C(C)=O)C(OCC)=O](2)</Checkpoint>

随后，由于乙酸酐是反应介质，羧基通过酸酐交换的方式被进一步活化为一个新的酸酐O=C(C1=CC=CC=C1C(OC(C)=O)=O)C(C(C)=O)C(OCC)=O。

<Checkpoint>[通过酸酐交换的方式进一步活化为一个新的酸酐O=C(C1=CC=CC=C1C(OC(C)=O)=O)C(C(C)=O)C(OCC)=O](2)</Checkpoint>

由于乙酰乙酸乙酯缩合后的次甲基碳原子还有一个活泼质子，其在三乙胺作用下脱除，生成的负离子再一次进行分子内关环，得到具有季碳原子的中间体O=C(C1=CC=CC=C1C2=O)C2(C(C)=O)C(OCC)=O。

<Checkpoint>[负离子再一次进行分子内关环](1)</Checkpoint>

<Checkpoint>[中间体O=C(C1=CC=CC=C1C2=O)C2(C(C)=O)C(OCC)=O](2)</Checkpoint>

这一中间体的季碳原子连接了四个羰基，有很强的脱除一个羰基并形成更大共轭体系的倾向。

<Checkpoint>[季碳原子连接了四个羰基，有很强的脱除一个羰基并形成更大共轭体系的倾向](1)</Checkpoint>

因此体系中的一分子乙酸进攻该中间体原来乙酰乙酸乙酯的乙酰基部分，脱除一分子乙酸酐，得到2-乙氧羰基1,3-茚二酮O=C(C1=CC=CC=C1C2=O)C2C(OCC)=O，其存在烯醇式的互变异构体O=C1C(C(OCC)=O)=C(O)C2=CC=CC=C21。

<Checkpoint>[生成2-乙氧羰基1,3-茚二酮O=C(C1=CC=CC=C1C2=O)C2C(OCC)=O](1)</Checkpoint>

<Checkpoint>[2-乙氧羰基1,3-茚二酮存在烯醇式的互变异构体O=C1C(C(OCC)=O)=C(O)C2=CC=CC=C21](1)</Checkpoint>

这一烯醇式中间体被体系中过量的乙酸酐捕获就能得到产物O=C1C(C(OCC)=O)=C(OC(C)=O)C2=CC=CC=C21，即$\mathbf{M}$。说明我们的推断是合理的。

<Checkpoint>[烯醇式中间体被体系中过量的乙酸酐捕获就能得到产物$\mathbf{M}$](1)</Checkpoint>

在推出目标产物后，由于2.64的甲基来自乙酰基，1.38的质子来自于乙基的末端甲基，因此其间连接方式为H-C-C-O-C-C-C-O-C-C-H，于是$a = 10$。

<Checkpoint>[连接方式为H-C-C-O-C-C-C-O-C-C-H，于是$a = 10$](2)</Checkpoint>

195.5的碳对应于五元环上的酮羰基碳，4.42的质子来自于乙基的亚甲基部分，故其间连接方式为C-C-C-O-C-H，于是$b = 5$。

<Checkpoint>[连接方式为C-C-C-O-C-H，于是$b = 5$](2)</Checkpoint>

由于$\mathbf{M}$可以视作乙酰基茚二酮衍生物，苯环和$\mathrm{C_3O_2}$片段应当共平面；

<Checkpoint>[苯环和$\mathrm{C_3O_2}$片段应当共平面](1)</Checkpoint>

由于酯基醇部分的氧原子仍与酯羰基共轭，因此酯的醇部分也可与羰基共轭。于是，烯醇酯部分的重原子可以共平面；

<Checkpoint>[烯醇酯部分的重原子可以共平面](1)</Checkpoint>

乙酯基团的所有重原子也可以共平面。

<Checkpoint>[乙酯基团的所有重原子也可以共平面](1)</Checkpoint>

于是，体系中所有重原子都可以共平面，使$c = 19$。因此全部正确的选项只有A。

<Checkpoint>[体系中所有重原子都可以共平面](1)</Checkpoint>

<Checkpoint>[$c = 19$](1)</Checkpoint>

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