<MultiModal>![This image describes several steps of organic synthesis. The substrate is $\bf{B}$ with an unknown structure, which reacts with $H_2, Pd/C$ to produce $\bf{A}$. $\bf{B}$ reacts with $O_3, Zn/H_2O$ to produce $\bf{C}$. $\bf{C}$ reacts with $Ag_2O$ to produce $\bf{D}$. $\bf{D}$ reacts with $I_2, OH^-$ to produce $\bf{E}$. $\bf{E}$ under heating conditions yields the final product CC1CCC(CC1)=O.](/media/uploads/1ae01d8c-706d-4748-8d97-3d9d64e4cfb3.png)</MultiModal>

Regarding the organic reaction in the above image, it is known that:
1. The molecular formula of $\bf{A}$ is $\rm{C_9H_{18}}$.
2. The molecular formula of $\bf{C}$ is $\rm{C_9H_{16}O_2}$.
3. The molecular formula of $\bf{E}$ is $\rm{C_8H_{14}O_4}$.

Which of the following descriptions about $\bf{A}$ is correct:
<MultiModal>![本图描述了几步有机合成过程。底物为未知结构的$\bf{B}$，与$H_2, Pd/C$反应生成$\bf{A}$，$\bf{B}$与$O_3, Zn/H_2O$反应生成$\bf{C}$，$\bf{C}$与$Ag_2O$反应生成$\bf{D}$，$\bf{D}$与$I_2, OH^-$反应生成$\bf{E}$，$\bf{E}$加热条件得到最终产物CC1CCC(CC1)=O。](/media/uploads/1ae01d8c-706d-4748-8d97-3d9d64e4cfb3.png)</MultiModal>

关于上图的有机反应，已知：
1. $\bf{A}$的分子式为$\rm{C_9H_{18}}$。
2. $\bf{C}$的分子式为$\rm{C_9H_{16}O_2}$。
3.  $\bf{E}$的分子式为$\rm{C_8H_{14}O_4}$。

下列关于$\bf{A}$的描述正确的是：
All other options are incorrect
$\bf{A}$ must be chiral
$\bf{A}$ contains a six-membered ring
$\bf{A}$ containing a five-membered ring
$\bf{A}$ containing a four-membered ring
$\bf{A}$ Contains a triple bond
$\bf{A}$ contains the bonding relationship $\rm{H_3C-C-C-CH_3}$
$\bf{A}$ contains an ethyl group.
$\bf{A}$ does not contain tertiary carbons
$\bf{A}$ containing isopropyl
$\bf{A}$ Containing terminal alkene
其他选项均不正确
$\bf{A}$一定具有手性
$\bf{A}$含有六元环
$\bf{A}$含有五元环
$\bf{A}$含有四元环
$\bf{A}$含有三键
$\bf{A}$中存在键联关系$\rm{H_3C-C-C-C-CH_3}$
$\bf{A}$含有一个乙基
$\bf{A}$不含有三级碳
$\bf{A}$含有异丙基
$\bf{A}$含有末端烯烃
This question examines the ability of retrosynthetic analysis.

The final product is 4-methylcyclohexanone, which is generated by heating, and the molecular formula of $\bf{E}$ is $\rm{C_8H_{14}O_4}$, indicating the loss of one molecule of $\rm{CO_2}$ and two molecules of $\rm{H_2O}$. Therefore, the first reaction should be decarboxylation, and $\bf{E}$ contains a carboxyl group;

<Checkpoint>[Heating loses one molecule of $\rm{CO_2}$ and two molecules of $\rm{H_2O}$](1)</Checkpoint>

<Checkpoint>[$\bf{E}$ contains a carboxyl group](1)</Checkpoint>

The loss of water produces a ketone carbonyl group, and the final product does not have a carbon-carbon double bond. The six-membered ring is entirely composed of carbon atoms, indicating that the loss of one molecule of water is only the dehydration of a hydrated ketal;

<Checkpoint>[The loss of one molecule of water is only the dehydration of a hydrated ketal](1)</Checkpoint>

The degree of unsaturation of $\bf{E}$ is calculated to be 2, and it contains a carboxyl group. Therefore, $\bf{E}$ either has a ring or a double bond; if $\bf{E}$ has a ring, then there is no double bond, and the two oxygen atoms other than the carboxyl group can only exist in the form of alcohols; but if there are alcohols, the final product cannot be without a carbon-carbon double bond, so $\bf{E}$ must not have a ring, and there must be another carbonyl group besides the carboxyl group;

<Checkpoint>[$\bf{E}$ must not have a ring, and there must be another carbonyl group besides the carboxyl group](1)</Checkpoint>

Combined with the fact that $\bf{E}$ loses one molecule of water in addition to the dehydration of a hydrated ketal, and the molecular formula of $\bf{E}$ contains four oxygen atoms, consider that $\bf{E}$ contains two carboxyl groups;

$\bf{E}$ must not have a ring, but the final product is a six-membered ring, so the newly formed carbon-carbon single bond on the ring can only be formed through a decarboxylation reaction; then at this time, we can think of the six-membered ring transition state decarboxylation process, as shown in the figure below:

<MultiModal>![This figure is a schematic diagram of the six-membered ring transition state decarboxylation; the left side is the state before decarboxylation, where two terminal carboxyl groups form a six-membered ring through the C=O1 of one carboxyl group and the C1-C2-O2-H2 of the other carboxyl group. C1 represents the α position of the carboxyl group, and C2 represents the carbon atom of the carboxyl group itself; a dashed line is drawn between C and C1, and a dashed line is drawn between H2 and O1. This transition state undergoes a reaction to generate the product on the right side of the figure, where C and C1 are connected, the single bond between C1 and C2 becomes a double bond, O2-H2 breaks, and H2 forms a bond with O1 of the other carboxyl group, forming an O=C=O structure and [H]OC(C[R])(O)[R]. In the figure, R represents other groups connected to the carboxyl group.](/media/uploads/9575a15d-a4b6-4d77-a5c5-0b08ea567486.png)</MultiModal>

<Checkpoint>[The newly formed carbon-carbon single bond on the ring can only be formed through a decarboxylation reaction](1)</Checkpoint>

<Checkpoint>[The decarboxylation reaction goes through a six-membered ring transition state, eventually yielding a hydrated ketal](1)</Checkpoint>

This process removes one molecule of water and one molecule of carbon dioxide, generating a six-membered ring and eventually yielding a hydrated ketal, which is consistent with the previous deduction, so $\bf{E}$ contains two terminal carboxyl groups. According to the position of the methyl group in the six-membered ring, the structure of $\bf{E}$ is deduced to be CC(CCC(O)=O)CCC(O)=O.

<Checkpoint>[$\bf{E}$ contains two terminal carboxyl groups](1)</Checkpoint>

<Checkpoint>[The structure of $\bf{E}$ is CC(CCC(O)=O)CCC(O)=O](1)</Checkpoint>

$\bf{E}$ is generated by the oxidation of $\bf{D}$ with iodine element under basic conditions, and the molecular formula of $\bf{C}$ has 1 more carbon atom than $\bf{E}$, which is easily thought of as the oxidation process of methyl ketone. Methyl ketone undergoes a haloform reaction to generate carboxylic acid. One carboxylic acid in the structure of $\bf{E}$ is produced by the oxidation of methyl ketone, so the structure of $\bf{D}$ is CC(CCC(C)=O)CCC(O)=O.

<Checkpoint>[One carboxylic acid in the structure of $\bf{E}$ is produced by the oxidation of methyl ketone via iodoform reaction](1)</Checkpoint>

<Checkpoint>[The structure of $\bf{D}$ is CC(CCC(C)=O)CCC(O)=O](1)</Checkpoint>

At this time, the information may not be able to deduce, look back and observe other reactions: the reaction of $\bf{B}$ generating $\bf{C}$ is ozonolysis, which cleaves the double bond to generate two carbonyl groups; therefore, $\bf{B}$ contains an alkene.

The reaction from $\bf{B}$ to $\bf{A}$ is a hydrogenation reaction, $\bf{B}$ contains an alkene, so the chemical formula of $\bf{B}$ can be deduced to be $\rm{C_9H_{16}}$; thus, the degree of unsaturation of $\bf{B}$ is 2, and there is an alkene, indicating that $\bf{B}$ contains a ring; the chemical formula of $\bf{B}$ generating $\bf{C}$ only has two more oxygen atoms, so it is judged that $\bf{B}$ has an alkene on the ring, and ozonolysis generates two carbonyl groups.

<Checkpoint>[$\bf{B}$ has an alkene on the ring, and ozonolysis generates two carbonyl groups](1)</Checkpoint>

The condition for $\bf{C}$ to generate $\bf{D}$ is silver oxide, which is a typical reaction of aldehyde being oxidized to carboxylic acid; therefore, the structure of $\bf{C}$ is CC(CCC(C)=O)CCC=O.

<Checkpoint>[$\bf{C}$ generating $\bf{D}$ aldehyde is oxidized to carboxylic acid](1)</Checkpoint>

<Checkpoint>[The structure of $\bf{C}$ is CC(CCC(C)=O)CCC=O](1)</Checkpoint>

According to the ozonolysis reaction, the structure of $\bf{B}$ can be deduced to be CC1CCC=C(C)CC1.

<Checkpoint>[The structure of $\bf{B}$ is CC1CCC=C(C)CC1](1)</Checkpoint>

The reaction from $\bf{B}$ to $\bf{A}$ is a hydrogenation reaction, and the structure of $\bf{A}$ is CC1CCCC(C)CC1.

<Checkpoint>[The structure of $\bf{A}$ is CC1CCCC(C)CC1](1)</Checkpoint>

According to the structure of $\bf{A}$, options B-I are all incorrect.

In summary, option A is correct.

<MultiModal>![This figure shows the structures of the unknown substances $\bf{A-E}$ involved in this question. The structure of $\bf{A}$ is CC1CCCC(C)CC1; the structure of $\bf{B}$ is CC1CCC=C(C)CC1; the structure of $\bf{C}$ is CC(CCC(C)=O)CCC=O; the structure of $\bf{D}$ is CC(CCC(C)=O)CCC(O)=O; the structure of $\bf{E}$ is CC(CCC(O)=O)CCC(O)=O.](/media/uploads/bf2fe5ff-fa10-4195-b586-7efef5d96c6f.png)</MultiModal>
本题考查逆合成推断能力。

最终产物为4-甲基环己酮，经过加热生成且$\bf{E}$分子式为$\rm{C_8H_{14}O_4}$，说明失去了一分子$\rm{CO_2}$和两分子$\rm{H_2O}$。则该步反应首先应为脱羧反应，$\bf{E}$中存在羧基；

<Checkpoint>[加热失去了一分子$\rm{CO_2}$和两分子$\rm{H_2O}$](1)</Checkpoint>

<Checkpoint>[$\bf{E}$中存在羧基](1)</Checkpoint>

失去水后产生酮羰基，且最终产物没有碳碳双键，六元环全都为碳原子构成，说明有一分子水的失去只是水合缩酮脱水；

<Checkpoint>[有一分子水的失去只是水合缩酮脱水](1)</Checkpoint>

计算$\bf{E}$不饱和度为2且存在羧基，因此$\bf{E}$要么存在环要么存在双键；如果$\bf{E}$存在环，则不存在双键，除羧基外的两个氧原子只能以醇的形式存在；但存在醇最终产物不可能没有碳碳双键，因此$\bf{E}$一定不存在环，则一定存在除羧基外的另一个羰基；

<Checkpoint>[$\bf{E}$一定不存在环，则一定存在除羧基外的另一个羰基](1)</Checkpoint>

结合$\bf{E}$除水合缩酮脱水之外还有一分子水的失去，且$\bf{E}$分子式中含有四个氧原子，考虑$\bf{E}$中含有两个羧基；

$\bf{E}$一定不存在环但最终产物为六元环，故新生成的环上的碳碳单键只能通过脱羧反应形成；则此时可以想到六元环过渡态脱羧过程，如下图所示：

<MultiModal>![本图为六元环过渡态脱羧示意图；左边为脱羧前的状态，两个末端羧基通过一个羧基的C=O1和另一个羧基的C1-C2-O2-H2构成六元环，C1代表羧基的α位，C2代表羧基的本身的碳原子；C与C1之间画有虚线，H2与O1之间画有虚线。该过渡态发生反应生成图中右边的产物，C与C1连接，C1与C2的单键变为双键，O2-H2断开，H2与另一个羧基的O1成键，形成O=C=O结构和[H]OC(C[R])(O)[R]。图中R代表连接羧基的其他基团。](/media/uploads/9575a15d-a4b6-4d77-a5c5-0b08ea567486.png)</MultiModal>

<Checkpoint>[新生成的环上的碳碳单键只能通过脱羧反应形成](1)</Checkpoint>

<Checkpoint>[脱羧反应经历六元环过渡态，最终得到水合缩酮](1)</Checkpoint>

该过程脱去一分子水和一分子二氧化碳，生成了六元环且最终得到水合缩酮，跟之前的推断符合，所以$\bf{E}$中含有两个末端羧基。根据六元环甲基位置，推出$\bf{E}$结构为CC(CCC(O)=O)CCC(O)=O。

<Checkpoint>[$\bf{E}$中含有两个末端羧基](1)</Checkpoint>

<Checkpoint>[$\bf{E}$结构为CC(CCC(O)=O)CCC(O)=O](1)</Checkpoint>

$\bf{E}$通过碱性条件碘单质氧化$\bf{D}$生成，且$\bf{C}$的分子式比$\bf{E}$多1个碳原子，很容易想到为甲基酮的氧化过程，甲基酮发生卤仿反应生成羧酸。$\bf{E}$结构有一个羧酸是甲基酮氧化产生的，因此$\bf{D}$结构为CC(CCC(C)=O)CCC(O)=O。

<Checkpoint>[$\bf{E}$结构有一个羧酸是甲基酮发生碘仿反应氧化产生的](1)</Checkpoint>

<Checkpoint>[$\bf{D}$结构为CC(CCC(C)=O)CCC(O)=O](1)</Checkpoint>

此时的信息可能无法推断，回头观察其他反应：$\bf{B}$生成$\bf{C}$的反应为臭氧化反应，断裂双键生成两个羰基；因此$\bf{B}$含有烯烃。

$\bf{B}$到$\bf{A}$为加氢反应，$\bf{B}$含有烯烃，可倒推$\bf{B}$的化学式为$\rm{C_9H_{16}}$；从而$\bf{B}$的不饱和度为2，存在烯烃，说明$\bf{B}$含有环；$\bf{B}$生成$\bf{C}$化学式只多了两个氧原子，因此判断$\bf{B}$存在环上的烯烃，臭氧化后生成两个羰基。

<Checkpoint>[$\bf{B}$存在环上的烯烃，臭氧化后生成两个羰基](1)</Checkpoint>

$\bf{C}$生成$\bf{D}$的条件为氧化银，是典型的醛基被氧化为羧酸的反应；因此$\bf{C}$的结构为CC(CCC(C)=O)CCC=O。

<Checkpoint>[$\bf{C}$生成$\bf{D}$醛基被氧化为羧酸](1)</Checkpoint>

<Checkpoint>[$\bf{C}$的结构为CC(CCC(C)=O)CCC=O](1)</Checkpoint>

根据臭氧化反应，可倒退$\bf{B}$的结构为CC1CCC=C(C)CC1。

<Checkpoint>[$\bf{B}$的结构为CC1CCC=C(C)CC1](1)</Checkpoint>

$\bf{B}$到$\bf{A}$为加氢反应，$\bf{A}$的结构为CC1CCCC(C)CC1。

<Checkpoint>[$\bf{A}$的结构为CC1CCCC(C)CC1](1)</Checkpoint>

根据$\bf{A}$的结构可知，选项B-I均不正确。

综上，选项A正确。

<MultiModal>![本图为本题涉及的未知物质结构$\bf{A-E}$。$\bf{A}$的结构为CC1CCCC(C)CC1；$\bf{B}$的结构为CC1CCC=C(C)CC1；$\bf{C}$的结构为CC(CCC(C)=O)CCC=O；$\bf{D}$结构为CC(CCC(C)=O)CCC(O)=O；$\bf{E}$结构为CC(CCC(O)=O)CCC(O)=O。](/media/uploads/bf2fe5ff-fa10-4195-b586-7efef5d96c6f.png)</MultiModal>