The following two substances can interconvert under certain conditions. $\bf{A}$ can be converted to $\bf{B}$ under condition $\bf X$, and $\bf{B}$ can be converted to $\bf{A}$ under condition $\bf Y$. Conditions $\bf X,Y$ are one of (a) heating or visible light irradiation; (b) ultraviolet light irradiation. The transformation from $\bf{A}$ to $\bf{B}$ has two intermediates, successively $\bf{M,N}$, one of which contains a three-membered ring.

<MultiModal>![[R]C1=NNC([R])=C2C1C(C=CC=C3)=C3C(C4=C2C=CC=C4)=O, substance code is $\bf{A}$](/media/uploads/ea36e7da-9e6c-40c6-a5b0-57b4cdded28e.png)</MultiModal>

<MultiModal>![O=C1C2=C(C=CC=C2)C3(C=C([R])NN=C3[R])C4=C1C=CC=C4, substance code is $\bf{B}$](/media/uploads/5346568a-af64-45b6-ad63-faf335debd7c.png)</MultiModal>

The following statements are made:

1. The one containing a three-membered ring among $\bf M,N$ is $\bf N$.

2. At least one of $\bf M,N$ contains a $\rm N-H$ bond.

3. Condition $\bf X$ is (a), and condition $\bf Y$ is (b).

Select all the correct statements.
如下两种物质可以在一定条件下相互转化。$\bf{A}$在条件$\bf X$下可转化为$\bf{B}$，$\bf{B}$在条件$\bf Y$下可转化为$\bf{A}$。条件$\bf X,Y$是 (a)加热或可见光照射；(b)紫外光照射 中的一者。$\bf{A}$到$\bf{B}$的转变过程中有两个中间体，依次为$\bf{M,N}$，其中一个含有一个三元环。

<MultiModal>![[R]C1=NNC([R])=C2C1C(C=CC=C3)=C3C(C4=C2C=CC=C4)=O，物质代号为$\bf{A}$](/media/uploads/ea36e7da-9e6c-40c6-a5b0-57b4cdded28e.png)</MultiModal>

<MultiModal>![O=C1C2=C(C=CC=C2)C3(C=C([R])NN=C3[R])C4=C1C=CC=C4，物质代号为$\bf{B}$](/media/uploads/5346568a-af64-45b6-ad63-faf335debd7c.png)</MultiModal>

有如下说法：

1.$\bf M,N$中含有三元环的是$\bf N$。

2.$\bf M,N$中至少一者存在 $\rm N-H$ 键。

3.条件$\bf X$是(a)，条件$\bf Y$是(b)。

选出所有的正确说法。
1,2,3
1,2
1,3
2,3
1
2
3
All other statements are incorrect.
1,2,3
1,2
1,3
2,3
1
2
3
其他所有说法均不对
The conjugation range of $\bf A$ is larger than that of $\bf B$, and the $\rm LUMO-HOMO$ energy gap of $\bf A$ is smaller than that of $\bf B$, absorbing light with a longer wavelength. Therefore, condition $\bf X$ is (a) heating or visible light irradiation, and condition $\bf Y$ is (b) ultraviolet light irradiation. Statement 3 is correct.

<Checkpoint>[The conjugation range of $\bf A$ is larger than that of $\bf B$](1)</Checkpoint>

<Checkpoint>[Condition $\bf X$ is (a), condition $\bf Y$ is (b)](1)</Checkpoint>

$\bf A$ to $\bf B$ undergoes two intermediates $\bf M,N$. If $\bf M$ contains a three-membered ring, the three-membered ring of $\bf M$ will open directly to obtain $\bf B$ in the next step, and there will be no other intermediate. In addition, the direct formation of a three-membered ring from $\bf A$ requires enamine to attack the carbonyl-linked aromatic ring. The nucleophilicity and electrophilicity of the two are not strong, and it is difficult to occur under heating or visible light irradiation. Therefore, the first step of the reaction from $\bf A$ to $\bf B$ is to prepare a suitable structure for the formation of a three-membered ring.

According to this idea, it can be deduced that under the electron-donating and electron-withdrawing effects of enamine and ketone, $\bf A$ tautomerizes to form $\bf M$, and the structure is as follows.

<MultiModal>![[R]C1=NN=C([R])C2=C3C(C=CC=C3)=C(O)C4=C(C=CC=C4)C12，物质代号为$\bf M$](/media/uploads/38204614-1fd4-4210-b0b8-c1499b723ca4.png)</MultiModal>

<Checkpoint>[Under the electron-donating and electron-withdrawing effects of enamine and ketone, $\bf A$ tautomerizes to form $\bf M$](1)</Checkpoint>

$\bf M$ to $\bf N$ undergoes $6\pi$ electrocyclization, changing the seven-membered ring into a hexa-fused structure, and the structure of $\bf N$ is as follows.

<MultiModal>![[R]C1=NN=C([R])C23C1C(C=CC=C4)2C4=C(O)C5=C3C=CC=C5，物质代号为$\bf N$](/media/uploads/f1c29e5d-270c-40c4-a3a3-afecd150bb61.png)</MultiModal>

<Checkpoint>[$\bf M$ to $\bf N$ undergoes $6\pi$ electrocyclization](1)</Checkpoint>

Under the electron-donating and electron-withdrawing effects of enol and imine, $\bf N$ tautomerizes to form $\bf B$.

<Checkpoint>[Under the electron-donating and electron-withdrawing effects of enol and imine, $\bf N$ tautomerizes to form $\bf B$](1)</Checkpoint>

Therefore, $\bf N$ contains a three-membered ring in $\bf M,N$, and statement 1 is correct. Both $\bf M,N$ have an azo structure and no $\rm N-H$ bond, statement 2 is wrong.

<Checkpoint>[$\bf N$ contains a three-membered ring, $\bf M$ does not contain a three-membered ring](1)</Checkpoint>

<Checkpoint>[Neither $\bf M$ nor $\bf N$ contains an $\rm N-H$ bond](1)</Checkpoint>

In summary, the correct statements are 1 and 3, and the answer is option C.
$\bf A$的共轭范围比$\bf B$更大，$\bf A$的$\rm LUMO-HOMO$能差比$\bf B$更小，吸收波长更长的光。所以条件$\bf X$是(a)加热或可见光照射，条件$\bf Y$是(b)紫外光照射，说法3正确。

<Checkpoint>[$\bf A$的共轭范围比$\bf B$更大](1)</Checkpoint>

<Checkpoint>[条件$\bf X$是(a)，条件$\bf Y$是(b)](1)</Checkpoint>

$\bf A$到$\bf B$经历两个中间体$\bf M,N$。若$\bf M$含有三元环，则下一步$\bf M$的三元环打开后就直接得到$\bf B$，不会有另一个中间体。此外，由$\bf A$直接形成三元环需要烯胺进攻连有羰基的芳香环，二者亲核性和亲电性不强，在加热或可见光照射条件下很难发生。所以$\bf A$到$\bf B$反应的第一步是为形成三元环准备合适结构。

根据此思路，可以推出在烯胺和酮的推电子和吸电子效应下$\bf A$互变异构生成$\bf M$，结构如下。

<MultiModal>![[R]C1=NN=C([R])C2=C3C(C=CC=C3)=C(O)C4=C(C=CC=C4)C12，物质代号为$\bf M$](/media/uploads/38204614-1fd4-4210-b0b8-c1499b723ca4.png)</MultiModal>

<Checkpoint>[在烯胺和酮的推电子和吸电子效应下$\bf A$互变异构生成$\bf M$](1)</Checkpoint>

$\bf M$到$\bf N$经历 $6\pi$ 电环化，将七元环变为六并三结构，$\bf N$结构如下。

<MultiModal>![[R]C1=NN=C([R])C23C1C(C=CC=C4)2C4=C(O)C5=C3C=CC=C5，物质代号为$\bf N$](/media/uploads/f1c29e5d-270c-40c4-a3a3-afecd150bb61.png)</MultiModal>

<Checkpoint>[$\bf M$到$\bf N$经历 $6\pi$ 电环化](1)</Checkpoint>

在烯醇和亚胺的推电子和吸电子效应下$\bf N$互变异构生成$\bf B$。

<Checkpoint>[在烯醇和亚胺的推电子和吸电子效应下$\bf N$互变异构生成$\bf B$](1)</Checkpoint>

所以$\bf M,N$中含有三元环的是$\bf N$，说法1正确。$\bf M,N$中都是偶氮结构，没有 $\rm N-H$ 键，说法2错误。

<Checkpoint>[$\bf N$含三元环，$\bf M$不含三元环](1)</Checkpoint>

<Checkpoint>[$\bf M,N$中都没有 $\rm N-H$ 键](1)</Checkpoint>

综上所述，正确的说法为1和3，答案为选项C。