The following compound **A** can alleviate the repulsion between lone pair electrons of nitrogen atoms when binding to a proton, and simultaneously form intramolecular hydrogen bonds. Therefore, **A** has a strong basicity (the $\mathrm{p} K_{\mathrm{a}}$ of its conjugate acid is $12.1$), and is called a "proton sponge":

<MultiModal>![Fig 1: The SMILES code of this reaction is: CN(C)C1=CC=CC2=C1C(N(C)C)=CC=C2>[H+]>CN(C)C3=CC=CC4=C3C([N+](C)([H])C)=CC=C4](/media/uploads/5c86a49c-a03f-40b3-bf94-6efe34d72e6f.png)</MultiModal>

Further studies have found that the introduction of methoxy groups at the ortho position of dimethylamino can greatly enhance its basicity. For example, the $\mathrm{p} K_{\mathrm{a}}$ of the conjugate acid of **B** in the following compounds is $16.1$, while the $\mathrm{p} K_{\mathrm{a}}$ of the conjugate acid of **C** is $13.9$:

<MultiModal>![The SMILES code of B is: CN(C)C1=C(OC)C=CC2=C1C(N(C)C)=C(OC)C=C2，C is：CN(C)C1=CC=C(OC)C2=C1C(N(C)C)=CC=C2OC](/media/uploads/5ced9030-60eb-4bfb-bdb1-10a39d62baf4.png)</MultiModal>

Which of the following statements are correct:

$1$. The nitrogen atom in **B** is close to $sp^{2}$ hybridization.

$2$. The methoxy group has an electron-donating inductive effect, which increases the electron cloud density on the nitrogen atom, making the nitrogen atom easier to bind to a proton.

$3$. In the comparison of the basicity of **A** and **B**, the electronic effect is a more important factor affecting the basicity of **B** than the steric effect.

$4$. The difference in basicity between **B** and **C** mainly comes from the steric hindrance effect provided by the methoxy group. The methoxy group and dimethylamino group are relatively crowded, resulting in the two dimethylamino groups of **B** being closer, the repulsion of lone pair electrons on the nitrogen atom being greater, and the combination with a proton can alleviate more repulsion.
以下化合物 **A** 结合质子时能够缓解氮原子孤对电子之间的排斥力, 同时形成分子内氢
键，因此 **A** 具有很强的碱性(其共轭酸的 $\mathrm{p} K_{\mathrm{a}}=12.1$ )， 被称为 “质子海绵”:

<MultiModal>![Fig 1: 该反应的SMILES编码为：CN(C)C1=CC=CC2=C1C(N(C)C)=CC=C2>[H+]>CN(C)C3=CC=CC4=C3C([N+](C)([H])C)=CC=C4](/media/uploads/5c86a49c-a03f-40b3-bf94-6efe34d72e6f.png)</MultiModal>

进一步研究发现， 在二甲氨基邻位引入甲氧基能使得其碱性大大增强, 如以下化合物中 **B**
的共轭酸的 $\mathrm{p} K_{\mathrm{a}}=16.1$, 而 **C** 的共轭酸的 $\mathrm{p} K_{\mathrm{a}}=13.9$ :

<MultiModal>![B 的SMILES编码为：CN(C)C1=C(OC)C=CC2=C1C(N(C)C)=C(OC)C=C2，C为：CN(C)C1=CC=C(OC)C2=C1C(N(C)C)=CC=C2OC](/media/uploads/5ced9030-60eb-4bfb-bdb1-10a39d62baf4.png)</MultiModal>

下列说法正确的有：

$1$.**B** 中的氮原子接近于$sp^{2}$杂化。

$2$.甲氧基具有给电子诱导效应,增大了氮原子上的电子云密度,使得氮原子更易结合质子。

$3$. **A**、**B** 碱性对比而言，电子效应是比空间效应更主要的影响 **B** 碱性的因素。

$4$. **B**、**C** 碱性差异主要来源于甲氧基提供的位阻效应，甲氧基与二甲氨基之间较为拥挤,导致 **B** 的两个二甲氨基距离更近,氮原子上孤对电子排斥力更大,结合质子能缓解更多的排斥力。
All other options are incorrect
$1$.
$2$.
$3$.
$4$.
$1、2$.
$1、3$.
$1、4$.
$2、3$.
$2、4$.
$3、4$.
$1、2、3$.
$1、2、4$.
$1、3、4$.
$2、3、4$.
$1、2、3、4$.
其他选项均不正确
$1$.
$2$.
$3$.
$4$.
$1、2$.
$1、3$.
$1、4$.
$2、3$.
$2、4$.
$3、4$.
$1、2、3$.
$1、2、4$.
$1、3、4$.
$2、3、4$.
$1、2、3、4$.
The conjugate acid of compound **A** has a $\mathrm{p} K_{\mathrm{a}}=12.1$, while the conjugate acid of **C** has a $\mathrm{p} K_{\mathrm{a}}=13.9$.

The difference between the two arises from the para-methoxy group to the dimethylamino group. Since the para-methoxy group cannot produce steric hindrance to the dimethylamino group, the difference in $\mathrm{p} K_{\mathrm{a}}$ of its conjugate acid is entirely due to the electron-donating resonance effect of the methoxy group, which is approximately $1.8$ orders of magnitude in the $\mathrm{p} K_{\mathrm{a}}$ of the conjugate acid.

<Checkpoint>[The difference in basicity between **A** and **C** arises from the electron-donating resonance effect of the methoxy group.](1)</Checkpoint>

For a conjugated system, adding one or a few double bonds does not significantly weaken the resonance effect of the substituent. Therefore, the electron-donating resonance effect of the methoxy group in **B** and **C** is somewhat reduced when it is in the para position compared to the ortho position, but the impact is not significant.

The conjugate acid of compound **B** has a $\mathrm{p} K_{\mathrm{a}}=16.1$, while the conjugate acid of **C** has a $\mathrm{p} K_{\mathrm{a}}=13.9$. Therefore, the difference in basicity between **B** and **C** can be largely attributed to the steric hindrance effect brought about by the ortho-methoxy group.

<Checkpoint>[The position of the methoxy group, whether ortho or para, has little effect on the electron-donating resonance effect.](1)</Checkpoint>

<Checkpoint>[The difference in basicity between **B** and **C** can be largely attributed to the steric hindrance effect brought about by the ortho-methoxy group.](1)</Checkpoint>

The methoxy group and the dimethylamino group are relatively crowded, resulting in the two dimethylamino groups of **B** being closer to each other, and the repulsion of lone pair electrons on the nitrogen atom being greater, which can alleviate more repulsion when binding to a proton.

<Checkpoint>[Binding a proton can alleviate more repulsion, $4$ is correct.](1)</Checkpoint>

At the same time, due to this repulsive effect, the two methyl groups of the dimethylamino group, when the nitrogen atom adopts $sp^{3}$ hybridization, will be repelled by the ortho substituent or another dimethylamino group regardless of the conformation. Therefore, in this case, it will tend to $sp^{2}$ hybridization to some extent.

<Checkpoint>[When the nitrogen atom adopts $sp^{3}$ hybridization, the two methyl groups of the dimethylamino group will be repelled by the ortho substituent or another dimethylamino group. $1$ is correct.](1)</Checkpoint>

Methoxy group has an electron-donating resonance effect. Among them, oxygen has an electron-withdrawing inductive effect due to its large electronegativity, but the overall electron-donating resonance is greater than the electron-withdrawing inductive effect and dominates.

<Checkpoint>[The electron-donating resonance effect of the methoxy group is greater than the electron-withdrawing inductive effect, $2$ is incorrect.](1)</Checkpoint>

Returning to the comparison of **A** and **B**, it is not difficult to know that the magnitude of the effect of the electron-donating resonance effect on the conjugate acid $\mathrm{p} K_{\mathrm{a}}$ is slightly greater than $1.8$, and the rest are the effects of steric hindrance. Therefore, the steric hindrance effect is greater than the electronic effect.

<Checkpoint>[The steric hindrance effect is greater than the electronic effect, $3$ is incorrect.](1)</Checkpoint>

In summary, the answers are $1, 4$.
化合物 **A** 
的共轭酸的 $\mathrm{p} K_{\mathrm{a}}=12.1$, 而  **C**  的共轭酸的 $\mathrm{p} K_{\mathrm{a}}=13.9$.

二者的差异来源于二甲氨基对位的甲氧基。而对位甲氧基无法对二甲氨基产生位阻效应，故其共轭酸的$\mathrm{p} K_{\mathrm{a}}$差异完全由甲氧基的给电子共轭效应给出，大概对共轭酸$\mathrm{p} K_{\mathrm{a}}$的影响是$1.8$个数量级。

<Checkpoint>[**A**、**C**碱性差异来源于甲氧基给电子共轭效应。](1)</Checkpoint>

对于共轭体系而言，增加一个或少数几个双键对于取代基的共轭效应削弱影响不大，故 **B**、**C** 中甲氧基位于对位相较于位于邻位，给电子共轭效应有所减小但影响不大。

化合物  **B** 
的共轭酸的 $\mathrm{p} K_{\mathrm{a}}=16.1$, 而  **C**  的共轭酸的 $\mathrm{p} K_{\mathrm{a}}=13.9$.故**B**、**C**碱性差异可基本归结于邻位甲氧基带来的位阻效应。

<Checkpoint>[甲氧基处于邻位或对位对于给电子共轭效应影响不大](1)</Checkpoint>

<Checkpoint>[**B**、**C**碱性差异可基本归结于邻位甲氧基带来的位阻效应。](1)</Checkpoint>

甲氧基与二甲氨基之间较为拥挤,导致 **B**  的两个二甲氨基距离更近,氮原子上孤对电子排斥力更大,结合质子能缓解更多的排斥力。

<Checkpoint>[结合质子能缓解更多的排斥力，$4$ 正确。](1)</Checkpoint>

同时，由于该排斥作用，导致二甲氨基的两个甲基在氮原子采取$sp^{3}$杂化时，不管处于怎样的构象都会与邻位取代基或另一个二甲氨基有所排斥，故在此情况下，会一定偏向于$sp^{2}$杂化。

<Checkpoint>[氮原子采取$sp^{3}$杂化时，二甲氨基的两个甲基会与邻位取代基或另一个二甲氨基有所排斥。$1$ 正确。](1)</Checkpoint>

甲氧基为给电子共轭效应，其中氧由于电负性大，具有吸电子诱导效应，但总体给电子共轭大于吸电子诱导效应，占主导。

<Checkpoint>[甲氧基给电子共轭大于吸电子诱导效应，$2$ 错误。](1)</Checkpoint>

再次回到 **A**、**B** 的比较，不难知道给电子共轭效应对共轭酸$\mathrm{p} K_{\mathrm{a}}$的影响的量级略大于$1.8$，剩下均为位阻效应的影响，故位阻效应大于电子效应。

<Checkpoint>[位阻效应大于电子效应，$3$ 错误。](1)</Checkpoint>

综上，答案为$1、4$