In the potential energy diagram of an organic molecule $\mathbf{X}$ rotating around one of its bonds, the points where the first derivative of the potential energy curve is 0 correspond to the following conformations $\mathbf{A\sim D}$:

<MultiModal>![Newman projection, the front carbon is connected to an alkenyl group $\mathrm{=CH_2}$ and a methyl group $\mathrm{CH_3}$ substituent, while the back carbon is connected to a methyl group $\mathrm{CH_3}$ substituent and two hydrogens $\mathrm{H}$, where in $\mathbf{A}$ the alkenyl group of the front carbon overlaps with the methyl group of the back carbon, in $\mathbf{B}$ the alkenyl group of the front carbon overlaps with the hydrogen of the back carbon, in $\mathbf{C}$ the methyl group of the front carbon overlaps with the hydrogen of the back carbon, and in $\mathbf{D}$ the methyl group of the front carbon overlaps with the methyl group of the back carbon](/media/uploads/df6b56cd-d046-4729-878a-c05ac2fc1e64.png)</MultiModal>

The points where the first derivative of the potential energy curve is 0 correspond to four different energies $1\sim 4$ (with the lowest energy conformation as the zero point, unit $\mathrm{kcal}$): 0.00, +0.06, +1.39, +2.68.

Please match the conformation numbers with the energy numbers, and select the correct result from the following options.
一个有机分子 $\mathbf{X}$ 围绕其中一根键旋转的势能图中，其势能曲线一阶导数为 0 的点对应有以下几个构象 $\mathbf{A\sim D}$：

<MultiModal>![纽曼式投影图，前面的碳连接有一个烯基 $\mathrm{=CH_2}$ 与一个甲基 $\mathrm{CH_3}$ 取代基，而后面的碳连接有一个甲基 $\mathrm{CH_3}$ 取代基与两个氢 $\mathrm{H}$，其中 $\mathbf{A}$ 中前面的碳的烯基与后面碳的甲基的重叠，$\mathbf{B}$ 中前面的碳的烯基与后面碳的氢重叠，$\mathbf{C}$ 中前面的碳的甲基与后面碳的氢重叠，$\mathbf{D}$ 中前面碳的甲基与后面碳的甲基重叠](/media/uploads/df6b56cd-d046-4729-878a-c05ac2fc1e64.png)</MultiModal>

势能曲线一阶导数为 0 的点对应有四种不同的能量 $1\sim 4$（以能量最低的构象为零点，单位 $\mathrm{kcal}$）：0.00、+0.06、+1.39、+2.68。

请将构象序号与能量序号相对应，从以下选项中选择正确的结果。
All other options are incorrect
A-1 B-2 C-3 D-4
A-3 B-2 C-1 D-4
A-3 B-1 C-2 D-4
A-2 B-1 C-3 D-4
A-2 B-3 C-1 D-4
A-1 B-2 C-4 D-3
A-1 B-3 C-4 D-2
A-1 B-4 C-3 D-2
A-1 B-4 C-2 D-3
其他选项均不正确
A-1 B-2 C-3 D-4
A-3 B-2 C-1 D-4
A-3 B-1 C-2 D-4
A-2 B-1 C-3 D-4
A-2 B-3 C-1 D-4
A-1 B-2 C-4 D-3
A-1 B-3 C-4 D-2
A-1 B-4 C-3 D-2
A-1 B-4 C-2 D-3
First, it is necessary to determine the structure of the organic molecule $\mathbf{X}$. According to the image, it is a Newman projection. The front carbon is connected to an alkene group $\mathrm{=CH_2}$ and a methyl $\mathrm{CH_3}$ substituent, while the rear carbon is connected to a methyl $\mathrm{CH_3}$ substituent and two hydrogens $\mathrm{H}$. Therefore, the SMILES of molecule $\mathbf{X}$ can be represented as `[CH2:1]=[CH:2]([CH3:5])[CH2:3][CH3:4]`, which is 2-methyl-1-butene.

<Checkpoint>[The SMILES of molecule $\mathbf{X}$ can be represented as `[CH2:1]=[CH:2]([CH3:5])[CH2:3][CH3:4]`, which is 2-methyl-1-butene](1)</Checkpoint>

The figure presents four different Newman conformations of SMILES [CH2:1]=[CH:2]([CH3:5])[CH2:3][CH3:4] rotated along the `[C:2][C:3]` bond, where in $\mathbf{A}$ `[C:1]` overlaps with `[C:4]`, in $\mathbf{B}$ `[C:1]` overlaps with the hydrogen on `[C:3]`, in $\mathbf{C}$ `[C:5]` overlaps with the hydrogen on `[C:3]`, and in $\mathbf{D}$ `[C:5]` overlaps with `[C:4]`.

<Checkpoint>[In $\mathbf{A}$ `[C:1]` overlaps with `[C:4]`, in $\mathbf{B}$ `[C:1]` overlaps with the hydrogen on `[C:3]`, in $\mathbf{C}$ `[C:5]` overlaps with the hydrogen on `[C:3]`, and in $\mathbf{D}$ `[C:5]` overlaps with `[C:4]`](1)</Checkpoint>

Furthermore, it is necessary to examine the relative magnitude of the steric hindrance effects of each conformation to determine the relative energy levels. This mainly involves A(1,3) strain and eclipsing steric hindrance, where the A(1,3) strain of the alkene group and the methyl group has less impact than the eclipsing steric hindrance of the two methyl groups.

<Checkpoint>[The A(1,3) strain of the alkene group and the methyl group has less impact than the eclipsing steric hindrance of the two methyl groups](0.5)</Checkpoint>

It is not difficult to find that, for $\mathbf{B}$, the two methyl groups in the front and back are in a staggered configuration and there is no eclipsing steric hindrance. The A(1,3) strain is only two hydrogens, which is the most stable conformation. Therefore, $\mathbf{B}$ should correspond to energy $1$.

<Checkpoint>[$\mathbf{B}$ corresponds to energy $1$](1)</Checkpoint>

For $\mathbf{A}$, there is A(1,3) strain between the methyl group and the alkene group, and there is no staggered conformation between the two methyl groups, so its energy will be slightly higher than $\mathbf{B}$, corresponding to energy $2$.

<Checkpoint>[$\mathbf{A}$ corresponds to energy $2$](1)</Checkpoint>

For $\mathbf{C/D}$, there is eclipsing strain between methyl and hydrogen or methyl, so its energy is higher than $\mathbf{A/B}$, and the eclipsing strain between methyl and methyl is greater. Therefore, the energy of $\mathbf{D}$ is higher than $\mathbf{C}$, $\mathbf{D}$ corresponds to energy $4$, and $\mathbf{C}$ corresponds to energy $3$.

<Checkpoint>[$\mathbf{C}$ corresponds to energy $3$](1)</Checkpoint>

<Checkpoint>[$\mathbf{D}$ corresponds to energy $4$](1)</Checkpoint>

Finally, select option E.
首先，需要确定有机分子 $\mathbf{X}$ 的结构，根据图象可得，这是一个纽曼式投影图，前面的碳连接有一个烯基 $\mathrm{=CH_2}$ 与一个甲基 $\mathrm{CH_3}$ 取代基，而后面的碳连接有一个甲基 $\mathrm{CH_3}$ 取代基与两个氢 $\mathrm{H}$，因此，分子 $\mathbf{X}$ 的SMILES可以表示为 `[CH2:1]=[CH:2]([CH3:5])[CH2:3][CH3:4]`，即2-甲基-1-丁烯。

<Checkpoint>[分子 $\mathbf{X}$ 的SMILES可以表示为 `[CH2:1]=[CH:2]([CH3:5])[CH2:3][CH3:4]`，即2-甲基-1-丁烯](1)</Checkpoint>

图中呈现了SMILES为[CH2:1]=[CH:2]([CH3:5])[CH2:3][CH3:4]的四种沿 `[C:2][C:3]` 键旋转的不同纽曼式构象，其中 $\mathbf{A}$ 中 `[C:1]` 与 `[C:4]` 重叠， $\mathbf{B}$ 中 `[C:1]` 与 `[C:3]` 上的氢重叠， $\mathbf{C}$ 中 `[C:5]` 与 `[C:3]` 上的氢重叠，  $\mathbf{D}$ 中 `[C:5]` 与` [C:4]` 重叠。

<Checkpoint>[$\mathbf{A}$ 中 `[C:1]` 与 `[C:4]` 重叠， $\mathbf{B}$ 中 `[C:1]` 与 `[C:3]` 上的氢重叠， $\mathbf{C}$ 中 `[C:5]` 与 `[C:3]` 上的氢重叠，  $\mathbf{D}$ 中 `[C:5]` 与` [C:4]` 重叠](1)</Checkpoint>

进一步，需要考察各个构象的位阻效应的相对大小，来决定能量的相对高低。其中主要涉及A(1,3)张力与重叠式的位阻，其中烯基与甲基的A(1,3)张力影响小于两个甲基的重叠式位阻的影响。

<Checkpoint>[烯基与甲基的A(1,3)张力影响小于两个甲基的重叠式位阻的影响](0.5)</Checkpoint>

不难发现，对于 $\mathbf{B}$ 而言，其前后两个甲基处于交叉式、不存在重叠式的位阻，而A(1,3)张力仅为两个氢，这是最稳定的构象。因此，$\mathbf{B}$ 应该对应能量 $1$。

<Checkpoint>[$\mathbf{B}$ 对应能量 $1$](1)</Checkpoint>

对于 $\mathbf{A}$ 而言，存在甲基与烯基的A(1,3)张力，而不存在两个甲基的交叉式构象，因此其能量会略高于 $\mathbf{B}$，对应能量 $2$。

<Checkpoint>[$\mathbf{A}$ 对应能量 $2$](1)</Checkpoint>

对于 $\mathbf{C/D}$ 而言，存在甲基与氢或甲基的重叠式张力，因此其能量高于 $\mathbf{A/B}$，而又以甲基与甲基的重叠式张力更大，因此，$\mathbf{D}$ 的能量高于 $\mathbf{C}$，$\mathbf{D}$ 对应能量 $4$，$\mathbf{C}$ 对应能量 $3$。

<Checkpoint>[$\mathbf{C}$ 对应能量 $3$](1)</Checkpoint>

<Checkpoint>[$\mathbf{D}$ 对应能量 $4$](1)</Checkpoint>

最终，选择选项 E。