Identify Substituents in Newman Projection of 2,2-Dimethylbutane

To determine the substituents $\mathbf{X}$ and $\mathbf{Y}$ in the given Newman projection of 2,2-dimethylbutane, we analyze the molecular structure and the projection diagram.

1. Molecular Structure of 2,2-Dimethylbutane

The IUPAC structure and numbering of 2,2-dimethylbutane is:
$\overset{1}{\text{C}}\text{H}_3 - \overset{2}{\text{C}}(\text{CH}_3)_2 - \overset{3}{\text{C}}\text{H}_2 - \overset{4}{\text{C}}\text{H}_3$

2. Analysis of the Given Newman Projection Diagram

From the visual representation:

• Front Carbon (represented by the central vertex where three bonds meet):
  It has two hydrogen atoms ($-\text{H}$) and one substituent $\mathbf{X}$ attached to it. Thus, its substituents are:
  $(\mathbf{X}, \text{H}, \text{H})$
• Back Carbon (represented by the circle):
  It has two methyl groups ($-\text{CH}_3$) and one substituent $\mathbf{Y}$ attached to it. Thus, its substituents are:
  $(\text{CH}_3, \text{CH}_3, \mathbf{Y})$

Since the back carbon has two methyl groups attached to it, it must correspond to $\text{C}_2$ of 2,2-dimethylbutane. The front carbon, being directly bonded to $\text{C}_2$, can be either $\text{C}_1$ or $\text{C}_3$.

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3. Case Analysis

Case 1: Viewing along the $\text{C}_1-\text{C}_2$ bond (Front Carbon = $\text{C}_1$, Back Carbon = $\text{C}_2$)

• Front Carbon ($\text{C}_1$): Being a terminal methyl group, $\text{C}_1$ is bonded to three hydrogen atoms ($-\text{H}, -\text{H}, -\text{H}$).
  Comparing with $(\mathbf{X}, \text{H}, \text{H})$, we get:
  $\mathbf{X} = \text{H}$

• Back Carbon ($\text{C}_2$): Apart from the $\text{C}_1-\text{C}_2$ bond, $\text{C}_2$ is bonded to two methyl groups ($-\text{CH}_3, -\text{CH}_3$) and an ethyl group ($-\text{CH}_2\text{CH}_3 = -\text{C}_2\text{H}_5$).
  Comparing with $(\text{CH}_3, \text{CH}_3, \mathbf{Y})$, we get:
  $\mathbf{Y} = \text{C}_2\text{H}_5$

Thus, $\mathbf{X} = \text{H}$ and $\mathbf{Y} = \text{C}_2\text{H}_5$, which corresponds to Option B.

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Case 2: Viewing along the $\text{C}_3-\text{C}_2$ bond (Front Carbon = $\text{C}_3$, Back Carbon = $\text{C}_2$)

• Front Carbon ($\text{C}_3$): $\text{C}_3$ is bonded to two hydrogen atoms and one methyl group ($-\text{C}_4\text{H}_3$).
  Comparing with $(\mathbf{X}, \text{H}, \text{H})$, we get:
  $\mathbf{X} = \text{CH}_3$

• Back Carbon ($\text{C}_2$): Apart from the $\text{C}_2-\text{C}_3$ bond, $\text{C}_2$ is bonded to three methyl groups (two attached to $\text{C}_2$ and one from $\text{C}_1$).
  Comparing with $(\text{CH}_3, \text{CH}_3, \mathbf{Y})$, we get:
  $\mathbf{Y} = \text{CH}_3$

Thus, $\mathbf{X} = \text{CH}_3$ and $\mathbf{Y} = \text{CH}_3$, which corresponds to Option D.

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Conclusion

Therefore, $\mathbf{X}$ and $\mathbf{Y}$ can respectively be $\text{H}$ and $\text{C}_2\text{H}_5$ or $\text{CH}_3$ and $\text{CH}_3$.

Correct Options: B and D