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(3pts) with your model set construct butane, c₄h₁₀. draw the expanded, …

Question

(3pts) with your model set construct butane, c₄h₁₀. draw the expanded, condensed and line-angle formulas. upload an image of your drawings here.

Explanation:

Brief Explanations
  1. Expanded Formula: Show all atoms and bonds. For butane ($C_4H_{10}$), the expanded structure has 4 carbon atoms in a chain, each bonded to hydrogen atoms. The first carbon (C₁) is bonded to 3 H atoms and 1 C (C₂). C₂ is bonded to 2 H atoms and 2 C atoms (C₁ and C₃). C₃ is bonded to 2 H atoms and 2 C atoms (C₂ and C₄). C₄ is bonded to 3 H atoms and 1 C (C₃). So the expanded formula is: $H - \overset{H}{\overset{|}{C}} - \overset{H}{\overset{|}{C}} - \overset{H}{\overset{|}{C}} - \overset{H}{\overset{|}{C}} - H$ (with each internal C having 2 H atoms, so more accurately: $H_3C - CH_2 - CH_2 - CH_3$ expanded would be showing each bond: $H - C(H)(H) - C(H)(H) - C(H)(H) - C(H)(H)(H)$? Wait, no, butane is $CH_3CH_2CH_2CH_3$. So expanded: each C - C bond is single, and each C has enough H to satisfy valency (4 bonds). So:

$H\quad H\quad H\quad H$

$|\quad |\quad |\quad |$

$H - C - C - C - C - H$

$|\quad |\quad |\quad |$

$H\quad H\quad H\quad H$ (Wait, no, the middle carbons (C₂ and C₃) have 2 H each, and the end carbons (C₁ and C₄) have 3 H each. So correct expanded:

$H_3C - CH_2 - CH_2 - CH_3$ expanded is:

$H\quad H\quad H$

$|\quad |\quad |$

$H - C - C - C - C - H$

$\quad |\quad |\quad |$

$\quad H\quad H\quad H$

Wait, no, let's do it properly. Carbon has 4 bonds. So C₁: bonded to 3 H and 1 C (C₂). C₂: bonded to 1 C (C₁), 1 C (C₃), and 2 H. C₃: bonded to 1 C (C₂), 1 C (C₄), and 2 H. C₄: bonded to 1 C (C₃) and 3 H. So the expanded structure is:

$H\quad H\quad H$

$|\quad |\quad |$

$H - C - C - C - C - H$

$\quad |\quad |\quad |$

$\quad H\quad H\quad H$

Wait, no, the first C (leftmost) has 3 H and 1 C (down? No, in a straight chain, it's a horizontal chain. So:

$H\quad H\quad H$

$|\quad |\quad |$

$H - C - C - C - C - H$

$\quad |\quad |\quad |$

$\quad H\quad H\quad H$

Wait, the middle two carbons (C₂ and C₃) have 2 H each, so between C₁ - C₂: C₁ has 3 H, C₂ has 2 H (and bonds to C₁ and C₃). C₃ has 2 H (bonds to C₂ and C₄), C₄ has 3 H. So the expanded formula is:

$H_3C - CH_2 - CH_2 - CH_3$ written with all bonds:

$H - C(H)(H) - C(H)(H) - C(H)(H) - C(H)(H)(H)$? No, that's condensed with parentheses. Wait, expanded formula shows each atom and each bond. So:

$H\quad H\quad H$

$|\quad |\quad |$

$H - C - C - C - C - H$

$\quad |\quad |\quad |$

$\quad H\quad H\quad H$

Yes, that's the expanded structure: each carbon is connected in a chain, with the end carbons having 3 H each and the middle carbons having 2 H each.

  1. Condensed Formula: Groups the atoms. For butane, it's $CH_3CH_2CH_2CH_3$ or $C_4H_{10}$ (molecular formula), but condensed structural formula is $CH_3(CH_2)_2CH_3$ or $CH_3CH_2CH_2CH_3$.
  1. Line - Angle Formula (Skeletal Formula): In line - angle formulas, we represent carbon atoms as the vertices of the lines, and hydrogen atoms attached to carbons are not shown (since each carbon is assumed to have enough H to satisfy valency, i.e., 4 bonds). For butane, which is a straight - chain alkane, the line - angle formula is a straight line of 4 carbon atoms (represented as 4 vertices in a straight line). So it's a zig - zag? No, butane is a straight chain, so the line - angle formula is four carbon atoms in a straight line (since all C - C bonds are single, and in skeletal formula, each "corner" or end is a carbon. So for 4 carbons, it's a straight line with 4 segments? Wait, no: the line - angle formula for a straight - chain alkane with n carbons is a straight line of (n - 1) segments, with n carbon atoms (the ends and the joints). So for butane ($n = 4$), the line - angle formula is:

$
ule{2c…

Answer:

To solve this, we construct the formulas for butane ($C_4H_{10}$) as follows:

1. Expanded Formula

Draw 4 carbon atoms in a horizontal chain, each connected by single bonds. Attach 3 H atoms to the first and fourth carbons, and 2 H atoms to the second and third carbons (show all bonds):

$$\begin{matrix} \ce{H} & \ce{H} & \ce{H} \\ | & | & | \\ \ce{H - C - C - C - C - H} \\ & | & | & | \\ & \ce{H} & \ce{H} & \ce{H} \\ \end{matrix}$$
2. Condensed Formula

Group atoms by carbon chains:
$\boldsymbol{\ce{CH3CH2CH2CH3}}$ (or $\ce{CH3(CH2)2CH3}$)

3. Line - Angle (Skeletal) Formula

Draw a straight line (4 carbon “nodes” in a row, no H atoms shown, as valency is implied):
$\boldsymbol{\ce{C - C - C - C}}$ (represented as a horizontal line with 4 carbon positions, e.g., $
ule{2cm}{0.1pt}$ with 4 endpoints).

After drawing these, upload an image of your hand - drawn or digital renderings of these formulas.