Question

Draw a line-bond structure for buta-1,3-diene, \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\); indicate the hybridization of each carbon; and predict the value of each bond angle.

Short answer

The line-bond structure of buta-1,3-diene features alternating double bonds. All carbons are \(sp^2\) hybridized with bond angles of approximately \(120^\circ\).

Step-by-step solution

Identify the Structure

Buta-1,3-diene is an aliphatic hydrocarbon with the formula \( \mathrm{H}_{2}C=CH-CH=CH_{2} \). It has four carbon atoms with alternating double bonds starting from the first carbon. This makes it a conjugated diene.

Draw the Line-Bond Structure

In the line-bond structure of buta-1,3-diene, we begin by writing four carbon atoms in a linear arrangement. Between the first and second carbon, draw a double bond, and between the second and third carbon, draw a single bond, followed by a double bond between the third and fourth carbon. Each terminal carbon (first and fourth) is bonded to two hydrogen atoms, while each internal carbon (second and third) is bonded to one hydrogen atom. The structure is as follows: H2C=CH-CH=CH2.

Determine Hybridization

The hybridization of a carbon atom is based on the number of atoms it is directly bonded to and the type of bonds. In buta-1,3-diene:- The first carbon (\( \mathrm{H}_{2}C\)) and the fourth carbon (\( \mathrm{CH}_{2}\)) are \(sp^2\) hybridized because each is involved in one double bond and is attached to two other atoms (one carbon and two hydrogens).- The second and third carbons (\( \mathrm{CH}\)) are also \(sp^2\) hybridized for the same reason: each is involved in one double bond and bonded to two other atoms (two carbons and one hydrogen).

Predict Bond Angles

For \( sp^2 \) hybridization, the bond angles are approximately \( 120^\circ \). Therefore, each bond angle in buta-1,3-diene, formed between consecutive bonds (carbon-carbon and hydrogen-carbon) is approximately \( 120^\circ \) because the geometry around each \( sp^2 \) carbon is trigonal planar.

Key concepts

Line-Bond Structure of Buta-1,3-diene

The line-bond structure of buta-1,3-diene is a detailed representation showing how atoms are connected. This structure uses lines to depict bonds between atoms — single lines for single bonds and double lines for double bonds.
Buta-1,3-diene consists of four carbon atoms and alternating double bonds, which are crucial for its unique chemical properties.
Here's how you draw it: Start by aligning four carbon atoms in a horizontal line. Draw a double bond between the first and second carbons, a single bond between the second and third carbons, and another double bond between the third and fourth carbons.
The terminal carbons at each end are also connected to two hydrogen atoms each, while the inner carbons have one hydrogen atom each.

• First Carbon: H2C, double-bonded to the second carbon.
• Second Carbon: CH, single-bonded to both adjacent carbons.
• Third Carbon: CH, single-bonded to both adjacent carbons.
• Fourth Carbon: CH2, double-bonded to the third carbon.

This structured approach helps in visualizing the molecule's framework, providing insights into its reactivity and interactions.

Carbon Hybridization in Buta-1,3-diene

Understanding carbon hybridization is key to predicting how molecules like buta-1,3-diene behave. In buta-1,3-diene, each carbon atom is involved in a particular bond type due to its hybridization state.
Each carbon is hybridized to optimize overlap with adjacent atoms, maximizing the molecule's stability and reducing electron repulsion.
In this compound:

• Every carbon atom is \( sp^2 \) hybridized.
• The first and fourth carbon atoms participate in one double bond and attach to two hydrogen atoms each.
• The second and third carbons are involved in double bonds with their neighbors and a single hydrogen.

The \( sp^2 \) hybridization means each carbon atom forms three sigma bonds: two with carbon and one with hydrogen. The remaining p orbital on each carbon can then form double bonds (pi bonds) with adjacent carbons.
This hybridization is crucial for the flat, planar shape of buta-1,3-diene and its conjugated system.

Bond Angles in Buta-1,3-diene

In buta-1,3-diene, the bond angles are a direct consequence of the carbon's hybridization state. Given that every carbon is \( sp^2 \) hybridized, you can predict that the geometry around these carbons will be trigonal planar.
This specific geometry allows for bond angles of about \( 120^{\circ} \). Such angles facilitate the optimal overlap of orbitals and maintain the stability of the planar structure.
To break it down further:

• The angle between the hydrogen atom and the carbon it is attached to, relative to the adjacent carbons, is generally \( 120^{\circ} \).
• The carbon-carbon-carbon (C-C-C) angles within the molecule are close to \( 120^{\circ} \) as well.

These consistent angles ensure the molecule remains flat, a necessary feature for proper \( \pi\) orbital overlap.

Conjugated Diene Characteristics in Buta-1,3-diene

Buta-1,3-diene is a prime example of a conjugated diene, which means it has alternating single and double bonds. This alternation allows for significant resonance, a key concept in organic chemistry that leads to increased stability.
The structure of buta-1,3-diene enables the \( \pi\) electrons to be delocalized across multiple atoms. This electron delocalization helps lower the overall energy of the molecule, granting it greater stability than regular dienes, which lack such conjugation.
Benefits of this conjugated system include:

• Enhanced stability compared to non-conjugated dienes.
• The ability to participate in unique reactions involving \( \pi\) electrons.
• Contributed to extended flexibility in potential chemical reactions, such as Diels-Alder reactions.

This conjugation is crucial, not only for understanding reactivity but also for appreciating the physical and chemical properties of buta-1,3-diene.