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Draw an atomic-orbital picture of 1,3 -dichloropropadiene, \(\mathrm{ClCH}=\mathrm{C}=\mathrm{CHCl}\) Examine the structure carefully and predict how many stereoisomers are possible for this structure. What kind of stereoisomers are these?

Short Answer

Expert verified
1,3-Dichloropropadiene has 4 \(E/Z\) stereoisomers.

Step by step solution

01

Understand the Structure

1,3-Dichloropropadiene has the molecular formula \( \mathrm{C}_3\mathrm{H}_2\mathrm{Cl}_2 \). It consists of a linear chain of three carbon atoms with two chlorine atoms attached at the terminal carbon atoms and two double bonds between carbons: \(\mathrm{ClCH} = \mathrm{C} = \mathrm{CHCl}\).
02

Visualizing Atomic Orbitals

Each carbon in the structure is \(sp\)-hybridized due to the presence of two adjacent double bonds, allowing for the formation of \(\pi\) bonds. The terminal carbons have \(sp^2\) orbitals for \(\sigma\)-bonding with hydrogen and the adjacent carbon. The central carbon uses two \(p\) orbitals to form the \(\pi\) bonds with the terminal carbons.
03

Identifying Stereocenters

Examine the structure for stereoisomers, which are possible where there are double bonds that can have \(cis/trans\) (or \(E/Z\)) configurations. In this molecule, the central \(\mathrm{C} = \mathrm{C} = \mathrm{C}\) linkage is linear, thus it does not provide stereochemical diversity, but each \(C = C\) part can allow for stereoisomerism.
04

Predicting Stereoisomers

The \(C_1\) and \(C_3\) double bonds each can exhibit \(E/Z\) isomerism (due to different groups on each carbon atom). Hence, two double bonds can lead to \(2^2 = 4\) stereoisomers: \(EE\), \(EZ\), \(ZE\), \(ZZ\).
05

Identifying the Type of Stereoisomers

These are \(E/Z\) stereoisomers, which are types of geometric isomers resulting from restricted rotation around the double bonds and differing in spatial arrangement of substituents around the double bond.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Atomic orbital theory
Atomic orbital theory is fundamental for understanding how atoms combine to form molecules. It involves the concept of orbitals, which are regions around the nucleus where electrons are likely to be found. Electrons occupy these orbitals based on their energy levels. In chemistry, orbitals are often considered as clouds with different shapes, such as spherical for s orbitals, and dumbbell-shaped for p orbitals. These shapes determine how atoms bond with one another.

When considering a molecule like 1,3-dichloropropadiene, one must consider the overlap of atomic orbitals. Each carbon atom in the molecule is bonded to hydrogen or chlorine atoms, and double bonds link the carbons. The shape and orientation of the atomic orbitals influence the molecule's geometry, contributing to its chemical properties and reactivity. This theory helps visualize the molecule's structure in terms of electron density and the formation of bonds within these regions.
Hybridization in organic molecules
Hybridization is a concept that explains the mixing of atomic orbitals to form new hybrid orbitals, which are suitable for pairing electrons to form chemical bonds.

In 1,3-dichloropropadiene, the carbon atoms exhibit different types of hybridization depending on their bonding. For example, terminal carbon atoms are often thought to be sp-hybridized in such chains due to their involvement in multiple bonds, meaning they use one s and one p orbital for bonding. This configuration allows for linearity and a specific angle between bonds.

Understanding hybridization is crucial in predicting molecule shapes and bond angles. It also explains the distribution of electrons in versatile organic structures and helps visualize how molecules like 1,3-dichloropropadiene are constructed, contributing to their chemical and physical properties.
Geometric isomerism
Geometric isomerism, also known as cis-trans isomerism or E/Z isomerism, arises due to the restricted rotation around double bonds. This restriction occurs because of the \(\pi \) bonds present in the structure.

In the context of 1,3-dichloropropadiene, each carbon-carbon double bond can adopt different geometries based on the spatial arrangement of its substituents, such as hydrogen and chlorine, across the double bonds. The potential for such isomers in each double bond results in multiple stereoisomers.

Geometric isomers are categorized into forms like "E" (entgegen, German for "opposite") and "Z" (zusammen, German for "together") depending on the priority of substituent groups. In 1,3-dichloropropadiene, these isomers fundamentally affect the molecule's physical properties and reactivity. Understanding and predicting such isomerism is critical in stereochemistry and organic synthesis.

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Most popular questions from this chapter

If methanal, \(\mathrm{H}_{2} \mathrm{C}=\mathrm{O}\), were protonated to give \(\mathrm{H}_{2} \mathrm{C}=\mathrm{OH}\), would you expect the \(\mathrm{C}=\mathrm{O}-\mathrm{H}\) angle to be closer to \(180^{\circ}, 120^{\circ}, 109^{\circ}\), or \(90^{\circ}\) ? Explain.

Examine the following structures and predict the most likely geometry, using concepts of orbital hybridization. State whether the molecule should be planar or nonplanar, and list the approximate values expected for the bond angles. a. \(\mathrm{SiCl}_{4}\) b. \(\mathrm{HCOO}^{\ominus}\) c. \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}\) d. \(\mathrm{F}_{2} \mathrm{C}=\mathrm{C}=\mathrm{CF}_{2}\) e. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{O}^{\oplus}\)

Suggest why the molecule \(\mathrm{Be}_{2}\) apparently is so unstable that it has not been observed. Explain why Be with an outer-shell electronic configuration of \((2 s)^{2}\) forms \(\mathrm{BeCl}_{2}\), whereas He with the configuration \((1 s)^{2}\) does not form \(\mathrm{HeCl}_{2}\).

Write electron-pair structures including bonding and unshared pairs for each of the following compounds. Predict the preferred shape of the molecule as linear, angular, planar and triangular, tetrahedral, or pyramidal. a. \(\oplus \mathrm{NO}_{2}\) b. \(\mathrm{CS}_{2}\) c. \(\mathrm{O}=\mathrm{C}=\mathrm{C}=\mathrm{O}\) d. \(\mathrm{H}_{2} \mathrm{C}=\mathrm{NH}\) e. \(\mathrm{HN}=\mathrm{NH}\) f. \(\mathrm{CH}_{3}\) g. ClNO h. \(\mathrm{NH}_{2}\) i. \(\mathrm{BH}_{4}^{-}\)

Write electron-pair structures for each of the following. Include both bonding and nonbonding pairs and predict the preferred shape of the molecule or ion as linear, triangular (planar), angular, tetrahedral, or pyramidal. a. \(\mathrm{CO}_{2}\) b. \(\mathrm{N} \equiv \mathrm{C}-\mathrm{O}^{\ominus}\) c. \(\mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}\) d. \(\mathrm{CH}_{3}^{+}\) e. \(\mathrm{F}_{2} \mathrm{C}=\mathrm{CH}_{2}\) f. \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{N}\) g. \(\mathrm{SiF}_{4}\) h. \(\mathrm{HCOOH}\) i. \(\mathrm{H}_{3} \mathrm{O}^{\ominus}\) j. \(\mathrm{CH}_{3} \mathrm{SH}\) \(\mathbf{k} . * \mathrm{SO}_{3}\)

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