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A gaseous compound of formula \(\mathrm{C}_{2} \mathrm{H}_{4}\) reacts with liquid bromine \(\left(\mathrm{Br}_{2}\right)\) to give a single \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Br}_{2}\) compound. The \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Br}_{2}\) so formed gives only one \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Br}_{3}\) substitution product. Deduce the structure of \(\mathrm{C}_{2} \mathrm{H}_{4}\) and the bromo compounds derived from it. (This was a key problem for the early organic chemists.)

Short Answer

Expert verified
Ethylene is \( \text{H}_2\text{C}=\text{CH}_2 \), forms \( \text{BrCH}_2\text{CH}_2\text{Br} \), and with more bromine \( \text{Br}_2\text{CHCH}_2\text{Br} \).

Step by step solution

01

Analyze the Ethylene Structure

Start by considering the structure of ethylene, \( \mathrm{C}_2\mathrm{H}_4 \). Ethylene, also known as ethene, is a hydrocarbon that contains a carbon-carbon double bond and is represented as \( \text{H}_2\text{C}=\text{CH}_2 \). This double bond is capable of reacting with bromine to form addition products.
02

Understand the Reaction Type

Ethylene reacts with liquid bromine (\( \mathrm{Br}_2 \)) in an addition reaction. The double bond between the carbon atoms opens up to form a single bond, allowing each carbon to form a new covalent bond with one bromine atom.
03

Deduce the Structure of the Dibromo Compound

From the reaction of ethylene with bromine, the compound that forms is \( \mathrm{C}_2\mathrm{H}_4\mathrm{Br}_2 \). Given the symmetrical nature of the bromine addition, the 1,2-dibromoethane structure forms: \( \text{BrCH}_2\text{CH}_2\text{Br} \). This product confirms the structure as the addition occurs on either side of the double bond.
04

Predict Further Bromination

Now consider the further bromination of \( \mathrm{C}_2\mathrm{H}_4\mathrm{Br}_2 \) to form \( \mathrm{C}_2\mathrm{H}_3\mathrm{Br}_3 \). The most likely substitution site for additional bromine is one of the hydrogen atoms on either carbon, leading to a compound such as \( \text{Br}_2\text{CHCH}_2\text{Br} \).
05

Conclude the Structure

Given that the final product \( \mathrm{C}_2\mathrm{H}_3\mathrm{Br}_3 \) results in only one substitution product, the initial compound must be symmetrical such that substitution leads to an indiscernible variation. Thus, the structure for the followed substitution product is consistent with bromine addition to any ethylene hydrogen position resulting in \( \text{Br}_2\text{CHCH}_2\text{Br} \).

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Key Concepts

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

Reaction Mechanism
When dealing with organic chemistry, understanding **reaction mechanisms** is crucial for predicting and explaining chemical reactions. A reaction mechanism refers to the step-by-step sequence through which reactants transform into products. In the case of ethylene (\(C_2H_4\)), when it reacts with bromine (\(Br_2\)), the mechanism follows what is known as an **addition reaction**. This consists of the double bond in ethylene opening up to accommodate new atoms. Here, bromine atoms are added across the carbon atoms of the initially present double bond.
  • First, as the ethylene approaches the bromine molecule, the electron-rich double bond induces a dipole in the \(Br_2\) molecule.
  • This results in the formation of cyclic bromonium ion and the opening of the double bond.
  • Then, the bromonium ion is attacked by another bromide ion, leading to the final dibrominated product.

Ultimately, understanding this mechanism allows for a deeper insight into how organic reactions proceed at the molecular level.
Chemical Structure
The **chemical structure** of a molecule provides vital information about its physical and chemical properties. Ethylene (\(C_2H_4\)) is a small molecule with a simple structure, composed of two carbon atoms double-bonded to each other and each bonded to two hydrogen atoms. The structure can be represented as \( \text{H}_2\text{C} = \text{CH}_2 \), highlighting its planar geometry.
  • This double bond is the reactive site that participates in various chemical reactions, particularly addition reactions.
  • The symmetry within the ethylene molecule also has implications on the type of products formed in reactions.
  • When bromine is added, each carbon atom of the former double bond gains a bromine atom, forming 1,2-dibromoethane: \( \text{BrCH}_2\text{CH}_2\text{Br} \).

The structure of this addition product is important to understand as it features saturated carbon atoms, which changes how the molecule can further react, such as in subsequent bromination.
Addition Reaction
An **addition reaction** is a common type of chemical reaction in organic chemistry where two or more molecules combine to form a larger molecule. These reactions usually involve compounds with unsaturated bonds, like triple or double bonds, that can 'open up' to allow new atoms to bond. In the reaction between ethylene and bromine, the \(C=C\) double bond of ethylene opens to form two new \(C-Br\) bonds, transforming into dibromoethane.
  • This process effectively saturates the carbon atoms, moving from a double bond to single bonds due to the new bromine-induced connections.
  • It follows the general pattern of an electrophilic addition, where the electron-rich \(C=C\) double bond donates electrons to the electron-poor bromine, leading to the symmetrical addition across the bond.

This addition is crucial as it introduces functionality into the molecule, altering its reactivity and opening pathways for further chemical transformations.
Bromo Compounds
**Bromo compounds** are organic compounds containing carbon bonded to bromine. They are a fascinating group of chemicals often used in synthetic chemistry due to bromine's reactivity. When ethylene reacts with bromine, the resulting molecule, dibromoethane, exemplifies a class of these compounds.
  • In particular, bromo compounds like dibromoethane are significant because of their potential in further chemical modifications, such as additional bromination.
  • The versatility of bromo compounds stems from the relatively stable yet reactive nature of the \(C-Br\) bond.
  • Dibromoethane can undergo a series of transformations, including further substitution reactions, which can introduce more bromine atoms, resulting in three bromines in the case of the given problem.

Thus, bromo compounds serve as important intermediates in many synthetic pathways due to their accessibility and the richness they provide to the molecular framework.

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Most popular questions from this chapter

(This problem is in the nature of review of elementary inorganic chemistry and may require reference to a general chemistry book.) Write Lewis structures for each of the following compounds. Use distinct, correctly placed dots for the electrons. Mark all atoms that are not neutral with charges of the proper sign. a. ammonia, \(\mathrm{NH}_{3}\) b. ammonium bromide, \(\mathrm{NH}_{4} \mathrm{Br}\) c. hydrogen cyanide, \(\mathrm{HCN}\) d. ozone \(\left(\angle \mathrm{O}-\mathrm{O}-\mathrm{O}=120^{\circ}\right)\) e. carbon dioxide, \(\mathrm{CO}_{2}\) f. hydrogen peroxide, \(\mathrm{HOOH}\) g. hydroxylamine, \(\mathrm{HONH}_{2}\) h. nitric acid, \(\mathrm{HNO}_{3}\) i. hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\) j. boron trifluoride, \(\mathrm{BF}_{3}\)

Why is the boiling point of water \(\left(100^{\circ}\right)\) substantially higher than the boiling point of methane \(\left(-161^{\circ}\right)\) ?

How many different isomers are there of \(\mathrm{CH}_{2} \mathrm{Br}_{4}\) ? (Assume free-rotating tetrahedral carbon and univalent hydrogen and bromine.) How could one determine which of these isomers is which by the substitution method?

An acid \((\mathrm{HA})\) can be defined as a substance that donates a proton to a base, for example water. The protondonation reaction usually is an equilibrium reaction and is written as $$ \mathrm{H}: \mathrm{A}+\mathrm{H}: \ddot{\mathrm{O}}: \mathrm{H} \rightleftarrows \mathrm{H}: \stackrel{\mathrm{H}}{\mathrm{O}} \cdot \stackrel{\oplus}{\mathrm{H}}+: \AA^{\ominus} $$ Predict which member of each of the following pairs of compounds would be the stronger acid. Give your reasons. a. \(\mathrm{LiH}, \mathrm{HF}\) b. \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{O}\) c. \(\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{H}_{2} \mathrm{O}\) d. \(\mathrm{CH}_{4}, \mathrm{CF}_{3} \mathrm{H}\)

There are a large number of known isomers of \(\mathrm{C}_{5} \mathrm{H}_{10}\), and some of these are typically unsaturated, like ethene, while others are saturated, like ethane. One of the saturated isomers on bromine substitution gives only one compound of formula \(\mathrm{C}_{5} \mathrm{H}_{9} \mathrm{Br}\). Work out a structure for this isomer of \(\mathrm{C}_{5} \mathrm{H}_{10}\) and its monobromo substitution product.

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