Problem 7
Question
Treatment of 2 -(4-hydroxyphenyl)ethyl bromide with basic alumina produces a white solid, \(\mathrm{mp}, 40-43^{\circ} \mathrm{C}\); IR \(1640 \mathrm{~cm}^{-1}\); UV \(_{\max } 282 \mathrm{~nm}\) in \(\mathrm{H}_{2} \mathrm{O}\); NMR two singlets of equal intensity at \(1.69\) and \(6.44 \mathrm{ppm}\) from TMS; anal: C \(79.98 \% \mathrm{H}\), \(6.71 \%\). Suggest a possible structure for this product and explain how it could be formed.
Step-by-Step Solution
Verified Answer
The product is likely styrene (vinyl benzene), formed via dehydrohalogenation.
1Step 1: Understanding the Task
We are given a compound, 2-(4-hydroxyphenyl)ethyl bromide, and information about a white solid product formed when this compound is treated with basic alumina. We need to deduce the possible structure of this product using the provided spectral data and elemental analysis.
2Step 2: Analyzing Spectral Data - IR Spectroscopy
The IR spectrum of the product shows a significant peak around 1640 cm⁻¹, which typically indicates the presence of a carbon-carbon double bond (C=C). This suggests that the product might be unsaturated, likely forming an alkene.
3Step 3: Analyzing Spectral Data - UV Spectroscopy
The UV spectrum shows a maximum at 282 nm, indicative of a conjugated system, often a phenyl ring with possible extended conjugation. This provides evidence that the aromatic ring is intact, and other conjugated systems might be present.
4Step 4: Analyzing NMR Data - 1H NMR
The NMR spectrum shows two singlets at 1.69 ppm and 6.44 ppm. The singlet at around 6.44 ppm likely belongs to protons on an unsaturated carbon (possibly vinyl or aryl protons), while the singlet at 1.69 ppm could correspond to alkyl protons, possibly indicating a methyl or similarly substituted group.
5Step 5: Examining Elemental Analysis
The elemental composition is about 79.98% C and 6.71% H. We compare these percentages with the expected percentages for a hydrocarbon with one oxygen (based on 100% total) to imply limited oxygen presence, suggesting a simple structure without multiple functional groups like additional -OH, -COOH, or -CHO groups.
6Step 6: Proposing a Reaction Pathway
Given that basic alumina (a dehydrating agent) and the observations, it's plausible the reaction involved a dehydrohalogenation to form a double bond. From 2-(4-hydroxyphenyl)ethyl bromide, elimination of HBr can lead to the formation of styrene by removing HBr from adjacent positions.
7Step 7: Determining Possible Product Structure
Based on all the above, the likely product of the reaction is styrene (vinyl benzene), characterized by the alkene double bond peak in the IR spectrum, the UV maximum due to its conjugated system, and the NMR signals corresponding to its hydrogen environment.
Key Concepts
IR SpectroscopyNMR SpectroscopyUV SpectroscopyElemental AnalysisDehydrohalogenation Reaction
IR Spectroscopy
IR Spectroscopy is a technique used to identify functional groups in a molecule based on the absorption of infrared light, which causes molecular vibrations. Each type of vibration corresponds to a specific bond within the molecule.
In the original exercise, the IR spectrum reveals a peak at 1640 cm⁻¹. Typically, this indicates the presence of a carbon-carbon double bond. Such double bonds absorb infrared light at this frequency, causing the characteristic vibration noted in the spectrum.
This information is essential for the proposed structure of styrene in the original exercise, as alkenes like styrene exhibit a C=C stretch that matches this specific frequency. Understanding these peaks helps chemists identify possible unsaturated compounds in a mixture.
In the original exercise, the IR spectrum reveals a peak at 1640 cm⁻¹. Typically, this indicates the presence of a carbon-carbon double bond. Such double bonds absorb infrared light at this frequency, causing the characteristic vibration noted in the spectrum.
This information is essential for the proposed structure of styrene in the original exercise, as alkenes like styrene exhibit a C=C stretch that matches this specific frequency. Understanding these peaks helps chemists identify possible unsaturated compounds in a mixture.
NMR Spectroscopy
NMR Spectroscopy is a powerful tool that provides information about the structure of organic compounds by observing the behavior of nuclei in a magnetic field. In this case, we are concerned with proton (\(^1\)H) NMR spectroscopy.
In the exercise, two singlet peaks are observed at 1.69 ppm and 6.44 ppm. Singlets occur when protons do not split due to interaction with adjacent protons, indicating an isolation or symmetry in their environment.
The peak at 6.44 ppm in the product might correspond to vinylic or aryl protons often associated with unsaturated carbons, like the protons on a styrene or benzene ring. The peak at 1.69 ppm can be linked to methyl or methylene protons typically located further downfield due to less shielding.
This spectral analysis helps confirm the proposed structure of the product as styrene, where such singlet peaks are anticipated.
In the exercise, two singlet peaks are observed at 1.69 ppm and 6.44 ppm. Singlets occur when protons do not split due to interaction with adjacent protons, indicating an isolation or symmetry in their environment.
The peak at 6.44 ppm in the product might correspond to vinylic or aryl protons often associated with unsaturated carbons, like the protons on a styrene or benzene ring. The peak at 1.69 ppm can be linked to methyl or methylene protons typically located further downfield due to less shielding.
This spectral analysis helps confirm the proposed structure of the product as styrene, where such singlet peaks are anticipated.
UV Spectroscopy
UV Spectroscopy deals with the absorption of UV light that excites electrons to higher energy levels, which is mainly used to study conjugated systems. UV spectroscopy indicates the presence of conjugated pi systems, such as aromatic rings.
For the compound in the exercise, a UV maximum is noted at 282 nm. This wavelength absorption suggests a system of conjugated pi bonds, typical for compounds with aromatic segments connected by double bonds.
Since the product proposed is styrene, a molecule like this would show UV absorption consistent with a phenyl group, verifying that the aromatic section remains intact following the reaction. Therefore, the UV data corroborates the preservation of the benzene ring in styrene.
For the compound in the exercise, a UV maximum is noted at 282 nm. This wavelength absorption suggests a system of conjugated pi bonds, typical for compounds with aromatic segments connected by double bonds.
Since the product proposed is styrene, a molecule like this would show UV absorption consistent with a phenyl group, verifying that the aromatic section remains intact following the reaction. Therefore, the UV data corroborates the preservation of the benzene ring in styrene.
Elemental Analysis
Elemental Analysis is used to determine the percentage composition of elements within a compound. This analysis is crucial for validating the molecular formula of an unknown substance.
In the exercise, the analysis results show approximately 79.98% carbon and 6.71% hydrogen. This ratio closely matches the composition expected in a hydrocarbon like styrene, taking into account the single oxygen atom initially present in the starting material.
Such composition excludes the presence of additional functional groups and aligns with the minimalistic structure of styrene. Elemental analysis thus supports the identification of styrene as the reaction product, fitting within the likely molecular framework given the data.
In the exercise, the analysis results show approximately 79.98% carbon and 6.71% hydrogen. This ratio closely matches the composition expected in a hydrocarbon like styrene, taking into account the single oxygen atom initially present in the starting material.
Such composition excludes the presence of additional functional groups and aligns with the minimalistic structure of styrene. Elemental analysis thus supports the identification of styrene as the reaction product, fitting within the likely molecular framework given the data.
Dehydrohalogenation Reaction
A Dehydrohalogenation Reaction involves the removal of a hydrogen halide from a substrate, resulting in an alkene through the formation of a carbon-carbon double bond. This reaction is highly relevant in organic synthesis for altering molecular frameworks.
In the reaction described in the exercise, basic alumina acts as a dehydrating agent, facilitating the elimination of HBr from 2-(4-hydroxyphenyl)ethyl bromide to form styrene.
The process signifies an E2 elimination mechanism, where the base removes a proton adjacent to the halogen-containing carbon, resulting in the creation of a double bond.
Understanding dehydrohalogenation helps students connect reactions to changes in molecular structure efficiently, reinforcing their comprehension of the mechanism forming the suggested product, styrene.
In the reaction described in the exercise, basic alumina acts as a dehydrating agent, facilitating the elimination of HBr from 2-(4-hydroxyphenyl)ethyl bromide to form styrene.
The process signifies an E2 elimination mechanism, where the base removes a proton adjacent to the halogen-containing carbon, resulting in the creation of a double bond.
Understanding dehydrohalogenation helps students connect reactions to changes in molecular structure efficiently, reinforcing their comprehension of the mechanism forming the suggested product, styrene.
Other exercises in this chapter
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