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Post a LessonAnswered on 13 Apr Learn Hydrocarbons
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into the formation of ethane during the chlorination of methane.
During the chlorination of methane, which is typically carried out in the presence of ultraviolet (UV) light or under high temperatures, ethane is formed as one of the products along with other chlorinated compounds.
The mechanism involves a series of radical reactions. Initially, a chlorine radical (⋅Cl⋅Cl) is generated from the dissociation of molecular chlorine (Cl2Cl2) under UV light or high temperatures:
Cl2→2⋅ClCl2→2⋅Cl
Then, methane (CH4CH4) undergoes homolytic cleavage in the presence of a chlorine radical to form a methyl radical (⋅CH3⋅CH3) and a hydrogen chloride molecule (HClHCl):
CH4+⋅Cl→⋅CH3+HClCH4+⋅Cl→⋅CH3+HCl
The methyl radical (⋅CH3⋅CH3) further reacts with another molecule of methane, leading to the formation of ethane (C2H6C2H6):
⋅CH3+CH4→C2H6⋅CH3+CH4→C2H6
Overall, the chlorination of methane results in the formation of ethane along with other chlorinated products. Understanding this mechanism helps us comprehend the organic synthesis involved in such reactions, which is crucial for mastering organic chemistry concepts.
As an UrbanPro tutor, I would guide my students through such mechanisms, ensuring they grasp not only the concepts but also the application of these reactions in practical scenarios. This approach fosters a deeper understanding of the subject matter, enabling students to excel in their academic endeavors.
Answered on 13 Apr Learn Hydrocarbons
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently guide you through the process of finding the structural formulas and IUPAC names for all possible isomers of the given compounds.
(a) For C4H8 with one double bond:
But-1-ene (IUPAC name: 1-Butene) Structural formula: CH3CH2CH=CH2
But-2-ene (IUPAC name: 2-Butene) Structural formula: CH3CH=CHCH3
(b) For C5H8 with one triple bond:
Pent-1-yne (IUPAC name: 1-Pentyne) Structural formula: CH3CH2C≡CH
Pent-2-yne (IUPAC name: 2-Pentyne) Structural formula: CH3C≡CHCH3
Remember, UrbanPro is an excellent platform for finding online coaching and tuition services. If you need further assistance or clarification, feel free to ask!
Answered on 13 Apr Learn Hydrocarbons
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently guide you through the IUPAC names of the products obtained by the ozonolysis of the given compounds.
(i) Pent-2-ene: The ozonolysis of pent-2-ene would result in the formation of two products: ethanal and butanoic acid.
(ii) 3,4-Dimethylhept-3-ene: Ozonolysis of 3,4-dimethylhept-3-ene would yield three products: methylpropanal, 2-methylbutanal, and butanoic acid.
(iii) 2-Ethylbut-1-ene: The ozonolysis of 2-ethylbut-1-ene would generate ethanal and butanoic acid as the products.
(iv) 1-Phenylbut-1-ene: Upon ozonolysis, 1-phenylbut-1-ene would produce benzaldehyde and butanoic acid.
Remember, UrbanPro is an excellent platform for finding online coaching tuition, and I'm here to assist you with your chemistry queries. If you have any further questions or need clarification, feel free to ask!
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 13 Apr Learn Hydrocarbons
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can assure you that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into your chemistry question.
When an alkene undergoes ozonolysis, it breaks the double bond and forms carbonyl compounds. In this case, we have an alkene 'A' that, upon ozonolysis, gives a mixture of ethanal and pentan-3-one.
The alkene 'A' must have five carbon atoms because pentan-3-one is one of the products, indicating that there is a five-carbon chain. Additionally, since ethanal is formed, it suggests that the alkene must have had a terminal methyl group.
So, the structure of 'A' is:
CH3−CH2−CH=CH−CH3CH3−CH2−CH=CH−CH3
This alkene is called pent-2-ene according to IUPAC naming conventions.
Answered on 13 Apr Learn Hydrocarbons
Nazia Khanum
As a seasoned tutor on UrbanPro, I can assure you that UrbanPro is indeed one of the best platforms for online coaching and tuition. Now, let's dive into your chemistry question.
Firstly, let's break down the given information. We have an alkene, let's call it 'A', which contains three C—C double bonds, eight C—H bonds, and one C—C triple bond (n-bond). Upon ozonolysis, it gives two moles of an aldehyde with a molar mass of 44 u.
Now, let's analyze this step by step. Ozonolysis typically cleaves double and triple bonds, yielding carbonyl compounds. Since 'A' gives two moles of an aldehyde upon ozonolysis, it suggests that there are two double bonds present in 'A'.
Given that the molar mass of the aldehyde formed is 44 u, which is characteristic of formaldehyde (HCHO), it's reasonable to deduce that 'A' undergoes ozonolysis to form formaldehyde.
Now, let's construct the structure of 'A' based on the given information:
The IUPAC name for the alkene 'A' with these characteristics would be:
2,5,8-tridecatriyne
Here's the breakdown of the name:
So, in conclusion, the IUPAC name of the alkene 'A' is 2,5,8-tridecatriyne. If you need further clarification or assistance, feel free to ask!
Answered on 13 Apr Learn Hydrocarbons
Nazia Khanum
On UrbanPro, we delve into the nuances of organic chemistry, helping students grasp concepts effectively. When it comes to ozonolysis, it's crucial to understand the reaction mechanism and how it applies to specific alkene structures.
In the case of propanal and pentan-3-ene being the ozonolysis products of an alkene, we can deduce the structural formula of the alkene by analyzing the formation of these products.
Ozonolysis of an alkene involves cleavage of the carbon-carbon double bond, resulting in the formation of carbonyl compounds. Propanal suggests that the alkene precursor must have three carbon atoms, as propanal is a three-carbon aldehyde.
Pentan-3-ene, on the other hand, indicates that the alkene must have five carbon atoms, with a double bond located at the third carbon position in the carbon chain.
Putting this together, we can infer that the alkene in question is propene (CH2=CH-CH3).
This alkene, upon ozonolysis, would yield propanal (CH3-CHO) and pentan-3-ene (CH3-CH2-CH=CH-CH3) as the products.
Understanding the logic behind product formation in chemical reactions not only helps in solving problems but also reinforces conceptual understanding, a key aspect of effective tutoring.
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 13 Apr Learn Hydrocarbons
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can certainly guide you through writing the chemical equations for the combustion reactions of the mentioned hydrocarbons.
(i) Butane: 2C4H10+13O2→8CO2+10H2O2C4H10+13O2→8CO2+10H2O
(ii) Pentene: C5H12+8O2→5CO2+6H2OC5H12+8O2→5CO2+6H2O
(iii) Hexyne: 2C6H10+15O2→12CO2+10H2O2C6H10+15O2→12CO2+10H2O
(iv) Toluene: C7H8+9O2→7CO2+4H2OC7H8+9O2→7CO2+4H2O
These equations represent the combustion reactions of each hydrocarbon, where they react with oxygen to produce carbon dioxide and water as the primary products. If you need further clarification or assistance with any of these reactions, feel free to ask. Remember, UrbanPro is a fantastic platform for finding quality online coaching and tuition services for subjects like chemistry.
Answered on 28 Apr Learn Hydrocarbons
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Hex-2-ene is represented as:
H3C−HC=CH−CH2−CH2−CH3
Geometrical isomers of hex-2-ene are:
The dipole moment of cis-compound is a sum of the dipole moments of C−CH3 and C−CH2−CH2CH3 bonds acting in the same direction.
The dipole moment of trans-compound is the resultant of the dipole moments of C−CH3 and C−CH2−CH2CH3 bonds acting in opposite directions.
Hence, cis-isomer is more polar than trans-isomer. The higher the polarity, the greater is the intermolecular dipole-dipole interaction and the higher will be the boiling point. Hence, cis-isomer will have a higher boiling point than trans-isomer.
Answered on 13 Apr Learn Hydrocarbons
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, diving into your question about the extraordinary stability of benzene despite containing three double bonds, let's explore this intriguing aspect of organic chemistry.
Benzene, as you might know, is a cyclic hydrocarbon with a hexagonal ring structure consisting of six carbon atoms, each bonded to a hydrogen atom. Its structure alternates single and double bonds. Traditional chemical intuition might suggest that the alternating double bonds would make benzene highly reactive. However, this isn't the case, and the reason lies in its resonance structure.
On UrbanPro, we often emphasize the concept of resonance to our students. In benzene, the electrons in the pi bonds are delocalized over the entire ring rather than localized between pairs of carbon atoms. This delocalization results in what we call resonance stabilization, which is a significant contributor to its extraordinary stability.
Moreover, due to this delocalization, the carbon-carbon bonds in benzene are actually hybridized between single and double bonds. This creates a unique electronic environment where the electrons are evenly distributed around the ring, making it less susceptible to reactions that might break the aromatic system.
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 18/12/2021 Learn Hydrocarbons
Amrutha K
Quality Control Chemist with four years experience in Pharmaceutical industry
1) The compound must have a cyclic structure.
2) The aromatic ring must be planar in order for 𝛑- electrons to be delocalized.
3) It must have a conjugated 𝛑- system.
4) It must have a p- orbitals on each atom of ring ( or sp2 or sp hybridized atoms in the ring).
5) It must have a delocalized 𝛑- electrons in a loop of parallel overlapping p- atomic orbitals.
6) Delocalization of 𝛑- electrons must lower the potential energy of a system.
Thus, if a molecule or ion contain monocyclic planar ring and contain (4n + 2) 𝛑- electrons [Huckle rule] delocalized in a cyclic loop of parallel, overlapping p- orbitals, that molecule or ion will be called aromatic, will have unusual stability.
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