Only glyoxal CHO - CHO is obtained by the ozonolysis of (a) Cyclobutene (b) 1,3 -butadiene (c) ethyne (d) 2 -butene

Short Answer

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Answer: 1,3-butadiene

Step by step solution

01

Understanding Ozonolysis

Ozonolysis is a chemical reaction used to break down alkenes or alkynes using ozone (O3). In this process, the double or triple bond in the molecule is cleaved and divided into two smaller molecules, forming carbonyl compounds like aldehydes, ketones or a mixture of them as the products. The general reaction for ozonolysis of alkene is as follows: R1-CH=CH-R2 + O3 → R1-CHO + R2-CHO (if R1 and R2 are hydrogen or alkyl groups) Now, let's analyze each option given in the exercise.
02

Analyzing Option (a) - Cyclobutene

Cyclobutene is a cyclic alkene with the molecular formula C4H6. The structure of cyclobutene is: ``` CH2=CH | | CH2---CH2 ``` Upon ozonolysis, it will form two molecules of ethanal (CH3CHO), which is not glyoxal (CHO-CHO). So, option (a) is not the correct answer.
03

Analyzing Option (b) - 1,3-butadiene

1,3-butadiene is an alkene with the molecular formula C4H6 and has the following structure: ``` CH2=CH-CH=CH2 ``` Upon ozonolysis of 1,3-butadiene, it will form glyoxal (CHO-CHO), as shown in the following reaction: ``` CH2=CH-CH=CH2 + O3 → CHO-CHO ``` Since glyoxal is the required product, option (b) is the correct answer.
04

Analyzing Option (c) - Ethyne

Ethyne, also known as acetylene, is an alkyne with the molecular formula C2H2. The structure of ethyne is: ``` HC≡CH ``` Ozonolysis of ethyne produces oxalic acid and hydrogen gas, as shown in the following reaction: ``` HC≡CH + O3 → H2C2O4 + H2 ``` Since glyoxal is not formed in this reaction, option (c) is not the correct answer.
05

Analyzing Option (d) - 2-butene

2-butene is an alkene with the molecular formula C4H8. Its structure is: ``` CH3-CH=CH-CH3 ``` Upon ozonolysis, 2-butene will form two molecules of ethanal (CH3CHO), which is not glyoxal (CHO-CHO). So, option (d) is not the correct answer. After analyzing each option, we can conclude that:
06

Conclusion

Only glyoxal (CHO-CHO) is obtained by the ozonolysis of 1,3-butadiene (option b).

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Key Concepts

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

Alkene Ozonolysis
Alkene ozonolysis is a vital organic reaction, especially for students preparing for competitive exams like IIT JEE. It involves the cleavage of carbon-to-carbon double bonds using ozone, a strong oxidizing agent. In this process, ozone reacts with an alkene to form a molozonide, which is an unstable intermediate and quickly rearranges to a more stable ozonide. Finally, the ozonide is reduced to give the corresponding carbonyl compounds (aldehydes or ketones).
Ozonolysis is important in structure determination of alkenes as it can identify the position of double bonds within the molecule. When ozonolysis results in the formation of glyoxal (CHO-CHO), it suggests that the original alkene had conjugated double bonds, as in the case of 1,3-butadiene. This is because the cleavage occurs at both ends of the conjugated system, leading to the smallest possible aldehyde, glyoxal. It's crucial to grasp this mechanism, as questions about product identification are common in exams like IIT JEE.
Organic Reaction Mechanisms
Understanding organic reaction mechanisms is fundamental to mastering organic chemistry. A reaction mechanism provides the step-by-step sequence of elementary reactions by which a chemical change occurs. It illustrates how the reactants are converted into products and which intermediates are formed during the course of the reaction. Such knowledge enables prediction of the outcome of the chemical reactions.
For ozonolysis, the key mechanism stages are the formation of the molozonide and its subsequent rearrangement to the ozonide, before finally being reduced to carbonyl compounds. Learning mechanisms not only helps with solving textbook exercises but also aids in visualizing the changes that occur at a molecular level. Breaking down complex reactions into simpler steps makes it easier for students to grasp, and it is particularly crucial for IIT JEE preparation where complex organic reactions are a significant part of the syllabus.
IIT JEE Chemistry Preparation
Students preparing for the IIT JEE must develop a robust understanding of concepts and the ability to apply them to solve problems efficiently. For organic chemistry, this means familiarizing oneself with a variety of reactions, including ozonolysis, and their mechanisms.

Study Strategies for IIT JEE

  • Conceptual Clarity: Make sure all underlying principles and reaction mechanisms are well understood, not just memorized.
  • Regular Practice: Consistently solve problems and refer to textbook solutions to clarify doubts.
  • Mock Tests: Take timed practice exams to improve speed and accuracy.
  • Analysis: After solving problems or taking tests, analyze mistakes to avoid repeating them.
Incorporating ozonolysis into study routines is one step further towards mastering organic chemistry. By understanding its application in problem-solving, students can significantly uplift their performance potential for IIT JEE Chemistry.

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Most popular questions from this chapter

Which of the following is the correct statement? (a) Acetylene is less reactive than ethylene in electrophilic addition reaction. (b) The hydrolysis of \(\mathrm{Mg}_{2} \mathrm{C}_{3}\) (Magnesium carbide) produces acetylene. (c) Ozonolysis of acetylene produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). (d) Reduction of dialkyl acetylene using Lindlar's catalyst gives a trans olefin.

The correct decreasing order of the acidity of hydrogen in acetylene, phenol, alcohol, ethylene and acetic acid is (a) Acetic acid \(>\) Phenol \(>\) alcohol \(>\) acetylene \(>\) ethylene (b) Phenol \(>\) acetic acid \(>\) alcohol \(>\) acetylene \(>\) ethylene (c) Acetic acid \(>\) alcohol \(>\) phenol \(>\) acetylene \(>\) ethylene (d) Acetic acid \(>\) acetylene \(>\) phenol \(>\) alcohol \(>\) ethylene

Identify the correct statement among the following. (a) A hydrocarbon of M.F. \(\mathrm{C}_{6} \mathrm{H}_{12}\) on reductive ozonolysis gave only one product which fails to react with Tollens reagent. The hydrocarbon could be 2,3 dimethyl but- 2 -ene. (b) Addition of bromine to trans-2-butene gives a racemic mixture of two optically active enantiomers. (c) But-1-yne and but-2-yne can be distinguished using ammoniacal cuprous chloride. (d) Ozonolysis of acetylene gives glyoxal.

Directions: This section contains 3 paragraphs. Based upon the paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. Passage I The three main classes of hydrocarbons are alkanes, alkenes and alkynes. Alkanes having cyclic structure are called cycloalkanes. They exhibit chain and ring chain isomerism. Alkenes and alkynes are more reactive than alkanes. A certain hydrocarbon has the molecular formula \(\mathrm{C}_{5} \mathrm{H}_{8} .\) Which of the following is not a structural possibility for this hydrocarbon? (a) a cycloalkane (b) contains a ring and a double bond (c) contains two double bond and no ring (d) an alkyne

Complete the following equations showing the stereo chemistry of the product wherever applicable. (i) Cis-2-pentene \(\stackrel{\mathrm{H}_{2} / \mathrm{Pt}}{\longrightarrow}\) (ii) Cis-2-butene \(\stackrel{\mathrm{Br}_{2} / \mathrm{CCl}_{4}}{\longrightarrow}\) (iii) Cis-2-butene \(\stackrel{\mathrm{m}-\mathrm{CPBA}}{\longrightarrow}\) (iv)

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