A reaction that does not lead to an alkene is (a) Kolbe's electrolysis (b) \(\alpha\) -elimination (c) Wittig reaction (d) dehalogenation

Short Answer

Expert verified
Answer: \(\alpha\)-elimination

Step by step solution

01

Option (a): Kolbe's Electrolysis

Kolbe's electrolysis involves the electrolysis of a carboxylate solution, leading to the formation of alkanes and carbon dioxide. There is no formation of an alkene in this reaction. The general reaction can be represented as: 2 RCOO- + 2 H2O → 2 CO2 + 2 R-H + 2 OH-
02

Option (b): \(\alpha\) -elimination

In an \(\alpha\)-elimination reaction, two groups are lost from the same carbon atom, and it does not lead to the formation of an alkene. An example of an \(\alpha\)-elimination reaction is as follows: R-C(-X)(-Y) → X-Y + R
03

Option (c): Wittig Reaction

The Wittig reaction is an organic reaction used to convert aldehydes or ketones into alkenes. A phosphonium ylide is the reactive intermediate in this reaction. It reacts with the carbonyl compound to produce an alkene. The general reaction can be represented as: Ph\(_3\)P=CR\(_2\) + R\(_1\)CO → Ph\(_3\)P=O + R\(_1\)C=CR\(_2\)
04

Option (d): Dehalogenation

Dehalogenation is a reaction in which a halogen atom is removed from a substrate molecule. In the context of organic molecules, dehalogenation can lead to the formation of an alkene when two halogens are removed from adjacent carbons. This is also known as a double dehydrohalogenation reaction. The general reaction is as follows: R-CH(X)-CH(Y)-R' → R-CH=CH-R' + HX + HY From the analysis of each option, we can conclude that:
05

Conclusion

Option (b) \(\alpha\)-elimination is the reaction that does not lead to the formation of an alkene.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

How is 1,3 -butadiene made from (i) 1 -butene (ii) acetylene

Identify the correct statements among the following (a) Chlorination of \(\mathrm{n}\) -butane involves the attack by a free radical. (b) 2,3-Dimethylbutane can form only two monochloro derivatives (c) Between \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) (I) and \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{CH}\) (II), only (II) will react with \(\mathrm{CH}_{3} \mathrm{MgI}\). (d) When a mixture of ethane, ethylene and ethyne is passed into a solution of ammoniacal \(\mathrm{AgNO}_{3}\), the gases bubbling out will be a mixture of ethane and ethylene

Identify the product in the given reaction. \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{NOCl} \rightarrow(\mathrm{A})\) (a) \(\mathrm{CH}_{3}-\mathrm{CHCl}-\mathrm{CH}_{2} \mathrm{NO}\) (b) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClNO}\) (d) \(\mathrm{CH}_{3} \mathrm{NOCH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\)

Which of the following will have zero dipole moment? (a) Cis 1, 2 dichloroethene (b) 1,1 dichloroethene (c) Cis-2-butene (d) trans-1, 2 dichloroethene

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.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free