The best method of preparation of an alkane with odd number of carbons from two alkylhalides is (a) Wurtz reaction (b) Kolbe's electrolysis (c) Corey-House synthesis (d) Ullmann reaction

Short Answer

Expert verified
A. Wurtz reaction B. Kolbe's electrolysis C. Corey-House synthesis D. Ullmann reaction Answer: C. Corey-House synthesis Explanation: The Corey-House synthesis is the best method for preparing an alkane with an odd number of carbons from two alkylhalides, as it allows the formation of both symmetric and asymmetric alkanes. In contrast, the Wurtz reaction and Kolbe's electrolysis typically yield alkanes with an even number of carbons, and the Ullmann reaction produces biaryl compounds instead of alkanes.

Step by step solution

01

1. Understand the four reaction methods

Before finding the best method, we need to understand each of the four given methods: (a) Wurtz reaction: A coupling reaction where two alkylhalides react in the presence of sodium metal to form a symmetrical alkane with an even number of carbons. (b) Kolbe's electrolysis: An electrochemical decarboxylation reaction involving the conversion of a carboxylic acid (or its salts) into an alkane, usually with an even number of carbons. (c) Corey-House synthesis: A two-step reaction involving organocuprates to form new carbon-carbon bonds, allowing the preparation of both symmetric and asymmetric alkanes. (d) Ullmann reaction: A copper-catalyzed coupling reaction involving the reaction between an aryl halide and an amine to form a biaryl compound.
02

2. Compare the reaction methods for odd-numbered alkanes

Comparing these methods: - The Wurtz reaction and Kolbe's electrolysis generally yield alkanes with an even number of carbons, which does not fulfill our requirement. - The Ullmann reaction yields biaryl compounds and not alkanes, so it is not suitable for the given condition. However, the Corey-House synthesis has the flexibility to create both symmetric and asymmetric alkanes, which means it can be used to prepare an alkane with an odd number of carbons.
03

3. Determine the best method

Based on the analysis above, the best method of preparation of an alkane with an odd number of carbons from two alkylhalides is: (c) Corey-House synthesis

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

The reactions of alkenes are mainly (a) nucleophilic additions (b) nucleophilic substitutions (c) free radical reactions (d) electrophilic additions

Identify the incorrect statement among the following (b) The dipole moment of \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) is greater than that of \(\mathrm{CH}_{3} \mathrm{Cl}\). (c) The cis isomers become increasingly less stable as the size of the \(\mathrm{R}\) groups increase. (d) Alkenes with two identical groups on the same doubly bonded carbon have no geometrical isomers.

1,6 Dibromohexane can be converted to cyclohexane, when treated with \(\mathrm{Na}\) or \(\mathrm{Zn}\), in the presence of a catalyst. This reaction is known as (a) Wurtz coupling (b) Dieckmann cyclization (c) Freund's method (d) Diels Alder reaction

\(\mathrm{HC} \equiv \mathrm{N}\) is more acidic than \(\mathrm{HC} \equiv \mathrm{CH}\) because (a) HCN is an acid whereas, \(\mathrm{HC} \equiv \mathrm{CH}\) is a hydrocarbon. (b) HCN has a pK of \(9.2\). (c) In addition to greater s character of the hybrid orbitals, the presence of more electronegative nitrogen further stabilizes the anion - CN. (d) of the triple bond nature of the \({ }^{-} \mathrm{C} \equiv \mathrm{N}\) bond.

HI does not exhibit peroxide effect because (a) HI is a reducing agent. (b) the iodine free radicals formed recombine readily to form the iodine molecule. (c) the bond energy of HI is very high that it does not undergo homolytic fission to give iodine radicals. (d) Free radical reactions work well when both the propagation steps are endothermic.

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