Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Reaction of one molecule of HBr with one molecule of 1,3 -butadiene at \(40^{\circ} \mathrm{C}\) gives predominantly: (a) 3-bromobutene under kinetically controlled conditions (b) 1 -bromo-2-butene under thermodynamically controlled conditions (c) 3-bromobutene under thermodynamically controlled conditions (d) 1-bromo-2-butene under kinetically controlled conditions.

Short Answer

Expert verified
(a) 3-bromobutene under kinetically controlled conditions.

Step by step solution

01

Understanding the Reaction

When 1,3-butadiene reacts with HBr, the reaction can proceed via different pathways leading to different products. The nature of the product depends on the reaction conditions, which can be either kinetic control (leading to the product formed fastest) or thermodynamic control (leading to the most stable product).
02

Identify Kinetic Product

Under kinetically controlled conditions, the reaction favors the formation of the product that forms the fastest. In the reaction of HBr with 1,3-butadiene, the kinetic product is 3-bromobutene. This is because the intermediate allylic cation leading to 3-bromobutene forms faster than others due to lower activation energy needed for its transition state.
03

Identify Thermodynamic Product

Under thermodynamically controlled conditions, the reaction favors the formation of the most stable product. In this case, the most stable product is 1-bromo-2-butene due to its more substituted alkene configuration, which is more stable according to Zaitsev's rule.
04

Analyze Given Conditions

The problem states that the reaction is occurring at 40°C, which suggests kinetic control may be predominant at this relatively low temperature. This implies that the reaction would predominantly produce the kinetic product, 3-bromobutene.
05

Match Conditions with Options

The question asks which product is predominantly formed under specific conditions. We have determined that under kinetically controlled conditions at 40°C, 3-bromobutene is formed predominantly. This matches option (a), which corresponds with the given conditions.

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!

Key Concepts

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

Kinetically Controlled Reactions
In the realm of organic chemistry, two pivotal types of control dictate the outcome of reactions: kinetic control and thermodynamic control. Kinetically controlled reactions are governed by how quickly a product can form.
When a reaction is under kinetic control, the major product is the one that forms the fastest. This "faster" product is often the one with the lower activation energy barrier, since less energy is required for the transition to occur.
The transition state leading to this product is accessed more readily, thus preferentially forming this product under conditions that favor kinetic control. These typically include lower temperatures and higher concentrations of reactants, as these conditions help to avoid the reaction proceeding over a longer period where equilibrium considerations might predominate.
Thermodynamically Controlled Reactions
Unlike kinetic control, thermodynamically controlled reactions focus on the stability of the products rather than the speed at which they form.
Under thermodynamic control, the major product is the one that is most stable—often, this is due to having the lowest overall energy or being the most substituted compound.
For example, in many reactions involving alkenes, more substituted alkenes are more thermodynamically stable. Higher temperatures and longer reaction times can allow the reaction to reach equilibrium, at which point the thermodynamically favored product predominates.
This is because the system can overcome higher energy barriers, reforming intermediates until the products settle into the most stable state possible.
Allylic Cations
Allylic cations are key intermediates in many organic mechanisms, especially reactions involving alkenes. An allylic cation is a positively charged species that has the charge localized on a carbon atom adjacent to a double bond.
This configuration is beneficial because the positive charge can be stabilized via resonance. The electrons from the double bond can delocalize, spreading out the positive charge across multiple atoms.
This kind of charge distribution not only stabilizes the intermediate but also influences which products are favored in both kinetically and thermodynamically controlled reactions. For instance, in the reaction of 1,3-butadiene with HBr, an allylic cation forms, leading to the products 3-bromobutene under kinetic control and 1-bromo-2-butene under thermodynamic control.
Zaitsev's Rule
Understanding Zaitsev's Rule helps predict the major product in elimination reactions, particularly when considering thermodynamic outcomes. According to Zaitsev's Rule, the most stable alkene is typically the most substituted one.
In other words, when more carbon atoms are bonded to the double-bonded carbon atoms, the system is generally more stable.
  • This is because additional alkyl groups donate electrons, helping to stabilize the double bond through hyperconjugation and inductive effects.
  • For example, in the reaction between 1,3-butadiene and HBr, the formation of 1-bromo-2-butene under thermodynamically controlled conditions is favored, as it is more substituted than its counterpart, 3-bromobutene.
Thus, Zaitsev's Rule is an essential guideline in predicting the outcome of reactions under thermodynamic control, guiding chemists in understanding which alkene product will be preferred.

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

Which is the decreasing order of strength of bases: \(\mathrm{OH}^{-}, \mathrm{NH}_{2}^{-}, \mathrm{HC} \equiv \mathrm{C}^{-}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2}^{-} ?\) (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{HC} \equiv \mathrm{C}^{-}>\mathrm{OH}^{-}\) (b) \(\mathrm{HC} \equiv \mathrm{C}^{-}>\mathrm{CH}_{3} \mathrm{CH}_{2}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{OH}^{-}\) (c) \(\mathrm{OH}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{HC} \equiv \mathrm{C}^{-}>\mathrm{CH}_{3} \mathrm{CH}_{2}^{-}\) (d) \(\mathrm{NH}_{2}^{-}>\mathrm{HC} \equiv \mathrm{C}^{-}>\mathrm{OH}^{-}>\mathrm{CH}_{3} \mathrm{CH}_{2}^{-}\)

A hydrocarbon of molecular formula, \(\mathrm{C}_{8} \mathrm{H}_{18}\) gives only one monochloro derivative. The hydrocarbon is: (a) 2-methyl heptane (b) Octane (c) \(2,2,3,3\) -tetramethyl butane (d) \(2,2,4\) -trimethyl pentane

In order to obtain trans \(\mathrm{Me}-\mathrm{CH}=\mathrm{CH}-\mathrm{Me}\) from \(\mathrm{MeC} \equiv \mathrm{CMe}\), the reagent(s) to be used would includes: (a) \(\mathrm{LiAlH}_{4}\) in \(\mathrm{EtO}_{2}\) (b) \(\mathrm{H}_{2} / \mathrm{Pd}, \mathrm{BaSO}_{4}\) (c) \(\mathrm{H}_{2} / \mathrm{Pd}, \mathrm{Pb}(\mathrm{OAc})_{2}\) (d) Sodium in liquid ammonia or sodium in \(\mathrm{EtOH}\)

Octane number of gasoline can be increased by the addition of BTX. BTX stands for: (a) Butane, TEL, xylene (b) Butane, tetramethyl lead, xylene (c) Benzene, toluene, xylene (d) Benzene, TEL, xylene

Which of the following is used for the conversion of 2-hexyne into trans-2-hexane? (a) \(\mathrm{H}_{2} / \mathrm{Pd} / \mathrm{BaSO}_{4}\) (b) \(\mathrm{H}_{2}, \mathrm{PtO}_{2}\) (c) \(\mathrm{NaBH}_{4}\) (d) \(\mathrm{Li}-\mathrm{NH}_{3} / \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)

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