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\(\left(\mathrm{H}_{3} \mathrm{C}\right)_{2} \mathrm{C}=\mathrm{CHCH}_{3}+\mathrm{NOBr} \longrightarrow \mathrm{A}\) The structure of the product (a)is given as (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{Br})-\mathrm{CH}(\mathrm{NO}) \mathrm{CH}_{3}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{NO})-\mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}\) (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{C}(\mathrm{NO})(\mathrm{Br}) \mathrm{CH}_{3}\) (d)

Short Answer

Expert verified
The product is (a) \((\text{CH}_3)_2\text{C(Br)}-\text{CH(NO)}\text{CH}_3\).

Step by step solution

01

Reaction Understanding

The reaction involves an alkene (\(\text{CH}_3)_2\text{C} = \text{CHCH}_3\) reacting with \(\text{NOBr}\). An electrophilic addition takes place where the double bond opens and adds \(\text{Br}\) and \(\text{NO}\).
02

Identify Major Product

In electrophilic addition, the more stable carbocation intermediate directs the product formation. The double bond breaks, electrons are used to form a carbocation on the more substituted carbon (tertiary), and the \(\text{Br}^-\) anion bonds with it. The \(\text{NO}^+\) cation will then bond to the less substituted carbon (secondary).
03

Drawing the Major Product Structure

Following Markovnikov's rule, the product will most likely have \(\text{Br}\) attached to the more substituted carbon (tertiary), and \(\text{NO}\) attached to the less substituted carbon. This follows a stability trend with the carbocation that forms after the addition of \(\text{Br}^-\).
04

Match with Provided Structure Options

Compare the derived structure from the steps above to the given options. The derived structure is \((\text{CH}_3)_2\text{C(Br)}-\text{CH(NO)}\text{CH}_3\), which matches with option (a).

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Key Concepts

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

Alkene Reactivity
Alkenes are hydrocarbons containing carbon-carbon double bonds, which make them highly reactive. This reactivity is thanks to the presence of \(\pi\) bonds, which are less stable than \(\sigma\) bonds. The \(\pi\) electrons are held more loosely, making them a target for chemical reactions. Electrophilic addition is a common reaction mechanism for alkenes.
In electrophilic additions, the \(\pi\) bond is broken when reacting with an electrophile—an electron-deficient species that seeks out electrons to form new bonds. During the reaction, alkenes become alkanes or other saturated hydrocarbons.
When \(\text{NOBr}\) reacts with an alkene, the electrophilic nitric oxide (NO^+) and Bromide ions (Br^-) are added across the double bond in the alkene. The reaction rate is significantly determined by the nature of substituents attached to the double-bonded carbons. The more substituted the alkene, the faster the reaction, due to the hyperconjugation and inductive effects that increase alkene reactivity.
Carbocation Stability
Carbocation stability is a fundamental concept when understanding chemical reactions involving carbocations as intermediates. Carbocations are positively charged ions that occur when a bond breaks unevenly, leaving a carbon atom with one fewer electron than normal. The stability of these ions greatly influences the path and outcome of chemical reactions.
Stability of carbocations follows this general order:
  • Tertiary carbocations (R3C^+) are more stable than
  • Secondary carbocations (R2HC^+), which are more stable than
  • Primary carbocations (RH2C^+).
This stability is due to the greater degree of positive charge delocalization allowed by additional neighboring alkyl groups. These groups provide electron-donating inductive effects and hyperconjugation, helping to stabilize the positively charged carbon.
In the context of the electrophilic addition reaction with \(\text{NOBr}\), the structure of the alkene favors the formation of a tertiary carbocation as an intermediate, which promotes the addition of the Br^- ion to the more substituted carbon.
Markovnikov's Rule
Markovnikov's rule is crucial for predicting the product of electrophilic addition reactions. It states that when an electrophile is added to an unsymmetrical alkene, the electrophile will attach to the carbon with the most substituents. Conversely, the nucleophile will attach to the carbon with fewer substituents.
This rule is a reflection of carbocation stability, as described earlier. By forming the more stable carbocation, the reaction proceeds in a direction that requires lower activation energy. Therefore, the major product reflects these trends.
In the reaction between \(\left(\text{H}_3\text{C}\right)_2\text{C} = \text{CHCH}_3\) and \(\text{NOBr}\), Markovnikov's rule tells us that the bromine ion (Br^-) will attach to the tertiary carbon, forming a stable tertiary carbocation. As a result, the nitric oxide ion (NO^+) will attach to the adjacent secondary carbon. This insight helps understand why the major product looks like \(\left(\text{CH}_3\right)_2\text{C}\text{(Br)}-\text{CH(NO)} \text{CH}_3\).
Understanding and applying Markovnikov's rule enables chemists to predict and control the outcomes of chemical reactions more accurately.

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Most popular questions from this chapter

Which of the following represent the correct order of nucleophillic addition for (I) HCHO, (II) \(\mathrm{CH}_{3} \mathrm{COCH}_{3^{+}}\) (III) \(\mathrm{CH}_{3} \mathrm{CHO}\), (IV) \(\mathrm{CH}_{3} \mathrm{COC}_{2} \mathrm{H}_{5}\) (a) \(\mid>\mathrm{II}>\mathrm{III}>\mathrm{IV}\) (b) \(\mathrm{I}>\mathrm{III}>\mathrm{II}>\mathrm{IV}\) (c) \(\mathrm{IV}>\mathrm{II}>\mathrm{III}>\mathrm{I}\) (d) \(\mathrm{I}>\mathrm{IV}>\mathrm{III}>\mathrm{II}\)

In which case, the intermediate involved is incorrect? (a) Pinacol - pinacolone rearrangement - carbocation (b) Hofmann's bromamide reaction \(-\) Nitrene (c) Aldol condensation - carbocation (d) Anti Markonikoff's addition of HBr to propene Free radical.

Free radicals are involved in mechanism of (a) Kharash effect (b) Cracking of alkanes (c) Halogenation of alkanes (d) all of these

Arrange the following compounds in order of increasing dipole moment. \(\mathrm{Cl}\) Clc1cccc(Cl)c1Cl Clc1ccc(Cl)cc1 Clc1ccc(Cl)c(Cl)c1 (I) (II) (III) ClC=C=C=CCl Clc1cc(Cl)cc(Cl)c1 (IV) (V) (a) \(\mathrm{II}=\mathrm{IV}<\mathrm{V}<\mathrm{I}<\mathrm{III}\) (b) \(\mathrm{IV}<\mathrm{III}<\mathrm{I}\) (c) \(\mathrm{II}<\mathrm{IV}<\mathrm{III}<\mathrm{I}<\mathrm{V}\) (d) \(\mathrm{V}=\mathrm{II}=\mathrm{IV}\)

Consider the following compounds: (1) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{NH}_{2}\) (2) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{NH}-\mathrm{C}_{6} \mathrm{H}_{5}\) (3) \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \mathrm{~N}\) (4) \(\mathrm{CH}_{1}-\mathrm{CH}_{2}-\mathrm{NH}_{2}\) Arrange these compounds in decreasing order of their basicity. (a) \(1>2>3>4\) (b) \(2>3>1>4\) (c) \(3>2>1>4\) (d) \(4>1>2>3\)

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