When 2-bromo-2-methylbutane is reacted with hot alcoholic potassium hydroxide, one major product K and a minor product J are formed. Both K and J decolorize bromine dissolved in carbon tetrachloride. When product J is reacted with HBr in the presence of benzoyl peroxide or hydrogen peroxide, product H is formed. When H is treated with aqueous sodium hydroxide, compound I is formed. When compound I is treated with acidified potassium dichromate (VI), a compound L is formed. When compound I is treated with Lucas reagent (ZnCl/HCl), the solution remains colorless unless it is subjected to heat. The solution forms an oily layer when heated. A solution of L turns litmus red. When a solution of L is added to a saturated solution of sodium bicarbonate, there is effervescence and carbon dioxide is evolved. 2.1 Write down a chemical equation for the reaction that results in the formation of compounds K and J. [2] 2.2 Give the IUPAC names of compound J. [1] 2.3 Give a reason why K is the major product when 2-bromo-2-methylbutane is reacted with hot alcoholic potassium hydroxide. [1] 2.4 Name and illustrate the reaction mechanism through which K is formed. Your illustration should have arrows showing electron movements as well as a clearly identified rate-determining step. [5] 2.5 Write an equation showing the reaction of J with HBr in the presence of benzoyl peroxide or hydrogen peroxide to form H. [2] 2.6 Draw the full structural formula of H and give its IUPAC name. [2] 2.7 Give the name of the reaction by which compound I is formed from the treatment of H with aqueous sodium hydroxide. [1] 2.8 Name and illustrate the mechanism of the reaction of H with aqueous sodium hydroxide. Your illustration should have arrows showing electron movements as well as a clearly identified rate-determining step. [5] 2.9 Give the identity of compound L and draw its line structure. [2] 2.10 What product would have been formed if the reaction between J and HBr was performed in the absence of benzoyl peroxide or hydrogen peroxide? Draw the structure of this product. [2] 2.11 What would you observe if the product you identified in 2.10 above was treated with dilute sulfuric acid followed by reaction with the Lucas reagent? Give a reason for your answer. [2] 3. Title_with_topic: Organic Chemistry Reactions and Mechanisms
The Correct Answer and Explanation is:
Let’s break down the questions and explain each part:
2.1 Reaction that results in the formation of K and J:
When 2-bromo-2-methylbutane is reacted with hot alcoholic potassium hydroxide, the reaction undergoes an elimination reaction. In this case, the reaction is a β-elimination (E2) reaction, where a proton (H⁺) is removed from the carbon adjacent to the carbon bonded to the bromine atom, and the bromine is eliminated. This results in the formation of alkenes.
The products K and J are alkenes that form as a result of the elimination of HBr. The two possible products come from removing a hydrogen from different adjacent carbon atoms.
For the reaction:
- K (major product): 2-methylbut-2-ene
- J (minor product): 2-methylbut-1-ene
The equation is:C5H11Br+KOH (alc)→C5H10+HBr\text{C}_5\text{H}_{11}\text{Br} + \text{KOH (alc)} \rightarrow \text{C}_5\text{H}_{10} + \text{HBr}C5H11Br+KOH (alc)→C5H10+HBr
Where C₅H₁₁Br is 2-bromo-2-methylbutane and C₅H₁₀ represents the alkenes K and J.
2.2 IUPAC Name of Compound J:
Compound J is 2-methylbut-1-ene. This is because the double bond is located between the first and second carbon atoms in the chain.
2.3 Reason for K being the Major Product:
The major product K, 2-methylbut-2-ene, is formed because it is more stable. The stability of the alkene product is higher when the double bond is in the more substituted position (i.e., the second carbon is more substituted than the first). This is known as Zaitsev’s rule, which states that in an elimination reaction, the more substituted alkene is typically the major product.
2.4 Reaction Mechanism for Formation of K (E2 Mechanism):
The mechanism is E2 (bimolecular elimination), where the base (OH⁻) abstracts a proton from the carbon adjacent to the carbon bonded to the leaving group (Br), and the leaving group (Br⁻) simultaneously leaves. The electron pairs from the C-H bond form the double bond.
- Rate-determining step: The simultaneous breaking of the C-H and C-Br bonds.
Mechanism diagram:
- Base (OH⁻) abstracts a proton (H⁺) from carbon adjacent to the carbon bearing the bromine.
- The electron pair from the C-H bond forms the double bond.
- The bromine leaves as Br⁻.
2.5 Reaction of J with HBr in Presence of Benzoyl Peroxide:
In the presence of benzoyl peroxide, the reaction undergoes radical substitution, where the Br radical adds to the more substituted carbon in J, forming an alkyl bromide.
The equation:C5H10+HBr→benzoyl peroxideC5H11Br\text{C}_5\text{H}_{10} + \text{HBr} \xrightarrow{\text{benzoyl peroxide}} \text{C}_5\text{H}_{11}\text{Br}C5H10+HBrbenzoyl peroxideC5H11Br
This results in 2-bromo-2-methylbutane (compound H).
2.6 Structural Formula of H and IUPAC Name:
- The structure of H is 2-bromo-2-methylbutane. Its IUPAC name is 2-bromo-2-methylbutane.
2.7 Name of the Reaction for Formation of I:
The reaction of H with aqueous sodium hydroxide is a hydrolysis reaction, where the bromine is replaced by a hydroxyl group (OH). This is a nucleophilic substitution (SN1) reaction.
2.8 Mechanism of the Reaction of H with Aqueous NaOH:
- SN1 Mechanism: The bromine leaves first, forming a carbocation. The hydroxide ion (OH⁻) then attacks the carbocation.
- Rate-determining step: The formation of the carbocation.
Mechanism diagram:
- C-Br bond breaks, forming a carbocation.
- OH⁻ attacks the carbocation, resulting in the formation of 2-methylbutan-2-ol (compound I).
2.9 Identity and Line Structure of Compound L:
Compound L is 2-methylbutanoic acid (carboxylic acid). The line structure is:
scssCopyEdit CH3-CH(CH3)-COOH
2.10 Product Formed if the Reaction of J with HBr is Without Benzoyl Peroxide:
Without benzoyl peroxide, the reaction between J and HBr would not undergo radical substitution. Instead, Markovnikov’s rule would apply, leading to the addition of the hydrogen to the carbon with the fewest hydrogens. The product would be 2-bromo-1-methylbutane.
2.11 Observation with Lucas Reagent:
When the product from 2.10 (2-bromo-1-methylbutane) is treated with dilute sulfuric acid and then with Lucas reagent, there would be no immediate reaction unless heat is applied. Upon heating, SN1 substitution occurs, forming a tertiary carbocation. The product would be tert-butyl alcohol.
The solution would form an oily layer upon heating due to the formation of a more hydrophobic product.
In summary, the key reactions involve E2 elimination, radical substitution, nucleophilic substitution, and carboxylic acid formation based on the reactivity of the compounds involved. Each step can be explained through mechanisms such as E2 for elimination, SN1 for substitution, and radical mechanisms for addition.
