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Chapter 8: Problem 48

Under certain reaction conditions, 2,3 -dibromobutane reacts with two equivalents of base to give three products, each of which contains two new \(\pi\) bonds. Product A has two sp hybridized carbon atoms, product \(\mathbf{B}\) has one sp hybridized carbon atom, and product \(\mathbf{C}\) has none. What are the structures of \(\mathbf{A}, \mathbf{B}\), and \(\mathbf{C} ?\)

Short Answer

Expert verified
A: 2-butyne; B: 1-butyne; C: 1,3-butadiene.

Step by step solution

01

Identify the structure of 2,3-dibromobutane

Start with the structure of 2,3-dibromobutane, which has the chemical formula C4H6Br2. The molecule is a four-carbon chain (butane) with two bromine atoms attached to the second and third carbon atoms.
02

Understand the reaction mechanism

The reaction involves removing two equivalents of hydrogen bromide (HBr) from 2,3-dibromobutane. This is done by the base abstracting protons from adjacent carbon atoms, leading to the formation of double bonds (alkenes).
03

Formulate Product A

Product A is an alkyne with two sp hybridized carbon atoms, meaning it must have triple bonds between two carbons. After removing two equivalents of HBr, we get 2-butyne with the structure: \( CH_3-C\equiv C-CH_3 \).
04

Formulate Product B

Product B has one sp hybridized carbon atom, indicating the presence of a triple bond but only on one carbon. Thus, 1-butyne is the result, with the structure: \( CH\equiv C-CH_2-CH_3 \).
05

Formulate Product C

Product C has no sp hybridized carbon atoms, so it must only contain double bonds. The logical structure that fits is 1,3-butadiene: \( CH_2=CH-CH=CH_2 \).
06

Verify the products

Check the hybridization: 2-butyne (A) has two sp hybridized carbons, 1-butyne (B) has one, and 1,3-butadiene (C) has none. This confirms our proposed structures.

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Key Concepts

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

2,3-Dibromobutane
2,3-Dibromobutane is a type of organic compound that features a four-carbon chain, known as butane, with two bromine atoms attached at specific positions. The chemical formula for this compound is C\(_4\)H\(_6\)Br\(_2\). The '2,3' designation specifies that the bromine atoms are located on the second and third carbon atoms. By understanding this structure, you begin to see how the molecule can undergo transformations to form different products. This specific arrangement of bromine atoms makes it a good candidate for base-induced elimination reactions.
Hybridization
Hybridization is a concept that helps us understand the shapes and properties of molecules, especially those involving carbon. In carbon compounds, hybridization can vary based on the number of bonds:
  • sp hybridization is seen in carbon atoms involved in triple bonds, like those found in alkynes.
  • sp\(^2\) hybridization happens in carbon atoms that form double bonds, typical of many alkenes.
  • sp\(^3\) hybridization is common in single-bonded carbon atoms, filling four orbitals through the overlap of one s and three p orbitals.
By determining the type of hybridization, you can predict the molecule鈥檚 shape and reactivity, which is crucial when looking at the products A, B, and C in this context.
Reaction Mechanism
A reaction mechanism describes the detailed process that occurs during a chemical reaction. For 2,3-dibromobutane, when treated with a base, a mechanism known as base-induced elimination occurs. The reaction primarily involves the removal of two molecules of hydrogen bromide (HBr) from the compound:
  • The base abstracts protons from the adjacent carbon atoms attached to bromine.
  • This leads to the formation of bonds between these carbon atoms, creating alkenes or alkynes due to the multiple elimination reactions.
Understanding the steps in the mechanism allows us to predict the formation of the final products, which, in our problem, include an alkyne and alkenes.
Alkenes
Alkenes are hydrocarbons that contain at least one carbon-carbon double bond. These bonds are characterized by sp\(^2\) hybridized carbon atoms. In the reaction involving 2,3-dibromobutane, the elimination of hydrogen bromide can also lead to the formation of alkenes, depending on the number of hydrogen and bromine atoms removed and their positions.An example from the reactions we've discussed is 1,3-butadiene. This product forms with conjugated double bonds, which stabilize the molecule through resonance. Such products without any sp hybridized carbons are significant for their reactivity and potential for further polymerization.
Base-Induced Elimination
Base-induced elimination is a common reaction pathway for removing small molecules like hydrogen halides (e.g., HBr) from organic compounds. In the context of 2,3-dibromobutane:
  • A strong base abstracts a proton adjacent to a bromine atom, leading to the elimination of HBr.
  • The resulting molecule features unsaturated bonds (double or triple) as the base triggers the formation of such bonds by repeated elimination steps.
Base-induced elimination reactions are essential for forming alkenes and alkynes, driving the conversion of saturated compounds to unsaturated ones. This technique is instrumental for chemically manipulating hydrocarbons to suit industrial and research needs.

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Most popular questions from this chapter

Rank the alkyl halides in each group in order of increasing reactivity in an E2 reaction. a. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{Br}) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}\) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}\) b. ClCCC1CCCCC1 CC1(Cl)CCCCC1 CC1CCCC(Cl)C1

Explain why \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{2} \mathrm{CH}_{3}\) reacts faster than \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}\) in an \(\mathrm{E} 2\) reaction, even though both alkyl halides are \(2^{\circ}\).

What alkenes are formed from each alkyl halide by an \(\mathrm{E} 1\) reaction? Use the Zaitsev rule to predict the major product. a. CCC(C)(C)Cl b. CCCC1CCCC1C

Draw an E1 mechanism for the following reaction. Draw the structure of the transition state for each step. $$ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{Cl}) \mathrm{CH}_{2} \mathrm{CH}_{3}+\mathrm{CH}_{3} \mathrm{OH} \quad \longrightarrow \quad\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCH}_{3}+\mathrm{CH}_{3} \mathrm{OH}_{2}+\mathrm{Cl}^{-} $$

Label the \(\alpha\) and \(\beta\) carbons in each alkyl halide, and draw all possible elimination products in each reaction. a. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2}-\mathrm{Cl} \quad \stackrel{\mathrm{K}^{+}-\mathrm{OC}\left(\mathrm{CH}_{3}\right)_{3}}{\longrightarrow}\) c. CCC(Cl)CC CO b. CCCCC(C)Br d. CC1(Br)CCCCC1 \(\mathrm{K}^{+-\mathrm{OH}}\)
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