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Chapter 26: Problem 161

1, 2 dibromopropane on treatment with \(X\) moles of \(\mathrm{NaNH}_{2}\) followed by treatment with ethyl bromide gave a pentyne, the value of \(\mathrm{X}\) is

Short Answer

Expert verified
The value of \(X\) is 2.

Step by step solution

01

Identify the structure of 1,2-dibromopropane

1,2-dibromopropane has a propane backbone with bromine atoms attached to the first and second carbon atoms. Its structure can be represented as \( ext{CH}_2Br-CHBr-CH_3\).
02

Understand the role of NaNH鈧�

Sodium amide (\( ext{NaNH}_2\)) is a strong base and is often used for elimination reactions to produce alkynes. In this case, \( ext{NaNH}_2\) will remove HBr from 1,2-dibromopropane, converting it into an alkyne.
03

Identify the elimination process

Treating 1,2-dibromopropane with \( ext{NaNH}_2\) will remove two equivalents of HBr, forming propyne (\( ext{CH}_3-C\equiv C-H\)). This requires \(2\) moles of \( ext{NaNH}_2\) for the complete elimination of both bromines.
04

Understand the reaction with ethyl bromide

After forming propyne, reacting it with ethyl bromide (an alkyl halide) in the presence of a strong, nucleophilic base will lead to nucleophilic substitution, extending the carbon chain to form pentyne. This step does not affect the value of \(X\).
05

Determine the value of X required

Because two moles of \( ext{NaNH}_2\) are required to eliminate both equivalents of HBr from 1,2-dibromopropane, the value of \(X\) is \(2\).

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Key Concepts

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

Elimination Reactions
Elimination reactions are a fundamental transformation in organic chemistry that involve the removal of elements from a molecule, typically forming a multiple bond. In these reactions, two substituents are removed from a molecule, resulting in the formation of a double or triple bond, depending on the substrate and reaction conditions.

For example:
  • When a single elimination occurs, it typically transforms saturated single-bonded compounds into unsaturated compounds with one double bond: an alkane into an alkene.
  • In a double elimination, which was showcased in this specific problem, it results in the formation of a triple bond, converting a dihaloalkane into an alkyne. This is often needed when synthesizing more complex organic molecules with varied functionalities.
Elimination reactions are often initiated by strong bases, like sodium amide. These reactions must be carefully controlled to obtain the desired product, such as an alkyne from a dihaloalkane through the removal of both bromine atoms as presented in this exercise.
Alkynes
Alkynes are hydrocarbons characterized by the presence of a triple bond between two carbon atoms. This triple bond gives alkynes unique properties and reactivity patterns compared to other hydrocarbons.

Some key features of alkynes include:
  • The carbon atoms in the triple bond are sp hybridized, resulting in a linear geometry with a bond angle of approximately 180 degrees.
  • The triple bond consists of one sigma bond and two pi bonds, making alkynes generally more reactive than alkanes and alkenes.
In the step from the original exercise, eliminating two bromine atoms from 1,2-dibromopropane using sodium amide leads to the formation of propyne, an alkyne, demonstrating a critical transformation from a saturated molecule to an unsaturated one.
Sodium Amide (NaNH2)
Sodium amide (NaNH鈧�) is a powerful, nucleophilic base widely used in organic chemistry, especially in the formation of alkynes via elimination reactions. NaNH鈧� plays a pivotal role due to its ability to abstract protons and facilitate the removal of hydrogen halides, aiding the formation of multiple bonds.

Its characteristics include:
  • Strong basicity: This allows it to effectively deprotonate even weakly acidic protons adjacent to a leaving group, such as a halogen in haloalkanes.
  • Nucleophilic nature: While its primary use in these contexts is as a base, NaNH鈧� can also participate in nucleophilic substitution reactions.
In the problem, NaNH鈧� removes two equivalents of HBr from 1,2-dibromopropane, facilitating the conversion to an alkyne by removing the bromine atoms and forming a triple bond. This requires precisely two moles of NaNH鈧�, as calculated in the solution, to achieve the complete double elimination necessary to form propyne.

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Most popular questions from this chapter

2-bromopentane is heated with potassium ethoxide in ethanol. The major product obtained is (a) 2 -ethoxypentane (b) 1 -pentane (c) cis-2-pentane (d) trans-2-pentane

\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Cl} \stackrel{\text { alc. } \mathrm{KOH}}{\longrightarrow}(\mathrm{X}) \stackrel{\mathrm{HBr}}{\longrightarrow}(\mathrm{Y}) \stackrel{\mathrm{Na} / \mathrm{cther}}{\longrightarrow}(\mathrm{Z})\) In the above sequence of reaction, the product \((Z)\) is (a) propane (b) hexane (c) 2,3 -dimethylbutane (d) allyl bromide

\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr} \stackrel{\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C} \overline{\mathrm{O}} \mathrm{K}^{+}}{\longrightarrow}(\mathrm{A}) \stackrel{\mathrm{NBS}}{\longrightarrow}\) (B) \(\frac{\text { Cold alk. }}{\mathrm{KMnO}_{4}}\) (C). The total number of stereisomers possible for the compound (C) are?

Identify the set of reagents/reaction conditions 'X' and ' \(\mathrm{Y}\) ' in the following set of transformations: CCCBr (a) \(\mathrm{X}=\) dilute aqueous \(\mathrm{NaOH}, 20^{\circ} \mathrm{C}\) \(\mathrm{Y}=\mathrm{HBr} /\) acetic acid, \(20^{\circ} \mathrm{C}\) (b) \(\mathrm{X}=\) concentrated alcoholic \(\mathrm{NaOH}, 80^{\circ} \mathrm{C}\) \(\mathrm{Y}=\mathrm{HBr} /\) acetic acid, \(20^{\circ} \mathrm{C}\) (c) \(\mathrm{X}=\) dilute aqueous \(\mathrm{NaOH}, 20^{\circ} \mathrm{C}\) \(\mathrm{Y}=\mathrm{Br}_{2} / \mathrm{CHCl}_{3}, 0^{\circ} \mathrm{C}\) (d) \(\mathrm{X}=\) concentrated alcoholic \(\mathrm{NaOH}, 80^{\circ} \mathrm{C}\) \(\mathrm{Y}=\mathrm{Br}_{2} / \mathrm{CHCl}_{3}, 0^{\circ} \mathrm{C}\)

Match the following: List I (Reactants) 1\. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{OH} \stackrel{\mathrm{NaBr}, \mathrm{H}_{2} \mathrm{SO}_{4}, \Delta}{\longrightarrow}\) 2\. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH} \frac{\text { Conc. } \mathrm{HCl}}{\text { Roser temp }}\) 3\. \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3} \stackrel{\mathrm{PB}_{3}}{\longrightarrow}\) 4\. \(\mathrm{Me}_{2} \mathrm{CHCH}_{2} \mathrm{OH} \stackrel{\mathrm{soc} \mathrm{h}}{\longrightarrow}\) List II (Alkyl halides) A. \(\mathrm{CH}_{3} \mathrm{CHBr}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3}\) B. \(\mathrm{Me}_{2} \mathrm{CHCH}_{2} \mathrm{Cl}\) C. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) D. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{Br}\) The correct matching is \(\begin{array}{lllll} & 1 & 2 & 3 & 4 \\ \text { (a) } \mathrm{C} & \mathrm{D} & \mathrm{B} & \mathrm{A}\end{array}\) (b) \(\mathrm{C}\) \(\mathrm{D} \quad \mathrm{A} \quad \mathrm{B}\) (c) D \(\mathrm{C} \quad \mathrm{A} \quad \mathrm{B}\) (d) D C \(\mathrm{A}\)
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