Question

Using your reaction roadmap as a guide, show how to convert 1-propanol into 2-hexyne. You must use 1-propanol as the source of all carbon atoms in the target molecule. Show all reagents needed and all molecules synthesized along the way.

Short answer

Question: Describe the sequence of reactions required to convert 1-propanol into 2-hexyne. Answer: To convert 1-propanol to 2-hexyne, we need to perform the following reactions: 1. Convert 1-Propanol to 1-Bromopropane using HBr. 2. Double the carbon chain via two consecutive Grignard reactions involving magnesium (Mg) and ethyl bromide (C₂H₅Br). 3. Brominate pentane to form 2-Bromopentane. 4. Form 2-Pentyne via an elimination reaction with NaNH₂. 5. Form 2-Hexyne via another elimination reaction with NaNH₂.

Step-by-step solution

Convert 1-Propanol to 1-Bromopropane

To start the synthesis, we need to convert the alcohol group in 1-propanol to a better leaving group. We can achieve this by converting 1-propanol into 1-bromopropane. This can be done via an S_N2 reaction using hydrobromic acid (HBr). The reaction is as follows: 1-Propanol + HBr → 1-Bromopropane + H₂O

Double the Carbon Chain via Grignard Reaction

Now, we need to extend the carbon chain. To do this, we will first convert 1-bromopropane into a Grignard reagent by reacting it with magnesium metal (Mg) in diethyl ether as the solvent. The reaction is as follows: 1-Bromopropane + Mg → Propylmagnesium bromide Next, react propylmagnesium bromide with ethyl bromide (C₂H₅Br) via another Grignard reaction. This will add two carbon atoms to the chain. The reaction is as follows: Propylmagnesium bromide + C₂H₅Br → Pentane (C₅H₁₂)

Bromination of Pentane to form 2-Bromopentane

To introduce the triple bond and obtain 2-hexyne, we first need a good leaving group at the desired position. We can achieve this by brominating the pentane, specifically 2-bromopentane as the major product. The reaction is as follows: Pentane + Br₂ → 2-Bromopentane + HBr

Formation of 2-Pentyne via Elimination Reaction

Now, we'll introduce a double bond in the 2-bromopentane by performing an E2 elimination reaction using sodium amide (NaNH₂). The reaction is as follows: 2-Bromopentane + NaNH₂ → 2-Pentyne + NaBr + NH₃

Formation of 2-Hexyne via Another Elimination Reaction

Finally, we will perform another E2 elimination reaction with 2-pentyne and sodium amide (NaNH₂) to form the triple bond in 2-hexyne. The reaction is as follows: 2-Pentyne + NaNH₂ → 2-Hexyne + NaBr + NH₃ In conclusion, the sequence of reactions to convert 1-propanol to 2-hexyne is as follows: 1. Convert 1-Propanol to 1-Bromopropane using HBr. 2. Double the carbon chain via two consecutive Grignard reactions involving magnesium (Mg) and ethyl bromide (C₂H₅Br). 3. Bromination of pentane to form 2-Bromopentane. 4. Formation of 2-Pentyne via an elimination reaction with NaNH₂. 5. Formation of 2-Hexyne via another elimination reaction with NaNH₂.

Key concepts

Grignard Reaction

The Grignard reaction is a cornerstone of organic synthesis. It involves the formation of a Grignard reagent, which is a type of organomagnesium compound, and is used to form carbon-carbon bonds.

A Grignard reagent is prepared by reacting an alkyl halide with magnesium metal in an anhydrous solvent, typically diethyl ether. In our example, 1-bromopropane reacts with magnesium to form propylmagnesium bromide. This reaction is highly sensitive to moisture; even trace amounts of water can quench the reagent, rendering it ineffective.

Once formed, the Grignard reagent can act as a nucleophile and attack electrophiles, such as a carbon atom within a carbonyl group. In the conversion of 1-propanol to 2-hexyne, the propylmagnesium bromide reacts with ethyl bromide. This step effectively doubles the carbon chain, expanding it from a three-carbon 1-propanol to a five-carbon chain, setting the stage for later steps to introduce unsaturation and reach the intended target molecule, 2-hexyne.

Being able to perform a Grignard reaction is essential for chemists, particularly when synthesizing complex organic molecules from smaller, simpler starting materials.

SN2 Reaction

The SN2 reaction is a type of nucleophilic substitution where a nucleophile attacks an electrophilic center with simultaneous departure of a leaving group, all occurring in a single, concerted step. The '2' in SN2 denotes bimolecular, meaning the reaction rate depends on the concentration of both the nucleophile and the electrophilic substrate.

In the context of the exercise, the initial step involves converting 1-propanol into 1-bromopropane. This transformation employs an SN2 mechanism where the hydroxyl group of 1-propanol is a poor leaving group and must first be converted to a better one, such as a bromide, to facilitate the displacement. Hydrobromic acid provides the bromide ion, which acts as a good nucleophile and displaces the hydroxyl group from 1-propanol, resulting in 1-bromopropane and water.

This reaction is characterized by its 'backside attack', where the nucleophile approaches from the opposite side of the leaving group, leading to an inversion of configuration at the carbon center. Understanding SN2 mechanisms is crucial for predicting the outcomes of synthesis exercises, including those involving the formation of different types of organohalides.

E2 Elimination Reaction

An E2 reaction is a bimolecular elimination process where a base removes a proton from a β-carbon, and the electrons from the broken C-H bond form a double bond as the leaving group departs. This reaction type is typically one step, with the bond formations and breakages occurring simultaneously.

In our exercise, an E2 reaction is employed twice. Initially, it is used to create 2-pentyne from 2-bromopentane. Sodium amide (NaNH₂) acts as a strong base and abstracts a proton that is β to the bromine atom in 2-bromopentane. The adjacent C-H and C-Br bonds break and form, respectively, resulting in the formation of 2-pentyne and the expulsion of bromide and ammonia.

The second E2 reaction in the exercise further extends this process by removing an additional β-hydrogen from 2-pentyne, generating the triple bond necessary for the final product, 2-hexyne. Mastery of E2 mechanisms is important for students as it enables them to perform dehydrohalogenations and other key elimination reactions crucial in the synthesis of alkenes and alkynes from alkyl halides.