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Another important pattern in organic synthesis is the construction of \(\mathrm{C}-\mathrm{C}\) bonds. Using your roadmap as a guide, show how to convert propane into hex-1-en-4-yne. You must use propane as the source of all of the carbon atoms in the hex-1-en-4-yne product. Show all reagents needed and all molecules synthesized along the way.

Short Answer

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Question: Outline the steps involved in converting propane into hex-1-en-4-yne using organic synthesis methods. Answer: The conversion of propane into hex-1-en-4-yne can be achieved through the following steps: 1) Convert propane into prop-1-yne by first converting propane into propene through elimination with a strong base like potassium tert-butoxide (t-BuOK), then converting propene into prop-1-yne using another strong base like NaNH2 in liquid ammonia (NH3). 2) Form a C-C bond to create hex-3-yne by performing a double prop-1-yne coupling reaction using Glaser's coupling method with copper(I) chloride (CuCl) as the catalyst, oxygen (O2) as the oxidizing agent, and an amine (e.g., triethylamine). 3) Partially reduce the triple bond in hex-3-yne to form hex-1-en-4-yne using the Lindlar catalyst (Pd/CaCO3/Pb) with hydrogen gas (H2), which selectively reduces alkynes to alkenes without attacking existing double bonds.

Step by step solution

01

Convert propane into prop-1-yne

To begin the synthesis, we will first convert propane into prop-1-yne. This process involves two steps: (1) converting propane into propene through elimination and (2) converting propene into prop-1-yne through further elimination. To accomplish the first step, we can utilize a strong base like potassium tert-butoxide (t-BuOK) to induce the elimination of a proton and a leaving group (hydrogen gas) to form propene. To accomplish the second step, we can convert propene to prop-1-yne using a strong base like NaNH2 in liquid ammonia (NH3).
02

Form a C-C bond to create hex-3-yne

Now that we have prop-1-yne, we can use it as a building block to form hex-3-yne. We do this by performing a double prop-1-yne coupling reaction through Glaser's coupling, which involves using copper(I) chloride (CuCl) as the catalyst and oxygen (O2) as the oxidizing agent in the presence of an amine (e.g., triethylamine). This reaction will couple two prop-1-yne molecules together, forming a triple bond between the terminal carbons and creating hex-3-yne.
03

Partially reduce the triple bond to form hex-1-en-4-yne

Finally, we will selectively reduce one of the triple bonds in hex-3-yne to form hex-1-en-4-yne. This can be achieved via the Lindlar catalyst (Pd/CaCO3/Pb) with hydrogen gas (H2). The Lindlar catalyst is a heterogeneous catalyst that selectively reduces triple bonds (alkynes) to double bonds (alkenes) without attacking double bonds already present. With this reaction, we will successfully convert propane into hex-1-en-4-yne, using propane as the only source of carbon atoms in the product.

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Key Concepts

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

Alkyne Formation
Alkyne formation is a fundamental part of organic chemistry, particularly when it comes to synthesizing complex molecules. Alkynes are hydrocarbons containing at least one carbon-carbon triple bond, characterized by the general formula CnH2n-2. To form an alkyne from a simple alkane like propane, a multistep process is necessary.

Firstly, an elimination reaction is used to remove hydrogen atoms and create a double bond, resulting in an alkene. This can be accomplished using a strong base. The alkene is then further dehydrohalogenated to form an alkyne, such as prop-1-yne from propene. Strong bases like sodium amide (NaNH2) are commonly involved in this second elimination step as they are capable of removing a proton from the carbon adjacent to the double bond, thus forming the desired triple bond.

The formation of alkynes is crucial in synthetic chemistry as it allows for the construction of more complex molecules through the introduction of highly reactive triple bonds which can be further manipulated to form various functional groups.
C-C Bond Formation
In organic synthesis, the construction of carbon-carbon (C-C) bonds is essential for building the skeletons of organic molecules. C-C bond formation enables the creation of larger and more complex structures from simpler precursors. There are various methods to form C-C bonds, and one common approach involves using coupling reactions.

Diverse strategies exist for coupling, such as metal-catalyzed cross-coupling reactions, which have become a staple in the chemist's toolkit. In the exercise mentioned, the C-C bond formation occurs through Glaser coupling, a method specifically designed to join together two terminal alkynes, resulting in a di-alkyne. C-C bond formation reactions are crucial in organic synthesis because they help expand the diversity of available organic compounds, providing the backbone for advanced materials, pharmaceuticals, and many more complex organic products.
Glaser Coupling
Glaser coupling is a specialized chemical reaction used in organic synthesis for the direct coupling of two terminal alkyne units to form a symmetrical di-alkyne. This reaction is named after the German chemist Carl Glaser who first reported it in 1869. Glaser coupling typically requires a copper(I) catalyst, like copper(I) chloride (CuCl), and an oxygen source to oxidize the copper which facilitates the formation of the carbon-carbon triple bond between the alkyne units.

The presence of an amine, such as triethylamine, is also important as it acts as a base to deprotonate the terminal alkyne, making it more reactive. In the context of the exercise, using Glaser coupling allows for the dimerization of prop-1-yne molecules, seamlessly creating the longer hex-3-yne chain, which is a crucial step in the synthetic route towards hex-1-en-4-yne.
Lindlar Catalyst Reduction
The Lindlar catalyst, a finely powdered palladium deposited on calcium carbonate and modified by the addition of various forms of lead, is specifically designed to catalyze the partial reduction of alkynes to alkenes. It provides a way to stop the reduction at the alkene stage, avoiding the full hydrogenation to an alkane.

When the Lindlar catalyst is used with molecular hydrogen (H2), the reaction is selective for the triple bond, reducing it to a cis-alkene. This selective hydrogenation is incredibly useful for synthesizing alkenes where the stereochemistry is important. In the proposed exercise, the Lindlar catalyst serves as a key reagent to selectively reduce one of the triple bonds in hex-3-yne, resulting in the target molecule hex-1-en-4-yne without affecting other double or triple bonds in the molecule. By using this catalyst, chemists can finely tune their synthetic pathways to obtain the desired level of saturation in organic molecules.

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Most popular questions from this chapter

Suggest a product of the following reaction. HI is a very strong acid. $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}+2 \mathrm{HI} \longrightarrow $$

The reaction of 1-bromopropane and sodium hydroxide in ethanol occurs by an \(\mathrm{S}_{\mathrm{N}} 2\) mechanism. What happens to the rate of this reaction under the following conditions? (a) The concentration of \(\mathrm{NaOH}\) is doubled. (b) The concentrations of both \(\mathrm{NaOH}\) and 1-bromopropane are doubled. (c) The volume of the solution in which the reaction is carried out is doubled.

Draw structural formulas of all chloroalkanes that undergo dehydrohalogenation when treated with KOH to give each alkene as the major product. For some parts, only one chloroalkane gives the desired alkene as the major product. For other parts, two chloroalkanes may work.

When cis-4-chlorocyclohexanol is treated with sodium hydroxide in ethanol, it gives mainly the substitution product trans-1,4-cyclohexanediol (1). Under the same reaction conditions, trans-4-chlorocyclohexanol gives 3-cyclohexenol (2) and the bicyclic ether (3).

1-Chloro-4-isopropylcyclohexane exists as two stereoisomers: one cis and one trans. Treatment of either isomer with sodium ethoxide in ethanol gives 4 -isopropylcyclohexene by an E2 reaction. CC(C)C1CC=CCC1 1-Chloro-4- 4-Isopropylcyclohexene isopropylcyclohexane The cis isomer undergoes E2 reaction several orders of magnitude faster than the trans isomer. How do you account for this experimental observation?

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