Question

If you work in the health profession, you might want to design a polymer for an artificial hip. Perhaps you decide to study the polymer of \((E)\)-2-pentene. But your lab is only stocked with 1-pentanol. Show the reactions you would need to use in order to convert 1-pentanol into \((E)\)-2-pentene. Hint: This conversion can be done cleanly in four steps.

Short answer

1-pentanol to 1-bromopentane (use HBr), 1-bromopentane to pent-1-ene (use KOtBu), pent-1-ene to pentane (use H2/Pd), pentane to (E)-2-pentene (use Cr_2O_3 catalyst).

Step-by-step solution

- Convert 1-pentanol to 1-bromopentane

The first step is to convert 1-pentanol to 1-bromopentane. This can be done by reacting 1-pentanol with hydrobromic acid \((HBr)\). The reaction that occurs is: \( \text{CH}_3\text{(CH}_2\text{)}_4\text{OH} + HBr \rightarrow \text{CH}_3\text{(CH}_2\text{)}_4\text{Br} + H_2O \).

- Convert 1-bromopentane to pent-1-ene

Next, 1-bromopentane needs to be converted to pent-1-ene. This can be achieved by using a strong base like potassium tert-butoxide (\(KOtBu\)). The reaction proceeds through an E2 elimination mechanism: \( \text{CH}_3\text{(CH}_2\text{)}_4\text{Br} + \text{KOtBu} \rightarrow \text{CH}_2= \text{CH-}\text{(CH}_2\text{)}_3\text{CH}_3 \).

- Convert pent-1-ene to pentane

The third step involves converting pent-1-ene to pentane. This can be done by hydrogenation, using hydrogen gas \((H_2)\) and a palladium catalyst \((Pd)\): \( \text{CH}_2= \text{CH-}\text{(CH}_2\text{)}_3\text{CH}_3 + H_2 \rightarrow \text{CH}_3\text{(CH}_2\text{)}_4\text{CH}_3 \).

- Convert pentane to \( (E) \)-2-pentene

In the final step, pentane must be dehydrogenated to form \( (E) \)-2-pentene. This step can be challenging, but heating pentane in the presence of a suitable dehydrogenation catalyst (like Cr_2O_3) can achieve the desired product: \( \text{CH}_3\text{(CH}_2\text{)}_4\text{CH}_3 \rightarrow \text{CH}_3\text{CH}= \text{CH-}\text{(CH}_2\text{)}_2\text{CH}_3 \).

Key concepts

1-pentanol conversion

Converting 1-pentanol to \(E\)-2-pentene involves a sequence of chemical reactions. To start, 1-pentanol (CH3(CH2)4OH) must be converted into 1-bromopentane (CH3(CH2)4Br). This initial step sets the stage for further transformations. By following a well-thought-out sequence of reactions, it is possible to obtain the final desired product.

hydrobromic acid reaction

The first step in the conversion of 1-pentanol is the reaction with hydrobromic acid (HBr). This reaction is an example of a substitution reaction, where the -OH group in 1-pentanol is replaced by a bromine atom, forming 1-bromopentane and water as a by-product.

The reaction can be represented as:
$$\text{CH}_3\text{(CH}_2\text{)}_4\text{OH} + HBr \rightarrow \text{CH}_3\text{(CH}_2\text{)}_4\text{Br} + H_2O$$

elimination mechanism

Next, 1-bromopentane undergoes an E2 elimination reaction to yield pent-1-ene. This reaction is facilitated by a strong base such as potassium tert-butoxide (KOtBu). During this process, a hydrogen atom and a bromine atom are removed, forming a double bond and resulting in pent-1-ene.

The reaction equation for the E2 elimination is:
$$\text{CH}_3\text{(CH}_2\text{)}_4\text{Br} + \text{KOtBu} \rightarrow \text{CH}_2=\text{CH-}\text{(CH}_2\text{)}_3\text{CH}_3$$

hydrogenation

Hydrogenation is the process of adding hydrogen (H2) to an unsaturated compound, such as an alkene, to convert it into a saturated compound. In this case, pent-1-ene is converted into pentane by reacting it with hydrogen gas in the presence of a palladium catalyst (Pd).

The hydrogenation reaction is represented as:
$$\text{CH}_2=\text{CH-}\text{(CH}_2\text{)}_3\text{CH}_3 + H_2 \rightarrow \text{CH}_3\text{(CH}_2\text{)}_4\text{CH}_3$$

dehydrogenation catalysis

The final step involves the dehydrogenation of pentane to obtain \(E\)-2-pentene. Dehydrogenation is the removal of hydrogen from a molecule, forming a double bond. This step can be achieved by heating pentane in the presence of a suitable dehydrogenation catalyst, such as chromium oxide (Cr2O3).

The reaction is shown as:
$$\text{CH}_3\text{(CH}_2\text{)}_4\text{CH}_3 \rightarrow \text{CH}_3\text{CH}=\text{CH-}\text{(CH}_2\text{)}_2\text{CH}_3$$

This step requires careful control of reaction conditions to ensure the desired \(E\) configuration of the product.