Question

Show how to convert 1-butyne to each of these compounds. (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C}=\mathrm{C}^{-} \mathrm{Na}^{+}$ (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C}=\mathrm{CD}\) (c) *C=C(C)C (d) C=C(C)C

Short answer

Question: Show how to convert 1-butyne into the following four compounds: (a) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C}=\mathrm{C}^{-}\mathrm{Na}^{+}\), (b) \(\mathrm{CH}_3 \mathrm{CH}_2\mathrm{C}=\mathrm{CD}\), (c) *C=C(C)C*, and (d) C=C(C)C. Answer: Compound (a): Perform a controlled reduction using Lindlar's catalyst (Pd/CaCO\(_3\)) and quinoline to convert the triple bond into a cis double bond, then deprotonate the product using a strong base such as sodium amide (NaNH\(_2\)). Compound (b): Perform a controlled reduction using Lindlar's catalyst to obtain a cis double bond, then perform a deuterium exchange using a deuterated acid like DCl (deuterium chloride). Compound (c): Perform a full reduction using sodium, liquid ammonia, and a catalytic amount of an electron carrier like naphthalene to obtain a trans double bond. Compound (d): Treat 1-butyne with a strong base such as potassium tert-butoxide (t-BuOK) in the presence of tert-butanol, causing the terminal hydrogen atom to migrate to the internal carbon and creating an internal double bond.

Step-by-step solution

Identify the starting material - 1-butyne

1-butyne has the following structure: \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}\). Its IUPAC name is but-1-yne, implying that it is an alkyne with four carbon atoms and a triple bond between the first and second carbon atoms.

Conversion to Compound (a)

To convert 1-butyne to compound (a) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C}=\mathrm{C}^{-}\mathrm{Na}^{+}\), we must change the triple bond into a double bond and add a sodium cation to form a naftyde ion. This can be achieved using a two-step process. 1. Perform a controlled reduction using Lindlar's catalyst (Pd/CaCO\(_3\)) and quinoline, which will convert the triple bond into a cis double bond, giving us cis-2-butene: \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C}=\mathrm{CH}_2\). 2. Deprotonation of cis-2-butene: Treat it with a strong base, such as sodium amide (NaNH\(_2\)), which will deprotonate the terminal proton of the alkene and form the desired product: \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C}=\mathrm{C}^{-}\mathrm{Na}^{+}\).

Conversion to Compound (b)

To convert 1-butyne to compound (b) \(\mathrm{CH}_3 \mathrm{CH}_2\mathrm{C}=\mathrm{CD}\), we must reduce the triple bond to a double bond and replace one of the terminal hydrogens with a deuterium atom. This can be achieved using the following two-step process: 1. As in Step 2, perform a controlled reduction using Lindlar's catalyst to obtain cis-2-butene: \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C}=\mathrm{CH}_2\). 2. Perform a deuterium exchange using a deuterated acid, such as DCl (deuterium chloride). This will replace one of the terminal hydrogens with a deuterium atom, giving the desired product: \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C}=\mathrm{CD}\).

Conversion to Compound (c)

To convert 1-butyne to compound (c) (SMILES code: *C=C(C)C*), first realize that this molecule is actually trans-2-butene. To convert 1-butyne to trans-2-butene, follow these steps: 1. Perform a full reduction of the alkyne using sodium, liquid ammonia, and a catalytic amount of an electron carrier like naphthalene. This will turn the triple bond into a trans double bond and yield the desired trans-2-butene: \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C}=\mathrm{C}\mathrm{(CH}_3\)).

Conversion to Compound (d)

To convert 1-butyne to compound (d) (SMILES code: C=C(C)C), we must change the terminal triple bond to an internal double bond. This can be achieved in a single step using an isomerization technique. 1. Treat 1-butyne with a strong base, such as potassium tert-butoxide (t-BuOK), in the presence of an appropriate solvent like tert-butanol. This base will cause the terminal hydrogen atom to migrate to the internal carbon, creating an internal double bond and forming the desired product: \(\mathrm{CH}_2=\mathrm{C}\mathrm{CH}_2\mathrm{CH}_3\).