Question

Draw structural formulas for the major product(s) formed by reaction of 3 -hexyne with each of these reagents. (Where you predict no reaction, write NR.) (a) \(\mathrm{H}_{2}\) (excess)/ \(\mathrm{Pt}\) (b) \(\mathrm{H}_{2}\) /Lindlar catalyst (c) \(\mathrm{Na}\) in \(\mathrm{NH}_{3}(l)\) (d) \(\mathrm{BH}_{3}\) followed by \(\mathrm{H}_{2} \mathrm{O}_{2} / \mathrm{NaOH}\) (e) \(\mathrm{BH}_{3}\) followed by \(\mathrm{CH}_{3} \mathrm{COOH}\) (f) \(\mathrm{BH}_{3}\) followed by \(\mathrm{CH}_{3} \mathrm{COOD}\) (g) \(\mathrm{Cl}_{2}\) ( \(\left.1 \mathrm{~mol}\right)\) (h) \(\mathrm{NaNH}_{2}\) in \(\mathrm{NH}_{3}(l)\) (i) \(\mathrm{HBr}\) (1 mol) (j) \(\mathrm{HBr}(2 \mathrm{~mol})\) (k) \(\mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{HgSO}_{4}\)

Short answer

(a) \(\mathrm{H}_{2}\) (excess)/ \(\mathrm{Pt}\): hexane (b) \(\mathrm{H}_{2}\) / Lindlar catalyst: (Z)-3-hexene (c) \(\mathrm{Na}\) in \(\mathrm{NH}_{3}(l)\): (E)-3-hexene (d) \(\mathrm{BH}_{3}\) followed by \(\mathrm{H}_{2} \mathrm{O}_{2} / \mathrm{NaOH}\): 3-hexanone (e) \(\mathrm{BH}_{3}\) followed by \(\mathrm{CH}_{3} \mathrm{COOH}\): 3-hexanoic acid methyl ester (f) \(\mathrm{BH}_{3}\) followed by \(\mathrm{CH}_{3} \mathrm{COOD}\): 3-hexanoic acid methyl ester (\(\mathrm{OD}\)) (g) \(\mathrm{Cl}_{2}\) ( \(\left.1 \mathrm{~mol}\right)\): 3,4-Dichloro-3-hexene (h) \(\mathrm{NaNH}_{2}\) in \(\mathrm{NH}_{3}(l)\): No reaction (NR) (i) \(\mathrm{HBr}\) (1 mol): (E)-1-bromo-3-hexene (j) \(\mathrm{HBr}(2 \mathrm{~mol})\): 1,2-Dibromo-3-hexane (k) \(\mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{HgSO}_{4}\): 2-hexanone

Step-by-step solution

(a) \(\mathrm{H}_{2}\) (excess)/ \(\mathrm{Pt}\)

The catalyst platinum with an excess of hydrogen gas will cause complete hydrogenation of the triple bond, turning it into an alkane. Major Product: hexane

(b) \(\mathrm{H}_{2}\) / Lindlar catalyst

The Lindlar catalyst will turn the triple bond into a cis-double bond without further reduction, creating an alkene. Major Product: (Z)-3-hexene

(c) \(\mathrm{Na}\) in \(\mathrm{NH}_{3}(l)\)

Sodium in liquid ammonia will perform a trans-elimination, leading to the formation of a trans-double bond. Major Product: (E)-3-hexene

(d) \(\mathrm{BH}_{3}\) followed by \(\mathrm{H}_{2} \mathrm{O}_{2} / \mathrm{NaOH}\)

Hydroboration-oxidation reaction will occur, adding an \(\mathrm{OH}\) group to the less substituted carbon of the triple bond, creating an enol. The enol can tautomerize to give a ketone. Major Product: 3-hexanone

(e) \(\mathrm{BH}_{3}\) followed by \(\mathrm{CH}_{3} \mathrm{COOH}\)

The reaction part is the same as in (d), the only difference is that the final product will be a carboxylic acid derivative. Major Product: 3-hexanoic acid methyl ester

(f) \(\mathrm{BH}_{3}\) followed by \(\mathrm{CH}_{3} \mathrm{COOD}\)

The reaction is similar to (e), except the alcohol will have a deuterium (D) atom instead of a hydrogen atom in the final product. Major Product: 3-hexanoic acid methyl ester (\(\mathrm{OD}\))

(g) \(\mathrm{Cl}_{2}\) ( \(\left.1 \mathrm{~mol}\right)\)

One equivalent of chlorine will react with the terminal alkyne to form a vicinal dichloride compound. Major Product: 3,4-Dichloro-3-hexene

(h) \(\mathrm{NaNH}_{2}\) in \(\mathrm{NH}_{3}(l)\)

Sodium amide will perform an elimination reaction, removing the terminal hydrogen and yield a compound with a terminal triple bond. Major Product: No reaction (NR)

(i) \(\mathrm{HBr}\) (1 mol)

One equivalent of HBr will add across the triple bond to form a 1-bromoalkene. Major Product: (E)-1-bromo-3-hexene

(j) \(\mathrm{HBr}(2 \mathrm{~mol})\)

Two equivalents of HBr will add across the triple bond, followed by the double bond, forming a dibromoalkane. Major Product: 1,2-Dibromo-3-hexane

(k) \(\mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{HgSO}_{4}\)

This reaction will perform oxymercuration-demercuration, turning the alkyne into a ketone. The terminal alkyne will go through a hydration reaction producing an enol that will tautomerize into a ketone. Major Product: 2-hexanone

Key concepts

Hydrogenation of Alkynes

Hydrogenation of alkynes is a chemical reaction in which the triple bond character of an alkyne is reduced to a double or single bond through the addition of hydrogen (H₂). This reaction allows chemists to selectively synthesize alkenes or alkanes from alkynes.

• When an alkyne is hydrogenated using a metal catalyst like platinum (Pt) with an excess of hydrogen, the triple bond is completely saturated, resulting in an alkane. This process is termed complete hydrogenation or full hydrogenation.
• In partial hydrogenation, a catalyst such as the Lindlar catalyst is employed to add hydrogen across only one pi bond of the triple bond, creating a cis-alkene.

The choice of catalyst and reaction conditions dictates whether the alkyne will be fully or partially hydrogenated.

Lindlar Catalyst

The Lindlar catalyst is pivotal in the gentle partial hydrogenation of alkynes to alkenes. It consists of palladium deposited on calcium carbonate and is poisoned, often by lead acetate and quinoline or a similar compound, to control the hydrogenation process.

The poisons moderate the catalyst's activity, ensuring that the alkyne is hydrogenated to the alkene stage without further reduction to an alkane. Specifically, the Lindlar catalyst facilitates the formation of the cis isomer of the alkene, preserving the geometry of the double bond due to the syn addition of hydrogen atoms.

Sodium Amide Reaction

Sodium amide (NaNH₂) in liquid ammonia is utilized for the deprotonation of alkynes, a reaction that is particularly important when dealing with terminal alkynes.

In the context of alkyne reactivity, sodium amide induces the elimination of hydrogen from the terminal carbon, resulting in the generation of a more reactive acetylide ion. This ion could itself be a nucleophile for further reactions. However, in some cases, there may be no reaction if the terminal proton isn't present or if steric hindrance prevents the deprotonation.

Hydroboration-Oxidation of Alkynes

Hydroboration-oxidation is a two-step reaction process that converts alkynes to aldehydes or ketones, depending on their structure.

During this process, borane (BH₃) adds across the triple bond of the alkyne in a syn addition manner. The resulting boron-containing intermediate is then oxidized with hydrogen peroxide (H₂O₂) in a basic aqueous solution, often sodium hydroxide (NaOH). This step replaces the -BH₂ group with an -OH group, leading to the formation of an enol. The enol immediately tautomerizes to the more stable carbonyl compound, such as a ketone or an aldehyde. Notably, in this reaction, the hydroxyl group attaches to the less substituted carbon, following Markovnikov's rule.

Vicinal Dibromide Formation

Vicinal dibromide formation involves the addition of bromine (Br₂) to the alkene or alkyne to form a compound with halogens on adjacent carbons.

This halogenation occurs via the anti addition mechanism, wherein the bromine atoms add on opposite sides of the double or triple bond. With alkynes, adding one equivalent of bromine will result in a trans-dibromide derivative because of the anti addition, which in the case of a terminal alkyne, like 3-hexyne, will yield a vicinal dibromide alkene such as 3,4-dichloro-3-hexene.