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) \(\mathrm{Cl}_{2}(1 \mathrm{~mol})\) (h) \(\mathrm{NaNH}_{2}\) in \(\mathrm{NH}_{3}(l)\) (i) \(\mathrm{HBr}(1 \mathrm{~mol})\) (j) \(\mathrm{HBr}\) (2 mol) (k) \(\mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{HgSO}_{4}\)

Step by step solution

01

(a) Hydrogenation with excess H2 and Pt catalyst

In this reaction, an alkyne undergoes hydrogenation with excess H2 and Pt as a catalyst. The triple bond of the alkyne gets reduced to a single bond, forming an alkane. Product: \(\displaystyle\mathrm{CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}CH_{3}}\) (hexane)

02

(b) Partial hydrogenation with Lindlar catalyst

The Lindlar catalyst is a heterogeneous catalyst used for the hydrogenation of alkynes to cis-alkenes (also called syn-hydrogenation). Product: \(\displaystyle\mathrm{CH_{3}CH_{2}CH=CHCH_{2}CH_{3}}\) (cis-3-hexene)

03

(c) Dissolving metal reduction

When sodium is used in liquid ammonia as a catalyst, it performs a dissolving metal reduction on the alkyne to yield trans-alkene. This reaction is also called anti-hydrogenation. Product: \(\displaystyle\mathrm{CH_{3}CH_{2}CH=CHCH_{2}CH_{3}}\) (trans-3-hexene)

04

(d) Hydroboration-oxidation

BH3, followed by H2O2/NaOH, is a hydroboration-oxidation reaction. It results in an anti-Markovnikov addition of water across the alkyne, turning it into an enol. The enol then tautomerizes into a more stable product, a ketone. Product: \(\displaystyle\mathrm{CH_{3}CH_{2}COCH_{2}CH_{2}CH_{3}}\) (3-hexanone)

05

(e) Hydroboration with carboxylic acid

When an alkyne reacts with BH3, followed by a carboxylic acid (CH3COOH), it undergoes hydroboration with an anti-Markovnikov addition, similar to (d) but instead, the product is an enol which tautomerizes to an aldehyde. Product: \(\displaystyle\mathrm{CH_{3}CH_{2}CHOCH_{2}CH_{2}CH_{3}}\) (2-hexanal)

06

(f) Deuterium-labeled hydroboration

With BH3 followed by CH3COOD, it's a similar reaction to (e). Instead of a regular hydrogen atom, deuterium is incorporated into the product. Product: \(\displaystyle\mathrm{CH_{3}CH_{2}CHDCH_{2}CH_{2}CH_{3}}\) (2-hexanal-d1)

07

(g) Halogen addition

Cl2 (1 mol) reacts with the alkyne in a halogen addition reaction. It adds Cl atoms onto the carbon atoms of the triple bond, yielding a trans-dihalide alkene. Product: \(\displaystyle\mathrm{CH_{3}CH_{2}CHCl=CHCH_{2}Cl}\) (trans-3,4-dichlorohexene)

08

(h) Deprotonation by NaNH2 in liquid NH3

An alkyne reacts with NaNH2 in liquid NH3 to form an acetylide ion. In this case, the terminal hydrogen is removed, and a negative charge remains on the carbon. Product: \(\displaystyle\mathrm{CH_{3}CH_{2}C{\equiv}C^{-}CH_{2}CH_{2}CH_{3}}\) (hex-3-yn-1-ide)

09

(i) Hydrohalogenation with 1 mol HBr

The addition of HBr (1 mol) to an alkyne results in Markovnikov's hydrohalogenation of the triple bond, forming a vinyl halide. Product: \(\displaystyle\mathrm{CH_{3}CH_{2}CHBr=CHCH_{2}CH_{3}}\) (3-bromo-1-hexene)

10

(j) Hydrohalogenation with 2 mol HBr

With 2 moles of HBr, a second reaction will occur on the newly formed double bond, and the product will be a dihalide aliphatic compound. Product: \(\displaystyle\mathrm{CH_{3}CH_{2}CHBrCHBrCH_{2}CH_{3}}\) (3,4-dibromohexane)

11

(k) Hydration with H2O in acidic medium

This is a hydration reaction of an alkyne, where the triple bond is converted into a carbonyl group in the presence of water, H2SO4 and HgSO4. This reaction follows Markovnikov's rule. Product: \(\displaystyle\mathrm{CH_{3}CH_{2}COCH_{2}CH_{2}CH_{3}}\) (3-hexanone)

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!