Chapter 9, Problem 24P

Write the structure of the major organic product isolated from the reaction of with

Hydrogen ( ), platinum

Hydrogen ( ), Lindlar palladium

Sodium amide in liquid ammonia

Product in part (c) treated with 1-bromobutane

Product in part (c) treated with tert-butyl bromide

Hydrogen chloride ( )

Hydrogen chloride ( )

Chlorine ( )

Chlorine ( )

Aqueous sulfuric acid, mercury sulfate

Ozone followed by hydrolysis

Interpretation Introduction

Interpretation:

The structure of the major product isolated from the reaction of with the given reagents is to be written.

Concept introduction:

In the presence a metal catalyst such as platinum, palladium, nickel, or rhodium, two molar equivalents of hydrogen get added across the triple bond of an alkyne to yield an alkane. In hydrogenation reaction, two hydrogen atoms get added to each carbon atom across the triple bond.

Hydrogenation reaction of alkynes to alkenes using the Lindlar catalyst is stereoselective. In this reaction only the cis alkene is produced.

Terminal alkynes react with to give product of the type .

Addition of hydrogen halides to alkynes gives alkenyl halides. The regioselectivity of this reaction follows Markovnikov’s rule. The hydrogen atom from hydrogen halide adds to the carbon in the alkyne that has the more number of hydrogens, and the halide adds to the carbon with the lower number of hydrogens.

In presence of excess of hydrogen halide, the sequential addition of two molecules of hydrogen halide to the carbon-carbon triple bond yields geminal dihalides.

The reaction of chlorine and bromine with alkynes yields tetrahaloderivatives. In this reaction, two molecules of the halogen get added to the triple bond. During the reaction, a dihaloalkene is get formed. When alkyne and the halogen are present in equimolar quantity, the dihaloalkene intermediate get formed and isolated. The stereochemistry of addition reaction is anti.

Alkynes, when subjected to ozonolysis followed by hydrolysis, produce carboxylic acids.

Answer to Problem 24P

Solution:

a)

b)

c)

d)

e)

f)

g)

h)

i)

j)

k)

Explanation of Solution

a) The reaction of with hydrogen (), platinum.

In the presence a metal catalyst such as platinum, palladium, nickel, or rhodium, two molar equivalents of hydrogen get added across the triple bond of an alkyne to yield an alkane. In case of , two molar equivalents of hydrogen add to the triple bond to produce hexane. Two hydrogen atoms get attached to each of the triple bonded carbon atoms of the .

b) The reaction of with hydrogen (), Lindlar palladium.

Reduction over Lindlar palladium catalyst with one mole of hydrogen partially reduces an alkyne to the corresponding cis alkene. has a terminal triple bond. Reduction of this triple bond with Lindlar palladium catalyst reduces it to a double bond. Since the double bond is terminal, there are no cis-trans isomers. The product is .

c) The reaction of with sodium amide in liquid ammonia.

Terminal alkynes react with to give species of the type . The hydrogen atom attached to the terminal carbon atom in is abstracted by the base to produce the corresponding acetylide anion.

d) The reaction of with product in part (c) treated with .

In the reaction of alkyl halide and sodium acetylide, the acetylide ion acts as a nucleophile. It displaces halide from carbon and forming a new carbon-carbon bond. This reaction follows mechanism.

The product formed in part (c) is the anion formed by . When this anion is treated with , the bromine atom is displaced, and a new bond is formed between the carbon atom of and the anion.

The product formed is .

e) The reaction of with product in part (c) treated with .

Acetylide anions are much more basic than hydroxide. Acetylide anions react with secondary and tertiary alkyl halides by elimination. Tert-butyl bromide is a tertiary alkyl halide and does not react by the mechanism.

The product formed in part (c) is the anion formed by . When this anion is treated with tert-butyl bromide, the bromine atom is eliminated along with a hydrogen atom attached to its adjacent carbon atom. The products formed are and .

f) The reaction of with product in part (c) treated with hydrogen chloride ().

Hydrogen halides add to alkynes to form alkenyl halides. When is treated with of hydrogen chloride, hexynyl chloride is formed. The regioselectivity of this addition reaction follows Markovnikov’s rule. Therefore, the hydrogen from hydrogen halide adds to the carbon in the alkyne that has the more number of hydrogens , and the chlorine adds to the carbon with the lower number hydrogens . The major product of the given addition reaction is .

g) The reaction of with product in part (c) treated with hydrogen chloride ().

In presence of excess of hydrogen halide, the sequential addition of two molecules of hydrogen halide to the carbon-carbon triple bond yields geminal dihalides. The second mole addition of hydrogen halide to the initially formed alkenyl halide is done in accordance with Markovnikov’s rule. Overall, both protons get attached to the same carbon and both halogens to the adjacent carbon atom. In , two hydrogen atoms get attached to the carbon atom while two chlorine atom gets attached to the carbon atom. The major product of this addition reaction is .

h) The reaction of with chlorine ().

When one mole of halogen molecule is added to an alkyne, the product formed is a dihaloalkene. One halogen atom gets attached to each of the triple bonded carbon atoms, and it is reduced to a double bond. , when reacted with one mole of , forms as the major product. One chlorine atoms gets attached to each and carbon atoms. The triple bond changes to a double bond. The two chlorine atoms are trans to each other.

i) The reaction of with chlorine ().

Alkynes react with two moles of chlorine to yield tetrachloroalkanes. Two molecules of the chlorine add to the triple bond. The stereochemistry of addition is anti. , when reacted with two mole of , forms as the major product. Two chlorine atoms get attached to each and carbon atoms. The triple bond changes to a single bond.

j) The reaction of with aqueous sulfuric acid, .

Hydration of alkynes employs aqueous sulfuric acid as the reaction medium and mercury(II) sulfate as a catalyst. The product formed is an enol which tautomerizes to the corresponding ketone. Markovnikov’s rule is followed in hydration of alkynes. Terminal alkynes yield methyl substituted ketones.

is a terminal alkyne. It undergoes hydration with aqueous sulfuric acid and to form as the major product.

k) The reaction of with ozone followed by hydrolysis.

Alkynes, when subjected to ozonolysis followed by hydrolysis, produce carboxylic acids. If carbonic acid is one of the products, it dissociates into carbon dioxide and water.

, when subjected to ozonolysis, produces pentanoic acid and carbonic acid as products. The carbonic acid dissociates into carbon dioxide and water.