 Bonds in Ethylene Molecules Figure 2-4.  Bonds in Ethylene Molecules

Example: CH 2 =CH - CH 2 - CH 2 - CH=CH 2

hex-1,5-diene

- Alkadienes have two consecutive double bonds.

Example: CH 2 =C = CH 2 CH 3 - CH = C = CH 2

prop-1,2-dylene but-1,2-dylene

- Alkadienes have two double bonds alternating with a single bond (conjugated alkadienes). For example: CH 2 =CH - CH = CH 2 CH 2 =C - CH = CH 2

CH 3

but-1,3-diene 2-methylbut-1,3-diene (isopren)

- Alkynes are unsaturated, open-chain hydrocarbons with a triple bond in the molecule. The general formula is C n H 2n -2 (n  2). They form a homologous series, at the top of the homologous series.

The chemical formula is acetylene (C 2 H 2 ).

2.1. Naming and isomerism

a. Common name

This naming method only applies to simple unsaturated, acyclic hydrocarbons with one double bond. The corresponding alkane is named and the suffix “ane” is replaced with “ylene”.

Example:

CH 2 =CH 2

CH 3 -CH=CH 2

CH 3 -C=CH 2

ethylene

propylene

CH 3

isobutylene

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Some alkadienes have the following systematic names:

2 2

CH=C=CH

CH 2 =CH-CH=CH 2

CH 2 =CH-CH 2 -CH 2 -CH=CH 2

CH 2 =C-CH=CH 2

CH 3

Allen

vinyl

diallyl

isoprene

For alkynes, acetylene is taken as the parent substance, and homologues of acetylene are considered substituted derivatives of acetylene when the hydrogen atom in acetylene is replaced by an alkyl radical and named as follows:

Example:

Alkyl radical + acetylene

CH 3 -C  CH CH 3 -C  C-CH 3

methyl acetylene dimethyl acetylene

b. IUPAC name

To name unsaturated acyclic hydrocarbons according to IUPAC nomenclature, the order of steps is similar to that for naming alkanes, but with the following difference:

- Choose the main chain as the longest carbon chain containing double bonds.

- Number the carbons on the main chain starting from the side closest to the double bond.

- The main chain has the name of the corresponding alkane without the suffix "ane", the number indicating the position of the double bond and the suffix "ene", "dien" or "ine".

Example:

CH 3 -CH=CH-CH 3 CH 2 =C-CH=CH 2 CH 3 –C  C-CH-CH 3

CH 3 CH 3

but-2-ene 2-methylbut-1,3-dien 4-methyl pent-2-ine

The radicals containing double bonds are collectively called alkenyls. Note the common names of some of the following alkenyls:

CH 2 =CH- CH 2 =CH-CH 2 - CH 2 =C-

CH 3

vinyl allyl iso- propenyl

CH 2 = CH 3 -CH=

The monovalent radical of an alkyne is called an alkyl.

CH=C - CH=C -CH 2 -

etinyl prop-2-inyl

c. Isomerism

Unsaturated hydrocarbons have the following isomers:

- Isomerism in carbon chain structure

Example:

CH 3 -CH-CH=CH 2 CH 3 -CH 2 -CH 2 -CH=CH 2CH 3

3-methylbut-1-ene pent-1-ene

(isopentylene) (pentylene)

- Isomerism in double bond position

CH 3 -CH 2 -C=CH CH 3 -C=C -CH 3

but-1-in but-2-in

- Geometric isomerism

CH 3

C = C

H CH 3 CH 2 -CH 3C = C

H CH 2 -CH 3 HH

trans -pent-2-en cis -pent-2-en

2.2. Preparation method

a. From halogen derivatives

Unsaturated hydrocarbons can be prepared by eliminating the HX of halogen derivatives in an alkaline/alcoholic medium:

3 3

CH -CH -CH -CH KOH

fall

CH 3 - CH =C -CH 3 + HBr

Br CH 3

CH 3

3 2 2

CH -CH -CHCl KOH

fall

CH 3 - C = C + 2 HCl

2 2 2 2

CH -CH -CH -CH KOH

fall

CH 2 = CH -CH =CH 2 + 2 HBr

Br Br

b. From alcohol

Unsaturated hydrocarbons can be prepared by dehydration of alcohol in the presence of concentrated H 2 SO 4 as a dehydrating agent at temperatures above 170 C.

CH-CH

-OH

H 2 SO 4 CH

= CH

+ HO

3 2 >170 0 C 2 2 2

CH-CH-CH

-OH

H 2 SO 4 CH

- C = CH + 2H O

CH-CH

-CH

>170 0 C 3 2

H 2 SO 4 solution

2 2 2

2 >170 0 C CH 2 = CH - CH =CH 2 + 2H 2 O

When Al2O3 and suitable temperature are present, but - 1,3 -diene can be synthesized from ethyl alcohol.

2 CH

-CH

-OH

Al2O3

= CH - CH =CH

+ 2H O + H

3 2 t 0 2

2 2 2

c. Specific methods for preparing acetylene

CaC 2 + 2 H 2 O Ca(OH) 2 + C 2 H 2

2 CH 4

1500 0 C

CH =CH + 3 H 2

2.3. Physical properties

Alkenes from C2H4 to C4H8 are gases, from C5H10 to C16H32 are liquids, and from C17H34 and above are solids. Alkenes have lower boiling and melting points than the corresponding alkanes. The boiling point of the cis isomer is higher than that of the trans isomer , and the melting point is the opposite.

The physical properties of alkynes are similar to those of alkenes. As the molecular weight of alkynes increases, the boiling point, melting point, and density also increase.

2.4. Structure

a. Structure of alkenes

The two carbon atoms in the double bond are in the sp2 hybridized state . The three sp2 hybridized orbitals of the carbon atom in the double bond form three  bonds , including one  bond.(CC) is formed by the overlap of sp 2 - sp 2 orbitals and two  bonds .other may be  bond(CH) due to sp2 -s orbital overlap or  bonding(CC) due to the hybridization of sp2 orbitals of carbon in the double bond and sp3 orbitals of carbon. Each carbon atom in the double bond has one 2pz orbital that has not participated in hybridization. Their axes are perpendicular to the plane containing the  bond , parallel to each other, have opposite spins, and they overlap to form a  bond. Therefore, the double bond consists of a  bond.and a  bond . Because the molecule contains a  bond.unstable, easily broken down to participate in addition, oxidation and polymerization reactions.

Figure 2-3.  linkin ethylene molecule Figure 2-4.  bondin ethylene molecule

b. Structure of alkadienes

Alkadienes have a similar structure to alkenes. Alkadienes conjugate electrons have combined to form a molecular orbital only release on the entire conjugate system.

Figure 2-5: Link in the but-1,3-diene molecule

c. Structure of alkynes

The two carbon atoms in the triple bond are in the sp hybridized state. The two sp hybridized orbitals form two  bonds, one of which is a  bond .(CC) is formed by the overlap of two sp hybrid orbitals, and a  bond , by the overlap of the sp hybrid orbital of carbon at the triple bond with the s orbital of hydrogen atom or formed by the overlap of two sp hybrid orbitals with sp 3 .

Each carbon atom at the triple bond has two unhybridized p orbitals, which are perpendicular to the plane containing the  bond , parallel to each other and have opposite spins. These p orbitals overlap each other in pairs to form two  bonds. The  bondsBecause of their poor stability, alkynes are chemically active compounds that easily participate in addition, oxidation, and polymerization reactions.

Figure 2-6 :  linksin acetylene molecule Figure 2-7:  bondsin acetylene molecule

2.5. Chemical properties

In the molecules of all unsaturated hydrocarbons there are  bonds.unstable, this bond is strongly polarized by electronic effects within the molecule, by the reactant and

Reaction conditions, so unsaturated hydrocarbons easily participate in addition reactions, oxidation reactions, polymerization reactions.

a. Addition reaction to double bonds

Addition reactions occur at  bonds , for example the addition reaction of alkenes occurs according to the following general diagram:



CC + A + B - C

Here AB can be X 2 , HOH, HX, HOX, H 2 SO 4

The reaction occurs by electrophilic addition mechanism in two stages:

- In the first stage, the electrophilic agent (cation A + ) combines with the negatively charged carbon atom in the double bond to form an intermediate carbocation. This is the slow, reversible stage that determines the reaction rate.

AB A + + B -



slow +

CC + A + C C

- In the next stage, the intermediate carbocation reacts with the anion B - to give products. In this stage, the reaction occurs quickly and in one direction.

CC + B -

fast

Here we consider some specific addition reactions:

- Halogen addition reaction

CC + X 2CC

X X

Here X 2 is F 2 , Cl 2 , Br 2 , I 2 .

Example:

CH 3 -CH =CH 2 + Br 2 CH 3 -CH -CH 2

Br Br

2 CH 2 =CH -CH =CH 2 +2Br 2 CH 2 -CH -CH =CH 2 + CH 2 -CH =CH -CH 2

Br Br Br

Br Br

CH 3 -C =CH

Br2

CH 3 -C =CH

Br2

CH-C-CH

Br Br Br

The Br2 addition reaction is used to identify double bonds.

- Hydrogen halide addition reaction (HX)

All unsaturated hydrocarbons easily combine with HX.

Example:



+ HXC

  Br

CH 3 -C = CH -CH 3 + HBr CH 3 -C -CH 2 -CH 3

CH 3

Electrophilic addition mechanism:

HBr H + + Br -



CH 3 -C = CH -CH 3 + H CH 3 -C -CH 2 -CH 3

CH 3

CH 3 -C -CH 2 -CH 3 + Br CH 3 -C -CH 2 -CH 3

CH 3 CH 3

Conjugated alkadienes with hydrogen halides always give two types of products: 1,2-addition and 1,4-addition.

Example:

2 CH 2 =C - CH =CH 2 +2 HBr CH 3 -C -CH =CH 2 + CH 3 -C =CH -CH 2

CH 3

Reaction mechanism:

HBr H + + Br -

CH 3

CH 3 Br

CH 2 C - CH CH 2 + H +

CH 3 - C CH CH 2

CH 3 CH 3

In the conjugated carbocation the positive charge is distributed over all three carbon atoms, but is more concentrated at C 2 and C 4 .

Br

2 CH 3 C CH CH 2 + 2B r - CH 3 - C CH CH 2 CH 3 - C CH CH 2

CH 3 CH 3 CH 3 Br

Hydrohalides added to alkyne will give the product alkane dihalogen gem. For example:

CH 3 -C=C -CH 3

+ HC l

CH 3 -C =CH -CH 3

+ HC l

CH 3 -C -CH 2 -CH 3

- National Assembly

Cl Cl

In acidic environments, alkenes and alkadienes combine with water to form alcohols.

Example:

+ HOH

H CC OH

CH 3 -CH =CH 2 + H 2 OH

CH 2 =CH -CH =CH 2 + H 2 O

CH 3 -CH - CH 3 OH

CH 3 -CH = CH -CH 2 -OH

2 4

In the presence of catalyst Hg2 + and strong acid (H SO), alkynes react with water to form enols, which are unstable and undergo isomerization to form carbonyl compounds. If the alkyne is acetylene, acetic aldehyde will be obtained, and if the alkyne is a homologue of acetylene, a ketone will be obtained.

Example:

CH = CH + H 2 O

Hg2 + HSO

CH =CH - OH displacement -CHO

2 4

transposition

2 CH 3

Hg 2+

CH 3 -C = CH + H 2 O

H SO

CH 3 -C= CH 2 CH 3 -C -CH 3

2 4 OH O

b. Oxidation reaction

Depending on the reaction conditions, unsaturated hydrocarbons can be oxidized to form different products: polyalcohols, aldehydes, ketones, acids, etc.

At normal temperature, unsaturated hydrocarbons are oxidized by KMnO 4 solution or H 2 O 2 solution in a weak acidic environment to form polyalcohols.

Example:

CH 3 - Ch = CH 2 +KM n O 4 +H 2 O CH 3 - CH -CH 2

oh oh

+ KOH + MnO2

CH 2 =Ch - CH = CH 2 +KM n O 4 +H 2 O CH 2 - CH -CH-CH 2

+ KOH + MnO2

oh oh oh oh

With strong oxidizing agents such as the mixture KMnO 4 + H 2 SO 4 or K 2 Cr 2 O 7 , ... unsaturated hydrocarbons are oxidized to produce carboxylic acids, ketones and carbon dioxide.

Example:

CH 3 - Ch =CH - CH 3

K 2 Cr 2 O 7 +H 2 SO 4

2 CH 3 - COOH + H 2 O

CH 2 =Ch - CH = CH 2

K 2 Cr 2 O 7 +H 2 SO 4

HOOC -COOH + 2CO 2 + H 2 O

Ch 3 - C = CH 2

CH 3

K 2 Cr 2 O 7 +H 2 SO 4 CH 3 - C -CH 3 + CO 2 + H 2 O