Chemistry ALKENES

### Topics to be covered

=> Alkenes
=> Structure of double bond
=> Nomenclature
=> Isomerism

### ALKENES

• Alkenes are unsaturated hydrocarbons containing at least one double bond.

• If there is one double bond between two carbon atoms in alkenes, they must possess two hydrogen atoms less than alkanes.

• Hence,general formula for alkenes is color{red}(C_nH_(2n)).

• Alkenes are also known as olefins (oil forming) since the first member, ethylene or ethene color{red}((C_2H_4)) was found to form an oily liquid on reaction with chlorine.

### Structure of Double Bond

• Carbon-carbon double bond in alkenes consists of one strong sigma color{red}((σ)) bond (bond enthalpy about color{red}(397 kJ mol^(–1))) due to head-on overlapping of color{red}(sp^2) hybridised orbitals and one weak pi color{red}((π)) bond (bond enthalpy about color{red}(284 KJ mol^(–1)) ) obtained by lateral or sideways overlapping of the two color{red}(2p) orbitals of the two carbon atoms

• . The double bond is shorter in bond length (134 pm) than the color{red}(C–C) single bond (154 pm).

• Because of the presence of the pi color{red}((π)) bond alkenes behave as sources of loosely held mobile electrons. Therefore, alkenes are easily attacked by reagents or compounds which are in search of electrons. Such reagents are called color{green}("𝐞𝐥𝐞𝐜𝐭𝐫𝐨𝐩𝐡𝐢𝐥𝐢𝐜 𝐫𝐞𝐚𝐠𝐞𝐧𝐭𝐬.")

• The presence of weaker color{red}(π)-bond makes alkenes unstable molecules in comparison to alkanes and thus, alkenes can be changed into single bond compounds by combining with the electrophilic reagents.

• Strength of the double bond (bond enthalpy, color{red}(681 kJ mol^(–1)) ) is greater than that of a carbon-carbon single bond in ethane (bond enthalpy, color{red}(348 kJ mol^(–1)) ).

• Orbital diagrams of ethene molecule are shown in Figs. 13.4 and 13.5.

### Nomenclature

• The longest chain of carbon atoms containing the double bond is selected.

• Numbering of the chain is done from the end which is nearer to the double bond.

• The suffix ‘𝐞𝐧𝐞’ replaces ‘𝐚𝐧𝐞’ of alkanes.

• IUPAC names of a few members of alkenes are given below :

tt(("𝐒𝐭𝐫𝐮𝐜𝐭𝐮𝐫𝐞" , "𝐈𝐔𝐏𝐀𝐂 𝐧𝐚𝐦𝐞") , (CH_3-CH=CH_2 , "Propene") , (CH_3-CH_2 - CH = CH_2 , "But – 1 - ene") , ( CH_3 - CH = CH - CH_3 , "But-2-ene") , (CH_2 = CH - CH = CH_2 , "Buta – 1,3 - diene") , ( CH_3 = underset(underset(CH_3)(|))C- CH_3 , "2-Methylprop-1-ene" ) , ( CH_2 = CH - underset(underset(CH_3)(|))CH - CH_3 , " 3-Methylbut-1-ene"))
Q 3284578457

Write IUPAC names of the following compounds:

Solution:

(ii) 1,3,5,7 Octatetraene;
(iii) 2-n-Propylpent-1-ene;
(iv) 4-Ethyl-2,6-dimethyl-dec-4-ene;
Q 3204678558

Calculate number of sigma (σ) and pi (π) bonds in the above structures (i-iv).

Solution:

σ bonds : 33, π bonds : 2
σ bonds : 17, π bonds : 4
σ bonds : 23, π bond : 1
σ bonds : 41, π bond : 1

### Isomerism

color{green}("𝐀𝐥𝐤𝐞𝐧𝐞𝐬 𝐬𝐡𝐨𝐰 𝐛𝐨𝐭𝐡 𝐬𝐭𝐫𝐮𝐜𝐭𝐮𝐫𝐚𝐥 𝐢𝐬𝐨𝐦𝐞𝐫𝐢𝐬𝐦 𝐚𝐧𝐝 𝐠𝐞𝐨𝐦𝐞𝐭𝐫𝐢𝐜𝐚𝐥 𝐢𝐬𝐨𝐦𝐞𝐫𝐢𝐬𝐦.")

color{green}("𝐒𝐭𝐫𝐮𝐜𝐭𝐮𝐫𝐚𝐥 𝐢𝐬𝐨𝐦𝐞𝐫𝐢𝐬𝐦 :") As in alkanes, ethene color{red}((C_2H_4)) and propene color{red}((C_3H_6)) can have only one structure but alkenes higher than propene have different structures. Alkenes possessing color{red}(C_4H_8) as molecular formula can be written in the following three ways:

Structures I and III, and II and III = color{green}("𝐜𝐡𝐚𝐢𝐧 𝐢𝐬𝐨𝐦𝐞𝐫𝐢𝐬𝐦")
Structures I and II = color{green}("𝐩𝐨𝐬𝐢𝐭𝐢𝐨𝐧 𝐢𝐬𝐨𝐦𝐞𝐫𝐬.")

color{green}("𝐆𝐞𝐨𝐦𝐞𝐭𝐫𝐢𝐜𝐚𝐥 𝐢𝐬𝐨𝐦𝐞𝐫𝐢𝐬𝐦 :") Doubly bonded carbon atoms have to satisfy the remaining two valences by joining with two atoms or groups. If the two atoms or groups attached to each carbon atom are different, they can be represented by color{green}("𝐘𝐗 𝐂 = 𝐂 𝐗𝐘") that can be represented in space in the following two ways :

In (a), the two identical atoms i.e., both the color{red}(X) or both the color{red}(Y) lie on the same side of the double bond but in (b) the two color{red}(X) or two color{red}(Y) lie across the double bond or on the opposite sides of the double bond. This results in different geometry of (a) and (b) i.e. disposition of atoms or groups in space in the two arrangements is different. Therefore, they are color{green}("𝐬𝐭𝐞𝐫𝐞𝐨𝐢𝐬𝐨𝐦𝐞𝐫𝐬.") They would have the same geometry if atoms or groups around color{red}(C=C) bond can be rotated but rotation around color{red}(C=C) bond is not free. It is restricted .The stereoisomers of this type are called color{green}("𝐠𝐞𝐨𝐦𝐞𝐭𝐫𝐢𝐜𝐚𝐥 𝐢𝐬𝐨𝐦𝐞𝐫𝐬.")

The isomer of the type (a), in which two identical atoms or groups lie on the same side of the double bond is called color{green}("𝐜𝐢𝐬 𝐢𝐬𝐨𝐦𝐞𝐫") and the other isomer of the type (b), in which identical atoms or groups lie on the opposite sides of the double bond is called color{green}("𝐭𝐫𝐚𝐧𝐬 𝐢𝐬𝐨𝐦𝐞𝐫") . Thus cis and trans isomers have the same structure but have different configuration (arrangement of atoms or groups in space). Due to different arrangement of atoms or groups in space, these isomers differ in their properties like melting point, boiling point, dipole moment, solubility etc. color{green}("𝐆𝐞𝐨𝐦𝐞𝐭𝐫𝐢𝐜𝐚𝐥 𝐨𝐫 𝐜𝐢𝐬-𝐭𝐫𝐚𝐧𝐬 𝐢𝐬𝐨𝐦𝐞𝐫𝐬") of but-2-ene are represented below :

Cis form of alkene is found to be more polar than the trans form. For example, dipole moment of cis-but - 2-ene is 0.33 Debye, whereas, dipole moment of the trans form is almost zero or it can be said that transbut- 2-ene is non-polar. This can be understood by drawing geometries of the two forms as given below from which it is clear that in the trans-but-2-ene, the two methyl groups are in opposite directions, Therefore, dipole moments of color{red}(C-CH_3) bonds cancel, thus making the trans form non-polar.

In the case of solids, it is observed that the trans isomer has higher melting point than the cis form. color{green}("𝐆𝐞𝐨𝐦𝐞𝐭𝐫𝐢𝐜𝐚𝐥 𝐨𝐫 𝐜𝐢𝐬-𝐭𝐫𝐚𝐧𝐬 𝐢𝐬𝐨𝐦𝐞𝐫𝐢𝐬𝐦") is also shown by alkenes of the types color{green}("𝐗𝐘𝐂 = 𝐂𝐗𝐙") and color{green}("𝐗𝐘𝐂 = 𝐂𝐙𝐖")
Q 3234778652

Write structures and IUPAC names of different structural isomers of alkenes corresponding to C_5 H_(10)

Solution:

(a) underset("Pent-1-ene")(CH_2 = CH – CH_2 – CH_2 – CH_3)

(b) underset("Pent-2-ene")(CH_3 – CH=CH – CH_2 – CH_3)

(c) underset("2-Methylbut-2-ene")( CH_3 – underset(underset(CH_3)(|))C = CH – CH_3)

(d) underset("3-Methylbut-1-ene")(CH_3 – underset(underset(CH_3)(|))CH – CH = CH_2)

(e) underset("2-Methylbut-1-ene")(CH_2 = underset(underset(CH_3)(|))C – CH_2 – CH_3)
Q 3274778656

Draw cis and trans isomers of the following compounds. Also write their IUPAC names :
(i) CHCl = CHCl
(ii) C_2H_5C CH_3 = C CH_3C_2H_5

Solution:

Q 3204778658

Which of the following compounds will show cis-trans isomerism?

(i) (CH_3)_2C = CH – C_2H_5

(ii) CH_2 = CBr_2

(iii) C_6H_5CH = CH – CH_3

(iv) CH_3CH = C Cl CH_3

Solution:

(iii) and (iv). In structures (i) and (ii), two identical groups are attached to one of the doubly bonded carbon atom.

### Preparation of alkenes:

color{green}("𝐅𝐫𝐨𝐦 𝐚𝐥𝐤𝐲𝐧𝐞𝐬 :") Alkynes on partial reduction with calculated amount of dihydrogen in the presence of palladised charcoal partially deactivated with poisons like sulphur compounds or quinoline give alkenes. color{green}("𝐏𝐚𝐫𝐭𝐢𝐚𝐥𝐥𝐲 𝐝𝐞𝐚𝐜𝐭𝐢𝐯𝐚𝐭𝐞𝐝 𝐩𝐚𝐥𝐥𝐚𝐝𝐢𝐬𝐞𝐝 𝐜𝐡𝐚𝐫𝐜𝐨𝐚𝐥 𝐢𝐬 𝐤𝐧𝐨𝐰𝐧 𝐚𝐬 𝐋𝐢𝐧𝐝𝐥𝐚𝐫’𝐬 𝐜𝐚𝐭𝐚𝐥𝐲𝐬𝐭.") Alkenes thus obtained are having cis geometry. However, alkynes on reduction with sodium in liquid ammonia form trans alkenes.

color{green}("𝐅𝐫𝐨𝐦 𝐚𝐥𝐤𝐲𝐥 𝐡𝐚𝐥𝐢𝐝𝐞𝐬:") Alkyl halides color{red}("(R-X")) on heating with alcoholic potash (potassium hydroxide dissolved in alcohol, say, ethanol) eliminate one molecule of halogen acid to form alkenes. This reaction is known as color{green}("𝐝𝐞𝐡𝐲𝐝𝐫𝐨𝐡𝐚𝐥𝐨𝐠𝐞𝐧𝐚𝐭𝐢𝐨𝐧") i.e., removal of halogen acid i.e. color{green}("β-𝐞𝐥𝐢𝐦𝐢𝐧𝐚𝐭𝐢𝐨𝐧 𝐫𝐞𝐚𝐜𝐭𝐢𝐨𝐧"), since hydrogen atom is eliminated from the β carbon atom (carbon atom next to the carbon to which halogen is attached).

• For halogens, the rate is: color{red}("𝐢𝐨𝐝𝐢𝐧𝐞 > 𝐛𝐫𝐨𝐦𝐢𝐧𝐞 > 𝐜𝐡𝐥𝐨𝐫𝐢𝐧𝐞")

• For alkyl groups it is : color{red}("𝐭𝐞𝐫𝐭 > 𝐬𝐞𝐜𝐨𝐧𝐝𝐚𝐫𝐲 > 𝐩𝐫𝐢𝐦𝐚𝐫𝐲.")

color{green}("𝐅𝐫𝐨𝐦 𝐯𝐢𝐜𝐢𝐧𝐚𝐥 𝐝𝐢𝐡𝐚𝐥𝐢𝐝𝐞𝐬 :") Dihalides in which two halogen atoms are attached to two adjacent carbon atoms are known as color{green}("𝐯𝐢𝐜𝐢𝐧𝐚𝐥") dihalides. Vicinal dihalides on treatment with zinc metal lose a molecule of color{red}(ZnX_2) to form an alkene. This reaction is known as color{green}("𝐝𝐞𝐡𝐚𝐥𝐨𝐠𝐞𝐧𝐚𝐭𝐢𝐨𝐧.")

color{red}(CH_2Br - CH_2Br +Zn → CH_2 = CH_2 + Zn Br_2) ..............(13.35)

color{red}(CH_3CHBr - CH_2Br + Zn → CH_3 CH = CH_2 + ZnBr_2) ............(13.36)

color{green}("𝐅𝐫𝐨𝐦 𝐚𝐥𝐜𝐨𝐡𝐨𝐥𝐬 𝐛𝐲 𝐚𝐜𝐢𝐝𝐢𝐜 𝐝𝐞𝐡𝐲𝐝𝐫𝐚𝐭𝐢𝐨𝐧 :") Alcohols on heating with concentrated sulphuric acid form alkenes with the elimination of one water molecule. Since a water molecule is eliminated from the alcohol molecule in the presence of an acid, this reaction is known as color{green}("𝐚𝐜𝐢𝐝𝐢𝐜 𝐝𝐞𝐡𝐲𝐝𝐫𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐚𝐥𝐜𝐨𝐡𝐨𝐥𝐬.") This reaction is also the example of β-elimination reaction since color{red}(–OH) group takes out one hydrogen atom from the β-carbon atom.