Alkenes

Question 1

General formula

Answer 1

CnH2n

Question 2

Characteristic

Answer 2

Unsaturated hydrocarbon with C=C present

Question 3

Nomenclature

Answer 3

Suffix: -ene
if there are more than one C=C write (di-ene/tri-ene)

Prefix Substituent (s) (state position(s) , if same then right di/tri)

Parent: Longest carbon chain

Question 4

Physical properties

Answer 4

Insoluble in water and soluble in non polar solvents
Less dense than water
Boiling point increases with increasing C number
Branching lowers boiling point

Question 5

cis and trans isomers of alkenes

Answer 5

may have different physical properties due to difference in polarity

trans isomers pack better in a crystal lattice leading to more IMF per unit volume. Hence they generally have higher melting point than their respective cis isomer

Question 6

Chemical properties/rxns

Answer 6

undergo
1. Elimination (alkene prep)
2.Combustion
3.Electrophilic addition
4.Redox : Reduction/Oxidation

Question 7

Elimination

Answer 7

Elimination of H and OH from alcohols
Reagant and condition: Excess concentrated H2SO4,heat / Al2O3(s),heat
results in formation of an alkene+H2O from a alcohol

Elimination of H and X from halogenoalkanes
Reagant and condition: Ethanolic KOH/NaOH, heat
results in the formation of alkene, K halide, H2O

Question 8

Stability of alkene isomers formed from elimination

Answer 8

The more highly substituted alkene is more stable and more readily formed in an elimination rxn

Question 9

Combustion

Answer 9

Alkenes are hydrocarbons and in complete combustion with excess O2, CO2 and H2O formed

Question 10

Reduction

Answer 10

Reagent and condition: H2(g) and Pt(s) catalyst/ H2(g) and Ni(s) catalyst and heat
results in formation of an alkane from alkene

Question 11

The susceptibility of alkenes to undergo rxns with electrophiles and addition rxns

Answer 11

C=C bond in alkene made up with one sigma head-on overlap between 2 2sp2 hybrid orbitals and one pi bond between the unhybridised p orbital from each carbon.

The struc of C=C results in the reactivity of Alkenes

Question 12

alkenes undergo rxns with electrophiles

Answer 12

The pi electrons found in regions above and below the plane of molecule. The less tightly bound pi electrons to the carbon nuclei causes a region of relatively high electron density at C=C

Question 13

alkenes undergo addition rxns

Answer 13

Less energy is required to break the pi bond in the C=C bond as pi bond less effective than sigma bond

Alkenes tend to undergo reactions involving the breaking of pi bond in C=C bond leaving sigma bond intact. The pi electrons are used to form 2 new sigma bonds with the electrophiles to form one product

Question 14

Electrophilic Addition Mechanism

Answer 14

Step 1 (slow): (rfr to notes for the eqn)
loosely bound pi electrons of C=C bond is a region of high electron density. As reagent molecule approaches the alkene. The pi electron cloud of C=C induces a dipole in the reagant mol. One atom in the reagent gains partial positive charge and reacts with alkene as an electrophile in slow step. And other gains partial negative charge.
Reagant molecule undergoes heterolytic fission producing a reactive carbocation intermediate and reagant ion (from partial negative atom)

Step 2 (fast): (refer to notes for eqn) negatively charged reagent acts as a nucleophile and attacks unstable positively charged carbocation forming product

rmbr curly arrows, lone pair and charge

Question 15

Electrophilic Addition of Halogen

Answer 15

Reagent and Condition
halogen dissolved in CCl4
( in the case of Br, orange-red is decolourised)

Question 16

Electrophilic Addition in bromine water

Answer 16

Reagent and Condition
Br2(aq)
Orange-red Br decolourised
results in the formation of a major (with H2O)and minor product (with Br- ion)

Question 17

Electrophilic Addition of Hydrogen Halides

Answer 17

Reagent and Condition
Gaseous HX (X is halogen)

Question 18

Electrophilic Addition of H2O (catalytic hydration)

Answer 18

Reagent and condition
Concentrated H2SO4 at rtp
H2O(l) and warm
OR
H2O(g), conc, H3PO4 catalyst ,heat and high pressure

Since H2SO4 is stronger acid than water it is the source of H+ (electrophile)
H2O behaves as the nucleophile and react with carbocation

Question 19

Markovnikov’s rule

Answer 19

For asymmetrical alkenes isomers are formed as product as major and minor products

The major product is produced more preferentially due to greater no.of alkyl groups attached to carbocation hence it is more stable leading to higher yield of that product

Question 20

Oxidation

Answer 20

Vigorous oxidation
Mild oxidation

pi bond is first cleaved

Question 21

Mild oxidation

Answer 21

cold, alkaline (NaOH) KMnO4 forming diols

pi bond cleaved by KMnO4 at a low temp and alkaline medium

Obs: Decolourisation of purple of KMnO4 brown ppt of MnO2 ppt formed
(MnO-4 to MnO2)

Question 22

Vigorous

Answer 22

hot acidified(H2SO4) KMnO4 C=C bond cleaved (sigma bond as well)

gives carbonyl and/or carboxylic acids as products
diol obtained from mild oxidation of C=C can be subjected to further oxidative cleavage to give similar products

Obs: Decolourisation of purple of KMnO4
(MnO-4 to Mn 2+)

Question 23

Types of product obtained from vigorous oxidation

Answer 23

depends on no.of H atoms attached to the C atom of the C=C

2 effervescence due to CO2 formed and decolourisation of KMnO4
Products: CO2 and H2O

1 and 0 decolourisation of purple KMnO4
Products:
Carboxylic acid and a ketone

the carboxylic acid is further oxidised by hot acidified KMnO4 to give CO2 and H2O