ALKENES

Question 1

why are alkenes more reactive to aq X2 than X2(l)

Answer 1

In polar solvent H2O, e- cloud of Br2 is polarised by polar H2O molecules, causes Br2 to have partial charge separation, causes Br atom to be more e- def and have a larger partial positive charge
Br molecule in H2O is more susceptible to Nu attack by alkane molecules

Question 2

Explain why __ is the major product according to markonikov rule

Answer 2

____ is the major pdt as the tertiary C+ int formed during E-add is more stable than the secondary C+ int that leads to the minor pdt ___

The tertiary C+ has more e- donating alkyl groups attached to the positively charged C+ thus the +ve charge is dispersed to a larger extent, ist is more stable and thus__ is formed faster and becomes the major pdt

Question 3

For E-add of HX to alkenes, explain the relative rate of rxn from HCL to HBr to HI

Answer 3

-HI > HBr > HCl, HI is the most reactive
-From Cl to I, atomic radius ↑
-effectiveness of orbital overlap ↓
- bond strength ↓, ↓amt of energy req to break H-X bonds

Question 4

Explain why the resulting mixture from E-add is optically inactive(product is chiral carbon)

Answer 4

• C+ int is trigonal planar, - the Cl/Br- Nu can attack the positively charged Carbon from either sides of the plane w/ equal probability
  -equal proportions of 2 enantiomers are formed, resulting in racemic mixt
  -the 2 enantiomers rotate PPL in opposite directions by the same extent
  -thus optically inactive

Question 5

Chemical test to distinguish alkene and alkane(4)

Answer 5

1) Br2 in CCl OR Br2(aq) to each sample @rtp
-> alkene undergoes E-Add and decolourises orange-red Br2
-> remains orange red for alkane

2)Cold dilute KMnO, dilute KOH
-> alkene undegoes [O], decolourises purple KMnO4, brown ppt MnO2 is formed
-> purple colour remains for alkane and no ppt is formed

3)KMnO4, dilute H2SO4, heat using water bath
-> alkene undergoes strong [O] and decolourises purple KMnO, effervescence of CO2 gas which forms white ppt in limewater
->purple colour remains for alkane and no effervescence formed

Question 6

Why can’t LiAl4 reduce alkene

Answer 6

[Li+Al H4]- forms H- hydride ion which repels from the e- rich C=Chi