Question

a.) Give the equation for the reaction of concentrated H₂SO₄ with propene.

b.) What is the formula of the product obtained from the reaction of dilute H₂SO₄ with propene?

Answer 1

SO3H a) con.H2SO4 он b) dil.H2SO4
1) H^+ hydrogen is attracted to the lone pairs on the oxygen atom of the OH. It then attacks the oxygen atom and forms -OH2^+ which is still attached to the carbon. The positive charge finds its way to the oxygen atom which is bonded to a carbon atom. Given the difference in electronegativities of carbon and oxygen, the bond between oxygen and carbon splices and oxygen acquires 2 electrons from the bond; thereby releasing a water molecule H2O. The carbon is left with one less electron, which oxygen stole away. But there's a C-H bond nearby on its neighboring carbon atom. The electrons of the C-H bond is turned into a second bond between C-C and now hydrogen is left with no electrons at all; its own electron that was once used to form the C-H bond is now in the C=C bond. So it goes away as H^+ and the reaction propagates itself.H^+ hydrogen is attracted to the lone pairs on the oxygen atom of the OH. It then attacks the oxygen atom and forms -OH2^+ which is still attached to the carbon. The positive charge finds its way to the oxygen atom which is bonded to a carbon atom. Given the difference in electronegativities of carbon and oxygen, the bond between oxygen and carbon splices and oxygen acquires 2 electrons from the bond; thereby releasing a water molecule H2O. The carbon is left with one less electron, which oxygen stole away. But there's a C-H bond nearby on its neighboring carbon atom. The electrons of the C-H bond is turned into a second bond between C-C and now hydrogen is left with no electrons at all; its own electron that was once used to form the C-H bond is now in the C=C bond. So it goes away as H^+ and the reaction propagates itself.

2) you end up with no double bond and a OH group on the most substituted carbon

its an alcohol synthesis by eletrophilic hydration
you can tell b/c you use dilute not concentrated sulfuric acid

Answer 2

SO3H a) con.H2SO4 он b) dil.H2SO4

Answer 3

CH3CH CH2 + H2S04 CH3CHCH3 OSO2OH

Answer 4

a)

in presence of conc.H2SO4

CH3CH CH2 + H2S04 CH3CHCH3 OSO2OH

b)
in presence of dil.H2SO4
CH3-CH=CH2 + H2SO4 -------> CH3-CH2-CH2-OH
                                     (propanol)

Answer 5

1)

CH3CH CH2 + H2S04 CH3CHCH3 OSO2OH

2) 2 C₃H₆ + 6 H₂SO₄ = 3 S₂ + 6 H₂CO₃ + 6 H₂O

Answer 6

a)

This is typical of the reaction with unsymmetrical alkenes. An unsymmetrical alkene has different groups at either end of the carbon-carbon double bond.

If sulphuric acid adds to an unsymmetrical alkene like propene, there are two possible ways it could add. You could end up with one of two products depending on which carbon atom the hydrogen attaches itself to.

However, in practice, there is only one major product.

CH3CH CH2 + H2S04 CH3CHCH3 OSO2OH

This is in line with Markovnikov's Rule which says:

When a compound HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already.

In this case, the hydrogen becomes attached to the CH2 group, because the CH2 group has more hydrogens than the CH group.

b) Product will be : CH3CH(OH)CH3