Alkane and alkene mechanisms
Can someone explain the steps in free radical substitution in (water, halogens, hydrogen halides and sulfuric) for alkanes. Also, the steps in working out Electrophilic Addition for alkanes? It would be much appreciated in simple terms as chemguide has confused me!
(edited 2 years ago)
Original post by alyxcts
Can someone explain the steps in free radical substitution in (water, halogens, hydrogen halides and sulfuric) for alkanes. Also, the steps in working out Electrophilic Addition for alkanes? It would be much appreciated in simple terms as chemguide has confused me!
To begin, a radical is an atom or molecule that has at least one unpaired electron. These radicals can take part in these 'free radical substitution' reactions. The mechanisms for these reactions use a 'half headed curly arrow' which means the movement of a single electron, in a normal mechanism you would use a full headed curly arrow which means the movement of a pair of electrons.
These reactions can be broken down into 3 steps:
• Initiation (generation of the radical species)
• Propagation (Chain reaction)
• Termination (Removal of the free radical species to stop the chain reaction) (There can be different termination reactions for any particular reaction, I have included all 3 for the example I have given).
As for electrophilic addition of alkenes, the rule is that the C=C bond is the nucleophile and the electrophile is the other species in the reaction (See second image attached). This is the reaction of HBr with a substituted alkene. The H-Br is the electrophile because the bromine is electronegative and is withdrawing electron density from the H atom in the H-Br bond, making the H delta positive and the Br delta negative. The H atom is now electrophilic and can be attacked by the nucleophile which is the C=C bond. This is the important part - the carbocation intermediate that you generate follows a preference for tertiary > secondary > primary. (See 3rd image attached) - I hope this has answered your question.
Original post by alyxcts
Can someone explain the steps in free radical substitution in (water, halogens, hydrogen halides and sulfuric) for alkanes. Also, the steps in working out Electrophilic Addition for alkanes? It would be much appreciated in simple terms as chemguide has confused me!
Try savemyexams... it helped me a lot.. their notes are very concise
Original post by alyxcts
Can someone explain the steps in free radical substitution in (water, halogens, hydrogen halides and sulfuric) for alkanes. Also, the steps in working out Electrophilic Addition for alkanes? It would be much appreciated in simple terms as chemguide has confused me!
Watch Eliot Rintoul, he's a God amongst men: https://www.youtube.com/channel/UCps4gUjfZsu6-b-7mwK41lg
Original post by Drew_Dittman
To begin, a radical is an atom or molecule that has at least one unpaired electron. These radicals can take part in these 'free radical substitution' reactions. The mechanisms for these reactions use a 'half headed curly arrow' which means the movement of a single electron, in a normal mechanism you would use a full headed curly arrow which means the movement of a pair of electrons.
These reactions can be broken down into 3 steps:
• Initiation (generation of the radical species)
• Propagation (Chain reaction)
• Termination (Removal of the free radical species to stop the chain reaction) (There can be different termination reactions for any particular reaction, I have included all 3 for the example I have given).
As for electrophilic addition of alkenes, the rule is that the C=C bond is the nucleophile and the electrophile is the other species in the reaction (See second image attached). This is the reaction of HBr with a substituted alkene. The H-Br is the electrophile because the bromine is electronegative and is withdrawing electron density from the H atom in the H-Br bond, making the H delta positive and the Br delta negative. The H atom is now electrophilic and can be attacked by the nucleophile which is the C=C bond. This is the important part - the carbocation intermediate that you generate follows a preference for tertiary > secondary > primary. (See 3rd image attached) - I hope this has answered your question.
These reactions can be broken down into 3 steps:
• Initiation (generation of the radical species)
• Propagation (Chain reaction)
• Termination (Removal of the free radical species to stop the chain reaction) (There can be different termination reactions for any particular reaction, I have included all 3 for the example I have given).
As for electrophilic addition of alkenes, the rule is that the C=C bond is the nucleophile and the electrophile is the other species in the reaction (See second image attached). This is the reaction of HBr with a substituted alkene. The H-Br is the electrophile because the bromine is electronegative and is withdrawing electron density from the H atom in the H-Br bond, making the H delta positive and the Br delta negative. The H atom is now electrophilic and can be attacked by the nucleophile which is the C=C bond. This is the important part - the carbocation intermediate that you generate follows a preference for tertiary > secondary > primary. (See 3rd image attached) - I hope this has answered your question.
thankyou! I'll try understand it thoroughly
Original post by Drew_Dittman
To begin, a radical is an atom or molecule that has at least one unpaired electron. These radicals can take part in these 'free radical substitution' reactions. The mechanisms for these reactions use a 'half headed curly arrow' which means the movement of a single electron, in a normal mechanism you would use a full headed curly arrow which means the movement of a pair of electrons.
These reactions can be broken down into 3 steps:
• Initiation (generation of the radical species)
• Propagation (Chain reaction)
• Termination (Removal of the free radical species to stop the chain reaction) (There can be different termination reactions for any particular reaction, I have included all 3 for the example I have given).
As for electrophilic addition of alkenes, the rule is that the C=C bond is the nucleophile and the electrophile is the other species in the reaction (See second image attached). This is the reaction of HBr with a substituted alkene. The H-Br is the electrophile because the bromine is electronegative and is withdrawing electron density from the H atom in the H-Br bond, making the H delta positive and the Br delta negative. The H atom is now electrophilic and can be attacked by the nucleophile which is the C=C bond. This is the important part - the carbocation intermediate that you generate follows a preference for tertiary > secondary > primary. (See 3rd image attached) - I hope this has answered your question.
These reactions can be broken down into 3 steps:
• Initiation (generation of the radical species)
• Propagation (Chain reaction)
• Termination (Removal of the free radical species to stop the chain reaction) (There can be different termination reactions for any particular reaction, I have included all 3 for the example I have given).
As for electrophilic addition of alkenes, the rule is that the C=C bond is the nucleophile and the electrophile is the other species in the reaction (See second image attached). This is the reaction of HBr with a substituted alkene. The H-Br is the electrophile because the bromine is electronegative and is withdrawing electron density from the H atom in the H-Br bond, making the H delta positive and the Br delta negative. The H atom is now electrophilic and can be attacked by the nucleophile which is the C=C bond. This is the important part - the carbocation intermediate that you generate follows a preference for tertiary > secondary > primary. (See 3rd image attached) - I hope this has answered your question.
What’s the role of the UV light?
Original post by alyxcts
What’s the role of the UV light?
UV light has just the correct energy to break the Cl-Cl bond.
The bond energy is 242 kJ mol-1
Energy per bond = 242000/6.02 x 1023 = 4.02 x 10-19 J
E = Hc/λ
λ = Hc/E = 6.63 x 10-34 x 2.99 x 108 /4.02 x 10-19 = 4.93 x 10-7 = 493 nm
So light of UV wavelength has enough energy to break the bond by homolytic fission (one electron going to each chlorine atom).
Cl2 ==> 2Cl·
(edited 2 years ago)
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