 
					
					
						Why don’t the other hydrogen halides behave in the same way?					
				 
				
					
						 المؤلف:  
						University of Missouri System
						 المؤلف:  
						University of Missouri System					
					
						 المصدر:  
						Organic Chemistry ii
						 المصدر:  
						Organic Chemistry ii					
					
						 الجزء والصفحة:  
						.................
						 الجزء والصفحة:  
						.................					
					
					
						 4-10-2020
						4-10-2020
					
					
						 1767
						1767					
				 
				
				
				
				
				
				
				
				
				
			 
			
			
				
				
Why don’t the other hydrogen halides behave in the same way?
The reason that hydrogen bromide adds in an anti-Markovnikov fashion in the presence of organic peroxides is simply a question of reaction rates. The free radical mechanism is much faster than the alternative electrophilic addition mechanism. Both mechanisms happen, but most of the product is the one from the free radical mechanism because that is working faster. With the other hydrogen halides, the opposite is true.
	- Hydrogen fluoride: The hydrogen-fluorine bond is so strong that fluorine radicals aren’t formed in the initiation step.
- Hydrogen chloride: With hydrogen chloride, the second half of the propagation stage is very slow. If you do a bond enthalpy sum, you will find that the following reaction is endothermic.


This is due to the relatively high hydrogen-chlorine bond strength.
	- Hydrogen iodide: In this case, the first step of the propagation stage turns out to be endothermic and this slows the reaction down. Not enough energy is released when the weak carbon-iodine bond is formed.


In the case of hydrogen bromide, both steps of the propagation stage are exothermic.
 
				
				
					
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