
Presence of peroxides affects the addition of
A. HBr
B. HCl
C. HI
D. All of these
Answer
349.5k+ views
Hint: In this question we have to use the concept of the “peroxide effect”. Peroxide reaction is the mechanism for free radical addition of certain substrates. Hydrogen peroxide is a chemical compound with the formula H2O2. It is a colorless liquid at room temperature with a bitter taste.
Complete Step by Step Answer:
The peroxide effect, also known as the anti-rule, Markownikoff's is the addition of HBr to unsymmetrical alkenes. This result caused the carbon atom of the unsymmetrical alkene's double bond that is carrying the most hydrogen atoms to become attached to the negative portion of the unsymmetrical attacking reagent. There is peroxide present during this process.
Normally, the hydrogen becomes connected to the carbon atom that already has the most hydrogens on it when a molecule of HX adds to a carbon-carbon double bond. The Markovnikov's Rule is what governs this. This is frequently referred to as the peroxide effect or anti-Markovnikov addition because the HBr adds on in the "wrong way around" when organic peroxides are present. Hydrogen bromide adds to propene using an electrophilic addition process in the absence of peroxides. This results in the outcome that Markovnikov's Rule expected.
CH3-CH=CH2 + HBr $\xrightarrow{peroxide}$ CH3-CH2-CH2-Br
Propene is converted to n-propyl bromide in presence of peroxide
The correct answer is HBr.
Additional Information: Because of how strong the hydrogen-fluorine connection is, fluorine radicals are not created during the initiation stage. The second phase of the propagation stage is extremely sluggish with hydrogen chloride. The subsequent reaction is endothermic, as shown by a bond enthalpy sum. This is because the hydrogen-chlorine connection is relatively strong. The first step of the propagation stage in the hydrogen iodide example turns out to be endothermic, which slows the reaction down. The weak carbon-iodine bond does not produce enough energy when it is created.
Note: Simply put, the issue of reaction rates is what causes hydrogen bromide to add in an anti-Markovnikov manner when organic peroxides are present. The alternate electrophilic addition mechanism is substantially slower than the free radical mechanism. Both mechanisms are active, but the free radical mechanism produces the majority of the product since it operates more quickly. It's the other way around for the other hydrogen halides.
Complete Step by Step Answer:
The peroxide effect, also known as the anti-rule, Markownikoff's is the addition of HBr to unsymmetrical alkenes. This result caused the carbon atom of the unsymmetrical alkene's double bond that is carrying the most hydrogen atoms to become attached to the negative portion of the unsymmetrical attacking reagent. There is peroxide present during this process.
Normally, the hydrogen becomes connected to the carbon atom that already has the most hydrogens on it when a molecule of HX adds to a carbon-carbon double bond. The Markovnikov's Rule is what governs this. This is frequently referred to as the peroxide effect or anti-Markovnikov addition because the HBr adds on in the "wrong way around" when organic peroxides are present. Hydrogen bromide adds to propene using an electrophilic addition process in the absence of peroxides. This results in the outcome that Markovnikov's Rule expected.
CH3-CH=CH2 + HBr $\xrightarrow{peroxide}$ CH3-CH2-CH2-Br
Propene is converted to n-propyl bromide in presence of peroxide
The correct answer is HBr.
Additional Information: Because of how strong the hydrogen-fluorine connection is, fluorine radicals are not created during the initiation stage. The second phase of the propagation stage is extremely sluggish with hydrogen chloride. The subsequent reaction is endothermic, as shown by a bond enthalpy sum. This is because the hydrogen-chlorine connection is relatively strong. The first step of the propagation stage in the hydrogen iodide example turns out to be endothermic, which slows the reaction down. The weak carbon-iodine bond does not produce enough energy when it is created.
Note: Simply put, the issue of reaction rates is what causes hydrogen bromide to add in an anti-Markovnikov manner when organic peroxides are present. The alternate electrophilic addition mechanism is substantially slower than the free radical mechanism. Both mechanisms are active, but the free radical mechanism produces the majority of the product since it operates more quickly. It's the other way around for the other hydrogen halides.
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