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Through which mechanism does HBr undergo...

Through which mechanism does HBr undergo reaction with unsymmetrical alkenes?

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Electrophilic addition mechanism.
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Alkenes undergo :

The alkenes have pi -bonds which make them highly reactive. These undergo electrophilic addition reactions, Addition of HX to unsymmetrical alkenes gives two products and follow Markovnikov rule and anti- Markovnikov rule. The addition occure through the formation of carbocation in Markovnikov rule and free radicals in anti-Markovnikov zule. The ozonolysis reaction of alkenes helps to locate the position of double bond. Predict the major product in the following reaction: CH_(3)-CH=CH_(2) underset((PhCOO)_(2))overset(HBr)to

Knowledge Check

  • Diborane does not undergo cleavage reaction with

    A
    trimethyle amine
    B
    `CO_(2)`
    C
    `CO`
    D
    ammonia

  • The carboxylic acid that does not undergo HVZ reaction is

    A
    ` CH_3COOH`
    B
    ` (CH_3)_2COOH`
    C
    ` CH_3CH_2CH_2CH_2COOH`
    D
    ` (CH_3)_3C C OOH`

  • The compound that does not undergo haloform reaction is

    A
    acetaldehyde
    B
    ethanol
    C
    acetone
    D
    propiophenone

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    The alkenes have pi -bonds which make them highly reactive. These undergo electrophilic addition reactions, Addition of HX to unsymmetrical alkenes gives two products and follow Markovnikov rule and anti- Markovnikov rule. The addition occure through the formation of carbocation in Markovnikov rule and free radicals in anti-Markovnikov zule. The ozonolysis reaction of alkenes helps to locate the position of double bond. Name the products formed when 2-methylpropene is treated with ozone followed by Zn, H_(2)O .

    The alkenes have pi -bonds which make them highly reactive. These undergo electrophilic addition reactions, Addition of HX to unsymmetrical alkenes gives two products and follow Markovnikov rule and anti- Markovnikov rule. The addition occure through the formation of carbocation in Markovnikov rule and free radicals in anti-Markovnikov zule. The ozonolysis reaction of alkenes helps to locate the position of double bond. Ozonolysis of an alkene 'A' followed by decomposition with water and a reducing agent gave a mixture of two isomers of the formula C_(3)H_(6)O . Give the structure of the alkene and its IUPAC name.

    The alkenes have pi -bonds which make them highly reactive. These undergo electrophilic addition reactions, Addition of HX to unsymmetrical alkenes gives two products and follow Markovnikov rule and anti- Markovnikov rule. The addition occure through the formation of carbocation in Markovnikov rule and free radicals in anti-Markovnikov zule. The ozonolysis reaction of alkenes helps to locate the position of double bond. Write the major product in the following reaction: CH_(3)- overset(CH_(3))overset(|)CH-CH=CH_(2) overset(HBr)to

    The alkenes have pi -bonds which make them highly reactive. These undergo electrophilic addition reactions, Addition of HX to unsymmetrical alkenes gives two products and follow Markovnikov rule and anti- Markovnikov rule. The addition occure through the formation of carbocation in Markovnikov rule and free radicals in anti-Markovnikov zule. The ozonolysis reaction of alkenes helps to locate the position of double bond. What product is obtained when propene reacts with CI_(2) in the presence of water?

    Why do alkynes undergo addition reactions while simple alkenes do not?

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