
What will be the product?
Answer
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Hint : The alkenes are the hydrocarbons consisting of only carbon and hydrogen atoms bonded by double covalent bonds. When the alkenes are treated with hydrogen bromide a markovnikov product, which means the negative part of the reagent is added to the carbon containing less number of hydrogen atoms.
Complete Step By Step Answer:
Note :
When alkenes are treated with hydrogen bromide in presence of peroxides, anti markovnikov’s product is formed, as the peroxide directs the free-radical mechanism it states that, a negative part of reagent attacks on the carbon with more number of hydrogen atoms, but in the given case markovnikov’s product is formed as it's not a free radical mechanism.
Complete Step By Step Answer:
Given a molecule is an alkene, an alkene is an unsaturated hydrocarbon consisting of only carbon and hydrogen atoms bonded through double covalent bonds. Out of which one bond is sigma and the other is pi-bond.
The given compound is $ 3,3 - diethyl1 - pentene $ . It has one double bond and the two ethyl groups as the substituent. The base or parent carbon chain is pentene.
When the given compound is treated with hydrogen bromide, the negative part of hydrogen bromide i.e.., bromide ion attacks on the double bonded carbon containing less number of hydrogen atoms, i.e.., the product is formed according to markovnikov’s rule.
Markovnikov’s rule states that the negative part of reagent attacks on the more substituted carbon, and the positive part of the reagent attacks on the less substituted carbon.
Thus, when $ 3,3 - diethyl1 - pentene $ is treated with hydrogen bromide forms $ 2 - bromo 3,3 - diethylpen\tan e $ .
The double bond can be reduced to substituted alkyl halide upon treatment with hydrogen bromide. Hydrogen bromide is an acid with the molecular formula of $ HBr $ .
Note :
When alkenes are treated with hydrogen bromide in presence of peroxides, anti markovnikov’s product is formed, as the peroxide directs the free-radical mechanism it states that, a negative part of reagent attacks on the carbon with more number of hydrogen atoms, but in the given case markovnikov’s product is formed as it's not a free radical mechanism.
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