Draw the product formed when the compound shown below undergoes a reaction with hbr in ch2cl2. interactive 3d display mode
2 answers:
Answer : The complete question is attached in the answer.
Explanation : In the given question an alkene is treated with alkylhalide in presence of methylene dichloride it gives a corresponding alkylbromide species.
The mechanism follows markonikov's addition rule, and the major product thus formed is bromine which is attached to least substituted carbon atom ad minor is bromine attached to highly substituted carbon atom.
Products formed follow markovnikov's rules:<u> 2.2 dimethyl 3 bromo butane</u>
<h3>Further explanation</h3>Alkenes are alkane-derived compounds
Alkene is an unsaturated hydrocarbon compound that has a double bond
<h3>-C = C- </h3>Alkene general formula: CnH2n
The nomenclature of the alkenes is generally the same as the alkanes
Alkenes including hydrocarbons are more reactive compared to alkanes because of the double bonds they have
One of the reactions that occur in alkenes is the addition reaction, which is generally the removal of the double bond due to the addition of another compound
In the addition reaction of HX gas (X = Cl, Br, or I) which is a halogen gas, Markovnikov rules apply,
1. If a carbon bonded double carbon atom binds to a different number of hydrogen atoms, then the halogen atom (X) will be bound to carbon atoms that hold fewer hydrogen atoms
2. If the number of carbon atoms in the double bond binds the same number of hydrogen atoms, then atom X will be bound to atom C which has the longest carbon chain
In the addition reaction of HBr to 3,3-dimethylbutene-1 in a solution of CH₂Cl₂ (CH₂Cl₂ does not react, is solvent) then the alkene compound 3,3-dimethylbutena-1 which has a double bond is found in atoms C no 1 and 2
In C-1 atoms, carbon atoms bind to 2 H atoms
In C-2 atoms, carbon atoms bind to 1 H atom
Then the Br Atom on HBr will bind to the C-2 atom that hold fewer hydrogen atoms, while the H atom on the HBr will bind to the C-2 atom
<h3>Learn more </h3>all correct statements concerning pentyne
the carbon-carbon double bond in eugenol
the statements about the properties of the carbon-carbon double bond are not true
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Ecell = 0.577 V
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Cr(s) ------------------ Cr+3(aq) + 3e- ] x 2 ...O
Sn+2(aq) + 2e- ------------ Sn(s) ] x 3 ...R
Reaction
2 Cr(s) + 3 Sn+2(aq) --------------- 2 Cr+3(aq) + 3 Sn(s)
Therefore
Eicell = - 0.14 - ( - 0.74)
Eicell = 0.60
In conclusion
Ecell = 0.577 V
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