The $cis,2-butene$ on reaction with $B{{r}_{2}}$ in $CC{{l}_{4}}$produces mainly
(A) $\text{1-bromo-2-butene}$
(B) $2,3-\text{dibromobutane}$
(C) $\text{meso-2,3-dibromobutane}$
(D) $(\pm )2,3-\text{dibromobutane}$
Answer
Verified
122.7k+ views
Hint: The electrophilic addition reaction of bromine molecules takes place with the alkene, known as the bromination process. It involves the breaking of the C-C double bond in the alkene given.
Complete step by step solution:
In the given reaction, between an alkene and bromine molecule in presence of carbon tetrachloride, it leads to an additional reaction.
-As in the bromine molecule, the bromine atoms have the same electronegativity and are nonpolar in nature. So, the $CC{{l}_{4}}$ having a polar C-Cl bond interacts with it and induces a partial positive charge on the closest bromine atom through the dipole-induced dipole interaction. Thus, making the bromine molecule partially charged.
-Then, the nucleophilic $C=C$ bond of the butene attacks the electrophilic centre, that is the $B{{r}^{+}}$ in the Br-Br molecule bond. This is the electrophilic addition step where the $B{{r}^{+}}$gets added to the $C=C$ bond. It forms a three-membered ring and hence, a bridged intermediate is formed known as the bromonium.
-The $B{{r}^{-}}$ left behind, now being highly nucleophilic, attacks the bromonium intermediate through nucleophilic addition. It attacks from the opposite side of the bridged ring as it may hinder its attack on the partially positive carbon atom. Therefore, the addition of the bromine molecule in the given reaction is anti-addiction, that is, both the bromine atoms are on opposite faces of the C-C bond.
-This leads to the formation of a racemic mixture of $2,3-\text{dibromobutane}$, that is, the two forms being $(2R,3R)-2,3-\text{dibromobutane}$and $(2S,3S)-2,3-\text{dibromobutane}$.They are enantiomers.
Therefore, the $cis,\,2-Butene$ on reaction with $B{{r}_{2}}$ in $CC{{l}_{4}}$produces mainly option (D)- $(\pm )2,3-\text{dibromobutane}$.
Note: It can be seen that the reaction is an anti-addition reaction and also stereospecific in nature. The carbon tetrachloride solvent only provides the opportunity for the reactant to get polarized and has no other effect on the reaction process.
Complete step by step solution:
In the given reaction, between an alkene and bromine molecule in presence of carbon tetrachloride, it leads to an additional reaction.
-As in the bromine molecule, the bromine atoms have the same electronegativity and are nonpolar in nature. So, the $CC{{l}_{4}}$ having a polar C-Cl bond interacts with it and induces a partial positive charge on the closest bromine atom through the dipole-induced dipole interaction. Thus, making the bromine molecule partially charged.
-Then, the nucleophilic $C=C$ bond of the butene attacks the electrophilic centre, that is the $B{{r}^{+}}$ in the Br-Br molecule bond. This is the electrophilic addition step where the $B{{r}^{+}}$gets added to the $C=C$ bond. It forms a three-membered ring and hence, a bridged intermediate is formed known as the bromonium.
-The $B{{r}^{-}}$ left behind, now being highly nucleophilic, attacks the bromonium intermediate through nucleophilic addition. It attacks from the opposite side of the bridged ring as it may hinder its attack on the partially positive carbon atom. Therefore, the addition of the bromine molecule in the given reaction is anti-addiction, that is, both the bromine atoms are on opposite faces of the C-C bond.
-This leads to the formation of a racemic mixture of $2,3-\text{dibromobutane}$, that is, the two forms being $(2R,3R)-2,3-\text{dibromobutane}$and $(2S,3S)-2,3-\text{dibromobutane}$.They are enantiomers.
Therefore, the $cis,\,2-Butene$ on reaction with $B{{r}_{2}}$ in $CC{{l}_{4}}$produces mainly option (D)- $(\pm )2,3-\text{dibromobutane}$.
Note: It can be seen that the reaction is an anti-addition reaction and also stereospecific in nature. The carbon tetrachloride solvent only provides the opportunity for the reactant to get polarized and has no other effect on the reaction process.
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE
How Electromagnetic Waves are Formed - Important Concepts for JEE
Electrical Resistance - Important Concepts and Tips for JEE
Average Atomic Mass - Important Concepts and Tips for JEE
Chemical Equation - Important Concepts and Tips for JEE
Concept of CP and CV of Gas - Important Concepts and Tips for JEE
Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main Chemistry Question Paper with Answer Keys and Solutions
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions
JEE Main Chemistry Online Mock Test for Class 12
Other Pages
NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes
NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions
NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs
NCERT Solutions for Class 12 Chemistry Chapter 7 Alcohol Phenol and Ether
NCERT Solutions for Class 12 Chemistry Chapter 8 Aldehydes Ketones and Carboxylic Acids