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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
119.4k+ 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.
![](https://www.vedantu.com/question-sets/b97f925f-11a6-4258-a047-c68250c8e0aa8139725201121089332.png)
-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.
![](https://www.vedantu.com/question-sets/1fd275b1-92f6-4f37-b685-289bb63bc2f43548842190800752515.png)
-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.
![](https://www.vedantu.com/question-sets/b97f925f-11a6-4258-a047-c68250c8e0aa8139725201121089332.png)
-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.
![](https://www.vedantu.com/question-sets/1fd275b1-92f6-4f37-b685-289bb63bc2f43548842190800752515.png)
-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.
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