Bromine Reagents: CCl4 vs H2O vs HCl | chemca

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Reactions of Bromine ($Br_2$)

The effect of Solvent ($CCl_4$ vs $H_2O$) and Added Nucleophiles ($Cl^-$) on Alkene Addition.

By chemca Team • Updated Jan 2026

The addition of bromine to alkenes proceeds via a Cyclic Bromonium Ion intermediate. The final product depends on which nucleophile opens this ring. The nucleophile comes from the reagent itself ($Br^-$), the solvent ($H_2O$), or added salts ($Cl^-$).

1. The General Mechanism

Formation of Cyclic Bromonium Ion

The alkene attacks $Br_2$, expelling $Br^-$ and forming a three-membered positively charged ring.

$$ >C=C< + Br_2 \rightarrow \underset{\text{Cyclic Bromonium Ion}}{>C(\overset{+}{Br})C<} + Br^- $$

Regiochemistry: The next step involves the attack of a nucleophile (Nu) from the backside (Anti-Addition) at the more substituted carbon (where the positive charge density is higher/more stable).

2. $Br_2$ in $CCl_4$ (Non-Polar Solvent)

Formation of Vicinal Dibromides

Since $CCl_4$ is non-nucleophilic, the only available nucleophile is the bromide ion ($Br^-$) generated in the first step.

$$ CH_2=CH_2 + Br_2 \xrightarrow{CCl_4} \underset{\text{1,2-Dibromoethane}}{Br-CH_2-CH_2-Br} $$

Stereochemistry: Anti-Addition.
Cis-alkene $\to$ Racemic mixture ($\pm$).
Trans-alkene $\to$ Meso compound.

Observation: The reddish-brown color of bromine disappears (Test for Unsaturation).

3. $Br_2$ in $H_2O$ (Bromine Water)

Formation of Halohydrins

Water is a polar protic solvent and a nucleophile. Since water is present in large excess compared to $Br^-$, it competes successfully to open the bromonium ring.

$$ CH_3-CH=CH_2 + Br_2 + H_2O \rightarrow \underset{\text{Bromohydrin}}{CH_3-CH(OH)-CH_2Br} + HBr $$

Regioselectivity (Markovnikov-like):

• The electrophile $Br^+$ adds first.
• The nucleophile ($H_2O$) attacks the more substituted carbon (which can better support the partial positive charge of the ring).
• Result: $-OH$ goes to the more hindered carbon, $-Br$ goes to the less hindered carbon.

4. $Br_2$ in presence of $NaCl$ or $HCl$ ($Cl^-$)

Formation of Mixed Halides

If excess chloride ions ($Cl^-$) are present (e.g., from added $NaCl$ or $HCl$), they act as nucleophiles competing with $Br^-$.

$$ CH_2=CH_2 + Br_2 \xrightarrow{NaCl (aq)} \underset{\text{1-Bromo-2-chloroethane (Major)}}{Br-CH_2-CH_2-Cl} + Br-CH_2-CH_2-Br $$

Why? The concentration of added $Cl^-$ is often higher than the $Br^-$ generated.
Note: $Cl^-$ cannot start the reaction (it's not an electrophile). The reaction must start with $Br^+$ (from $Br_2$). Hence, no 1,2-dichloro product is formed.

5. Solvent Effect on Phenol

Solvent	Reaction	Product
Non-polar ($CS_2, CCl_4$)	Low ionization of Phenol. Ring moderately activated.	Mono-substitution (o/p Bromophenol).
Polar ($H_2O$)	High ionization forms Phenoxide ion ($PhO^-$). Ring highly activated.	Poly-substitution (2,4,6-Tribromophenol).

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