Halogenoalkanes

//Halogenoalkanes// are compounds that contain a halogen (fluorine, chlorine, bromine or iodine) bonded to an alkyl group. Halogenoalkanes undergo substitution reactions. This is because the carbon atom that is bonded to the halogen contains a partial positive charge ( δ + ) due to the difference in electronegativity between the carbon and the halogen. Therefore negative ions with a lone pair, are attracted to the carbon atom and displace the halogen.  Iodoalkanes are the most reactive due to the weak C-I bond that can be easily broken. Eg of reaction: C 2 H 5 Br (aq) + NaOH (aq) ---> C 2 H 5 OH (aq) + NaBr (aq)

In this reaction the hydroxide ion (OH - ) is attracted to the carbon with the δ +  charge. The negative ion with the lone pair, which displaces the halogen, during the reaction is called a **nucleophile**. Therefore, in the above reaction OH -  is a nucleophile. <span style="font-family: Arial,sans-serif; font-size: 14pt;">Usually “curly arrows” are used to show the movement of an electron pair during such kind of a nucleophilic substitution reaction. (the end of the arrow always shows where the electrons move to).

<span style="font-family: Arial,sans-serif; font-size: 14pt;">Fluoroalkanes usually do not undergo substitution reactions due to the strong C-F bond. <span style="font-family: Arial,sans-serif; font-size: 14pt;">The mechanism of the substitution reaction depends on the type of halogenoalkane that is reacting. There are two reaction mechanisms, namely the S N <span style="font-family: Arial,sans-serif; font-size: 14pt;">1 and the S N <span style="font-family: Arial,sans-serif; font-size: 14pt;">2 mechanisms.

<span style="font-family: Arial,sans-serif; font-size: 14pt;">This is a two step mechanism. <span style="font-family: Arial,sans-serif; font-size: 14pt;">In the first step the carbon-to-halogen bond is broken to form a carbocation intermediate. <span style="font-family: Arial,sans-serif; font-size: 14pt;">In the second step, the carbocation reacts with the nucleophile to form the product. <span style="font-family: Arial,sans-serif; font-size: 14pt;">Eg: <span style="font-family: Arial,sans-serif; font-size: 14pt;">The reaction of 2-bromo-2methylpropane, with dilute aqueous sodium hydroxide solution to form the tertiary alcohol 2-methylpropan-2-ol. <span style="font-family: Arial,sans-serif; font-size: 14pt;">C(CH 3 <span style="font-family: Arial,sans-serif; font-size: 14pt;">) 3 <span style="font-family: Arial,sans-serif; font-size: 14pt;">Br (aq) + OH - <span style="font-family: Arial,sans-serif; font-size: 14pt;"> (aq) ---> C(CH 3 <span style="font-family: Arial,sans-serif; font-size: 14pt;">) 3 <span style="font-family: Arial,sans-serif; font-size: 14pt;">OH (aq) + Br - <span style="font-family: Arial,sans-serif; font-size: 14pt;"> (aq) <span style="font-family: Arial,sans-serif; font-size: 14pt;">1.In the first step the C-Br bond is broken to form a carbocation intermediate. <span style="font-family: Arial,sans-serif; font-size: 14pt;">
 * <span style="font-family: Arial,sans-serif; font-size: 14pt;">S ** N **<span style="font-family: Arial,sans-serif; font-size: 14pt;">1 (//substitution nucleophilic unimolecular//) <span style="font-family: Arial,sans-serif; font-size: 19px;">mechanism **

<span style="font-family: Arial,sans-serif; font-size: 14pt;">2.In the second step the carbocation intermediate reacts with a hydroxide ion to form the alcohol. <span style="font-family: Arial,sans-serif; font-size: 14pt;">


 * <span style="font-family: Arial,sans-serif; font-size: 14pt;">S ** N **<span style="font-family: Arial,sans-serif; font-size: 14pt;">2 (//substitution nucleophilic bimolecular//) <span style="font-family: Arial,sans-serif; font-size: 19px;">mechanism **

<span style="font-family: Arial,sans-serif; font-size: 14pt;">In this reaction the nuclophile is attracted to the carbon atom and begins to bond to the carbon atom, just as the halogen atom begins to leave the carbon atom. <span style="font-family: Arial,sans-serif; font-size: 14pt;">Thus a transition state is formed with the carbon atom having 5 different groups/molecules around it. The transition state is an ion as it contains a negative charge. <span style="font-family: Arial,sans-serif; font-size: 14pt;">The halogen then leaves as a halide ion and the product is formed. <span style="font-family: Arial,sans-serif; font-size: 14pt;">Eg: <span style="font-family: Arial,sans-serif; font-size: 14pt;">The reaction of bromoethane with sodium hydroxide, (ionic equation below): <span style="font-family: Arial,sans-serif; font-size: 14pt;"> <span style="font-family: Arial,sans-serif; font-size: 14pt;">C 2 <span style="font-family: Arial,sans-serif; font-size: 14pt;">H 5 <span style="font-family: Arial,sans-serif; font-size: 14pt;">Br (aq) + OH - <span style="font-family: Arial,sans-serif; font-size: 14pt;"> (aq) ---> C 2 <span style="font-family: Arial,sans-serif; font-size: 14pt;">H 5 <span style="font-family: Arial,sans-serif; font-size: 14pt;">OH (aq) + Br - <span style="font-family: Arial,sans-serif; font-size: 14pt;"> (aq)

<span style="font-family: Arial,sans-serif; font-size: 14pt;">The hydroxide ion is attracted to the carbon atom (bonded to bromine) and begins to form a bond to it just as the bromine begins to leave the carbon atom. Thus a transition state is created with 5 groups/atoms around the carbon. <span style="font-family: Arial,sans-serif; font-size: 14pt;">The bromine atom then leaves the carbon atom as a bromide ion and an alcohol is produced.

<span style="font-family: Arial,sans-serif; font-size: 14pt;">//**Primary** halogenoalkanes// react mostly via the **S** ** N <span style="font-family: Arial,sans-serif; font-size: 14pt;">2 mechanism ** <span style="font-family: Arial,sans-serif; font-size: 14pt;">//**Tertiary** halogenoalkanes// react mostly via the **S** ** N <span style="font-family: Arial,sans-serif; font-size: 14pt;">1 mechanism ** <span style="font-family: Arial,sans-serif; font-size: 14pt;">//**Secondary**// halogenoalkanes react via both mechanisms.
 * <span style="font-family: Arial,sans-serif; font-size: 14pt;">Note: **

<span style="font-family: Arial,sans-serif; font-size: 14pt;">The breaking of the carbon-to-carbon bond for primary, secondary and tertiary halogenoalkanes involves heterolytic fission as the halogen atom always leaves as the halide ion.

By the end of this lesson you should be able to: __**<span style="color: #0000ff; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Next up- Reaction pathways **__
 * use equations to demonstrate the substitution reaction of halogenoalkanes with sodium hydroxide.
 * explain the substitution reactions of halogenoalkanes with sodium hydroxide using the S N 1 and S N 2 mechanisms.
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