`=>` When a haloalkane with `color{red}(β)`-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from `color{red}(β)`-carbon and a halogen atom from the `color{red}(α)`-carbon atom.

● As a result, an alkene is formed as a product.

● Since `color{red}(β)`-hydrogen atom is involved in elimination, it is often called `color{red}(β)`-elimination. See fig.

`=>` If there is possibility of formation of more than one alkene due to the availability of more than one `color{red}(α)`-hydrogen atoms, usually one alkene is formed as the major product.

● Alexander Zaitsev (also pronounced as Saytzeff) in 1875 formulated a rule summarised as “In dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.”

● Thus, 2-bromopentane gives pent-2-ene as the major product.

`color{red}(undersettext{pent -2- ene (81 %) } (H_3C-CH_2-CH= CH- CH_3) overset{ overset(-) OH } ← undersettext(2- Bromopentane) ( H_3C - CH_2-CH_2- overset( overset(Br)(|))CH - underset(underset(H)(|))CH_2) overset{ overset(-)OH } → undersettext{Pent -1- ene (19 % ) } (H_3C- CH_2-CH_2-CH = CH_2))`

`=>` A chemical reaction is the result of competition; it is a race that is won by the fastest runner.

`=>` A collection of molecules tend to do, by and large, what is easiest for them.

`=>` An alkyl halide with `color{red}(α)`-hydrogen atoms when reacted with a base or a nucleophile has two competing routes : substitution (`color{red}(S_N1)` and `color{red}(S_N2)`) and elimination.

● Which route will be taken up depends upon the nature of alkyl halide, strength and size of base/nucleophile and reaction conditions.

● Thus, a bulkier nucleophile will prefer to act as a base and abstracts a proton rather than approach a tetravalent carbon atom (steric reasons) and vice versa.

● Similarly, a primary alkyl halide will prefer a `color{red}(S_N2)` reaction, a secondary halide - `color{red}(S_N2)` or elimination depending upon the strength of base/nucleophile and a tertiary halide - `color{red}(S_N1)` or elimination depending upon the stability of carbocation or the more substituted alkene.

`color{green}("Organometallic Compounds ")` : Most organic chlorides, bromides and iodides react with certain metals to give compounds containing carbon-metal bonds. Such compounds are known as organo-metallic compounds.

● An important class of organo-metallic compounds discovered by Victor Grignard in 1900 is alkyl magnesium halide, `color{red}(RMgX)`, referred as `color{green}("Grignard Reagents")`.

● These reagents are obtained by the reaction of haloalkanes with magnesium metal in dry ether.

`color{red}(CH_3CH_2 Br + Mg oversettext(dry ether )→ undersettext(Grignard reagent )(CH_3CH_2 MgBr))`

`=>` In the Grignard reagent, the carbon-magnesium bond is covalent but highly polar, with carbon pulling electrons from electropositive magnesium; the magnesium halogen bond is essentially ionic.

`color{red}(overset(delta-)R - overset(delta+)Mg overset(delta-)X)`

`=>` Grignard reagents are highly reactive and react with any source of proton to give hydrocarbons.

● Even water, alcohols, amines are sufficiently acidic to convert them to corresponding hydrocarbons.

`color{red}(RMgX +H_2O → RH + Mg (OH) X)`

`=>` It is therefore necessary to avoid even traces of moisture from a Grignard reagent.

`=>` On the other hand, this could be considered as one of the methods for converting halides to hydrocarbons.

Alkyl halides react with sodium in dry ether to give hydrocarbons containing double the number of carbon atoms present in the halide. This reaction is known as Wurtz reaction.

`color{red}(2RX + Na → R R +NaX)`