Question

Which of the following reactions is an elimination reaction?
(a) ${ CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 2 }-OH\xrightarrow { { PCl }_{ 3 } } { CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 2 }-Cl$
(b) ${ Br-CH }_{ 2 }-{ CH }_{ 2 }-Br\xrightarrow { Zn } { CH }_{ 2 }={ CH }_{ 2 }+{ ZnBr }_{ 2 }$
(c) ${ CH }_{ 3 }-{ CH }(Cl)-{ CH }_{ 3 }\xrightarrow { alc.\quad KOH } { CH }_{ 3 }-{ CH }={ CH }_{ 2 }$
(d) ${ CH }_{ 3 }-COOH\xrightarrow { { CH }_{ 3 }OH } { CH }_{ 3 }-C(O)-O{ CH }_{ 3 }$
(e) Both (b) and (c) are correct

Answer 1

Hint: Elimination reaction involves the loss of a small molecule and the formation of a product. The two most common elimination reactions are dehydrohalogenation and dehalogenation reactions.

Complete step by step answer:
Before solving this question, we need to understand the two basic types of organic reactions namely substitution reactions and elimination reactions.

We will look into substitution reactions briefly:
Substitution reactions involve the loss of a leaving group and the addition of a nucleophilic group. These reactions can follow the $ { S }_{ N }1$ mechanism or the $ { S }_{ N }2$ mechanism. The $ { S }_{ N }1$ mechanism involves the formation of enantiomers in a 50:50 ratio due to which the mixture of the products is not optically active (racemic mixture). The ${ S }_{ N }2$ mechanism involves Walden inversion.

Now, we will look into elimination reactions in more detail:
Elimination reaction involves the loss of a small molecule and the formation of a product. There are many different types of elimination reactions but we will look into only two types namely dehydrohalogenation and dehalogenation.

Dehydrohalogenation: This involves the formation of an alkene from a haloalkane which takes place due to the removal of a halogen halide (HCl, HBr, HI). The hydrogen will be removed from the $ \beta$-carbon and halogen from the $ \alpha $-carbon. The ease of dehydrohalogenation of alkyl halides having the same alkyl group but different halogens is: $ iodides > bromides > chlorides$ while for isomeric alkyl halides having the same halogen but different structures is: $ tertiary > secondary > primary$. If we have an alkyl halide such that there is $ \beta $-hydrogen on either side of the carbon atom bonded to the halogen, then two different alkene products are possible but their relative amounts will differ. It is the more highly substituted alkene which predominates and this rule is called Saytzeff rule. The reaction given below is an example of dehydrohalogenation reaction:
${ CH }_{ 3 }-{ CH }(Cl)-{ CH }_{ 3 }\xrightarrow { alc.\quad KOH } { CH }_{ 3 }-{ CH }={ CH }_{ 2 }$

Dehalogenation: It involves the removal of a molecule of halogen (${ Cl }_{ 2 },{ Br }_{ 2 },{ I }_{ 2 }$) from a dihaloalkane in which the two halogen atoms are present on adjacent carbon atoms (vicinal or 1,2-dihaloalkanes) to form an alkene when heated with zinc dust with methanol as solvent. 1,1-dihaloalkanes also undergo dehalogenation on heating with zinc dust. The reaction given below is an example of dehalogenation reaction:
${ Br-CH }_{ 2 }-{ CH }_{ 2 }-Br\xrightarrow { Zn } { CH }_{ 2 }={ CH }_{ 2 }+{ ZnBr }_{ 2 }$
So, the correct answer is “Option E”.

Note: Opposed to the Saytzeff elimination, there is also Hofmann elimination. Alkyl fluorides undergo dehydrohalogenation in accordance with Hofmann rule. According to Hofmann rule, the orientation of elimination is governed by the acidity of the $\beta $-hydrogen to be eliminated. The major product is formed by the elimination of the more acidic $\beta $-hydrogen.