The ease of dehydrohalogenation of alkyl halide with alcoholic KOH is:
A. \[{3^0} < {2^0} < {1^0}\]
B. \[{3^0} > {2^0} > {1^0}\]
C. \[{3^0} < {2^0} > {1^0}\]
D. \[{3^0} > {2^0} < {1^0}\]

Hint: First we should be aware of the term dehydrohalogenation of alkyl halides and Saytzeff`s which is used to find the ease of carbon compounds. Then arrange the ease of dehydrohalogenation of different alkyl halides having the same halogen in suitable order.

Complete answer:
The dehydrohalogenation of alkyl halides, another \[\beta \] elimination reaction, involves the loss of a hydrogen and a halide from an Alkyl halide (RX).
Saytzeff's rule states that “The alkene formed in the greatest amount is the one that corresponds to removal of the hydrogen from the alpha-carbon having the fewest hydrogen substituents”.
So, according to Saytzeff's rule, any alkyl halide that gives a more substituted (more stable) alkene undergoes dehydrohalogenation faster than one which gives a less highly substituted (less stable) alkene. Thus, the ease of dehydrohalogenation of different alkyl halides having the same halogen decreases in the order, \[{\text{tertiary }}\left( {{3^0}} \right) > {\text{secondary }}\left( {{2^0}} \right) > {\text{primary}}\left( {{1^0}} \right)\].

Hence the correct option is B. \[{3^0} > {2^0} > {1^0}\]

Note: Alkyl halides (also known as haloalkanes) are compounds in which one or more hydrogen atoms in an have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine).