Question

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. $$3^0>2^0<1^0$$

Answer 1

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Note: Alkyl halides (also known as haloalkanes) are compounds in which one or more hydrogen atoms in an have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine).