Question

The ease of dehydrohalogenation of alkyl halide with alcoholic koh is

Answer 1

Ease of dehydrohalogenation of alkyl halides is: tert > sec > pr

The alkyl halides, in which halogen is attached to a terminal carbon, yield a single alkene but alkyl halides in which the halogen atom is attached to a non-terminal carbon atom and both adjacent position have hydrogen atoms yield a mixture of alkenes. For example: KOH (alc)

CH3CH2CH2Cl CH3CH=CH2chloropropane propene

Cl | KOH (alc) CH3CH2CHCH3CH3CH=

CHCH3+ CH3CH2CH=CH22-chlorobutane 2-butene 1-butene (major) (minor) In the first reaction, chloropropane can lose hydrogen only from C2; therefore, it gives only one product, i.e., propene. However, in the second reaction, 2-chlorobutane can lose hydrogen from any of the two-β-carbon atoms and, hence, it gives a mixture of 2-butene (80%) and 1-butene (20%). Now you may ask why 2-butene is the major product? Dehydrohalogenation follows Saytzeff rule which says that the more highly substituted alkene is the dominant product.

Answer 2

Alkyl halide dehydrohalogenation with alcoholic KOH is a 3°>2°>1° instance.

tertiary> secondary > primary

Explanation:

• On the basis of the stability of the alkene created during the dehydrohalogenation of haloalkanes, this sequence of alkyl halides can be explained.
• Dehydrohalogenation of 3° alkyl halides produces more substituted alkenes, which are more stable and form at a faster rate, whereas dehydrohalogenation of primary alkyl halides produces the fewest substituted alkenes, which are less stable and form at a slower rate.
• The Saytzeff rule states that any alkyl halide that produces a more highly substituted (and hence more stable) alkene dehydrohalogenation more quickly than one that produces a less highly substituted (and therefore less stable) alkene.
• As a result, different alkyl halides with the same halogen are easier to dehydrohalogenate in decreasing order.

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