Product formed when ch3ch2ch(cl)ch(h)ch3 is heated with alcoholic KOH and explanation in detail:

When ch3ch2ch(cl)ch(h)ch3 is heated with alcoholic KOH, an elimination reaction occurs. Here's the detailed explanation:

Step 1: Formation of alkoxide ion

The alcoholic KOH dissociates to form KOH and alcohol. The alcohol acts as a solvent and KOH acts as a base. The base abstracts a proton from the α-carbon, leading to the formation of a carbanion intermediate. The carbanion intermediate then interacts with KOH to form an alkoxide ion.



Step 2: Dehydrohalogenation

The alkoxide ion attacks the β-hydrogen, leading to the formation of a double bond between the β and γ carbons. This process is called dehydrohalogenation, which means the elimination of a hydrogen atom and a halogen atom (in this case, Cl) from adjacent carbon atoms.



Step 3: Formation of the product

The final product is an alkene (propene) and HCl. The reaction is an example of an E2 (elimination bimolecular) reaction because it occurs in a single step and involves the simultaneous loss of two groups (H and Cl) from adjacent carbon atoms.



Conclusion:

Thus, when ch3ch2ch(cl)ch(h)ch3 is heated with alcoholic KOH, it undergoes an E2 elimination reaction to form propene and HCl.

Alkene will be formed as a major product.
An alcoholic solution of KOH contains RO` ion which is a strong base. Thus it can abstract hydrogen from bita carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.