Anorganic compound A (molecular formula `C_4H_10O`) reacts vigorously with acetyl chloride and responds to iodoform test. When passed over hot alumina, A is converted to another compound B (`C_4H_8`) which on ozonolysis gives only one aldehyde. Identify A and B with reasons.

Explanation

Since the given compound A has four carbons, the aldehyde formed is `CH_3CHO`. Removing the oxygen atoms of two acetaldehyde molecules and combining the carbons of C=Ogroup, the alkene B is
`CH_3 - underset(H)underset(|)C = underset(H)underset(|)C - CH_3` (But - 2- ene)
Alcohol which responds to iodoform test and gives but-2-ene on dehydration is:
`CH_3 - underset(OH)underset(|)(CH)(CH_2)CH_3` (Butan-2-ol)
Butan-2-ol on treatment with acetyl chloride gives the ester:

The reactions may be represented as: