Daily Chemistry Problems for JEE - Day 2
An optically active compound (A), forms a hydrochloride but dissolves in water to give a neutral solution. On heating with soda lime, (A) yields (B).
Both (A) and (B) react with and dilute HCl, the former yields a compound (C) , which on heating is converted to (D), while the latter yields (E), .
In (A) the type of carbon to which N is attached is . The value of z is:
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1 solution
Firstly, since the compound A dissolves in water to give a neutral solution, hence, we are looking at the presence of an acid as well as a basic group, i.e. a carboxylic group, and an − N H 2 group.
Secondly, since the compound's formula does not show any degree of un-saturation, hence it's a straight chain compound, and therefore, it can be either:
C H 3 − C H N H 2 − C O O H
Or
C H 2 N H 2 C H 2 − C O O H
Now, the presence of an − N H 2 can be confirmed by the next reaction, i.e. addition of N a N O 2 + H C l which is use in diazonium preparation.
Finally, since the compound A is an optically active compound, hence, it has to be the former of the two compounds, i.e.
C H 3 − C H N H 2 − C O O H
and therefore, the required answer is 2 as N in this compound is attached to a 2 o carbon.