# Daily Chemistry Problems for JEE - Day 2

Chemistry Level 2

An optically active compound (A), $$C_3H_7O_2N$$ forms a hydrochloride but dissolves in water to give a neutral solution. On heating with soda lime, (A) yields $$C_2H_7N$$ (B).

Both (A) and (B) react with $$NaNO_2$$ and dilute HCl, the former yields a compound (C) $$C_3H_6O_3$$, which on heating is converted to (D), $$C_6H_8O_4$$ while the latter yields (E), $$C_2H_6O$$.

In (A) the type of carbon to which N is attached is $$z^{0}$$. The value of z is:

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