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Organic reasoning in college chemistry
A is optically active, indicating that it contains chiral carbon and its molecules are asymmetric.

B is also optically active (ibid.)

B can form salts with both acids and bases: it refers to amino acids.

It can react with HNO2 to release gas: the prepared fatty primary amine reacts with nitrous acid to generate alcohol, and nitrogen is released: RNH2+HNO2→ROH+N2↑.

Therefore, B has the structure of H2NRCOOH (at least 2 carbons).

The interaction between c and Fecl3 _ 3 _ 3 is blue-purple: it can be inferred that c has phenolic hydroxyl groups (including benzene rings and at least 6 carbons).

The molecular composition of a is C9H 1 1O2N and it is hydrolyzed into b and C(B and c add up to 9 carbons, so it is just right).

Therefore, b is h2n (CH3) chcooh and c is phenol (C6H5OH).

A is c6h5occh (CH3) NH2.

The related reaction equation is: C6H5OOCCH (CH3) NH2+2NaOH → C6H5ONA+NaOCCH (CH3) NH2+H2O.

Acidification with weak acid (that is, it can't form salt with amino NH2): c6h5na+h +→ c6h5hoh naoocch (CH3) NH2+h +→ hoocch (CH3) NH2.

Interaction between B and HNO2: H2n (CH3) chcooh+HNO2 → hoochch (CH3) OH+N2 ↑.

The interaction between C and FeCl _ 3 is blue-purple: C6H5OH+FeCl _ 3 → blue-purple complex.