Question

For the determination of the structural formula of aromatic compound A, the 1 H-NMR spectrum was obtained, which is given below.

Compound A was subjected to the five (5) stage experiment, which follows. It is stated that in Stage I the yield was 45%, while for the remaining stages the yield was 100%.

Step I: Amount of 0.02 mol of Compound A after treatment with concentrate NaOH under the appropriate conditions, gave the organic products B and C.

Step II: To the reaction mixture of Step I, 2.4-DNPH was added so that the mixture was completely reacted to form a colored precipitate, which was removed by filtration.

Stage III: Extract is isolated from Compound B, which has a lower boiling point than Compound C.

Step IV: An aqueous solution of hydrochloric acid was added to the solution of compound C, so that the pH became slightly acidic. With cooling of the solution, white crystals of organic compound D were displaced.

Step V: The crystals of D, after being separated by filtration, suitably cleaned and dried, were found to have a mass of 0.612g.

a) Write the structural formulas of organic compounds A, B, C and D, using all the information and recording your course of work.

b) Write the structural formula of organic compound X, which under the appropriate conditions reacts in a (1) step and gives the organic compounds B and D.

c) Calculate the percentage (% by mass) of compound A which was converted to precipitate by 2,4-DNPH

30 2.0 2.0 1.0 8 (ppm)

Answer 1

The singlet at 10 ppm indicate presence of aldehyde functionality. The two doublets of 4 protons in the range 7 - 8 ppm indicate presence of aromatic ring and it is substituted at 1 and 4 positions. The 3 proton singlet at 2.25 ppm indicate a methyl gropup.

From these H-NMR data the compound is concluded as 4-Methylbenzaldehyde.

Stage I

In presence of conc NaOH aromatic aldehydes under go cannizzaro reaction.

OH Nao Conc NaOH Cannizzaro reaction В A

here, the reaction yield is only 45%. So, 55% of compound A will be remain unreacted.   

Step II

In this step addition of 2,4-DNPH (2,4-dinitrophenylhydrazine) gives colored product.

NO2 02N NH2 O2N NO2 A

Stage III. The Compound B is a benzyle alcohol and compound C is a 4-methylbenzoicacid sodium salt. so, the boiling point of B is less compared to C.

Stage IV.

Addition of HCl to C gives a precipitate under acidic pH. Sodium salt of 4-methylbenzoic acid gives 4-methylbenzoic acid after reacting with HCl.

NaO но НС1 NaC D

In step I the conversion yield is 45%. So the remaining 55% of compound A will be reacted with 2,4-DNPH.