>>  >>  >>  >>  >>

Problem 6:

An organic acid A, C3H4O3 is catalytically reduced in presence of ammonia to give B, C3H7NO2. B reacts with acetyl chloride, hydrochloric acid and alcohols. It can also react with nitrous acid to give another compound C, C3H6O3, along with the evolution of nitrogen. What are A, B and C. Give reasons?

Solution:

Compound A is acid having one –COOH group only, the remaining part C2H3O can be

on catalytic reduction keto group is converted into secondary alcohol which with ammonia will give amino acid i.e.,

O                                        OH                                  NH2
||              reduction              ||               HNH2             ||
CH3 – C – COOH ——————→ CH3 – CH – COOH ————→ CH3 – CH – COOH
–H2O

with nitrous acid, B, react to give

NH2                                       OH
|                                          |
CH3 – CH – COOH + HNO2 ——→ CH3 – CH – COOH + N2↑ + H2O
(B)                                       (C)

Problem 7:

Compound A (C6H12O2) on reduction yields two compounds B and C. The compound B on oxidation gave D, which on treatment with aqueous alkali and subsequent heating furnished E. The latter on catalytic hydrogenation gave C. The compound D was oxidised further to give F which was found to be a monobasic acid (mw = 60). Deduce the structures of A to E.

Solution:

From the molecular weight data F comes out to be CH3COOH. If we work back B is CH3CH2­OH and D is CH3CHO.

OH                                      H2/Pt
CH3CHO ————→ CH3 – HC = CH – CHO —————→ CH3CH2CH2CH2OH
(D)        Δ                    (E)                                       (C)

From here we can deduce that reduction of A with LiAlH4 gives two alcohols. So A must be an ester. Hence (A) comes out to be CH3CH2CH2COOC2H5.

Problem 8:

An organic compound (A) on treatment with acetic acid in the presence of sulfuric acid produces an ester (B). (A) on mild oxidation gives (C). (C) with concentrated KOH followed by acidification with dil. HCl regenerates (A) and produces (D). (D) with phosphorous pentachloride followed by reaction with ammonia gives (E). (E) on dehydration produces hydrocyanic acid. Identify compounds A, B, C, D and E.

Solution:

O
–H2O                                        ||
E ——————→ HCN So (E) must be H — C — NH2

O                                    O                                 O
||               PCl5               ||               NH3              ||
H — C — OH ————→ H — C — Cl ————→ H — C — NH2
(D)                                                                       (E)

(A) reacting with acetic acid is giving an ester (B). So (A) must be an alcohol. (A) alcohol on mild oxidation gives (C). Hence (C) is an aldehyde. (C) with conc. KOH followed by acidification is yielding (A) alcohol and  H — C — OH (D). So this reaction is cannizzaro reaction. So A must be CH3OH.

O
||
CH3OH       CH3 — C — O — CH3         HCHO        HCOOH       HCONH2
(A)                   (B)                        (C)             (D)             (E)

### IIT JEE Resources

Untitled Document
Stay connected