A. Write the structures of A, B, C and D in the following reactions:

B. Distinguish between:

(i) and

(ii) and HCOOH

C. Arrange the following in the increasing order of their boiling points:

a) Step 1: When CH₃CN is treated with SnCl₂ (stannous chloride) in the

presence of HCl, it gives CH₃CHO (A)

CH₃CN → CH₃CHO

The above reaction is called Stephen reaction.

Step 2: When CH₃CHO is treated with HCN, it gives cyanohydrin.

CH₃ CH₃ CH₃

(A) Cyanohydrin (D)

Step 3: When CH₃CHO (A) is treated with dil. NaOH, it undergoes aldol condensation to form β-hydroxyacetaldehyde (B).

⇒ OH ion removes the acidic α-hydrogen to form enolate ion

⇒ The enolate ion being a strong nucleophile attacks the carbonyl group of the second molecule of acetaldehyde (CH₃CHO) to anion.

⇒ Protonation of anion to form a β-hydroxy aldehyde (aldol)

Step 4: Now, when aldol(B) is heated with dilute acids undergo dehydration to from But-2-en-1-al(C).

Thus, A is CH₃CHO

B is CH₃—CH(OH)—CH₂—CHO

C is CH₃—CH=CH—CHO

D is CH₃—CH(CN)(OH)

b) (i) Iodoform test:

C₆H₅—CH=CH—COCH₃ being a methyl ketone on treatment with I₂/NaOH(NaOI) undergoes iodoform test to give yellow precipitate but C₆H₅—CH=CH—CO-CH₂CH₃ being ethyl ketone does not.

C₆H₅—CH=CH—COCH₃ + 3NaOI → C₆H₅—COONa + CHI₃ + 2NaOH

Iodoform

(yellow ppt)

C₆H₅—CH=CH—CO-CH₂CH + NaOI → No yellow ppt. of CHI₃

(ii) CH₃—CH₂—COOH and HCOOH can be distinguished by the following test:

⇒ Tollen’s reagent test: Formic acid contains a H-atom attached to >C=O group and hence can be regarded as an aldehyde.

⇒ It can be easily oxidized to CO₂ and H₂O but acetic acid cannot. Therefore, HCOOH (formic acid) behaves as a reducing agent whereas CH₃—CH₂—COOH (ethanoic acid) does not.

Formic acid reduces Tollen’s Reagent to metallic silver but ethanoic acid does not.

Tollens’ reagent is an ammoniacal solution of silver nitrate and is prepared by adding NH₄OH solution to AgNO₃ solution till the precipitate of Ag₂O first formed just redissolves. When propanal (aldehyde) is heated with Tollens’ reagent, the latter is reduced to metallic silver which deposits on the walls of the test tube as bright silver mirror.

During this reduction, the following reaction occur:

⇒ Formation of Tollens’ reagent:

2AgNO₃ + 2NH₄OH → Ag₂O + 2 NH₄NO₃ + H₂O

Ag₂O + 4NH₄OH → 2[Ag(NH₃)₂]⁺OH^- + 3H₂O

Tollens’ Reagent

⇒ Now, formic acid is treated with Tollens’ reagent:

HCOOH + 2Ag(NH₃)₂]⁺ + 3OH- →

2Ag + CO₂ + 4NH₃ + 2H₂O

The silver (Ag) thus deposited shines like a mirror. The formation of silver mirror confirms the presence of propanal.

On the other hand, CH₃—CH₂—COOH does not give this test. It does not form silver with Tollens’ reagent.

c) The increasing order of boiling points of the given organic compounds is

CH₃CHO < CH₃CH₂OH < CH₃COOH

Explanation:

The boiling point of carboxylic acids are much higher than those of aldehydes and alcohols of comparable molecular masses. This means that carboxylic acids form even stronger intermolecular H- bonding than alcohols and aldehyde.