Explain Aldol condensation reaction of ethanal.(b) Write four properties of interstitial compounds of transition elements.

Answer:

(a) Aldol condensation of ethanal: Aldehydes having an α-hydrogen, in the presence of dilute alkali (dil. NaOH), undergo self-addition: the carbanion formed from one molecule adds to the carbonyl group of another, giving a β-hydroxy aldehyde (aldol).

2CH₃CHO —(dil. NaOH)→ CH₃–CH(OH)–CH₂–CHO (3-hydroxybutanal, aldol)

On heating, the aldol loses a water molecule to give an α,β-unsaturated aldehyde:

CH₃–CH(OH)–CH₂–CHO —(Δ)→ CH₃–CH=CH–CHO + H₂O (but-2-enal, crotonaldehyde)

Mechanism in brief: OH^– removes an α-H from ethanal to form the carbanion ^–CH₂CHO; this attacks the carbonyl carbon of a second ethanal molecule; the alkoxide formed takes H⁺ from water to give aldol.

(b) Properties of interstitial compounds:

• They have high melting points, higher than those of the pure metals.
• They are very hard; some borides are nearly as hard as diamond.
• They retain metallic conductivity (electrical and thermal).
• They are chemically inert, and are often non-stoichiometric (e.g. TiC, Mn₄N, Fe₃H, VH_0.56).