To convert ethanal (CH₃CHO) to but-2-enal (CH₃CH=CHCHO), we can follow a series of organic reactions involving aldol condensation and dehydration.
1. Aldol Condensation:
Ethanal undergoes aldol condensation in the presence of a dilute base (e.g., NaOH) to form 3-hydroxybutanal (aldol).
$ 2 \, \text{CH}_3\text{CHO} \xrightarrow{\text{NaOH}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO} $
2. Dehydration:
The 3-hydroxybutanal intermediate is then dehydrated (elimination of water) in the presence of an acid (e.g., H₂SO₄) or heat to form but-2-enal (crotonaldehyde).
$ \text{CH}_3\text{CH(OH)CH}_2\text{CHO} \xrightarrow{\text{H}^+ \text{ or } \Delta} \text{CH}_3\text{CH=CHCHO} + \text{H}_2\text{O} $
Summary of the Conversion:
Ethanal → (Aldol Condensation) → 3-Hydroxybutanal → (Dehydration) → But-2-enal
Final Answer:
The conversion of ethanal to but-2-enal is achieved via $\boxed{\text{aldol condensation followed by dehydration}}$.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline
Account for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.