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 $\boxed{\text{aldol condensation followed by dehydration}}$.





Consider the following reaction of benzene. the percentage of oxygen is _______ %. (Nearest integer) 
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).