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Question
A compound 'A' of molecular formula C2H3OCl undergoes a series of reactions as shown below. Write the structures of A, B, C and D in the following reactions :
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Solution
Using the given molecular formula, compound A is Ethanoyl Chloride CH3COCl, which undergoes reaction with poisoned palladium.
On carrying hydrogenation of A in the presence of poisoned palladium, we get an aldehyde. Hence, B can be Ethanal, CH3CHO.
An aldehyde, on treating with dilute alkali, undergoes aldol condensation reaction. Hence, C can be CH3CH(OH)CH2CHO.
On heating an aldol product, it loses water to produce a double bond and we get CH3CH=CHCHO.
Hence, we have
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is treated with concentrated aqueous \[\ce{KOH}\] solution?
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(i) \[\ce{CH3 - CHO}\]
(ii) 
(iii) \[\begin{array}{cc}
\phantom{}\ce{O}\\
\phantom{}||\\
\ce{CH3 - C - CH3}
\end{array}\]
(iv) \[\begin{array}{cc}
\phantom{}\ce{CH3}\\
|\phantom{...}\\
\ce{CH3 - C - CHO}\phantom{..}\\
|\phantom{...}\\
\phantom{}\ce{CH3}\\
\end{array}\]
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Identify A and B from the following reaction:
\[\begin{array}{cc}
\ce{CH3}\phantom{.................}\\
|\phantom{....................}\\
\phantom{}\ce{2CH3 - C = O ->[Ba(OH)2] A ->[Δ] B + H2O}
\end{array}\]
\[\ce{CH3-CH2-CHO ->[dil][alkali] Product}\]
The product in the above reaction is:
