In the given reaction CH_3-CH_2-C(=O)-CH_2-CO | vedclass

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(B) The starting material contains both a ketone and an ester group.$LiAlH_4$ is a strong reducing agent that reduces both groups.To selectively reduc

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$HO-CH_2-CH_2-OH + H^{+}$

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(B) The starting material contains both a ketone and an ester group.$LiAlH_4$ is a strong reducing agent that reduces both groups.To selectively reduce the ester group to an alcohol while keeping the ketone group intact,the ketone must first be protected as a cyclic ketal using ethylene glycol $(HO-CH_2-CH_2-OH)$ in the presence of an acid catalyst $(H^{+})$.The subsequent acidic workup $((ii) H_2O/H^{+})$ after reduction removes the protecting group and restores the ketone functionality.

In the given reaction $CH_3-CH_2-C(=O)-CH_2-COOC_2H_5$ $\xrightarrow{[X]} (A)$ $\xrightarrow{(i) LiAlH_4, (ii) H_2O/H^{+}} CH_3-CH_2-C(=O)-CH_2-CH_2OH + C_2H_5OH$,$[X]$ will be:

Similar Questions

The structure of the main product $(B)$ of the following reaction will be:

Arrange the following compounds in the correct order of their reactivity towards nucleophilic addition reactions:
$I: CH_3CHO$
$II: CH_3CONH_2$
$III: HCHO$
$IV: CH_3COCH_3$

Assertion : Benzaldehyde is more reactive than ethanol towards nucleophilic attack.
Reason : The overall effect of $-I$ and $+R$ effect of phenyl group decreases the electron density on the carbon atom of $>C=O$ group in benzaldehyde.

Give the product from the following reaction sequence:

The product formed in Aldol condensation is

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