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Aldehydes, Ketones and Carboxylic Acids

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In the following reaction sequence, products I, J and L are formed. K represents a reagent.

Section: Chemistry

Question : 77 of 77

Marks: +1, -0

Solution:

The reaction sequence for formation of J and K is

\mathrm{CH_3-CH_2-C\equiv C-CH_3CH_2Br}

\xrightarrow{\text{Mg/ Ether}}

\mathrm{CH_3-CH_2-C\equiv C-CH_2CH_2MgBr}

\xrightarrow{+\mathrm{CO}_2}

\mathrm{CH_3-CH_2-C\equiv C-CH_2CH_2-CO-OMgBr}

\xrightarrow{\mathrm{H}^+ / \mathrm{H}_2\mathrm{O}}

\mathrm{CH_3-CH_2-C\equiv C-CH_2CH_2-CO-OH}

- (J)

\xrightarrow{\mathrm{SOCl}_2}

(K)

\mathrm{CH_3-CH_2-C\equiv C-CH_2CH_2-CO-Cl}

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