The conversion of CH_{3}CH_{2}COOH to the pro | vedclass

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Question

(A, D) The conversion of propanoic acid $(CH_{3}CH_{2}COOH)$ to the cyclic acetal $(CH_{3}CH_{2}CH(OCH_{2}CH_{2}O))$ involves the following steps:$1$.

Vedclass · Accepted Answer

$HCl, SnCl_{2},$ ethylene glycol

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(A, D) The conversion of propanoic acid $(CH_{3}CH_{2}COOH)$ to the cyclic acetal $(CH_{3}CH_{2}CH(OCH_{2}CH_{2}O))$ involves the following steps:$1$. Conversion of carboxylic acid to acid chloride using $SOCl_{2}$ or $PCl_{5}$.$2$. Reduction of acid chloride to aldehyde using $LiAlH_{4}$ (at low temperature) or Rosenmund reduction $(H_{2}/Pd-BaSO_{4})$.$3$. Protection of the aldehyde group by forming a cyclic acetal using ethylene glycol in the presence of an acid catalyst.Both options $(A)$ and $(D)$ describe valid pathways for this transformation, as $SnCl_{2}/HCl$ is also a standard reagent for the reduction of acid chlorides to aldehydes (Stephen reduction).

The conversion of $CH_{3}CH_{2}COOH$ to the product shown below can be accomplished by:
$CH_{3}CH_{2}CH(OCH_{2}CH_{2}O)$

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The conversion of $CH_{3}CH_{2}COOH$ to the product shown below can be accomplished by:$CH_{3}CH_{2}CH(OCH_{2}CH_{2}O)$