CH_3-CHO xrightarrow{HCN} (A) xrightarrow{H_3 | vedclass

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Question

(C) The reaction sequence is as follows:$1. CH_3-CHO + HCN \rightarrow CH_3-CH(OH)-CN (A)$ (Acetaldehyde cyanohydrin).$2. CH_3-CH(OH)-CN \xrightarrow{

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$CH_3-CHO$

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(C) The reaction sequence is as follows:$1. CH_3-CHO + HCN ightarrow CH_3-CH(OH)-CN (A)$ (Acetaldehyde cyanohydrin).$2. CH_3-CH(OH)-CN \xrightarrow{H_3O^{+}} CH_3-CH(OH)-COOH (B)$ (Lactic acid).$3. 2$ molecules of Lactic acid $(B)$ on heating $(\Delta)$ undergo intermolecular esterification to form a cyclic diester called Lactide $(C)$. Lactide shows geometrical isomerism (cis and trans).$4. 	ext{Lactide} (C) \xrightarrow{LiAlH_4} CH_3-CH(OH)-CH_2OH (D)$ (Propane$-1,2-$diol).$5. CH_3-CH(OH)-CH_2OH (D) \xrightarrow{HIO_4} CH_3-CHO (E) + HCHO$.Therefore,product $(E)$ is Acetaldehyde $(CH_3-CHO)$.

$CH_3-CHO$ $\xrightarrow{HCN} (A)$ $\xrightarrow{H_3O^{+}} (B)$ $\xrightarrow{\Delta} (C)$ $\xrightarrow{LiAlH_4} (D)$ $\xrightarrow{HIO_4} HCHO + (E)$. Compound $(C)$ can show geometrical isomerism. Product $(E)$ of the reaction will be:

Similar Questions

Predict the product when the given compound reacts with $LiAlH_4$.

The reactant is $3$-methylpentanal. The reaction with $NaCN$ followed by $H_2SO_4$ (cyanohydrin formation) is shown below:
$CH_3CH_2CH(CH_3)CH_2CHO \xrightarrow{NaCN/H^+} CH_3CH_2CH(CH_3)CH_2CH(OH)CN$
Products of the reaction are:

The reagent$(s)$ which can be used to bring about the following transformation is(are):
$(A)$ $LiAlH_4$ in $(C_2H_5)_2O$
$(B)$ $BH_3$ in $THF$
$(C)$ $NaBH_4$ in $C_2H_5OH$
$(D)$ Raney $Ni / H_2$ in $THF$

Identify the pair of compounds which can be distinguished by the iodoform test.

The product $(A)$ of the reaction is:

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