Which of the following reaction products is c | vedclass

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Question

(C) Let us analyze each reaction:$(A)$ $NaBH_4$ is a selective reducing agent that reduces aldehydes and ketones to alcohols but does not reduce ester

Vedclass · Accepted Answer

Phenylacetonitrile $\xrightarrow{(1) DIBAL-H, (2) H_3O^+}$ Phenylacetaldehyde

Vedclass · Answer

(C) Let us analyze each reaction:$(A)$ $NaBH_4$ is a selective reducing agent that reduces aldehydes and ketones to alcohols but does not reduce esters. The reaction shown reduces the ester group to an alcohol while keeping the ketone intact,which is incorrect.$(B)$ $H_2/Pd-C$ is a strong reducing agent that reduces both the $C=C$ double bond and the $C=O$ double bond. It would not stop at the alcohol stage with one equivalent; it would reduce the alkene as well.$(C)$ The reduction of nitriles $(R-CN)$ with $DIBAL-H$ followed by acid hydrolysis $(H_3O^+)$ is a standard method to produce aldehydes $(R-CHO)$. Thus,$C_6H_5-CH_2-CN \xrightarrow{(1) DIBAL-H, (2) H_3O^+} C_6H_5-CH_2-CHO$ is correct.$(D)$ The reduction of an internal alkyne with $Na/liq. NH_3$ (Birch reduction conditions) produces a trans-alkene,not a cis-alkene. Therefore,this is incorrect.

$A$. $C_6H_5COCl + H_2 \xrightarrow{Pd-BaSO_4} C_6H_5CHO$	$I$. Gattermann-Koch reaction
$B$. $CH_3CN + SnCl_2 + HCl \rightarrow CH_3CHO$	$II$. Etard reaction
$C$. $C_6H_5CH_3 + CrO_2Cl_2 \rightarrow C_6H_5CHO$	$III$. Stephen reaction
$D$. $C_6H_6 + CO + HCl \xrightarrow{AlCl_3} C_6H_5CHO$	$IV$. Rosenmund reaction

Which of the following reaction products is correct?

Similar Questions

In the following reaction,product $P$ is: $RCOCl \xrightarrow{H_2, Pd-BaSO_4} P$

The oxidation of toluene to benzaldehyde by chromyl chloride $(CrO_2Cl_2)$ is called:

$R-CN \xrightarrow[(ii) H_2O]{(i) DIBAL-H} R-Y$
Consider the above reaction and identify $Y$.

Which of the following sets of reagents convert toluene to benzaldehyde?
$A$. $Cl_2 / h\nu ; H_2O, \Delta$
$B$. $Cl_2 / Fe ; H_2O$
$C$. $KMnO_4 / OH^{-} ; H^{+}$
$D$. $CrO_2Cl_2 / CS_2 ; H_3O^{+}$

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