The reaction of benzene with CO and HCl in th | vedclass

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(C) The reaction of benzene with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3$ is the Gattermann-Koch reaction,which produces benzaldehyde $(C_

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Benzoyl chloride reacting with $H_2-Pd$ and $BaSO_4$

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(C) The reaction of benzene with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3$ is the Gattermann-Koch reaction,which produces benzaldehyde $(C_6H_5CHO)$. Thus,$X$ is benzaldehyde.Let us analyze the given options:$A$: Benzonitrile $(C_6H_5CN)$ on reduction with $LiAlH_4$ gives benzylamine $(C_6H_5CH_2NH_2)$.$B$: Toluene $(C_6H_5CH_3)$ on oxidation with $KMnO_4 | OH^-$ gives benzoic acid $(C_6H_5COOH)$.$C$: Benzoyl chloride $(C_6H_5COCl)$ on reduction with $H_2-Pd$ and $BaSO_4$ (Rosenmund reduction) gives benzaldehyde $(C_6H_5CHO)$.$D$: Benzyl alcohol $(C_6H_5CH_2OH)$ on oxidation with $CrO_3-H_2SO_4$ (Jones reagent) gives benzoic acid $(C_6H_5COOH)$.Therefore,benzaldehyde is obtained in reaction $C$.

Column - $I$ and $II$	Column - $III$
$(A).$ Rosenmund reaction - $(i).$ $H_2, Pd, BaSO_4$	$(p).$ Benzene to benzaldehyde
$(B).$ Stephen reaction - $(iii).$ $SnCl_2, HCl$	$(q).$ Benzoyl chloride to benzaldehyde
$(C).$ Etard reaction - $(iv).$ $CrO_2Cl_2, CS_2$	$(r).$ Toluene to chromium complex
$(D).$ Gatterman-Koch reaction - $(ii).$ $CO, HCl$ anhydrous $AlCl_3/CuCl$	$(s).$ $RCN \to RCHO$

The reaction of benzene with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3$ gives a compound $X$. $X$ can also be obtained from which of the following reactions?

Similar Questions

Identify the compound $C$ in the following sequence:
$(CH_3)_2CHCH_2C \equiv N$ $\xrightarrow{HCl, H_2O} \text{compound } A$ $\xrightarrow[2. H_2O]{1. LiAlH_4} \text{compound } B$ $\xrightarrow{PCC, CH_2Cl_2} \text{compound } C$

Benzaldehyde can be prepared by the oxidation of toluene by:

Which of the following reactions will give a carbonyl compound as the major product?

The product $A$ in the following reaction is:
$CH_3-CH=CH-CH_2-CN \xrightarrow[(2) H_3O^+]{(1) DIBAL-H, -78^{\circ}C} A$

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