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Question

 $$CH_3 OH \xrightarrow{HBr}  A  \xrightarrow{KCN} B \xrightarrow{H_2/Ni} C \xrightarrow{HNO_2} D$$

Consider the above reaction sequence and choose the correct option regarding the different products obtained in the above reaction sequence.


A
'D' is an alcohol consisting one carbon more than the starting alcohol
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B
Product 'B' is formed via SN2pathway
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C
'C' is an amine
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D
All of the above
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Solution

The correct option is D All of the above
$$CH_3OH\xrightarrow[S_N2]{HBr} CH_3Br\xrightarrow[S_n2]{KCN} \underset{(B)}{CH_3CN}\xrightarrow[Reduction]{H_2/Nl}\underset{(C)}{CH_3CH_2NH_2} \xrightarrow{HNO_2} \underset {(D)}{CH_3CH_2OH}$$  
                                                                                      
                                                     Diazotisation followed by hydrolysis

In this reaction sequence that 'D' is ethyl alcohol which consists of one carbon atom more than methyl alcohol, the starting alcohol.

Product 'B', i.e., $$CH_3CN$$ is a product formed vis $$S_N2$$ pathway on the treatment of $$CH_3Br$$ and KCN.

Product 'C' is ethylamine formed by the reduction of methyl cyanide. 

Hence, the correct option is $$\text{D}$$

Chemistry

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