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Match List -I with List -II.

List-I List-II
(A)

(I) Gatterman Koch reaction
(B) \mathrm{CH}_{3}-\mathrm{CN} \xrightarrow[H_{3}O^{+}]{Sncl_{2}/HCl}\mathrm{CH}_{3}-\mathrm{CHO}

(II) Etard reaction

(III) Stephen reaction

(IV) Rosenmund reaction

Choose the correct answer from the options given below :    

Option: 1

\begin{aligned} &(\mathrm{A})-(\mathrm{IV}),(\mathrm{B})-(\mathrm{III}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{I}) \\ \end{aligned}


Option: 2

(\mathrm{A})-(\mathrm{I}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{IV}) \\


Option: 3

(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{III}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{I}) \\


Option: 4

(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{IV})


Answers (1)

best_answer

\mathrm{(A) \: \rightarrow (IV) \, Rosenmund \: reaction}

\mathrm{(B) \: \rightarrow (III)\, Stephon\: reaction}

\mathrm{(C) \: \rightarrow (II) \, Etard \: reaction }

\mathrm{(D) \: \rightarrow (I) \: Gatterman \: Koch\: reaction}

So, option (1) is correct.


 

Posted by

Shailly goel

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