NEET · Chemistry · STD 11 - 9. Hydricarbon
In the reaction \(H - C\, \equiv CH\,\xrightarrow[{(ii)\,C{H_3}C{H_2}Br}]{{(i)\,NaN{H_2}\,/\,liq.\,N{H_3}}}\,X\) \(\xrightarrow[{(ii)\,C{H_3}C{H_2}Br}]{{(i)\,NaN{H_2}\,/\,liq.\,N{H_3}}}\,Y\) \(X\) and \(Y\) are
- A \(X= {2-}\)Butyne\(,\, Y = {2-}\)Hexyne
- B \(X= 1-\)Butyne\(,\, Y = {2-}\)Hexyne
- C \(X= 1-\)Butyne\(,\, Y = {3-}\)Hexyne
- D \(X= {2-}\)Butyne\(, \,Y = {3-}\)Hexyne.
Answer & Solution
Correct Answer
(C) \(X= 1-\)Butyne\(,\, Y = {3-}\)Hexyne
Step-by-step Solution
Detailed explanation
\(H-C\equiv C-H\xrightarrow[liq.\,N{{H}_{3}}]{Na\,N{{H}_{2}}/}HC\equiv C\,Na\) \(\xrightarrow[-NaBr]{C{{H}_{3}}C{{H}_{2}}-Br}CH\equiv C-C{{H}_{2}}-C{{H}_{3}}\) \(C{{H}_{3}}-C{{H}_{2}}-C\equiv CH\xrightarrow[liq.\,N{{H}_{3}}]{Na\,N{{H}_{2}}/}\) \(C{{H}_{3}}-C{{H}_{2}}-C\equiv C\,Na\xrightarrow[-NaBr]{C{{H}_{3}}C{{H}_{2}}-Br}\) \(\underset{3-Hexyne}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C\equiv C-C{{H}_{2}}-C{{H}_{3}}}}\,\)
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