A) \[HC\equiv C-C{{H}_{2}}-C\equiv CH\]
B) \[NaOH\]
C) \[{{104}^{o}}40\]
D) \[{{103}^{o}}\]
Correct Answer: B
Solution :
\[C=85.71%=\frac{85.71}{12}=7.14;\] \[\frac{7.14}{7.14}=1\] \[H=14.29%=\frac{14.29}{1}=14.29;\] \[\frac{14.29}{7.14}=2\] \[\therefore \] Empirical formula \[=C{{H}_{2}}\] and, empirical formula weight \[=12+2=14\] Again, molecular formula weight \[=2\times \] vapour density \[=2\times 14=28\] \[\therefore \] \[n=\frac{28}{14}=2\] \[\therefore \] Molecular formula \[={{(C{{H}_{2}})}_{2}}={{C}_{2}}{{H}_{4}}\] \[\underset{(A)}{\mathop{C{{H}_{2}}=C{{H}_{2}}}}\,+HOCl\xrightarrow{{}}\underset{(B)}{\mathop{\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C{{H}_{2}}-C{{H}_{2}}-Cl}}\,}}\,\] \[+KCN\xrightarrow{EtOH}\] \[\underset{OH}{\mathop{\underset{|}{\mathop{C{{H}_{2}}-C{{H}_{2}}-CN}}\,}}\,\xrightarrow{{{H}_{3}}{{O}^{+}}}\underset{(C)}{\mathop{\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C{{H}_{2}}-C{{H}_{2}}-COOH}}\,}}\,\]
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