What are Y and Z in the following sequence?

${\mathrm{CH}}_2=\mathrm{CHBr} \stackrel{{\mathrm{NaNH}}_2}{\to } \mathrm{Y} \underset{{\mathrm{H}}_2\mathrm{O}; 333 \mathrm{K}}{\overset{{\mathrm{Hg}}^{2+}/{\mathrm{H}}^{+}}{\to }}\mathrm{Z}$

The reaction of bromo ethylene with sodamine produces acetylene. The obtained 

acetylene upon reaction with mercury (II) in acidic medium forms ethanol also called as  acetaldehyde. The reaction sequence is as follows

 ${\mathrm{CH}}_2=\mathrm{CHBr} \stackrel{{\mathrm{NaNH}}_2}{\to } \mathrm{ethyne} \underset{{\mathrm{H}}_2\mathrm{O}; 333 \mathrm{K}}{\overset{{\mathrm{Hg}}^{2+}/{\mathrm{H}}^{+}}{\to }}{\mathrm{CH}}_3\mathrm{CHO}$  

Hence the correct option is (C).