Answer:
An aldehyde
'A' with molar mass 44u is\[C{{H}_{3}}CH=O\] (Ethanal). Since, two moles of ethanol
are obtained by ozonolysis of alkene 'A'.
\[\underset{Ethanal}{\mathop{C{{H}_{3}}CH=O}}\,+\underset{Ethanal}{\mathop{O=HC-C{{H}_{3}}}}\,\]
Now remove the oxygen atoms and join them by a double bond,
the structure of alkene 'A' is:
\[\underset{But-2-ene}{\mathop{C{{H}_{3}}CH=CHC{{H}_{3}}}}\,or\,H-\underset{\underset{\text{H}}{\mathop{|}}\,}{\overset{\overset{\text{H}}{\mathop{\text{
}\!\!|\!\!\text{
}}}\,}{\mathop{C}}}\,-\underset{\underset{\text{H}}{\mathop{|}}\,}{\mathop{C}}\,=\underset{\underset{\text{H}}{\mathop{|}}\,}{\mathop{C}}\,-\underset{\underset{\text{H}}{\mathop{|}}\,}{\overset{\overset{\text{H}}{\mathop{\text{
}\!\!|\!\!\text{ }}}\,}{\mathop{C}}}\,-\text{H}\]
It has three C-C, eight C-H \[\sigma \] bonds and
one C-C \[\pi \] bond.
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