Question

An organic compound A ( ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}{{\text{O}}_{\text{2}}}$ ) reacts with ${\text{B}}{{\text{r}}_{\text{2}}}$ in presence of phosphorus to produce B which can be resolved into enantiomers. B on dehydrobromination yields C ( ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{2}}}$ ) which does not show stereoisomerism. C on decarboxylation given an alkene which on further ozonolysis gives D and E. D gives Schiff’s test but E does not. The correct identity of these compounds are:
(A)- A is ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{CHC}}{{\text{H}}_{\text{2}}}{\text{COOH}}$
(B)- B is ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{CHCHBrCOOH}}$
(C)- C is ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{C = CHCOOH}}$
(D)- E is acetone

Answer 1

Hint Enantiomers are those pair of molecules which shows their mirror images and those compounds which have same molecular formula but have different atomic arrangement in space will show stereoisomerism like cis – trans isomerism.

Complete step by step solution:
All steps of the above given reaction is explained as follow:
-In the 2st step an organic compound A ( ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}{{\text{O}}_{\text{2}}}$ ) reacts with ${\text{B}}{{\text{r}}_{\text{2}}}$ in presence of phosphorus to produce B i.e. ${{\text{C}}_{\text{5}}}{{\text{H}}_9}{{\text{O}}_{\text{2}}}$ which can be resolved into enantiomers because it has one asymmetric or chiral carbon in it and reaction is shown as follow:
${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{CHC}}{{\text{H}}_{\text{2}}}{\text{COOH}}\xrightarrow{{{\text{P\& B}}{{\text{r}}_{\text{2}}}}}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{CHCHBrCOOH}}$
-Now the formed product in the above reaction B ( ${{\text{C}}_{\text{5}}}{{\text{H}}_9}{{\text{O}}_{\text{2}}}$ ) produces C ( ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{2}}}$ ) by the removal of bromine and hydrogen through dehydrobromination process and formed product of this step i.e. C ( ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{2}}}$ ) will not show stereoisomerism or cis – trans isomer as in the one double bonded carbon atom same substituents ( ${\text{ - C}}{{\text{H}}_{\text{3}}}$ ) are present, and reaction is shown as follow:
${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{CHCHBrCOOH}}\xrightarrow[{{\text{ - Br, - H}}}] {}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{C = CHCOOH}}$
-On decarboxylation of C ( ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{2}}}$ ) it produces an alkene ( ${{\text{C}}_{\text{4}}}{{\text{H}}_{\text{8}}}$ ) and reaction is shown as follow:
${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{C = CHCOOH}}\xrightarrow[{{\text{ - C}}{{\text{O}}_{\text{2}}}}] {}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{C = C}}{{\text{H}}_{\text{2}}}$
-Now these alkenes on further ozonolysis produce D (Formaldehyde) which gives Schiff’s test because it is an aldehyde and E (Acetone) which does not give Schiff’s test.
${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{C = C}}{{\text{H}}_{\text{2}}}\xrightarrow{{{\text{Ozonolysis}}}}\mathop {{\text{HCHO}}}\limits_{{\text{(Formaldehyde)}}} {\text{ + }}\mathop {{{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)}_{\text{2}}}{\text{CO}}}\limits_{{\text{(Acetone)}}}$

Hence, E is acetone i.e. option (D) is correct.

Note: Here some of you think that why acetone will not give Schiff’s test so the reason is that in acetone ketone group is present and Schiff’s test is only given by aldehydes that’s why acetone will not give Schiff’s test.