Question

Consider the reactions.
a) $C{H_3} - CH = C{H_2}\xrightarrow[{(ii){{{H_2}O} \mathord{\left/
 {\vphantom {{{H_2}O} {OH}}} \right.
 } {OH}}}]{{(i)B{H_3}/THF}}A\xrightarrow[{inC{H_2}C{l_2}}]{{PCC}}B$

b) $C{H_3} - CH = C{H_2}\xrightarrow[{(ii)NaB{H_4}OH}]{{(i)Hg{{\left( {OAc} \right)}_2}}}C\xrightarrow[{inC{H_2}C{l_2}}]{{PCC}}D$
Products B and D are:
A) Functional isomers
B) Positional isomers
C) Chain isomers
D) Metamers

Answer 1

Hint:The compound given to us is propene. This compound is undergoing various reactions to form B and D. In order to find B and D, we have to first understand the nature of the catalyst used and how they react with the given compound.

Complete step-by-step answer:It is given to us that Propene is undergoing various reactions to form products B and D. Let us find these products individually.
a) In this process, propene is reacting with $B{H_3}$ in presence of THF i.e. tetrahydrofuran. In this reaction $B{H_3}$ dissociates to form ${H^ + }$ and $B{H_2}^ - $. These ions react with propene by breaking its double bond. This reaction can be written as follows.
$C{H_3} - CH = C{H_2}\xrightarrow{{{{B{H_3}} \mathord{\left/
 {\vphantom {{B{H_3}} {THF}}} \right.
 } {THF}}}}C{H_3} - C{H_2} - C{H_2} - B{H_2}$
This product is further reacted with water. In this reaction the $ - B{H_2}$ molecule is substituted by $ - OH$ and this reaction can be written as follows.
$C{H_3} - C{H_2} - C{H_2} - B{H_2}\xrightarrow{{{{{H_2}O} \mathord{\left/
 {\vphantom {{{H_2}O} {O{H^ - }}}} \right.
 } {O{H^ - }}}}}C{H_3} - C{H_2} - C{H_2} - OH$
This is compound A. Compound A is then treated with PCC which is an oxidizing agent and hence propanol will be oxidized to propanal.
Reaction: $C{H_3} - C{H_2} - C{H_2} - OH\xrightarrow[{inC{H_2}C{l_2}}]{{PCC}}C{H_3} - C{H_2} - CHO$
Therefore, compound B is propanal.
b) Similarly for this process, propene is first treated with $Hg{\left( {OAc} \right)_2}$ which dissociates and adds to the propene molecule.
This reaction can be written as $C{H_3} - CH = C{H_2}\xrightarrow{{Hg{{\left( {OAc} \right)}_2}}}C{H_3} - CH\left( {OAc} \right) - C{H_2}\left( {HgOAc} \right)$
This on further reaction with $NaB{H_4}OH$ eliminates the mercury molecules to form a secondary alcohol.
Reaction: $C{H_3} - CH\left( {OAc} \right) - C{H_2}\left( {HgOAc} \right)\xrightarrow{{NaB{H_4}OH}}C{H_3} - CH\left( {OH} \right) - C{H_3}$
This is compound C. Compound C in reaction with PCC oxidizes to form a ketone.
Reaction: $C{H_3} - CH\left( {OH} \right) - C{H_3}\xrightarrow[{inC{H_2}C{l_2}}]{{PCC}}C{H_3} - CO - C{H_3}$
Hence, compound D is propanone.
Therefore, compounds B and D are functional isomers.

So,the correct option is (A).

Note:Compounds that have the same molecular formula but different functional groups are known as functional isomers. Here, the general molecular formula of both aldehyde and ketone is ${C_n}{H_{2n}}O$ but in aldehydes the Oxygen atom is attached to terminal carbons and in ketones, oxygen atom is attached to secondary carbons.