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Question: Consider the reactions. a) \(C{H_3} - CH = C{H_2}\xrightarrow[{(ii){{{H_2}O} \mathord{\left/ {\...

Consider the reactions.
a) CH3−CH=CH2→(ii)H2O/H2OOHOH(i)BH3/THFA→inCH2Cl2PCCBC{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) CH3−CH=CH2→(ii)NaBH4OH(i)Hg(OAc)2C→inCH2Cl2PCCDC{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

Explanation

Solution

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 BH3B{H_3} in presence of THF i.e. tetrahydrofuran. In this reaction BH3B{H_3} dissociates to form H+{H^ + } and BH2−B{H_2}^ - . These ions react with propene by breaking its double bond. This reaction can be written as follows.
CH3−CH=CH2→BH3/BH3THFTHFCH3−CH2−CH2−BH2C{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 −BH2 - B{H_2} molecule is substituted by −OH - OH and this reaction can be written as follows.
CH3−CH2−CH2−BH2→H2O/H2OOH−OH−CH3−CH2−CH2−OHC{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: CH3−CH2−CH2−OH→inCH2Cl2PCCCH3−CH2−CHOC{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(OAc)2Hg{\left( {OAc} \right)_2} which dissociates and adds to the propene molecule.
This reaction can be written as CH3−CH=CH2→Hg(OAc)2CH3−CH(OAc)−CH2(HgOAc)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 NaBH4OHNaB{H_4}OH eliminates the mercury molecules to form a secondary alcohol.
Reaction: CH3−CH(OAc)−CH2(HgOAc)→NaBH4OHCH3−CH(OH)−CH3C{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: CH3−CH(OH)−CH3→inCH2Cl2PCCCH3−CO−CH3C{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 CnH2nO{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.