Question

Consider the reactions.
a) $C{H}_3-CH=C{H}_2\underset{\left(ii\right)H_{2}O/OH}{\overset{\left(i\right)B{H}_3/THF}{\to }}A\underset{\left(inC{H}_{2}C{l}_2\right)}{\overset{PCC}{\to }}B$
b) $C{H}_3-CH=C{H}_2\underset{\left(ii\right)NaB{H}_4\overline{O}H}{\overset{\left(i\right)Hg(OAc{)}_2}{\to }}C\underset{\left(inC{H}_{2}C{l}_2\right)}{\overset{PCC}{\to }}D$
Product B and D are:

• A. Functional isomers
• B. Positional isomeres
• C. Chain isomeres
• D. Metamers

Answer 1

The correct option is A Functional isomers

$n_{3}c-CH=C{H}_2\stackrel{B{H}_3/THF}{\to }{n}_{3}c-C{H}_2-C{H}_2-B{H}_2$
$\downarrow H_2{O}_2/O{H}^{-}$
$n_{3}c-C{H}_2-C{H}_2-OH$
$\downarrow PCCin\left(C{H}_{2}C{l}_2\right)$
$n_{3}c-C{H}_2-\stackrel{O}{\stackrel{||}{C}}-H$ (ethanal)
$n_{3}c-CH=C{H}_2\stackrel{Hg(OAc{)}_2}{\to }{n}_{3}c-\underset{OH}{\underset{|}{CH}}-\underset{HgOAc}{\underset{|}{C{H}_2}}$
$\downarrow NaB{H}_4$
$n_{3}c-\underset{OH}{\underset{|}{CH}}-C{H}_3$
$\downarrow PCC$
$n_{3}c-\stackrel{O}{\stackrel{||}{C}}-C{H}_3$ (Acetone)

B and D are aldehyde and ketones so these are functional isomers.