Question

An alkene ‘A’ (Mol. formula \[\ce{C5H10}\]) on ozonolysis gives a mixture of two compounds ‘B’ and ‘C’. Compound ‘B’ gives positive Fehling’s test and also forms iodoform on treatment with \[\ce{I2}\] and \[\ce{NaOH}\]. Compound ‘C’ does not give Fehling’s test but forms iodoform. Identify the compounds A, B and C. Write the reaction for ozonolysis and formation of iodoform from B and C.

Answer 1

Compound B gives positive Fehling’s test. It shows that it is an aldehyde and gives iodoform test which shows it has \[\ce{^-COCH3}\] group. Compounds C is a ketone because it does not give Fehling’s test but gives iodoform test which shows it also has \[\ce{^-COCH3}\] groups.

Hence compound A is

\[\begin{array}{cc}
\ce{CH3CH = C - CH3}\\
\phantom{....}|\\
\phantom{.......}\ce{CH3}
\end{array}\]

(i)

\[\begin{array}{cc}
\ce{CH3 - CH = C - CH3 ->[(i) O3][(ii) Zn/H2O] \underset{Acetaldehyde}{CH3 - CHO} + O = C - H3}\\
\phantom{........}|\phantom{...................}\ce{\underset{(B)}{(Ethanal)}}\phantom{..........}|\phantom{}\\
\phantom{..........}\ce{\underset{(A)}{\underset{2-Methylbut-2-ene}{CH3}} \phantom{............................}\underset{(C)}{\underset{(Acetone)}{CH3}}}\phantom{...}
\end{array}\]

Other isomer of (A) will not give products corresponding to the given test.

\[\ce{(2NaOH + I2 -> NaOI + NaI + H2O)}\]

(ii) 

\[\ce{\underset{(B)}{CH3CHO} + \underset{hypoiodite}{\underset{Sodium}{3NaOI}} -> \underset{accetadehyde}{\underset{Tri-iodo}{CI3CHO}} + 3NaOH}\]

\[\ce{CH3\underset{Iodal}{CHO} + NaOH ->[Hydrolysis] \underset{Iodoform}{CHI3} + HCOONa}\]

(iii)

\[\begin{array}{cc}
\phantom{..}\ce{O}\phantom{............................}\ce{O}\phantom{............}\\
\phantom{.}||\phantom{.............................}||\phantom{...........}\\
\ce{CH3 - \underset{(C)}{C} - CH3 + 3NaOI -> \underset{Tri-iodoacetone}{CI3 - C - CH3} + 3NaOH}
\end{array}\]

\[\begin{array}{cc}
\ce{O}\phantom{.........................................}\\
||\phantom{.........................................}\\
\ce{CH3 - C - CH3 + NaOH ->[Hydrolysis]  \underset{Iodoform}{CHl3} + \underset{Sodium acetate}{CH3COONa}}
\end{array}\]