An alkene A with molecular formula \(\mathrm{C_5H_{10}}\) on ozonolysis gives a mixture of two compounds B and C. Compound B gives positive Fehling test and also reacts with iodine and NaOH solution. Compound C does not give Fehling solution test but forms iodoform. Identify the compounds A, B and C.
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Solution and Explanation
Ozonolysis breaks the C=C bond and gives two carbonyl compounds. The two products together hold all five carbons of A.
B gives the Fehling test, so B is an aldehyde. B also gives iodoform, so it has the \(\mathrm{CH_3CO-}\) group. The only aldehyde with this group is ethanal, \(\mathrm{CH_3CHO}\).
C does not give the Fehling test, so it is a ketone. C forms iodoform, so it is a methyl ketone. It must have the other three carbons, so C is propanone, \(\mathrm{CH_3COCH_3}\).
Step 2: Join the two pieces to get A:
Put the two carbonyl carbons back together as a double bond: \(\mathrm{CH_3CH=}\) from ethanal and \(\mathrm{=C(CH_3)_2}\) from propanone.
So A is \(\mathrm{CH_3CH=C(CH_3)_2}\), which is 2-methylbut-2-ene. Its formula is \(\mathrm{C_5H_{10}}\), as given.
Step 3: Write the reactions:
\[ \mathrm{(CH_3)_2C=CHCH_3 \xrightarrow{(i)\ O_3,\ (ii)\ Zn/H_2O} (CH_3)_2C=O + CH_3CHO} \]
Fehling test by B: \(\mathrm{CH_3CHO + 2Cu^{2+} + 5OH^- \to CH_3COO^- + Cu_2O\downarrow + 3H_2O}\) (red precipitate).
Iodoform test by B: \(\mathrm{CH_3CHO + 3I_2 + 4NaOH \to CHI_3\downarrow + HCOONa + 3NaI + 3H_2O}\)
Iodoform test by C: \(\mathrm{CH_3COCH_3 + 3I_2 + 4NaOH \to CHI_3\downarrow + CH_3COONa + 3NaI + 3H_2O}\) (yellow precipitate).
Final Answer:
A is 2-methylbut-2-ene, B is ethanal and C is propanone.
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