Question

Alkene 'X' (Molecular formula ${C_5}{H_{10}}$ ) on ozonolysis gives a mixture of two compounds 'Y' and 'Z'. Compound 'Y' gives positive Fehling's test and iodoform test. Compound 'Z' does not give Fehling's test but gives iodoform test. Compounds X, Y and Z are?

$$A.{\text{ }}X = {C_6}{H_5}COC{H_3},Y = C{H_3}CHO,Z = C{H_3}COC{H_3} \\\ B.{\text{ }}X = C{H_3} - CH = C - {\left( {C{H_3}} \right)_2},Y = C{H_3}CHO,Z = C{H_3}COC{H_3} \\\ C.{\text{ }}X = C{H_3}C{H_2}CH = C{H_2},Y = C{H_3}C{H_2}CHO,Z = HCHO \\\ D.{\text{ }}X = C{H_3} - CH = CH - C{H_3},Y = C{H_3}CHO,Z = C{H_3}CHO \\\$$

Answer 1

Hint- In order to deal with this question we will identify the compound with the help of property of test that given as if any compound gives positive Fehling test it means it is a aldehyde and if any compound gives iodoform test it means it has $- COC{H_3}$ group.

Complete step-by-step answer:
Given that compound X is ${C_5}{H_{10}}$
For Y and Z we will use the given question statement
Compound Y is also positive for Fehling. It shows it is an aldehyde and gives an iodoform check that shows it has a group $- COC{H_3}$ . Compound Z is a ketone since it does not measure Fehling but gives iodoform check that also reveals the $- COC{H_3}$ groups.
Here the compound X is $C{H_3} - CH = C - {\left( {C{H_3}} \right)_2}$
Now we will write the reaction which will clear our compounds
$C{H_3} - CH = C - {\left( {C{H_3}} \right)_2}\mathop \to \limits_{(ii)\;Zn/{H_2}O}^{(i){\text{ }}{O_3}} C{H_3}CHO + C{H_3} - \left( {CO} \right) - C{H_3}$
Here in the above reaction it is clear that 2-Methylbut-2-ene on ozonolysis gives two products which are acetaldehyde or ethanal and acetone so out of these two product one of them is Y and the other will be Z. on the basis of further reaction we will find them out.
Now let us move on with the ethanol for its Fehling’s and iodoform test respectively.

$$C{H_3}CHO + 3NaOI \to C{I_3}CHO + 3NaOH{\text{ }}\left[ {{\text{Fehling's Test}}} \right] \\\ C{H_3}CHO + 3{I_2} + 3NaOH{\text{ }}\mathop \to \limits^{Hydrolysis} {\text{ }}CH{I_3} + 3HI + HCOONa{\text{ }}\left[ {{\text{Iodoform Test}}} \right] \\\$$

From the above reaction we can see that ethanal gives positive Fehling's test and iodoform test. So compound Y is ethanol.
Now let us move on with the acetone for its Fehling’s and iodoform test respectively.

$$C{H_3}COC{H_3} + 3NaOI \to C{I_3}COC{H_3} + 3NaOH{\text{ }}\left[ {{\text{Fehling's Test}}} \right] \\\ C{H_3}COC{H_3} + 3{I_2} + 3NaOH{\text{ }}\mathop \to \limits^{Hydrolysis} {\text{ }}CH{I_3} + 3HI + C{H_3}COONa{\text{ }}\left[ {{\text{Iodoform Test}}} \right] \\\$$

From the above reaction we can see that acetone does not give Fehling's test but give iodoform test. So compound Z is acetone.
Hence, compounds X, Y and Z are: $X = C{H_3} - CH = C - {\left( {C{H_3}} \right)_2},Y = C{H_3}CHO,Z = C{H_3}COC{H_3}$
So, option B is the correct option.

Note- The Iodoform test is used to verify the existence of $R - COC{H_3}$ carbonyl compounds or $R - CH\left( {OH} \right)C{H_3}$ alcohols in a given unknown material. The iodine, a base, and a methyl ketone reaction gives a yellow precipitate together with a "antiseptic" fragrance.
The presence of aldehydes but not of ketones is observed in Fehling 's test by reducing the deep blue copper(II) solution to a red precipitate of insoluble copper oxide. The method is widely used for sugar reduction but is considered to be NOT aldehyde sensitive.