An alkene ' A ' (molecular formula $\mathrm{C}_5 \mathrm{H}_{10}$ ) 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 $\mathrm{I}_2$ and 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 .
Molecular formula $=\mathrm{C}_5 \mathrm{H}_{10}$
Degree of unsaturation $=\left(C_n+1\right)-\frac{H_n}{2}$
where, $\mathrm{C}_n=$ number of carbon atoms
$\mathrm{H}_n=$ number of hydrogen atoms
$$=(5+1)-\frac{10}{2}=1$$
Compound $A$ will be either alkene or cyclic hydrocarbon. Since, $A$ is undergoing ozonolysis hence $A$ must be an alkene.
Possible structures of alkene are
I. $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2$
II. $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$
III.
IV. 
Ozonolysis of structure I produces aldehyde only
Ozonolysis of structure Il produces aldehyde only
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3 \xrightarrow[\text { (i) } \mathrm{Zn} / \mathrm{H}_2 \mathrm{O}]{\left(\text { i) } \mathrm{O}_3\right.} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CHO}+\mathrm{CH}_3 \mathrm{CHO}$
After ozonolysis of each of structures I, II and III produces only aldehydes as both components. But as given in the question one compound doesn't give Fehling test but must give iodoform test. Hence, compound must be a ketone with 
group. Hence, correct structure is IV.
$$\mathrm{\mathop {C{H_3}CHO}\limits_{Acetaldehyde\,[B]} + 3{I_2} + 4NaOH\buildrel \Delta \over \longrightarrow \mathop {CH{I_3}}\limits_{Iodoform} + \mathop {HCOONa}\limits_{Sodium\,formate} + 3Nal + 3{H_2}O}$$
$$\mathrm{\mathop {C{H_3}COC{H_3}}\limits_{Acetone\,[C]} + 3{I_2} + 4NaOH\buildrel \Delta \over \longrightarrow \mathop {CH{I_3}}\limits_{Iodoform} + \mathop {C{H_3}COONa}\limits_{Sodium\,\,acetate} + 3Nal + 3{H_2}O}$$
An aromatic compound ' A ' (molecular formula $\mathrm{C}_8 \mathrm{H}_8 \mathrm{O}$ ) gives positive $2,4-$ DNP test. It gives a yellow precipitate of compound 'B' on treatment with iodine and sodium hydroxide solution. Compound 'A' does not give Tollen's or Fehling's test. On drastic oxidation with potassium permanganate it forms a carboxylic acid 'C' (molecular formula $\mathrm{C}_7 \mathrm{H}_6 \mathrm{O}_2$ ), which is also formed alongwith the yellow compound in the above reaction. Identify $\mathrm{A}, \mathrm{B}$ and C and write all the reactions involved.
$$\begin{array}{ll} \text { Molecular formula }=\mathrm{C}_8 \mathrm{H}_8 \mathrm{O} \\ \text { Degree of unsaturation } & =\left(\mathrm{C}_n+1\right)-\frac{\mathrm{H}_n}{2} \\ & =(8+1)-\frac{8}{2}=9-4=5 \end{array}$$
Degree of unsaturation $>5$ i.e., it may contain benzene ring having degree of unsaturation equal to 4 and one degree of unsaturation must be carbonyl group.
Thus, possible structures are
According to question, compound ' $A$ ' does not respond to Tollen's or Fehling's test, So, it is a ketone not aldehyde. Therefore, structure I is correct. Complete reaction sequence is as follows
Write down functional isomers of a carbonyl compound with molecular formula $\mathrm{C}_3 \mathrm{H}_6 \mathrm{O}$. Which isomer will react faster with HCN and why? Explain the mechanism of the reaction also. Will the reaction lead to the completion with the conversion of whole reactant into product at reaction conditions? If a strong acid is added to the reaction mixture what will be the effect on concentration of the product and why?
$$\begin{aligned} &\text { Functional isomers of } \mathrm{C}_3 \mathrm{H}_6 \mathrm{O} \text { containing carbonyl group are }\\ &\underset{\text { Propanal }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}} \text { and } \underset{\text { Propanone }}{\mathrm{CH}_3 \mathrm{COCH}_3} \end{aligned}$$
(a) Propanal, $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$ will react faster with HCN because there is less steric hindrance and electronic factors, which increases its electrophilicity.
(b) The reaction mechanism is as follow
The reaction does not lead to completion because it is a reversible reaction. Equilibrium is established.
(c) If a strong acid is added to the reaction mixture, the reaction is inhibited because production of $\overline{\mathrm{C}} \mathrm{N~ions}$ prevented.
When liquid ' $A$ ' is treated with a freshly prepared ammoniacal silver nitrate solution it gives bright silver mirror. The liquid forms a white crystalline solid on treatment with sodium hydrogen sulphite. Liquid ${ }^{\mathrm{B}}$ ' also forms a white crystalline solid with sodium hydrogen sulphite but it does not give test with ammoniacal silver nitrate. Which of the two liquids is aldehyde? Write the chemical equations of these reactions also.
Since the liquid A reduces ammoniacal silver nitrate, (Tollen's reagent), it 'A' is aldehyde.
