Assertion (A) The $\alpha$-hydrogen atom in carbonyl compounds is less acidic.
Reason (R) The anion formed after the loss of $\alpha$-hydrogen atom is resonance stabilised.
Assertion (A) Aromatic aldehydes and formaldehyde undergo Cannizzaro reaction.
Reason (R) Aromatic aldehydes are almost as reactive as formaldehyde.
Assertion (A) Aldehydes and ketones, both react with Tollen's reagent to form silver mirror.
Reason (R) Both, aldehydes and ketones contain a carbonyl group.
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
