Assertion (A) Compounds containing - $\mathrm{CH0}$ group are easily oxidised to corresponding carboxylic acids.
Reason (R) Carboxylic acids can be reduced to alcohols by treatment with $\mathrm{LiAlH}_4$.
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.
$$\mathrm{\mathop {C{H_3}CHO}\limits_{Acetaldehyde\,[B]} + 3{I_2} + 4NaOH\buildrel \Delta \over \longrightarrow \mathop {CH{I_3}}\limits_{Iodoform} + \mathop {HCOONa}\limits_{Sodium\,formate} + 3Nal + 3{H_2}O}$$
$$\mathrm{\mathop {C{H_3}COC{H_3}}\limits_{Acetone\,[C]} + 3{I_2} + 4NaOH\buildrel \Delta \over \longrightarrow \mathop {CH{I_3}}\limits_{Iodoform} + \mathop {C{H_3}COONa}\limits_{Sodium\,\,acetate} + 3Nal + 3{H_2}O}$$