$$ \underset{\mathrm{N}}{\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}} \frac{\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}}{\mathrm{H}_{2} \mathrm{SO}_{4}} \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O} \cdot \frac{\mathrm{I}_{2} / \mathrm{NaOH}}{\mathrm{Warm}}, \mathrm{CHI}_{3} $$ Here, N is

Given sequence of chemical equation is $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} \frac{\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}}{\mathrm{H}_{2} \mathrm{SO}_{4}}-\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O} \frac{\mathrm{I}_{2} / \mathrm{NaOH}}{\text { Warm }} \mathrm{CHI}_{3}$ $(\mathrm{N})$ Here N will be $=$ Butan- $2 \cdot 0$ l) Here, Step I It involves the oxidation of secondary alcohol to corresponding alcohol in the presence of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and $\mathrm{H}_{2} \mathrm{SO}_{4}$ Step II It involves the lodoform reaction. Ketone formed in step 1 have one methyl group linked to the carbonyl carbon atom are oxidised by sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less than that of carbonyl compound.