The product is:

Which of the following tetracarboxylic acids forms a di-anhydride upon heating?

An organic compound having molecular formula $C_2H_6O$ undergoes oxidation with $K_2Cr_2O_7 / H_2SO_4$ to produce $X$ which contains $40\,\%$ carbon,$6.7\,\%$ hydrogen and $53.3\,\%$ oxygen. The molecular formula of the compound $X$ is

How will you bring about the following conversions in not more than two steps?
$(i)$ Propanone to Propene
$(ii)$ Benzoic acid to Benzaldehyde
$(iii)$ Ethanol to $3-$Hydroxybutanal
$(iv)$ Benzene to $m-$Nitroacetophenone
$(v)$ Benzaldehyde to Benzophenone
$(vi)$ Bromobenzene to $1-$Phenylethanol
$(vii)$ Benzaldehyde to $3-$Phenylpropan$-1-$ol
$(viii)$ Benzaldehyde to $\alpha-$Hydroxyphenylacetic acid
$(ix)$ Benzoic acid to $m-$Nitrobenzyl alcohol

Draw the structures of the following compounds:
$(i)$ $3-$Methylbutanal
$(ii)$ $p-$Nitropropiophenone
$(iii)$ $p-$Methylbenzaldehyde
$(iv)$ $4-$Methylpent$-3-$en$-2-$one
$(v)$ $4-$Chloropentan$-2-$one
$(vi)$ $3-$Bromo$-4-$phenylpentanoic acid
$(vii)$ $p, p^{\prime}-$Dihydroxybenzophenone
$(viii)$ Hex$-2-$en$-4-$ynoic acid

The product $(C)$ of the above reaction is: