(A) (i) Define a homologous series of carbon compounds. (ii) Why is the melting and boiling points of $\mathrm{C_4H_8}$ higher than that of $\mathrm{C_3H_6}$ or $\mathrm{C_2H_4}$ ? (iii) Why do we NOT see any gradation in chemical properties of a homologous series compounds ? (iv) Write the name and structures of (i) aldehyde and (ii) ketone with molecular form $\mathrm{C_3H_6O}$ . OR (B) (i) Write the name and structure of an organic compound 'X' having two carbon atoms in its molecule and its name is suffixed with '-ol'. (ii) What happens when 'X' is heated with excess concentrated sulphuric acid at 443 K? Write chemical equation for the reaction stating the conditions for the reaction. Also state the role played by concentrated sulphuric acid in the reaction. (iii) Name and draw the electron dot structure of hydrocarbon produced in the above reaction.

(A) (i) A homologous series is a series of carbon compounds in which the same functional group substitutes for hydrogen in a carbon chain. Members of such a series have the same general formula and similar chemical properties.

(ii) The melting and boiling points of $C_4H_8$ are higher than $C_3H_6$ or $C_2H_4$ because molecular mass increases as we move down a homologous series. With the increase in molecular mass, the intermolecular forces of attraction (Van der Waals forces) become stronger, requiring more energy to overcome.

(iii) We do not see a gradation in chemical properties because chemical properties are determined solely by the functional group present in the molecule. Since all members of a homologous series have the identical functional group, their chemical reactivity remains similar. Gradation is only observed in physical properties.

(iv) For molecular formula $C_3H_6O$:

(i) Aldehyde: Propanal. Structure: $CH_3-CH_2-CHO$

(ii) Ketone: Propanone. Structure: $CH_3-CO-CH_3$

OR

(B) (i) The organic compound 'X' is Ethanol. Structure: $CH_3-CH_2-OH$.

(ii) When Ethanol is heated with excess concentrated sulphuric acid at $443 \text{ K}$, it undergoes dehydration to form ethene.

Chemical Equation:

$$CH_3-CH_2OH \xrightarrow{\text{Hot conc. } H_2SO_4, 443 \text{ K}} CH_2=CH_2 + H_2O$$

Role of concentrated $H_2SO_4$: It acts as a dehydrating agent, as it removes a water molecule from ethanol.

(iii) The hydrocarbon produced is Ethene.

Electron Dot Structure of Ethene ($C_2H_4$):

In the structure, each Carbon atom is bonded to two Hydrogen atoms by single covalent bonds, and the two Carbon atoms are linked by a double covalent bond ($C=C$).