Question

$CaC_2 + N_2 \xrightarrow{1100^\circ C} C + 'M'$

$M + H_2SO_4 \rightarrow$ white ppt. + 'N'

$N \xrightarrow{pH = 7-9}$ Dimer of 'N' $\xrightarrow{Pyrolysis}$ Aromatic compound (O)

If the number of σ-bonds in 'O' = a and number of lone pairs in 'O' = b

Then find the value of (a/b)

Answer 1

2.5

Answer 2

Solution: The given reactions are:

1. $CaC_2 + N_2 \xrightarrow{1100^\circ C} C + 'M'$ This is the Frank-Caro process for producing calcium cyanamide. $CaC_2 + N_2 \xrightarrow{1100^\circ C} CaCN_2 + C$ So, 'M' is $CaCN_2$ (Calcium cyanamide) and 'C' is Carbon.

2. $M + H_2SO_4 \rightarrow$ white ppt. + 'N' 'M' is $CaCN_2$. Calcium cyanamide reacts with acid to form cyanamide and a calcium salt. $CaCN_2 + H_2SO_4 \rightarrow CaSO_4 \downarrow + H_2NCN$ $CaSO_4$ (Calcium sulfate) is a white precipitate. So, 'N' is $H_2NCN$ (Cyanamide).

3. $N \xrightarrow{pH = 7-9}$ Dimer of 'N' $\xrightarrow{Pyrolysis}$ Aromatic compound (O) 'N' is $H_2NCN$ (Cyanamide). Cyanamide dimerizes at pH 7-9 to form dicyandiamide (cyanoguanidine). $2 H_2NCN \xrightarrow{pH=7-9} (H_2N)_2C=N-C \equiv N$ The dimer of 'N' is dicyandiamide. Pyrolysis of dicyandiamide yields melamine, which is an aromatic compound. $3 (H_2N)_2C=N-C \equiv N \xrightarrow{Pyrolysis} 2 C_3H_6N_6$ The aromatic compound 'O' is Melamine ($C_3H_6N_6$).

Melamine has the structure 1,3,5-triazine-2,4,6-triamine:

      NH₂
      / \
     C   N
    /     \
   N       C
  //       \\
 N           N
 \         /
  C---N---C
  |   |   |
 NH₂ NH₂ NH₂

Melamine is an aromatic compound (planar ring, cyclic delocalization of 6 $\pi$ electrons, follows Huckel's rule).

We need to find the number of $\sigma$-bonds (a) and the number of lone pairs (b) in Melamine (O).

Number of $\sigma$-bonds (a):

• Ring $\sigma$-bonds: The six-membered ring has 6 atoms (3 C and 3 N) connected by single and double bonds. Counting one bond from each double bond and all single bonds in the ring skeleton gives 6 $\sigma$-bonds in the ring.
• Exo-cyclic C-N $\sigma$-bonds: There are 3 amino groups ($NH_2$) attached to the carbon atoms of the ring. Each C-$NH_2$ bond is a single bond. So, 3 C-N $\sigma$-bonds.
• N-H $\sigma$-bonds: Each $NH_2$ group has 2 N-H single bonds. There are 3 $NH_2$ groups. So, $3 \times 2 = 6$ N-H $\sigma$-bonds.

Total number of $\sigma$-bonds (a) = 6 (ring) + 3 (C-N) + 6 (N-H) = 15.

Number of lone pairs (b):

• Ring Nitrogen atoms: There are 3 nitrogen atoms in the ring. Each ring nitrogen is bonded to two carbon atoms and has one lone pair contributing to the aromatic system. So, 3 lone pairs on ring nitrogens.
• Exo-cyclic Nitrogen atoms ($NH_2$ groups): There are 3 nitrogen atoms in the $NH_2$ groups. Each $NH_2$ nitrogen is bonded to one carbon and two hydrogen atoms, and has one lone pair. So, 3 lone pairs on $NH_2$ nitrogens.

Total number of lone pairs (b) = 3 (ring N) + 3 ($NH_2$ N) = 6.

We need to find the value of (a/b). a = 15 b = 6 (a/b) = 15 / 6 = 5 / 2 = 2.5

The final answer is 2.5.