Chemistry - CHEMICAL BONDING Question with Solution | TestHub

The problem asks to determine the value of $(x-y)$ based on intramolecular hydrogen bonding in a given list of compounds.

Here's a breakdown of the analysis for each compound:

1. Chloral hydrate (2,2,2-trichloroethane-1,1-diol):

Structure: $CC{l}_{3}CH(OH{)}_2$

Intramolecular H-bond: Forms a 5-membered ring between one -OH group and a Cl atom.

Ring members: O-H...Cl-C-C.

Number of compounds forming at least one 5-membered ring (y): 1 (Chloral hydrate).

$s{p}^2$ hybridized atoms in the ring (excluding oxygen): 0 (all carbons are $s{p}^3$).

2. o-Nitrophenol:

Structure: $C_6{H}_4(OH)(N{O}_2)$

Intramolecular H-bond: Forms a 6-membered ring between the -OH group and one oxygen of the nitro group.

Ring members: O-H...O=N-C-C-C.

$s{p}^2$ hybridized atoms in the ring (excluding oxygen): 3 (1 N atom and 2 C atoms from the benzene ring).

3. Salicylaldehyde:

Structure: $C_6{H}_4(OH)(CHO)$

Intramolecular H-bond: Forms a 6-membered ring between the -OH group and the carbonyl oxygen.

Ring members: O-H...O=C-C-C-C.

$s{p}^2$ hybridized atoms in the ring (excluding oxygen): 3 (3 C atoms from the benzene ring, including the carbonyl carbon).

4. Hydrogen Maleate anion:

Structure: $[HOOC-CH=CH-COO{]}^{-}$

Intramolecular H-bond: Forms a 7-membered ring between the carboxylic acid proton and the carboxylate oxygen.

Ring members: O-H...O=C-C=C-C.

$s{p}^2$ hybridized atoms in the ring (excluding oxygen): 4 (4 C atoms, two from the double bond and two from the carboxyl groups).

5. Chloral ($CC{l}_{3}CHO$): Does not form intramolecular hydrogen bonds.

Summary:

y = Number of compounds forming at least one 5-membered ring = 1 (Chloral hydrate).

x = Sum of $s{p}^2$ hybridized atoms (excluding oxygen) in the rings formed by intramolecular H-bonding:

Chloral hydrate: 0

o-Nitrophenol: 3

Salicylaldehyde: 3

Hydrogen Maleate anion: 4

Total $x=0+3+3+4=10$.

Finally, calculate $(x-y)$:

$x-y=10-1=9$.

The final answer is $\boxed{9}$.