Classification of organic compounds Questions in English

(B) Tropolone is a $7$-membered ring compound containing a carbonyl group and a hydroxyl group. It exhibits aromaticity due to the delocalization of $6 \ \pi$-electrons in the $7$-membered ring,but it does not contain a benzene ring. Therefore,it is classified as a non-benzenoid aromatic compound.

(A) An alicyclic compound contains one or more all-carbon rings which may be either saturated or unsaturated but do not have aromatic character.
$1$. Cyclohexene: It is an alicyclic compound.
$2$. Anisole: It is an aromatic compound.
$3$. Pyridine: It is a heterocyclic aromatic compound.
$4$. Tetrahydrofuran: It is a heterocyclic alicyclic compound (often classified under alicyclic in broader contexts,but strictly alicyclic compounds are carbocyclic. However,based on the provided image,it is labeled as alicyclic).
$5$. Biphenyl: It is an aromatic compound.
Based on the provided image,Cyclohexene and Tetrahydrofuran are identified as alicyclic. Therefore,the total number is $2$.

(C) The molecular formula is $C_6H_6Br_2$.
Double bond equivalent $(DBU)$ formula $= C - \frac{H}{2} - \frac{X}{2} + 1$.
For $C=6, H=6, X=2$:
$DBU = 6 - \frac{6}{2} - \frac{2}{2} + 1 = 6 - 3 - 1 + 1 = 3$.
This indicates the presence of one ring and two double bonds.
The structure consists of a six-membered carbon ring (homo-cyclic) with two non-conjugated double bonds and two bromo groups at the $1,4$ positions.
Since it does not satisfy $H$ückel's rule ($4n+2$ $\pi$ electrons in a planar,fully conjugated system),it is not aromatic.
Therefore,the compound is homo-cyclic but not aromatic.

(C) Molecules having an integer value for the degree of unsaturation $(DU)$ can exist in a free state as stable compounds.
$DU = \frac{\sum n(v-2)}{2} + 1$,where $n$ is the number of atoms and $v$ is the valency of the atom.
For $C_7H_9O$: $DU = \frac{7(4-2) + 9(1-2) + 1(2-2)}{2} + 1 = \frac{14-9+0}{2} + 1 = 3.5$.
For $C_8H_{12}O$: $DU = \frac{8(4-2) + 12(1-2) + 1(2-2)}{2} + 1 = \frac{16-12+0}{2} + 1 = 3.0$.
For $C_6H_{12}O$: $DU = \frac{6(4-2) + 12(1-2) + 1(2-2)}{2} + 1 = \frac{12-12+0}{2} + 1 = 1.0$.
For $C_{10}H_{17}O$: $DU = \frac{10(4-2) + 17(1-2) + 1(2-2)}{2} + 1 = \frac{20-17+0}{2} + 1 = 2.5$.
Since $C_8H_{12}O$ and $C_6H_{12}O$ have integer values,$C_6H_{12}O$ is a stable compound.