The reaction proceeds as follows:
Cyclohexane $\xrightarrow{\text{Br}_2}$ C$_6$H$_{10}$Br$_2$
1. Product A: Bromination followed by elimination with alcoholic KOH gives a conjugated diene:
\[\text{C}_6\text{H}_{10}\text{Br}_2 \xrightarrow{\text{alc. KOH (3 eq.)}} \text{C}_6\text{H}_6 \, (\text{benzene}).\]
- Benzene contains 6 $\pi$ electrons.
2. Product B: Bromination followed by reaction with sodium hydroxide gives an enolate ion:
\[\text{C}_6\text{H}_{10}\text{Br}_2 \xrightarrow{\text{Na}^+/\text{O}^-} \text{O-CH}_2\text{CH=CH}_2.\]
- This product contains 2 $\pi$ electrons.
Total $\pi$ electrons: $6 + 2 = 8$
Conc. HNO\(_3\)
Designate whether each of the following compounds is aromatic or not aromatic.
Find the IUPAC name of the compound.
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: