To identify chiral carbons, examine carbon atoms with four different substituents, which render them asymmetric:
Step 1. CH₃–CH₂–CH(NO₂)–COOH: The second carbon is chiral due to four distinct substituents.
Step 2. CH₃–CH₂–CHBr–CH₂–CH₃: The third carbon is chiral, as it has four different substituents.
Step 3. CH₃–CH(I)–CH₂–NO₂: The second carbon is chiral due to its four different groups.
Step 4. CH₃–CH₂–CH(OH)–CH₂OH: The third carbon is chiral, as it has four distinct substituents.
Step 5. CH₃–CH–CH(I)–C₂H₅: The second carbon is chiral due to four different substituents.
Thus, there are five compounds containing chiral carbons.
The Correct answer is: 5
How many different stereoisomers are possible for the given molecule?
Assertion (A): All naturally occurring \(\alpha\)-amino acids except glycine are optically active. Reason (R): Most naturally occurring amino acids have L-configuration.
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32