In this question, we are determining which compound is least likely to produce effervescence of CO\(_2\) when reacted with aqueous NaHCO\(_3\). Effervescence occurs when an acid reacts with NaHCO\(_3\), producing CO\(_2\).
Compound (1) contains a hydroxyl group (-OH) and nitro groups (-NO\(_2\)) which will likely result in an acidic environment and cause effervescence with NaHCO\(_3\).
Compound (2) contains a carboxyl group (-COOH), a strong acid, which will react with NaHCO\(_3\) and release CO\(_2\).
Compound (3) contains an amine group (-NH\(_3\)), which is basic and does not typically react with NaHCO\(_3\) to produce CO\(_2\).
Compound (4) contains a nitro group (-NO\(_2\)) but lacks a strongly acidic functional group that would promote CO\(_2\) production.
Thus, it is the least likely to produce CO\(_2\) effervescence in the presence of NaHCO\(_3\).
Therefore, the compound least likely to give effervescence of CO\(_2\) is compound (4).
A dipeptide, “x”, on complete hydrolysis gives “y” and “z”; “y” on treatment with aqueous HNO$_2$, produces lactic acid. On the other hand, “z” on heating gives the following cyclic molecule.
Based on the information given, the dipeptide X is:
If $ \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p $, then $ 96 \log_e p $ is equal to _______
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:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: