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:
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $