The major product P formed in the following sequence of reactions is: (i) SOCl2 (ii) R-NH2 (iii) LiAlH4 (iv) H3O+ (1) [Structure with OH and NHR] (2) [Structure with Cl and OH] (3) [Structure with CONHR] (4) [Structure with CH2NHR]
Explanation: 1. SOCl2: Converts the carboxylic acid (-COOH) to an acyl chloride (-COCl). 2. R-NH2: Reacts with the acyl chloride to form an amide (-CONHR). 3. LiAlH4: Reduces the amide to an amine (-CH2NHR). 4. H3O+: Acidic workup to protonate any intermediates. Therefore, the correct answer is (3).
Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq).
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: