The major product ‘P’ for the following sequence of reactions is :
When solving reaction sequences:
Analyze each reagent step-by-step and apply known reaction mechanisms.
Clemmensen reduction (Zn/Hg, HCl) is used to reduce carbonyl groups to methylene (\(-\text{CH}_2-\)).
LiAlH\(_4\) reduces amides to amines.
1. First Step: Clemmensen Reduction (Zn/Hg, HCl):
The Clemmensen reduction converts the ketone group (\(-\text{C}= \text{O}\)) into a methylene (\(-\text{CH}_2-\)) group.
Starting compound:
\[\text{Ph-CH}_2-\text{CO-CH}_2-\text{CH}_2-\text{NH}_2.\]
After Clemmensen reduction:
\[\text{Ph-CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{NH}_2.\]
2. Second Step: LiAlH\(_4\) Reduction:
LiAlH\(_4\) reduces amides (\(-\text{CONH}_2\)) to amines (\(-\text{CH}_2-\text{NH}_2\)). In this compound, no further reduction occurs as the ketone has already been reduced to \(-\text{CH}_2-\) in the first step.
3. Final Product:
The major product (P) is:
\[\text{Ph-CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{NH}_2.\]
Final Answer: \(\text{Ph-CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{NH}_2\).
The correct increasing order of stability of the complexes based on \( \Delta \) value is:
List I (Molecule) | List II (Number and types of bond/s between two carbon atoms) | ||
A. | ethane | I. | one σ-bond and two π-bonds |
B. | ethene | II. | two π-bonds |
C. | carbon molecule, C2 | III. | one σ-bonds |
D. | ethyne | IV. | one σ-bond and one π-bond |
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