The correct order of basicity for the following molecules is:
The correct order of basicity for the following molecules is:
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- \( \text{P}>\text{Q}>\text{R} \)
- \( \text{R}>\text{P}>\text{Q} \)
- \( \text{Q}>\text{P}>\text{R} \)
- \( \text{R}>\text{Q}>\text{P} \)
The Correct Option is D
Approach Solution - 1
To determine the correct order of basicity for the given molecules, we need to consider the electronic effects and structural factors affecting the nitrogen atom's ability to donate a lone pair of electrons.
- Basicity of a molecule is generally determined by the availability of the nitrogen lone pair for donation. More available the lone pair, the stronger the base.
- Looking at molecule (P), which is a N-methylpiperidin-2-one, the nitrogen is part of a structure that has some resonance stabilization due to the carbonyl group adjacent to it. This delocalization reduces the availability of the nitrogen's lone pair, reducing basicity.
- In molecule (Q), the nitrogen is part of a N-methylpyrrole-2-one structure. The nitrogen's lone pair is part of a conjugated system with the carbonyl, making it less available for donation due to resonance with the carbonyl group.
- Molecule (R) is a saturated alkaloid structure, specifically, delta-valerolactam. Here, the nitrogen's lone pair is most available for donation as it is least involved in resonance or conjugation compared to (P) and (Q).
Based on these observations:
- (R) is the most basic due to the highest availability of the nitrogen lone pair.
- (Q) is less basic than (R) but more than (P) because of the structure's partial conjugation.
- (P) is the least basic as the nitrogen's lone pair is involved in resonance with the adjacent carbonyl group.
Therefore, the correct order of basicity is \(R > Q > P\).
Approach Solution -2
The question asks for the correct order of basicity among the given molecules. The basicity of a molecule is primarily determined by the availability of the lone pair of electrons on the nitrogen atom for donation.
For the molecules (P), (Q), and (R):
- Molecule (P): This is an amide with a carbonyl group (C=O), which is highly electronegative. The presence of this carbonyl group decreases the electron availability on the nitrogen because of resonance. The lone pair on nitrogen is delocalized into the carbonyl group, making (P) less basic.
- Molecule (Q): Similar to (P), this compound also contains a carbonyl group adjacent to nitrogen. However, the presence of the double bond in the ring structure introduces a partial resonance effect that can slightly increase the availability of the lone pair on nitrogen for donation, compared to a purely saturated cycle like (P).
- Molecule (R): In contrast, (R) is a secondary amine without a carbonyl group in proximity. This means the nitrogen’s lone pair is more available for protonation, making (R) the most basic among the three due to the absence of any resonance effects involving a carbonyl group.
Therefore, the correct order of basicity is:
\(R > Q > P\)
This order is due to the increasing availability of the nitrogen's lone pair for protonation as we move from (P) to (R).
Top Questions on Amines
Which of the following amine(s) show(s) positive carbamylamine test?
- Identify correct statements: (A) Primary amines do not give diazonium salts when treated with \({NaNO}_2\) in acidic condition.
(B) Aliphatic and aromatic primary amines on heating with \({CHCl}_3\) and ethanolic \({KOH}\) form carbylamines.
(C) Secondary and tertiary amines also give carbylamine test.
(D) Benzenesulfonyl chloride is known as Hinsberg’s reagent.
(E) Tertiary amines react with benzenesulfonyl chloride very easily.
Choose the correct answer from the options given below: Amines are usually formed from amides, imides, halides, nitro compounds, etc. They exhibit hydrogen bonding which influences their physical properties. In alkyl amines, a combination of electron releasing, steric and H-bonding factors influence the stability of the substituted ammonium cations in protic polar solvents and thus affect the basic nature of amines. Alkyl amines are found to be stronger bases than ammonia. Amines being basic in nature, react with acids to form salts. Aryldiazonium salts, undergo replacement of the diazonium group with a variety of nucleophiles to produce aryl halides, cyanides, phenols and arenes.
How can you convert the following:
(i) Ethanoic acid to methanamine- Write short notes on the following:
(i) Carbylamine reaction
(ii) Hofmann–bromamide reaction
(iii) Diazotisation - The purification method based on the following physical transformation is: \[ {Solid} \xrightarrow{{Heat}} {Vapour} \xrightarrow{{Cool}} {Solid} \]
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