Organic Chemistry Quiz: Do You Know Amines?
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An organic compound ‘A’ on treatment with NH3 gives ‘B’ which on heating gives ‘C’, ‘C’ when treated with Br2 in the presence of KOH produces ethylamine. Compound ‘A’ is:
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Welcome to the Amines Quiz! This interactive quiz is designed to test your knowledge of amines, an essential functional group in organic chemistry. You'll encounter questions covering a wide array of topics, from basic structures and nomenclature to reactions, properties, and applications of primary, secondary, and tertiary amines.
Whether you're a student preparing for exams, a teacher seeking engaging classroom See morecontent, or a chemistry enthusiast eager to reinforce your understanding, this quiz provides a comprehensive assessment tailored to all levels. Instant feedback and detailed explanations after each question help reinforce learning and clarify key concepts.
Take the Amines Quiz today and discover how much you know about this fascinating class of organic compounds. Test yourself, learn, and improve your understanding!
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- 2.
Which of the following amines cannot be prepared by Gabriel’s synthesis:
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Butylamine
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Isopropylamine
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2-phenylethylamine
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N-Methylbenzylamine
Correct Answer
A. N-MethylbenzylamineExplanation
Gabriel's synthesis is a method used to prepare primary amines. It involves the reaction of phthalimide with an alkyl halide followed by hydrolysis and decarboxylation. However, N-Methylbenzylamine cannot be prepared by Gabriel's synthesis because it is a secondary amine, not a primary amine. In Gabriel's synthesis, only primary amines can be obtained.Rate this question:
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- 3.
Acetamide is treated with the following reagents separately. Which one of these yields methylamine?
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PCl5
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NaOH–Br2
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Sodalime
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Hot. Conc. H2SO4
Correct Answer
A. NaOH–Br2Explanation
NaOH-Br2 is the reagent that yields methylamine when treated with acetamide. This is because NaOH (sodium hydroxide) is a strong base that can deprotonate the amide group in acetamide, forming the corresponding carboxylate ion. Br2 (bromine) is an electrophilic reagent that can react with the carboxylate ion, replacing the oxygen with a bromine atom. This intermediate can then undergo a rearrangement reaction to form methylamine.Rate this question:
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- 4.
An organic compound (C3H9N) (A) when treated with nitrous acid, gave an alcohol, and N2 gas was evolved. (A) on warming with CHCl3 and caustic potash gave (C) which on reduction gave isopropylmethylamine. Predict the structure of (A):
Correct Answer
A.Explanation
Based on the given information, compound (A) is treated with nitrous acid to yield an alcohol and N2 gas. This suggests that (A) contains an amine group (NH2), which reacts with nitrous acid to form a diazonium salt. (A) is also warmed with CHCl3 and caustic potash to give (C), which is then reduced to isopropylmethylamine. This indicates that (A) contains an alkyl group that can be converted to an amine through reduction. Therefore, the structure of (A) is likely an alkyl amine with the formula C3H9N.Rate this question:
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- 5.
Which of the following compounds is most basic?
Correct Answer
A.Explanation
The most basic compound is the one that can easily accept a proton (H+) to form a stable conjugate acid. In general, amines are more basic than alcohols and ethers. Among the given compounds, ammonia (NH3) is the most basic because it has a lone pair of electrons on the nitrogen atom, which can readily accept a proton to form NH4+. The other compounds, water (H2O) and methanol (CH3OH), also have lone pairs of electrons, but they are less basic compared to ammonia.Rate this question:
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- 6.
Basicity of CH3CH2NH2 (I), CH3CONH2 (II) and C6H5CONH2 (III) follows the order:
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I > II > III
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I > III > II
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II > III > I
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III > II > I
Correct Answer
A. I > III > IIExplanation
The basicity of a compound is determined by its ability to donate a lone pair of electrons. In this case, CH3CH2NH2 (I) is an amine compound, which has a lone pair of electrons on the nitrogen atom. This lone pair can easily be donated, making it a strong base. C6H5CONH2 (III) is an amide compound, which also has a lone pair of electrons on the nitrogen atom. However, the presence of the phenyl group (C6H5) attached to the nitrogen atom reduces the availability of the lone pair, making it a weaker base compared to I. CH3CONH2 (II) is also an amide compound, but it does not have any electron-donating groups attached to the nitrogen atom, making it the weakest base among the three compounds. Therefore, the correct order of basicity is I > III > II.Rate this question:
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- 7.
Predict the product:
Correct Answer
A.Explanation
Reacting the given amine with sodium nitrite (NaNO2) and hydrochloric acid (HCl) results in the formation of a diazonium salt, which can subsequently be used to introduce various functional groups into the aromatic ring. This reaction sequence is known as diazotization followed by a Sandmeyer reaction.Rate this question:
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- 8.
Which among the following amines can be directly oxidized to the corresponding nitro compound by potassium permanganate?
Correct Answer
A.Explanation
Primary amines can be directly oxidized to the corresponding nitro compound by potassium permanganate. This is because primary amines have a hydrogen atom attached to the nitrogen atom, which can be easily oxidized by potassium permanganate. Secondary amines and tertiary amines do not have this hydrogen atom and therefore cannot be directly oxidized to nitro compounds by potassium permanganate.Rate this question:
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- 9.
The optically active compound (X) (X) on treatment with NaNO2/HCl gives:
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1° alcohol with retention of configuration
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2° alcohol with inversion of configuration
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Racemic mixture of 2° alcohols
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Racemic mixture of 1° alcohols
Correct Answer
A. Racemic mixture of 2° alcoholsExplanation
When the optically active compound (X) is treated with NaNO2/HCl, it undergoes a reaction called the Sandmeyer reaction. In this reaction, the compound is converted into a diazonium salt. The diazonium salt can then undergo various reactions depending on the conditions. In this case, it is converted into a 2° alcohol. However, the conversion of the diazonium salt to the alcohol involves the loss of the chiral center, resulting in a racemic mixture of 2° alcohols. This is because the reaction proceeds through a carbocation intermediate, which can undergo both front-side and back-side attacks, leading to the formation of both R and S configurations of the alcohol.Rate this question:
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- 10.
Which of the following compounds will dissolve in an alkali solution after it undergoes reaction with Hinsberg’s reagent?
Correct Answer
A.Explanation
In Hinsberg's test, secondary amines like (C2H5)2NH react with Hinsberg's reagent (benzene sulfonyl chloride, C6H5SO2Cl) to form sulfonamides that are insoluble in alkali, while primary amines form sulfonamides soluble in alkali due to their free -NH group. Tertiary amines do not react with the reagent. Thus, the secondary amine (C2H5)2NH forms an insoluble sulfonamide that remains undissolved in alkali.Rate this question:
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- 11.
Which of the following exists as zwitterions?
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P-aminophenol
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Salicylic acid
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Sulphanilic acid
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Ethanolamine
Correct Answer
A. Sulphanilic acidExplanation
Sulphanilic acid exists as zwitterions. Zwitterions are molecules that have both positive and negative charges, but overall they have a neutral charge. In the case of sulphanilic acid, it contains an amino group (-NH2) and a sulfonic acid group (-SO3H), which can both donate and accept protons. This allows the molecule to exist in a zwitterionic form, with the amino group being positively charged and the sulfonic acid group being negatively charged. This property makes sulphanilic acid useful in various chemical reactions and applications.Rate this question:
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- 12.
Which of the following is not the correct reaction of aryldiazonium salts?
Correct Answer
A.Explanation
Aryldiazonium salts can undergo various reactions, including diazotization, Sandmeyer reaction, and coupling reactions. However, one reaction that is not observed with aryldiazonium salts is the Cannizzaro reaction. The Cannizzaro reaction is a disproportionation reaction of aldehydes or ketones in the presence of a strong base, resulting in the formation of an alcohol and a carboxylic acid. Since aryldiazonium salts do not contain any carbonyl group, they cannot undergo the Cannizzaro reaction.Rate this question:
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- 13.
In the diazotization of arylamine, the use of nitrous acid is: It suppresses the hydrolysis of phenol
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It suppresses hydrolysis of phenol
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It is a source of electrophilic nitrosonium ion
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It neutralizes the base liberated
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All of the above
Correct Answer
A. It is a source of electrophilic nitrosonium ionExplanation
Nitrous acid (HNO2) is used in the diazotization of arylamine as it acts as a source of electrophilic nitrosonium ion (NO+). The nitrosonium ion is a highly reactive electrophile that can react with the arylamine to form a diazonium salt. This reaction is an important step in the synthesis of various organic compounds, such as azo dyes. The other options listed (suppressing hydrolysis of phenol and neutralizing the base liberated) are not the primary roles of nitrous acid in this context.Rate this question:
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- 14.
Aniline is treated with bromine water and the resulting product is treated with an aqueous solution of sodium nitrite in presence of dilute HCl. The compound so formed is converted into tetrafluoroborate which is subsequently heated dry. The final product is
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P-Bromofluorobenzene
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P-Bromoaniline
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2, 4, 6-Tribromofluorobenzene
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1, 3, 5-Tribromobenzene
Correct Answer
A. 2, 4, 6-TribromofluorobenzeneExplanation
When aniline is treated with bromine water, bromination occurs at the para position of the benzene ring, resulting in p-bromoaniline. When p-bromoaniline is treated with an aqueous solution of sodium nitrite in the presence of dilute HCl, a diazonium salt is formed. This diazonium salt can be converted into a tetrafluoroborate salt. Heating the tetrafluoroborate salt leads to the elimination of nitrogen gas and the formation of a fluorobenzene derivative. In this case, the bromine substituents on the benzene ring will be replaced by fluorine atoms, resulting in 2, 4, 6-tribromofluorobenzene as the final product.Rate this question:
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- 15.
Aniline in a set of the following reactions yielded a coloured product Y. The structure of Y would be
Correct Answer
A.Explanation
Reacting aniline C6H5NH2 with sodium nitrite NaNO2 and hydrochloric acid HCl at 0-5°C results in the formation of benzene diazonium chloride C6H5N2^+Cl^- , an intermediate diazonium salt. This process is known as diazotization. The diazonium salt then undergoes an electrophilic aromatic substitution reaction with N,N -dimethylaniline C6H5NCH32 via a coupling reaction. In this reaction, the diazonium group -N2^+ acts as an electrophile and couples with the activated aromatic ring of N, N -dimethylaniline to produce the azo compound 4-dimethylaminoazobenzene C6H5N=NC6H4NCH32, commonly known as Butter Yellow. This vibrant yellow azo dye is formed due to the extended conjugation in the azo group -N=N- and serves as an example of aromatic diazonium coupling with electron-rich amines.Rate this question:
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May 14, 2024Quiz Edited by
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Jan 29, 2014Quiz Created by
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