Amines Chemistry Quiz

The correct answer is CH3CH2CH2CH2NH2. This molecule contains an amino group (-NH2) which can form hydrogen bonds with other molecules. Hydrogen bonding is a strong intermolecular force that increases the boiling point of a substance. The other molecules listed do not contain a hydrogen bond donor or acceptor, so they do not have the same strong intermolecular forces and therefore lower boiling points.

The correct answer is sp3. In amines, the nitrogen atom is bonded to three other atoms and has one lone pair of electrons. This arrangement requires the nitrogen atom to undergo sp3 hybridization, where the s orbital and the three p orbitals of nitrogen combine to form four sp3 hybrid orbitals. These orbitals are then used to form sigma bonds with the other atoms, resulting in a tetrahedral geometry around the nitrogen atom.

The carbylamine test is given by primary amines. This test involves heating the amine with chloroform and potassium hydroxide, which results in the formation of isocyanides. The presence of isocyanides can be detected by their foul odor. Secondary and tertiary amines do not give this test because they do not have a hydrogen atom attached to the nitrogen atom, which is necessary for the reaction to occur. Therefore, the correct answer is primary amine.

When an imine is reduced, it undergoes a reaction called imine reduction. In this reaction, the double bond between the carbon and nitrogen in the imine is converted into a single bond, resulting in the formation of an amine. Therefore, the correct answer is amine.

Amines are classified as weak bases because they can accept protons (H+) from other molecules. This is due to the presence of a lone pair of electrons on the nitrogen atom, which can form a coordinate bond with a proton. However, compared to strong bases, amines have a lower tendency to accept protons and have a weaker ability to donate electrons. Therefore, they are classified as weak bases.

All of the given options are water soluble.

The correct answer is Benzene sulfonic acid. Hinsberg reagent is a reagent used in organic chemistry to distinguish between primary, secondary, and tertiary amines. It is commonly used in the Hinsberg test, where the amine is reacted with the reagent to form an insoluble sulfonamide. In this case, the Hinsberg reagent is benzene sulfonic acid, which reacts with the amine to form a benzene sulfonamide. This reaction is used to determine the type of amine present in a given compound.

Primary amines produce N2 on treatment with HONO because primary amines have a nitrogen atom bonded to only one carbon atom. When primary amines react with HONO (nitrous acid), a diazonium salt is formed. This diazonium salt is unstable and decomposes to release N2 gas, resulting in the formation of a nitroso compound. Tertiary and secondary amines do not produce N2 gas when treated with HONO because they do not have a hydrogen atom bonded directly to the nitrogen atom, which is necessary for the formation of the diazonium salt. Quaternary amines do not have a nitrogen-hydrogen bond, so they also do not produce N2 gas in this reaction.

When acetonitrile reacts with LiAlH4, it undergoes reduction to form ethylamine. LiAlH4 is a strong reducing agent and can reduce the nitrile group (-CN) in acetonitrile to an amine group (-NH2). Ethylamine is formed by replacing the nitrile group with an amino group in acetonitrile. Methylamine, dimethyltryptamine, and trimethylamine are not the correct products of this reaction.

When CH3CH2Cl reacts with (CH3)2NH, the chlorine atom in CH3CH2Cl is replaced by the (CH3)2NH group, resulting in the formation of a 3° amine. A 3° amine is characterized by having three alkyl groups bonded to the nitrogen atom. Therefore, the major product of this reaction is a 3° amine.

Diethylamine is the correct answer because it contains two ethyl groups attached to the nitrogen atom. When diethylamine reacts with nitrous acid, it forms a nitrosamine compound. The presence of two ethyl groups enhances the reactivity of diethylamine towards nitrous acid, allowing for the formation of the nitrosamine product.

Diethylamine is the most basic compound among the given options. Basicity is determined by the availability of lone pairs of electrons on the nitrogen atom. Diethylamine has two ethyl groups attached to the nitrogen, which increases the electron density on the nitrogen atom, making it more basic compared to the other compounds. Ammonia has only one methyl group, and ethylamine has one ethyl group attached to the nitrogen, making them less basic than diethylamine. Triethylamine has three ethyl groups, but the steric hindrance caused by the bulky groups decreases its basicity compared to diethylamine.