CBSE Class 12 Chemistry Amines NCERT Solutions

Solution: A tertiary (3°) amine is characterized by a nitrogen atom bonded to three carbon atoms (from alkyl or aryl groups). Let's examine the structures:

(a) 1-methylcyclohexylamine: The nitrogen is bonded to one methyl group and two carbon atoms of the cyclohexyl ring, making it a 3° amine. However, the provided solution indicates (b) is correct, let's re-evaluate based on common conventions and the source's likely intent.

(b) Triethylamine: This amine has the formula $(CH\_3CH\_2)\_3N$. The nitrogen atom is bonded to three ethyl groups, thus it is a tertiary (3°) amine.

(c) tert-butylamine: This amine has the formula $(CH\_3)\_3CNH\_2$. The nitrogen atom is bonded to one tertiary butyl group and two hydrogen atoms, making it a primary (1°) amine.

(d) N-methylaniline: This amine has the formula $C\_6H\_5NHCH\_3$. The nitrogen atom is bonded to one methyl group, one phenyl group, and one hydrogen atom, making it a secondary (2°) amine.

Based on the standard definitions, triethylamine is a tertiary amine. The source's diagrammatic representation, though somewhat unclear, supports this. Therefore, triethylamine is the correct answer.

Answer: (b)

Solution: To determine the IUPAC name, we need to identify the longest carbon chain containing the principal functional group (amine) and the double bond, and assign the lowest possible locants.

The given structure is $CH\_2\; =\; CHCH\_2NHCH\_3$. We can number the carbons in the chain attached to the nitrogen. The chain containing the double bond has three carbons. The double bond starts at carbon 1. The nitrogen atom is substituted with a methyl group. The parent chain is propene. The amine group is attached to the first carbon of the propyl chain, and the nitrogen is substituted with a methyl group.

The structure can be visualized as:

$$CH\_2\; =\; CH\; -\; CH\_2\; -\; NH\; -\; CH\_3$$

The longest chain containing the double bond is a three-carbon chain (propene). The double bond is at position 1. The amine group is attached to carbon 1. The nitrogen atom is substituted with a methyl group.

Therefore, the IUPAC name is N-methylprop-2-en-1-amine.

Answer: (d)

Solution: The basic strength of amines in aqueous medium depends on several factors, including the inductive effect (+I), resonance, and solvation of the conjugate acid.

Let's analyze each option:

(a) $CH\_3NH\_2$ (Methylamine): The methyl group has a +I effect, increasing electron density on the nitrogen atom, making it a stronger base than ammonia. Solvation of the conjugate acid ($CH\_3NH\_3^+$) is significant.

(b) $NCCH\_2NH\_2$ (2-cyanoethanamine): The cyano group (-CN) is strongly electron-withdrawing (-I effect), which decreases the electron density on the nitrogen atom, making it a weaker base.

(c) $(CH\_3)\_2NH$ (Dimethylamine): This is a secondary amine. It has two methyl groups, both exerting a +I effect, which significantly increases the electron density on the nitrogen. The conjugate acid ($(CH\_3)\_2NH\_2^+$) is also well-solvated. Generally, for lower aliphatic amines, secondary amines are stronger bases than primary or tertiary amines due to a balance of inductive effect and solvation.

(d) $C\_6H\_5NHCH\_3$ (N-methylaniline): This is a secondary amine where one substituent is an aryl group (phenyl). The lone pair on the nitrogen atom is delocalized into the benzene ring through resonance, significantly reducing its availability for protonation. This makes it a much weaker base compared to aliphatic amines.

Comparing the factors, dimethylamine ($(CH\_3)\_2NH$) has the strongest electron-donating effect from two methyl groups and good solvation of its conjugate acid, making it the strongest base among the given options in aqueous medium.

Answer: (c)

Solution: A Bronsted base is a substance that can accept a proton ($H^+$). The strength of a base depends on the availability of the lone pair of electrons on the atom (in this case, nitrogen) to accept the proton.

Let's analyze the options:

(a) Aniline ($C\_6H\_5NH\_2$): In aniline, the lone pair of electrons on the nitrogen atom is delocalized into the pi-electron system of the benzene ring through resonance. This delocalization makes the lone pair less available to accept a proton, rendering aniline a weak base.

(b) A secondary amine with electron-withdrawing groups: Electron-withdrawing groups decrease the electron density on the nitrogen atom, making the lone pair less available and thus reducing basicity.

(c) A tertiary amine with electron-donating groups: Electron-donating groups increase the electron density on the nitrogen atom, making the lone pair more available and increasing basicity. However, tertiary amines can sometimes be weaker bases than secondary amines in aqueous solution due to steric hindrance and reduced solvation of the conjugate acid.

(d) Ammonia ($NH\_3$): Ammonia is a base, but its basicity is generally lower than that of aliphatic primary and secondary amines due to the absence of electron-donating alkyl groups.

Comparing these, aniline is significantly weakened as a base due to resonance with the benzene ring. While option (b) describes a weak base, aniline is a classic example of a very weak base among amines due to this resonance effect. Therefore, aniline is the weakest Bronsted base among the typical choices presented in such questions.

Answer: (a)

Solution: The reaction described is the alkylation of benzylamine, and the question asks which alkyl halide is best suited for this reaction via an S_N1 mechanism.

An S_N1 (Substitution Nucleophilic Unimolecular) reaction proceeds through a carbocation intermediate. The rate-determining step is the formation of this carbocation from the substrate (alkyl halide in this case). Therefore, the alkyl halide that forms the most stable carbocation will react fastest via the S_N1 mechanism.

Let's consider the carbocations formed from each option:

(a) $CH\_3Br$: Forms a methyl carbocation ($CH\_3^+$), which is highly unstable.

(b) $C\_6H\_5Br$: Forms a phenyl carbocation ($C\_6H\_5^+$), which is extremely unstable due to the sp2 hybridization of the carbon and the lack of resonance stabilization for a positive charge directly on the ring in this manner.

(c) $C\_6H\_5CH\_2Br$: Forms a benzyl carbocation ($C\_6H\_5CH\_2^+$). This carbocation is resonance-stabilized by the adjacent benzene ring, making it relatively stable compared to simple alkyl carbocations.

(d) $C\_2H\_5Br$: Forms an ethyl carbocation ($C\_2H\_5^+$), which is more stable than a methyl carbocation due to hyperconjugation but less stable than a benzyl carbocation.

Since the S_N1 mechanism favors the formation of a stable carbocation, benzyl bromide ($C\_6H\_5CH\_2Br$) is the best choice because it yields the resonance-stabilized benzyl carbocation.

Answer: (c)