Class 12 Chemistry Chapter 9 – Amines NCERT Solutions

Intext Questions – Class 12 Chemistry Chapter 9 Amines NCERT Solutions

9.1

Ans – (i) 1°
(ii) 3°
(iii) 1°
(iv) 2°

9.2

Ans – Structures and IUPAC names of different isomeric amines:-

9.3

9.4

Ans – (i) Given each of the alkyl-based compounds C₂H₅NH₂, (C₂H₅)₂NH’s & NH₃ inductive action, they can be ranked in an ascending sequence of fundamental potency as mentioned below:

NH₃ < C₂H₅NH₂ < (C₂H₅)₂NH

Once more, NH₃ has more proton acceptance compared to C₆H₅NH₂. Consequently, we have the following characteristics: C₆H₅NH₂ < NH₃ < C₂H₅NH₂ < (C₂H₅)₂NH

The density of electrons on the N-atom surrounding C₆H₅CH₂NH₂ is greater compared to the one in NH₃, but much smaller than the one found on the N-atom within C₂H₅NH₂ because of the −I impact created by the C₆H₅− group. As a result, the compounds that come next can be ranked according to their fundamental attributes:C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH

(ii) By considering the fact of steric restriction alongside the inductive impact of the alkyl compounds C₂H₅NH₂ & (C₂H₅)₂NH along with their basic reaction can be described in the following manner: C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH

Once more, the overall electron density regarding the nitrogen atom present in C₆H₅NH₂ is significantly less when compared to that around the N atom in C₂H₅NH₂ due to the −R impact occurring inside the C₆H₅− band. Consequently, C₆H₅NH₂ consists of declined fundamental features as opposed to C₂H₅NH₂ compounds.

As a result, the compounds that come next can be grouped in ascending hierarchy based on their basic physical properties: C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH

(iii) By considering the fact of the steric constraint & inductive impact of alkyl divisions, the CH₃NH₂, (CH₃)₂NH & (CH₃)₃N compounds are capable of being ranked in ascending sequence based on their fundamental qualities in the following way:

(CH₃)₃N < (CH₃)₂NH < CH₃NH₂

N is joined straight with the benzene ring through C₆H₅NH₂. Consequently, the N-atom’s single set of electrons becomes dispersed across the benzene ring. N is not bonded completely with the benzene ring around C₆H₅CH₂NH₂. Its single couple is therefore not decentralized across the benzene ring.

As a result, C₆H₅CH₂NH₂ happens to be more base compared to C₆H₅NH₂ due to the electrons that exist on the atom of nitrogen, which is easier to acquire for protonation to occur in the former case.

Once more, the density of electrons on the N-atom in C₆H₅CH₂NH₂ is relatively lower compared to the one present on the N-atom in (CH₃)₃N due to the −I impact caused by the C₆H₅ molecular compound. Consequently, compared to C₆H₅CH₂NH₂, (CH₃)₃N remains more basic. As a result, the following is an arrangement for rising basic potency for the substances mentioned.

C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃N < (CH₃)₂NH < CH₃NH₂

9.5

9.6

9.7

Ans –

9.8

9.9

Exercise Questions – Class 12 Chemistry Chapter 9 Amines NCERT Solutions

9.1

Ans – (i) Propan-2-amine(1°)
(ii) Propan-1-amine (1°),
(iii) N-Methylpropan-2-amine (2°).
(iv) 2-Methylpropan-2-amine(l°)
(v) N-Methylbenzenamine or N-methylaniline(2°)
(vi) N-Ethyl-N-methylethanamine (3°)
(vii) 3-Bromobenzenamine or 3-bromoaniline (1°)

9.2

Ans – (i) Methylamine and methylamine can be distinguished by using the carbylamine test.

(ii) Secondary and tertiary amine can be distinguished by using the Liebermann’s nitrosamine test.

(iii) Ethylamine and aniline can be distinguished by using the azo test

(iv) Aniline and benzylamine can be distinguished by using the nitrous acid test.

(v) Aniline and N-methylaniline can be distinguished by using the carbylamine test.

9.3

Ans – (i) The previous N-atom’s singular set of electrons in order becomes decentralized across the benzene ring within the case of aniline. The nitrogen atoms of electrons therefore drop. In contrast, the −CH₃ group’s +I action in CH₃NH₂ raises the N-atom’s abundance of electrons. Consequently, aniline has a greater pK_b ratio compared to methylamine since it constitutes a less effective base material.

(ii) The H-bond between molecules causes ethylamine to dissociate in fluids. Nevertheless, aniline has become impermeable in water because of its significant hydrophobic portion, or hydrocarbon portion, which results in relatively little H-bonding.

(iii) Since methylamine is more fundamental compared to water, it may take a proton coming from it and release OH^– ions. When these OH^– ions mix together with the Fe⁺³ ions in H₂O, a brown compound of properly hydrated ferric oxide is created.

(iv) Typically, a combination of conc HNO³⁺ conc H₂SO₄ is used for nitration. The majority of the aniline is protonated to create the anilinium ion when acidic substances are present. Meanwhile, aniline alongside anilinium ions makes up the outcome of the solution when acidic substances are readily available in the compound.
At this point, the element that gained NH³⁺ ions within the anilinium-charged particles might tend to be eliminating & m-directing, whereas the −NH₂ compound around aniline gets activated alongside o, p-directing: Because of steric resistance at the o-region, aniline nitration mostly yields p-nitroaniline, while anilinium ion nitration yields m-nitroaniline.
In real life, p-nitroaniline & m-nitroaniline are mixed roughly around a 1:1 ratio. As a result of the chain of amino acids being protonated, nitration of aniline yields a significant quantity of m-nitroaniline.

(v) The study of organic chemistry method called the Friedel-Crafts reaction adds substituted molecules to aromatic ring structures. It’s also defined as an electrophilic aromatic replacement process (EAS). A substance that is salty when created due to the presence of aniline, a base gained from the Lewis formula gets merged with the Lewis acid AlCl₃.
It acts as a robust deactivating reagent for an electrophilic replacement mechanism, created when the N of aniline attains a positive value. Therefore, Aniline doesn’t experience the Friedel Crafts response.
C₆H₅NH₂ + AlCl₃ → C₆H₅NH₂⁺AlCl₃⁻

(vi) Because of resonant frequencies, the positive electrical charge around the benzene ring is dispersed, making the diazonium salts containing aromatic amines of greater stability as opposed to those containing aliphatic amines.

(vii) The Gabriel synthesis is a type of chemical process that converts primary alkyl halides to basic amines. Historically, potassium phthalimide can be employed in this process. In the field of organic chemistry, it’s a vital aspect of developing amine-based substances. Siegmund Gabriel, a German chemist, gave the process its name. Clean primary amines are produced using Gabriel’s phthalimide responses, free of both tertiary & secondary amine interference. As a result, it is recommended during the synthesis of primary amines.

9.4

Ans – (i) While there are two −C₂H₅ compounds present in (C₂H₅)₂NH, there is one sole pair of compounds available in C₂H₅NH₂. As a result, (C₂H₅)₂NH has a greater +I impact as opposed to C₂H₅NH₂ & (C₂H₅)₂NH consists of a maximum electron density throughout the entire N-atom compared to C₂H₅NH₂. Therefore, (C₂H₅)₂NH remains more basic in nature as compared to C₂H₅NH₂.

Additionally, because the only pair of molecules in C₆H₅NHCH₃ & C₆H₅NH₂ is delocalized, they are both more fundamental than (C₂H₅)₂NH & C₂H₅NH₂. Furthermore, due to the significant presence of the −CH₃ group’s +I action, C₆H₅NHCH₃ is expected to be relatively much easier compared to C₆H₅NH₂. Therefore, the below factor will be the sequence in which the basic features of supplied molecular compounds get increased:
C₆H₅NH₂ < C₆H₅NHCH₃ < C₂H₅NH₂ < (C₂H₅)₂NH

We are aware that pK_b readings will decrease while increasing the basic force.
C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH

(ii) (C₂H₅)₂NH tends to be highly basic as opposed to CH₃NH₂, mainly because of the larger +I impact of two separate −C₂H₅ molecules over a single −CH₃ unit.  The individual pair of charged particles that surround the N-atom become dispersed around the benzene ring throughout both C₆H₅NH₂ & C₆H₅N(CH₃)₂, but the +I impact of the 2 −CH₃ groupings makes C₆H₅N(CH₃)₂ a more fundamental compound.
(C₂H₅)₂NH > CH₃NH₂ > C₆H₅N(CH₃)₂ > C₆H₅NH₂

(iii) a) The fundamental stability of amines reduces when the electron-withdrawing NO₂ compound is present, but the electron-donating −CH₃ compound rises.
p−nitroaniline < aniline < p−toluidine

(iii) b) The element nitrogen is joined immediately with the benzene ring around C₆H₅NH₂ & C₆H₅NHCH₃. The N-atom’s single set of electrons is thus decentralized underneath the benzene ring. Consequently, compared to C₆H₅CH₂NH₂, both C₆H₅NH₂ & C₆H₅NHCH₃ are less effective bases, among others. With regard to the +1 impact of the −CH₃ category, C₆H₅NHCH₃ is considered a far more fundamental compound as opposed to C₆H₅NH₂.
C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂

(iv) The solvation consequences, especially the stability of the conjugate acids owing to H-bonding, cannot be observed in the gas state or in solvents that are not water-based like chlorobenzene, etc. As a result, the alkyl compounds’ +I action is the sole element influencing basic potency. The more pairs of alkyl groups there are the greater the +I impact. Therefore, in the gas period, the proper sequence for reducing basic intensity has become,
(C₂H₅)₃N > (C₂H₅)₂N > C₂H₅NH₂ > NH₃

(v) Alcohols generate more robust H-bonds compared to amines because oxygen has an increased electronegativity as opposed to nitrogen. Furthermore, the degree of H-bonding is higher in basic amines compared with secondary amines because it is dependent on the amount of H-atoms surrounding the N-atom.
(CH₃)₂NH₂ < C₂H₅NH₂ < C₂H₅OH

(vi) As amino acids’ molecular weight increases, their solubility reduces given that the water-resistant hydrocarbon portion is bigger and since there are fewer hydrogen atoms adjacent to the N-atom that form H-bonds.
C₆H₅NH₂ < (C₆H₅)₂NH < C₂H₅NH₂

9.5

Ans –

9.6

Ans – The Heinsberg method can be utilized to differentiate between all three categories of amines that exist. The procedure involves shaking the amine using benzene sulphonyl chloride (C₆H₅SO₂Cl) when there is too much aqueous NaOH or KOH present. An impenetrable substance is produced when a primary amine interacts to produce a transparent liquid that becomes more acidic.

An impervious chemical substance is created by a secondary amine and stays that way regardless of being acidified.

A ternary amine evaporates forming an acidic substance instead of reacting against the chemical reagents.

9.7

Ans – (i) Carbylamine reaction consists of the inclusion of amine in the middle reaction developed from the dehydrohalogenation of chloroform substances. This intermediate chain reaction is defined as dichlorocarbene. This process is called Hofmann isocyanide synthesis or carbylamine reaction. It’s a fundamental assessment to find the composition of primary amines-based compounds. When heated with chloroform & an alcohol-based KOH aqueous solution both aliphatic & fragrant basic amines yield isocyanides or carbylamines, which possess a highly disagreeable smell.
R−NH₂ + CHCl₃ + 3KOH(alc) → R−N≡C + 3KCl + 3H₂O

(ii) In 1858, Peter Griess, a German-based industrial scientist, became the first to record this sort of reaction. Diazotization is a technique by which a primary aromatic amino molecule is changed to produce a diazonium salt. These diazonium salts are often made by reacting an aromatic amine combined with nitrous acid when an additional acid is present.
This procedure involves mixing a water-based solution that comprises sodium nitrite into a principal existing component of aromatic amine (such as aniline) fluid that contains a substantial amount of HCl components around an approximate temperature lower than 5°C.

(iii) August Wilhelm Von Hoffmann has successfully presented Hoffmann’s explanation of the bromamide degradation process. The Hoffmann Rearrangement Reaction is another name given to this process, that is deployed to generate primary amines. To penetrate the amide compound & cause deprotonation followed by the resulting formation of an anion, the Hoffmann bromamide reaction process often involves the application of an alkali representing a powerful base.
During the creation, the basic amine consists of one less carbon atom, as opposed to the primary amide. The subsequent occurrence happens whenever an amide undergoes treatment alongside bromine under an alkali-based environment. This phenomenon is defined as Hoffmann’s bromamide degradation process.

(iv) A chemical-based event known as a coupling reaction occurs when a new molecular structure merges with two or more molecular components. In the field of organic chemistry, it’s an essential process that turns basic compounds into more complicated ones. Luminously coloured azo-based substances are created when arenediazonium salt combines alongside either phenol (in an alkaline solution) or aromatic amino acid compounds (in a slightly acidic environment).
In most cases, the chain reaction occurs within the para location that contains an amino or hydroxyl category in the protein. The ortho region is where the para slot is immobilized, and no interaction arises when both the ortho & para locations are engaged.

(v) Ammonolysis is a detailed reaction mechanism where ammonia is deployed in a process that incorporates chemicals to develop compounds such as amines or nitrides. This phenomenon happens only when an amino group (NH₂) gets linked along with ammonia (NH₃) under an ethanolic solution by replacing the presence of halogen atoms with an alkyl or benzyl halide compound.
Ammonolysis additionally refers to the method by which a group of amino substances replaces the hydroxyl group within alcoholic substances (or phenols) or the halogen atom around alkyl halides (or aryl halides). Alcoholic ammonia was the compound employed during the ammonolysis process. Tertiary, secondary & primary amines typically interact while producing a complex combination.
These methods can additionally employ hydroammonolysis, which uses the hydrogenation process as a trigger movement alongside an ammonia-hydrogen combination to seamlessly generate amines via carbonyl compounds without any external factors.

(vi) According to the internationally recognized IUPAC terminology, ethanoylation refers to the chemical process of acetylation. It explains the process by which a chemical molecule acquires an acetyl group with functional properties. Deacetylation, or the elimination of the acetyl category, is a reverse chemical process.
This can also be characterized when an acetyl group (CH₃C=O component) in a molecule is replaced with an atom of hydrogen. Acetoxy functional groups are commonly found in the results of acetylation processes. Acetylation is an approach in which one employs acetyl chloride or acetic anhydride to add another acetyl (CH₃CO−) compound to an existing molecule.

(vii) A German chemist named Siegmund Gabriel gave this process as Gabriel synthesis. It’s a mode of chemical process that converts primary alkyl halides to basic amines. Historically, potassium phthalimide is used in the process.
In the field of organic chemistry, it’s a crucial process for making amine substances. Pure aliphatic along with aralkyl amines that are primary may be prepared using this procedure. Potassium phthalimide is produced when a phthalimide substance has been treated using ethanolic KOH.
When heated with an appropriate alkyl or aralkyl halide molecular compound, N-substituted phthalimides are produced, which hydrolyse with either HCl or an alkaline solution to produce the principal amines.

9.8

9.9

Ans – (i) A = CH₃CH₂CN, B = CH₃CH₂CONH₂, C = CH₃CH₂NH₂ 
(ii) A = C₆H₅CN, B = C₆H₅COOH, C = C₆H₅CONH₂
(iii) A = CH₃CH₂CN, B = CH₃CH₂CH₂NH₂, C = CH₃CH₂CH₂OH 
(iv) A = C₆H₅NH₂, B = C₆H₅N₂Cl⁻, C = C₆H₅OH 
(v) A = CH₃CONH₂, B = CH₃NH₂, C = CH₃OH
(vi) A = C₆H₅NH₂, B = C₆H₅N₂⁺Cl⁻,

9.10

Ans – Component “B” should be an amide along with component “C” which is also amine similar to component C, which has the chemical structure C₆H₇N. This substance is produced from component B upon processing with Br₂ & KOH compounds. Aniline is the sole known amine with the chemical structure C₆H₇N, or C₆H₅NH₂.
Considering “C” represents an aniline, benzamide (C₆H₅CONH₂) must belong to the amide compound that is used when it comes into existence. Component “B” is therefore benzamide. Because component ‘B’ is created from a substance created using liquid ammonia by heating. Thus, component ‘A’ has to contain benzoic acid by definition.

9.11

9.12

Ans – The phthalimide anion turns nucleophilic when there is an assault against the organic compounds of halogen molecules is what makes Gabriel’s phthalimide synthesis successful. The Gabriel phthalimide process cannot be used to create arylamines, or aromatic primary amines, given that aryl halides are very difficult to serve as substitutes nucleophilically.

9.13

Ans – Between 273-278 K, aliphatic & aromatic main amines interact using HNO₂ to produce aliphatic & aromatic diazonium sodium chloride, correspondingly. Nevertheless, when exposed to this extremely low temperature, aliphatic diazonium salts become volatile and easily break down into a variety of chemical substances. The following is the reaction between HNO₂, highly aromatic & aliphatic principal amines.

9.14

Ans – (i) An amide ion is produced when an amine loses its proton charge, whereas an alkoxide ion is produced when a substance like alcohol loses its proton charge.
R−NH₂ → R−NH⁻ + H⁺
R−O−H → R−O− + H⁺
O is likely to draw positively charged substances with greater force to N due to the fact it happens to be more of an electronegative compound. RO is considered extremely stable as opposed to RNH⁻ as a result. Alcohols therefore have a higher acidity compared to amines. Alcoholic compounds are considerably more acidic instead of amines as a result.

(ii) The solution is that O is going to draw positive creatures far more effectively as opposed to N because the former happens to be more electronegative in nature. Meanwhile, RNH^– seems more volatile compared to RO. Alcohols therefore have a higher acidity rather than amines. On the other hand, alcoholic substances are more acidic in nature compared to amines.

(iii) For a few different causes, amines that are aromatic are significantly less basic as opposed to aliphatic & ammonia amines-based compounds. It includes,

(a) The isolated pair of electron particles present inside the nitrogen atom becomes decentralized around the benzene ring because of the resonating frequency experienced in aniline along with other aromatic amines, making it less readily accessible for protonation to occur Because of this consequence, aliphatic amines & ammonia are somewhat stronger bases compared to aromatic amines.
(b) Aromatic amine compounds have a lower basic compound as opposed to comparable protonated charged particles simply because they have a greater persistence while exhibiting an extremely small tendency to interact with another proton to generate similar protonated ionic substances.

Related Study Resources of Chapter 9 – Amines

Students can use the links below to get extra study materials for Class 12 Chemistry Chapter 5: Coordination Compounds

Sl No.	Related Links
1	Class 12 Chemistry Chapter 9 Amines – Important Questions
2	NCERT Official Textbook (PDF download)

Chapter Overview: Amines – Class 12 NCERT Chemistry

Class 12 NCERT Chemistry’s Chapter 9, “Amines,” teaches students a crucial class of organic molecules derived from ammonia. Both industrial and biological chemistry depend on these nitrogen-containing molecules in great part. A key component of the Class 12 Chemistry syllabus, knowledge of the structure, classification, preparation, and chemical behaviour of amines is particularly crucial for students getting ready for competitive tests like NEET and JEE as well as CBSE board exams.

Beginning with the classification of amines into primary, secondary, and tertiary kinds depending on the number of alkyl or aryl groups bonded to the nitrogen atom, the chapter opens A basic idea in organic chemistry, students also pick up the IUPAC nomenclature of amines. A thorough review of several techniques of amine preparation—including reduction of nitro compounds, ammonolysis of alkyl halides, Gabriel phthalimide synthesis, and Hoffmann bromamide degradation—follows here. Mechanisms help to describe these reactions so that a strong conceptual basis—which is essential for good Class 12 Chemistry board exam performance—may be developed.

The chapter also covers the physical and chemical characteristics of amines, together with their basic character, solubility, boiling temperatures, and interactions with acids and alkyl halides. Particularly important is the ability to differentiate among several kinds of amines using chemical testing, which frequently show up on board and admission tests. This chapter is not only theoretical but also useful in practical chemistry since it clarifies electronic effects, hydrogen bonding, and steric variables, therefore enabling students to grasp the behaviour of amines in various situations.

Students are first exposed to diazonium salts—especially benzene diazonium chloride—and their importance in organic synthesis at the close of the chapter. From an application point of view, the production and reactions of diazonium salts—including Sandmeyer and Gattermann reactions—are crucial and often questioned on CBSE exam papers. For any serious Chemistry student, the chapter on amines offers a thorough knowledge of one of the most crucial functional groups in organic chemistry, so it is absolutely essential.

Benefits of Using These NCERT Solutions

Using these Amines Class 12 NCERT Solutions gives students a disciplined method for learning one of the most crucial organic chemistry topics. Every answer is designed according to the most recent CBSE Class 12 Chemistry syllabus, therefore guaranteeing that pupils remain in line with what is anticipated on board tests. Our NCERT Solutions for Class 12 Chemistry Chapter 9 – Amines offer detailed step-by-step explanations to develop a strong conceptual basis whether your focus is on intext questions or back exercise problems.

For last-minute revision, these Amines Class 12 NCERT Solutions are very helpful since they succinctly and clearly outline the significant reactions, preparation techniques, and properties. Students can understand difficult subjects such Hoffmann degradation, Gabriel phthalimide synthesis, and electronic influences influencing basicity by means of thorough responses and well stated methods. Particularly while getting ready for competitive tests like NEET and JEE, students increase their writing speed, accuracy, and confidence by routinely using these Class 12 NCERT Chemistry Amines Solutions.

Furthermore carefully chosen by subject specialists, these NCERT Solutions for Amines Class 12 guarantee accuracy and clarity. Exam-oriented and plagiarism-free, the material guides students toward both theory-based and application-based answers. These Class 12 Amines NCERT Solutions are absolutely essential for your study if you want perfect grades in organic chemistry.