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NCERT Exemplar Class 12 Chemistry Solutions chapter 13 Amines

NCERT Exemplar Class 12 Chemistry Solutions chapter 13 Amines

Edited By sumit saini | Updated on Sep 17, 2022 11:10 AM IST | #CBSE Class 12th

NCERT exemplar Class 12 Chemistry solutions chapter 13 Anime - Do you wonder how medicines or fibres are synthesized? What does cough syrups like Benadryl contain? NCERT exemplar Class 12 Chemistry chapter 13 solutions, answers questions like these and in turn defines Amines as a vital group of organic compounds derived by replacing some hydrogen atoms in ammonia molecule with alkyl or aryl groups, which often occur naturally in vitamins, proteins, hormones, and alkaloids. Class 12 Chemistry NCERT exemplar solutions chapter 13 discusses the structure and geometry of amines, the classification of amines as primary, secondary, and tertiary, the naming of amines, methods to prepare them, the properties of amines, the process of formation of diazonium salts, and their importance. All of which are detailed in the NCERT exemplar Class 12 Chemistry solutions chapter 13 pdf download

This Story also Contains
  1. NCERT Exemplar Class 12 Chemistry Solutions Chapter 13: MCQ (Type 1)
  2. NCERT Exemplar Class 12 Chemistry Solutions Chapter 13: MCQ (Type 2)
  3. NCERT Exemplar Class 12 Chemistry Solutions Chapter 13: Short Answer Type
  4. NCERT Exemplar Class 12 Chemistry Solutions Chapter 13: Matching Type
  5. NCERT Exemplar Class 12 Chemistry Solutions Chapter 13: Assertion and Reason Type
  6. NCERT Exemplar Class 12 Chemistry Solutions Chapter 13: Long Answer Type
  7. Main Subtopics of NCERT Exemplar Solutions for Class 12 Chemistry Chapter 13 Amines
  8. NCERT Exemplar Class 12 Chemistry Solutions Chapter 13 Amines - Learning Outcome
  9. NCERT Exemplar Class 12 Chemistry Solutions
  10. Important Topics Covered in NCERT Exemplar Solutions For Class 12 Chemistry Chapter 13

NCERT Exemplar Class 12 Chemistry Solutions Chapter 13: MCQ (Type 1)

Question:1

Which of the following is a 3^{o} amine?
(i) 1-methylcyclohexylamine
(ii) Triethylamine
(iii) tert-butylamine
(iv) N-methyl aniline
Answer:

Option (ii) is the correct answer.

Question:2

The correct IUPAC name for CH_{2}=CHCH_{2} NHCH_{3} is
(i) Allylmethylamine
(ii) 2-amino-4-pentene
(iii) 4-aminopent-1-ene
(iv) N-methylprop-2-en-1-amine
Answer:

Option (iv) is the correct answer.

Question:3

Amongst the following, the strongest base in aqueous medium is ____________.
(i) CH_{3}NH_{2}
(ii) NCCH_{2}NH_{2}
(iii) (CH_{3})_{2}NH
(iv) C_{6}H_{5}NHCH_{3}
Answer:

Option (iii) is the correct answer.
Explanation: (CH_{3})_{2}NH is a 2^{o} amine and CH_{3}NH_{2} is a 1^{o} amine, therefore (CH_{3})_{2}NH is more basic than CH_{3}NH_{2}. And because of –I effect of CN group, NC-CH_{2}NH_{2} is less basic than CH_{3}NH_{2}. Moreover, C_{6}H_{5}NHCH_{3} is the least basic and less basic than CH_{3}NH_{2} and (CH_{3})_{2}NH; it is due to delocalisation of lone pair electrons that are present in nitrogen atom into the benzene ring. Therefore, the decreasing order of amines will be:
\left ( CH_{3} \right )_{2}NH>CH_{3}NH_{2}>C_{6}H_{5}NHCH_{3}>NC-CH_{2}NH_{2}

Question:4

Which of the following is the weakest Brönsted base?

Answer:

Option (A) is the correct answer.
Explanation: The lone pair of electrons on the N-atom has delocalised into the benzene ring; therefore C_{6}H_{5}NH_{2} s the weakest base.

Question:5

Benzylamine may be alkylated, as shown in the following equation:
C_{6}H_{5}CH_{2}NH_{2} + R-X \rightarrow C_{6}H_{5}CH_{2}NHR
Which of the following alkyl halides is best suited for this reaction through S_{N}1 mechanism?
(i) CH_{3}Br
(ii) C_{6}H_{5}Br
(iii) C_{6}H_{5}CH_{2}Br
(iv) C_{2}H_{5}Br
Answer:

Option (iii) is the correct answer.
Explanation:S_{N}1 is a two-step reaction. At first, R-X bond is broken, which produces carbocation. Then nucleophile attacks the carbocation. The more the stability of carbocation, the more will be the rate of reaction. Benzylic halides are highly reactive towards S_{N}1 reactions.

Question:6

Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine?
(i) \text {H}_{2} (excess)/Pt
(ii) \text {LiAIH}_{4} in ether
(iii) \text {Fe} and \text {HCl}
(iv) \text {Sn} and \text {HCl}
Answer:

Option (ii) is the correct answer.
Explanation: Using \text {LiAIH}_{4} in ether, Aryl-nitro compounds can't convert into an amine.

Question:7

In order to prepare a 1° amine from an alkyl halide with simultaneous addition of one \text {CH}_{2} group in the carbon chain, the reagent used as source of nitrogen is ___________.
(i) Sodium amide, \text {NaNH}_{2}
(ii) Sodium azide, \text {NaN}_{3}
(iii) Potassium cyanide, \text {KCN}
(iv) Potassium phthalimide, \text {C}_{6}\text {H}_{4}\left ( \text {CO}_{2} \right )\text N^{-}K^{+}

Answer:

Option (iii) is the correct answer.
Explanation: The number of carbon atoms increases with the help of \text {KCN}

Question:8

The source of nitrogen in Gabriel synthesis of amines is _____________.
(i) Sodium azide, \text {NaN}_{3}
(ii) Sodium nitrite, \text {NaNO}_{2}
(iii) Potassium cyanide, \text {KCN}
(iv) Potassium phthalimide, \text {C}_{6}\text {H}_{4}\left ( \text {CO} \right )_{2}\text N^-K^+
Answer:

Option (iv) is the correct answer.
Explanation: In Gabriel's synthesis, Potassium Phthalimide is the source of nitrogen.

Question:9

Amongst the given set of reactants, the most appropriate for preparing 2° amine is _____.
(i) 2° \text {R-Br+NH}_{3}
(ii) 2° \text {R-Br+NaCN} followed by \text {H}_{2}/\text {Pt}
(iii) 1° \text {R-NH}_{2}+\text {RCHO} followed by \text {H}_{2}/\text {Pt}
(iv) 1° \text {R-Br}\left ( \text {2 mol} \right )+ potassium phthalimide followed by \text {H}_{3}\text {O}^{+}/heat
Answer:

Option (iii) is the correct answer.

Question:10

The best reagent for converting 2–phenylpropanamide into 2-phenylpropanamine is _____.
(i) excess \text {H}_{2}
(ii) \text {Br}_{2} in aqueous \text {NaOH}
(iii) iodine in the presence of red phosphorus
(iv) \text {LiAlH}_{4} in ether
Answer:

Option (iv) is the correct answer.

Question:11

The best reagent for converting, 2-phenylpropanamide into 1- phenylethanamine is ____.
(i) excess \text {H}_{2}/\text {Pt}
(ii) \text {NaOH}/\text {Br}_{2}
(iii) \text {NaBH}_{4}/methanol
(iv) \text {LiAlH}_{4}/ether

Answer:

Option (ii) is the correct answer.

Question:12

Hoffmann Bromamide Degradation reaction is shown by __________.
(i) \text {ArNH}_{2}
(ii) \text {ArCONH}_{2}
(iii) \text {ArNO}_{2}
(iv) \text {ArCH}_{2}\text {NH}_{2}

Answer:

Option (ii) is the correct answer.

Question:13

The correct increasing order of basic strength for the following compounds is______________.

(i) II < III < I
(ii) III < I < II
(iii) III < II < I
(iv) II < I < III
Answer:

Option (iv) is the correct answer.
Explanation: The basic strength is decreased with electron-withdrawing groups, and it is increased with electron releasing groups of aniline.

Question:14

Methylamine reacts with \text {HNO}_{2} to form _________.
(i) \text {CH}_{3}\text {-O-N=O}
(ii) \text {CH}_{3}\text {-O-CH}_{3}
(iii) \text {CH}_{3}\text {OH}
(iv) \text {CH}_{3}\text {CHO}
Answer:

Option (iii) is the correct answer.
Explaination : (c)\; \text {CH}_{3}\text {NH}_{2}+\text {HNO}_{3}\overset{0^{o}-5^{o}C}{\rightarrow}\text {CH}_{3}\text {OH}+\text {N}_{2}+\text {H}_{2}\text {O}

Question:15

The gas evolved when methylamine reacts with nitrous acid is __________.
(i) \text {NH}_{3}
(ii) \text {N}_{2}
(iii) \text {H}_{2}
(iv) \text {C}_{2}\text {H}_{6}
Answer:

Option (ii) is the correct answer.
Explanation: The chemical reaction of methylamine with nitrous acid is as follows:

Question:16

In the nitration of benzene using a mixture of conc. \text {H}_{2}\text {SO}_{4} and conc. \text {HNO}_{3}, the species which initiates the reaction is __________.
(i) \text {NO}_{2}
(ii) \text {NO}^{+}
(iii) \text {NO}_{2}^+
(iv) \text {NO}_{2}^{-}
Answer:

Option (iii) is the correct answer.
Explanation: The process of nitration is initiated by \text {NO}_{2}^+ (Nitronium Ion) electrophile. It is then obtained as:
H_{2}SO_{4}\left ( conc. \right )\rightarrow H^{+}+HSO_{4}^{-}
H^{+}+HNO_{3}\rightarrow H_{2}NO_{3}^{+}
H_{2}NO_{3}^{+}\rightarrow NO_{2}^{+}+H_{2}O

Question:17

Reduction of aromatic nitro compounds using \text {Fe and HCl} gives __________.
(i) aromatic oxime
(ii) aromatic hydrocarbon
(iii) aromatic primary amine
(iv) aromatic amide
Answer:

Option (iii) is the correct answer.
Explanation :

Question:18

The most reactive amine towards dilute hydrochloric acid is ___________.

Answer:

Option (ii) is the correct answer.
Explanation: The reactivity towards dilute HCl is more if the strength of the base is more. Therefore,\left ( CH_{3} \right )_{2}NH has the highest basic strength.

Question:19

Acid anhydrides on reaction with primary amines give ____________.
(i) amide
(ii) imide
(iii) secondary amine
(iv) imine
Answer:

Option (i) is the correct answer.
Explanation:
\\(a)C_{2}H_{5}NH_{2}+\left ( CH_{3}CO \right )_{2}O\rightarrow CH_{3}CONHC_{2}H_{5}+CH_{3}COOH \\
N-Ethylacetamide

Question:20

The reaction ArN_{2}^+Cl^{-} \overset{Cu/HCl}{\rightarrow} ArCl + N_2+ CuCl is named as _________.
(i) Sandmeyer reaction
(ii) Gatterman reaction
(iii) Claisen reaction
(iv) Carbylamine reaction
Answer:

Option (ii) is the correct answer.
Explanation:

Question:21

The best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain is
(i) Hoffmann Bromamide reaction
(ii) Gabriel phthalimide synthesis
(iii) Sandmeyer reaction
(iv) Reaction with NH_{3}
Answer:

Option (ii) is the correct answer.
Explanation: To get primary amines from alkyl halide without changing the number of carbon atoms, Gabriel Phthalimide synthesis is used.

Question:22

Which of the following compound will not undergo an azo coupling reaction with benzene diazonium chloride.
(i) Aniline
(ii) Phenol
(iii) Anisole
(iv) Nitrobenzene
Answer:

Option (iv) is the correct answer.
Explanation: Diazonium Chloride is a very weak electrophile; therefore, it reacts with any electron rich compound which contains electron donating groups; -OH, -NO_{2}, -OCH_{3}. The compounds should not contain electron withdrawing groups like: -NO_{2}, etc.

Question:23

Which of the following compounds is the weakest Brönsted base?

Answer:

Option (iii) is the correct answer.
Explanation: All the amines have a tendency to accept protons, therefore are stronger Bronsted Bases than phenols and alcohols. And since phenol is much more acidic than alcohol, Phenols have fewer tendencies to accept any proton. Hence the weakest.

Question:24

Among the following amines, the strongest Brönsted base is __________.

Answer:

Option (iv) is the correct answer.
Explanation:NH_{3} is a stronger base than aniline because of the delocalisation of the lone pair of electrons at N-atom and into the Benzene Ring. And in Pyrrole, as the lone pair electrons on N-atom donate to aromatic sextet formation, it is not basic at all. Hence, Pyrrolidine accepts the proton readily while also being the strongest base.

Question:25

The correct decreasing order of basic strength of the following species is _______. H_{2}O, NH_{3}, OH^{-}, NH_{2}^{-}
(i)\; NH_{2}^{-}>OH^{-}>NH_{3}>H_{2}O
(ii)\; OH^{-}>NH_{2}^{-}>H_{2}O>NH_{3}
(iii)\;NH_{3}>H_{2}O>NH_{2}^{-}>OH^{-}
(iv)\;H_{2}O>NH_{3}>OH^{-}>NH_{2}^{-}
Answer:

Option (i) is the correct answer.
Explanation: The electronegativity of O is more than N atom; therefore, O-H bond is more polar than N-H bond. Therefore, O-H is more acidic than the N-H bond. NH_{2}^- and OH^-, both have a negative charge, and because of that, they are more basic than NH_{3} and H_{2}O.

Question:26

Which of the following should be most volatile?

(i) II
(ii) IV
(iii) I
(iv) III
Answer:

Option (ii) is the correct answer.
Explanation: The boiling points of 1° and 2° amines are higher than 3° amines because of intermolecular H-Bonding. They are also less volatile than 3° amines and hydrocarbons of similar molecular mass.

Question:27

Which of the following methods of preparation of amines will not give the same number of carbon atoms in the chain of amines as in the reactant?
(i) The reaction of nitrite with LiAlH_{4}.
(ii) The reaction of the amide with LiAlH_{4} followed by treatment with water.
(iii) Heating alkyl halide with potassium salt of phthalimide followed by hydrolysis.
(iv) Treatment of amide with bromine in the aqueous solution of sodium hydroxide.
Answer:

Option (iv) is the correct answer.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 13: MCQ (Type 2)

Question:28

Which of the following cannot be prepared by Sandmeyer's reaction?
(i) Chlorobenzene
(ii) Bromobenzene
(iii) Iodobenzene
(iv) Fluorobenzene
Answer:

Option (iii) and (iv) are the correct answers.
Explanation: Using Sandmeyer's reaction, Chloro and Bromo arenes are prepared. And by simply warming Diazonium salt solution with aqueous KI solution, Iodarenes can be prepared. Fluoroarenes are made using Balz-Schiemann reaction. All other reagents give off aniline.

Question:29

Reduction of nitrobenzene by which of the following reagent gives aniline?
(i) {Sn}/{HCl}
(ii) {Fe}/{HCl}
(iii) H_{2}-Pd
(iv) {Sn}/{NH_{4}OH}
Answer:

Option (i), (ii) and (iii) are the correct answers.

Question:30

Which of the following species are involved in the carbylamine test?
(i) R-NC
(ii) CHCl_{3}
(iii) COCl_{2}
(iv) NaNO_{2}+HCl
Answer:

Option (i) and (ii) are the correct answers.
Explanation: Amine when reacts with a mixture of \text {CHCl}_{3} and \text {KOH} gives out alkyl isocyanate. This reaction is called a Carbylamine reaction.
Here, \text {R-NH}_{2}+\text {CHCl}+\text {3KOH}\rightarrow \text {RNC}+\text {3KCl}+\text {3H}_{2}\text {O} Only \text {RNC} and \text {CHCl}_{3} are involved in carbylamine reaction.

Question:31

The reagents that can be used to convert benzene diazonium chloride to benzene are __________.
(i) {SnCl_{2}}/{HCl}
(ii) CH_{3}CH_{2}OH
(iii) H_{3}PO_{2}
(iv) LiAlH_{4}
Answer:

Option (ii) and (iii) are the correct answers.
Explanation :

Question:32

The product of the following reaction is __________.

Answer:

Option (A) and (B) is the correct answer.
Explanation :

Question:33

Arenium ion involved in the bromination of aniline is __________.

Answer:

Option (i), (ii) and (iii) are the correct answers.
Explanation: Bromination of aniline involves Arenium is as follows:

Question:34

Which of the following amines can be prepared by Gabriel synthesis.
\\(i) Isobutyl amine\\ (ii) 2-Phenylethylamine\\ (iii) N-methyl benzylamine\\ (iv) Aniline
Answer:

Option (i) and (ii) are the correct answers.
Explanation:(CH_{3})_{2}CH-CH_{2}NH_{2} and C_{6}H_{5}CH_{2}NH_{2} are the primary aliphatic amines that can be prepared by Gabriel synthesis. 2^{o} amines are C_{6}H_{5}CH_{2}NHCH_{3} (C) and 1^{o} amine, C_{6}H_{5}NH_{2} cannot be prepared by this process.

Question:35

Which of the following reactions are correct ?

Answer:

Options (i) and (iii) are the correct answers.
Explanation : CH_{3}CH_{2}NH_{2}+NH_{4}Cl

Question:36

Under which of the following reaction conditions, aniline gives p-nitro derivative as the major product?
(i) Acetyl chloride/pyridine followed by reaction with conc. H_{2}SO_{4} + conc. HNO_{3}
(ii) Acetic anhydride/pyridine followed by conc. H_{2}SO_{4} + conc. HNO_{3}
(iii) Dil. HCl followed by reaction with conc. H_{2}SO_{4} + conc. HNO_{3}
(iv) Reaction with conc. HNO_{3}+conc. H_{2}SO_{4}
Answer:

Option (i) and (ii) are the correct answers.
Explanation: When aniline reacts with acetyl chloride or acetic anhydride in the presence of pyridine produces N-acetyl aniline. This is a ortho, para directing group which produces p-nitroaniline on reaction with nitrating mixtures. As shown below:

Question:37

Which of the following reactions belong to electrophilic aromatic substitution?
(i) Bromination of acetanilide
(ii) Coupling reaction of aryldiazonium salts
(iii) Diazotisation of aniline
(iv) Acylation of aniline
Answer:

Option (i) and (ii) are the correct answers.
Explanation: Nucleophilic substitution reaction where -NH_{2} \; and\; H�NH_{2} \; and\; H atoms are replaced by acyl groups is known as Acylation. Diazotisation is also a type of nucleophilic substitution reaction.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 13: Short Answer Type

Question:38

What is the role of \text {HNO}_{3} in the nitrating mixture used for nitration of benzene?
Answer:

The 1:1 solution of \text {HNO}_{3} and \text {H}_{2}\text {SO}_{4} is a nitrating mixture, and it is used for nitration of organic compounds. This mixture acts as a base which provides electrophile in the nitration process of benzene.

Question:39

Why is \text {NH}_{2} group of aniline acetylated before carrying out nitration?
Answer:

The \text {NH}_{2} group of aniline is acetylated before carrying out the nitration for controlling the nitration reaction, carry oxidation products and nitro derivative products formation.
As the acetyl group is an electron-withdrawing group, it attracts the lone pair of the electron in the N atom towards the carbonyl group.

Question:40

What is the product when \text {C}_{6}\text {H}_{5}\text {CH}_{2}\text {NH}_{2} reacts with \text {HNO}_{2} ?
Answer:

\text {HNO}_{2} is reacted \text {C}_{6}\text {H}_{5}\text {CH}_{2}\text {NH}_{2} which forms unstable diazonium salt, and in turn, alcohol is given out. This is the reaction:
\text {C}_{6}\text {H}_{5}\text {CH}_{2}\text {NH}_{2}+\text {HNO}_{2}\rightarrow \text {C}_{6}\text {H}_{5}\text {CH}_{2}\text {OH}+\text {N}_{2}+\text {H}_{2}\text {O}

Question:41

What is the best reagent to convert nitrile to primary amine?

Answer:

\text {LiAlH}_{4} and Sodium/Alcohol are the best reagents for the conversion of nitrile to a primary amine. And nitriles can be converted into a corresponding primary amine using reduction.

Question:42

Give the structure of 'A' in the following reaction.

Answer:


3-Methylnitrobenzene is the product formed in the above reaction.

Question:43

What is Hinsberg reagent?

Answer:

\text {C}_{6}\text {H}_{5}\text {SOCl} aka Benzene sulphonylchloride is known as Hinsberg's reagent. It is used to differentiate between primary, secondary and tertiary amines.

Question:44

Why is benzene diazonium chloride not stored and is used immediately after its preparation?
Answer:

Benzene Diazonium is very stable at low temperatures and highly soluble in water at high temperatures. It is recommended to use it immediately after its preparation as it is unstable.

Question:45

Why does the acetylation of \text {-NH}_{2} group of aniline reduce its activating effect?
Answer:

The activating effect of \text {-NH}_{2} group of aniline is reduced by its acetylation. It happens because of the lone pair electrons on the nitrogen of acetanilide react with oxygen atom because of resonance.

Question:46

Explain why \text {MeNH}_{2} is a stronger base than \text {MeOH} ?
Answer:

\text {MeOH} is a weaker base than \text {MeNH}_{2} because \text {MeNH}_{2} has lower electronegativity and has lone pair electrons present on the nitrogen atom.

Question:47

What is the role of pyridine in the acylation reaction of amines?
Answer:

By acylation of \text {-NH}_{2} group with acetic anhydride in the presence of pyridine and after that carrying substitution followed with hydrolysis of the substituted amide with the substituted amine, the activating effect of \text {-NH}_{2} group can be controlled. \text {HCl} is a side product, which is removed with the help of pyridine as a base.

Question:48

Under what reaction conditions (acidic/basic), the coupling reaction of aryldiazonium chloride with aniline is carried out?
Answer:

The reaction is performed in a mild basic medium and known as electrophilic substitution reaction. Aniline and Aryldiazonium chloride react to form a yellow dye of p-Aminoazobenzene.

Question:49

Predict the product of the reaction of aniline with bromine in a non-polar solvent such as \text {CS}_{2}.
Answer:

When aniline and bromine react in a non-polar solvent, such as \text {CS}_{2}, non-polar products are formed that are 4-Bromoaniline and 2-Bromoaniline in which 4-Bromoaniline is present in the majority.

Question:50

Arrange the following compounds in increasing order of dipole moment.
\text {CH}_{3}\text {CH}_{2}\text {CH}_{3},\text {CH}_{3}\text {CH}_{2}\text {NH}_{2},\text {CH}_{3}\text {CH}_{2}\text {OH}
Answer:

\text {CH}_{3}\text {CH}_{2}\text {CH}_{3}<\text {CH}_{3}\text {CH}_{2}\text {NH}_{2}<\text {CH}_{3}\text {CH}_{2}\text {OH}
\text {CH}_{3}\text {CH}_{2}\text {OH} has dipole moment greater than \text {CH}_{3}\text {CH}_{2}\text {NH}_{2}. Whereas,\text {CH}_{3}\text {CH}_{2}\text {CH}_{3} has the least dipole moment amongst the three compounds. \text {CH}_{3}\text {CH}_{2}\text {CH}_{3} is an almost non-polar molecule.

Question:51

What is the structure and IUPAC name of the compound, allyl amine?
Answer:

The structure of allyl amine is and its IUPAC name is prop-2-en-1-amine.

Question:52

Write down the IUPAC name of

Answer:

IUPAC name of N, N-Dimethylbenzenamine

Question:53

A compound Z with molecular formula \text {C}_{3}\text {H}_{9}\text {N} reacts with \text {C}_{6}\text {H}_{5}\text {SO}_{2}\text {Cl} to give a solid, insoluble in alkali, identify Z.
Answer:

Solid, insoluble alkali is given off when compound Z with molecular formula \text {C}_{3}\text {H}_{9}\text {N} and is an aliphatic amine on treatment with \text {C}_{6}\text {H}_{5}\text {SO}_{2}\text {Cl}. And hence, the product does not have any replaceable hydrogen on its nitrogen atom. The amine (Z) should be a secondary amine which means Z is ethyl methylamine (\text {C}_{2}\text {H}_{5}\text {NHCH}_{3}).

Question:54

A primary amine, \text {RNH}_{2} can be reacted with \text {CH}_{3}-\text {X} to get secondary amine, \text {RNHCH}_{3}, but the only disadvantage is that 3^{o} amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where \text {RNH}_{2} forms only 2^{o} amine?
Answer:

\\\text {R}-\text {NH}_{2}+\text {CHCl}_{3}+\text {3KOH}\rightarrow \text {R}-\text {NC}\xrightarrow[\text {(Reduction)}]{\text {H}_{2}/\text {Pd}}\text {R}-\text {NH}-\text {CH}_{3}\\\; 1^{o}\text {Amine}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \text {Alkyl isocyanide} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 2^{o}\text {Amine}
Carbylamine reaction is shown by 1^{o} amine only which result in the replacement of two hydrogen atoms attached to N atom of -\text {NH}_{2} group by one carbon atom to form isocyanide. On catalytic reduction, the isocyanide will give a secondary amine with one methyl group.

Question:55

Complete the following reaction.

Answer:

Question:56

Why is aniline soluble in aqueous \text {HCl} ?
Answer:


Anilinium chloride salt is formed when aniline reacts with aqueous \text {HCl}, and it is soluble in water.

Question:57

Suggest a route by which the following conversion can be accomplished.

Answer:

Question:58

Identify A and B in the following reaction.

Answer:

Question:59

How will you carry out the following conversions?
(i) toluene \rightarrow p-toIuidine
(ii) p-toluidine diazonium chloride \rightarrow p-toluic acid
Answer:

Question:60

Write the following conversions:
(i) nitrobenzene \rightarrow acetanilide
(ii) acetanilide \rightarrow p-nitroaniline
Answer:

Question:61

A solution contains 1 g mol each of p-toluene diazonium chloride and p-nitrophenyl diazonium chloride. To this 1 g mol of alkaline solution of phenol is added. Predict the major product. Explain your answer.
Answer:

This reaction is an example of electrophilic aromatic substitution. Phenol forms phenoxide ion in an alkaline medium, which is more electron-rich than phenol and therefore more reactive in the electrophilic attack. Arydiazonium cation is electrophile in this reaction, and p-Nitrophenyldiazonium cation is a stronger electrophile than p-toluene diazonium cation. Hence, it pairs preferentially with phenol.

Question:62

How will you bring out the following conversion?

Answer:

Question:63

How will you carry out the following conversion?

Answer:

Question:64

How will you carry out the following conversion?

Answer:

Question:65

How will you carry out the following conversion ?

Answer:

NCERT Exemplar Class 12 Chemistry Solutions Chapter 13: Matching Type

Question:66

Match the reactions given in Column I with the statements given in Column II.

Column I

Column II

(i)

Ammonolysis

(a)

Amine with lesser number of carbon atoms

(ii)

Gabriel phthalimide synthesis

(b)

Detection test for primary amines

(iii)

Hofmann Bromamide reaction

(c)

Reaction of phthalimide with \text {KOH} and \text {R-X}

(iv)

Carbylamine reaction

(d)

Reaction of alkyl halides with \text {NH}_{3}


Answer:

(i \rightarrow d), (ii \rightarrow c), (iii \rightarrow a), (iv \rightarrow b)

Question:67

Match the compounds given in Column I with the items given in Column II.

Column I

Column II

(i)

Benzene sulphonyl Chloride

(a)

Zwitter ion

(ii)

Sulphanilic acid

(b)

Hinsberg reagent

(iii)

Alkyl diazonium salts

(c)

Dyes

(iv)

Aryl diazonium salts

(d)

Conversion to alcohols

Answer:

(i \rightarrow b), (ii \rightarrow a), (iii \rightarrow d), (iv \rightarrow c)

NCERT Exemplar Class 12 Chemistry Solutions Chapter 13: Assertion and Reason Type

Question:68

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Acylation of amines gives a monosubstituted product whereas alkylation of amines gives polysubstituted product.
Reason (R): Acyl group sterically hinders the approach of further acyl groups.
(i) Both assertion and reason are wrong.
(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.
(v) Assertion is wrong statement but reason is correct statement.

Answer:

(iii) Assertion is correct statement but reason is wrong statement.

Question:69

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Hoffmann's bromamide reaction is given by primary amines.
Reason (R): Primary amines are more basic than secondary amines.
(i) Both assertion and reason are wrong.
(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.
(v) Assertion is wrong statement but reason is correct statement.
Answer:

(iii) Assertion is correct statement but reason is wrong statement.

Question:70

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): N-Ethylbenzene sulphonamide is soluble in alkali.
Reason (R): Hydrogen attached to nitrogen in sulphonamide is strongly acidic.
(i) Both assertion and reason are wrong.
(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.
(v) Assertion is wrong statement but reason is correct statement.
Answer:


Ethylbenzene sulphonamide is soluble in alkali because it has acidic hydrogen.
Hence (iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.

Question:71

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): N, N-diethyl benzene sulphonamide is insoluble in alkali.
Reason (R): Sulphonyl group attached to the nitrogen atom is a strong electron-withdrawing group.
(i) Both assertion and reason are wrong.
(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.
(v) Assertion is wrong statement but reason is correct statement.
Answer:

N, N-diethyl benzene sulphonamide is not soluble in alkali as it there is no acidic hydrogen present on this. Sulphonyl group is an electron-withdrawing group which is attached to a nitrogen atom.
Hence, (ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.

Question:72

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Only a small amount of \text {HCl} is required in the reduction of nitro compounds with iron scrap and \text {HCl} in the presence of steam.
Reason (R):\text {FeCl}_{2} formed gets hydrolysed to release \text {HCl} during the reaction.
(i) Both assertion and reason are wrong.
(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.
(v) Assertion is wrong statement but reason is correct statement.
Answer:

\text {Fe}+\text {2HCl}\rightarrow \text {FeCl}_{2}+\text {2[H]}
Nacent hydrogen is reduced into nitro compounds.
\text {FeCl}_{2}+\text {H}_{2}\text {O}\text {(g)}\rightarrow \text {FeO}+\text {2HCl}
Hence, (iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.

Question:73

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Aromatic 1^{o} amines can be prepared by Gabriel phthalimide synthesis.
Reason (R): Aryl halides undergo nucleophilic substitution with anion formed by phthalimide.
(i) Both assertion and reason are wrong.
(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.
(v) Assertion is wrong statement but reason is correct statement.
Answer:

(i) Both assertion and reason are wrong

Question:74

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Acetanilide is less basic than aniline.
Reason (R): Acetylation of aniline results in the decrease of electron density on nitrogen.
(i) Both assertion and reason are wrong.
(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.
(v) Assertion is wrong statement but reason is correct statement.
Answer:

(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 13: Long Answer Type

Question:75

A hydrocarbon 'A', \left ( \text {C}_{4}\text {H}_{8} \right ) on reaction with \text {HCl} gives a compound 'B' \left ( \text {C}_{4}\text {H}_{9}\text {Cl} \right ), which on reaction with 1 mol of \text {NH}_{3} gives compound 'C',\left ( \text {C}_{4}\text {H}_{11}\text {N} \right ) . On reacting with \text {NaNO}_{2} and \text {HCl} followed by treatment with water, compound 'C' yields optically active alcohol, 'D'. Ozonolysis of 'A' gives 2 moles of acetaldehyde. Identify compounds' A' to 'D'. Explain the reactions involved.
Answer:

(i) Addition of \text {HCl} to compound 'A' shows that compound 'A' is alkene. Compound 'B' is \text {C}_{4}\text {H}_{9}\text {Cl}
(ii) Compound 'B' reacts with \text {NH}_{2}, it forms amine 'C'.
\\\text {C}_{4}\text {H}_{8}\overset{\text {HCl}}{\rightarrow}\text {C}_{4}\text {H}_{9}\text {Cl}\overset{\text {NH}_{3}}{\rightarrow}\text {C}_{4}\text {H}_{11}\text {N}\; \text {Or}\; \text {C}_{4}\text {H}_{9}\text {NH}_{2}\\\text {(A)}\; \; \; \; \; \; \; \; \; \; \; \; \text {(B)}\; \; \; \; \; \; \; \; \; \; \; \; \; \text {(C)}
(iii) 'C' gives diazonium salt with \text {NaNO}_{2}/\text {HCl}, which yields an optically active alcohol. So, 'C' is aliphatic amine.
(iv) 'A' on Ozonolysis produces 2 moles of \text {CH}_{3}\text {CHO}. So, 'A' is \text {CH}_{3}-\text {CH}=\text {CH}-\text {CH}_{3}\left ( \text {But-2-ene} \right )
Reactions :


Question:76

A colourless substance 'A' \left ( \text {C}_{6}\text {H}_{7}\text {N} \right ) is sparingly soluble in water and gives a water-soluble compound 'B' on treating with a mineral acid. On reacting with \text {CHCl}_{3} and alcoholic potash 'A' produces an obnoxious smell due to the formation of compound 'C'. The reaction of 'A' with benzene sulphonyl chloride gives compound 'D' which is soluble in alkali. With \text {NaNO}_{2} and \text {HCl}, 'A' forms compound 'E' which reacts with phenol in alkaline medium to give an orange dye 'F' Identify compounds' A' to 'F'.

Answer:

Question:77

Predict the reagent or the product in the following reaction sequence.

Answer:

Also, check - NCERT solutions for class 12, other subjects

The NCERT exemplar Class 12 Chemistry chapter 13 solutions given on this page would help you to prepare well for the 12 boards and competitive exams like NEET and JEE Main by brushing over the important concepts and, in turn, making revision easy . By Utilising the NCERT exemplar Class 12 Chemistry solutions chapter 13 pdf download function students can make learning even more convenient .

Main Subtopics of NCERT Exemplar Solutions for Class 12 Chemistry Chapter 13 Amines

  • Structure of Amines
  • Classification
  • Nomenclature
  • Preparation of Amines
  • Reduction of Nitro Compounds
  • Ammonolysis of Alkyl Halides
  • Reduction of nitrates
  • Reduction of Amides
  • Gabriel phthalimide synthesis
  • Hoffmann bromamide degradation reaction
  • Physical Properties
  • Chemical Reactions
  • Basic character of amines
  • Alkylation
  • Acylation
  • Carbylamine reaction
  • Reaction with nitrous acid
  • Reaction with aryl sulphonyl chloride
  • Electrophilic substitution
  • Method of Preparation of Diazonium Salts
  • Physical Properties
  • Chemical Reaction
  • Reactions involving the displacement of nitrogen
  • Reactions involving retention of the diazo group
  • Importance of Diazonium Salts in the synthesis of Aromatic Compounds
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NCERT Exemplar Class 12 Chemistry Solutions Chapter 13 Amines - Learning Outcome

  • NCERT exemplar solutions for Class 12 Chemistry chapter 13 are completely authentic and will guide you in the right direction.

  • It helps in understanding the concepts of animes in an easy way. Animes are derivatives of ammonia, obtained by replacing the Hydrogen atom.

  • Detailed explanation can be found in the NCERT exemplar Class 12 Chemistry solutions chapter 13.

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NCERT Exemplar Class 12 Chemistry Solutions

Important Topics Covered in NCERT Exemplar Solutions For Class 12 Chemistry Chapter 13

  • Class 12 Chemistry NCERT exemplar solutions chapter 13 has detailed that Structure of Amines, Classification, Nomenclature, Preparation, Physical Properties, Chemical Reactions are important topics which students should pay extra attention to.

  • NCERT exemplar Class 12 Chemistry chapter 13 solutions tries to explain all the chemical concepts by proper visualization in tables and structures to make understanding more comprehensible and easier.

NCERT Exemplar Class 12 Solutions

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Frequently Asked Questions (FAQs)

1. 1. Why should one pay attention to this chapter and NCERT exemplar Class 12 Chemistry solutions chapter 13?

Class 12 Chemistry NCERT exemplar solutions chapter 13 covers the topic of amines which is a crucial part of organic chemistry. Those who are planning to take up engineering or biochemistry should not skip this chapter.

2. 2. How are these NCERT exemplar Class 12 Chemistry chapter 13 solutions helpful?

Students can make the most of these questions and NCERT exemplar solutions for Class 12 Chemistry chapter 13 by understanding the topic better and also these solutions will help in answering the questions asked in the exam.

3. 3. What all topics are included in this Class 12 Chemistry NCERT exemplar solutions chapter 13?

The entire chapter covers the topic of amines, its properties, types, classification, applications and uses, and the chemical reactions.

4. 4. Are these NCERT exemplar Class 12 Chemistry solutions chapter 13 useful in board exams?

-        Yes, these NCERT exemplar Class 12 Chemistry chapter 13 Solutions are helpful for board exams as they help in understanding topics better and also in learning the pattern of the paper.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer
Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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