NCERT Solutions for Class 12 Chemistry Chapter 13 Amines

 

NCERT Solutions for Class 12 Chemistry Chapter 13 Amines -  You will find it interesting that amines are one of the important organic molecules which are present in proteins, vitamins, and hormones, etc. In this chapter, you will deal with organic compounds of amines and learn about amines and diazonium salts. You may also have some doubts which can be resolved by going through CBSE NCERT solutions for class 12 chemistry chapter 13 Amines.  Amines are also considered as derivatives of ammonia. In this chapter, there are 14 questions in the exercise. The NCERT solutions for class 12 chemistry chapter 13 Amines are prepared in a very detailed manner which has all the answers to questions. Chemistry chapter 13 Amines holds four marks in the class 12 CBSE board exam. These NCERT solutions help students in their preparation of class 12 CBSE Board exam as well as various competitive exams like JEE Mains, NEET, etc.   

In solutions of NCERT class 12 chemistry chapter 13 Amines, there are ten sub-topics from which questions are generally asked that covers important concepts of chemistry such as the structure, nomenclature, and classification of amines and also covers its chemical and physical properties and method of preparation of diazonium salts. After studying CBSE NCERT solutions for class 12 chemistry chapter 13 Amines you will be able to describe amines as derivatives of ammonia; classify amines as primary, secondary and tertiary; explain IUPAC nomenclature of amines; describe methods of preparation of amines and properties of amines. Scroll down to get solutions of NCERT class 12 chemistry chapter 13 Amines.

Amines are derivatives of ammonia and they are obtained by the replacement of H-atoms with aryl or alkyl groups. Based on the number of H-atoms replaced from ammonia, amines are classified into three groups: Primary, Secondary and Tertiary amines. Structure of primary amines isR-NH_2, the structure of secondary amines are R_2NH or R-NHR^' and the structure of tertiary amines are  R_3N,\:\: R_2NHR^' \:\:  or RNR^'R''.

Topics and Sub-topics of NCERT Class 12 Chemistry Chapter 13 Amines-

13.1 Structure of Amines

13.2 Classification

13.3 Nomenclature

13.4 Preparation of Amines

13.5 Physical Properties

13.6 Chemical Reactions

13.7 Method of Preparation of Diazonium Salts

13.8 Physical Properties

13.9 Chemical Reactions

13.10 Importance of Diazonium Salts in the Synthesis of Aromatic  Compounds

NCERT Solutions for Class 12 Chemistry Chapter 13 Amines- Exercise Questions

Solutions to In-Text Questions Ex 13.1  to 13.9

Question 13.1   Classify the following amines as primary, secondary or tertiary:

                (i)    

Answer:

N atom is directly connected with only one C atom, so it is a primary aromatic amine

 

Question 13.1   Classify the following amines as primary, secondary or tertiary:

                (ii) 

Answer:

In this compound N atom directly connected with 3 carbon atoms. So, it is a tertiary aromatic amine

 

Question 13.1   Classify the following amines as primary, secondary or tertiary:

             (iii)    (C_{2}H_{5})_{2}CHNH_{2}

Answer:

(C_{2}H_{5})_{2}CHNH_{2}

Here N atom directly connected with only one C atom. So it is a primary aliphatic amine.

 

Question 13.1     Classify the following amines as primary, secondary or tertiary:

                (iv)   (C_{2}H_{5})_{2}NH

Answer:

(C_{2}H_{5})_{2}NH

In structure, we can clearly see that N is directly connected with 2 carbon atom. so, it is a secondary amine.

 

Question 13.2   (i) Write structures of different isomeric amines corresponding to the molecular formula,

                   C_{4}H_{11}N

Answer:

different isomeric amines corresponding to the molecular formula, C_{4}H_{11}N

(i )      CH_{3}CH_{2}CH_{2}CH_{2}NH_{2}                                   (v)  CH_{3}CH_{2}CH_{2}CH_{2}NHCH_{3}

(ii)                                          (vi)   

(iii)                                            (vii)        

(iv)                                                    (viii)          

                   

Question 13.2   (ii) Write IUPAC names of all the isomers.

Answer:

IUPAC name of all the isomers-

CH_{3}CH_{2}CH_{2}CH_{2}NH_{2}

Butanamine

Butan-2-amine

2-Methylpropanamine

N-Methylpropan-2-amine

2-Methylpropan-2-amine

N,N-Dimethylethanamine

CH_{3}CH_{2}CH_{2}CH_{2}NHCH_{3}

N-Methylpropanamine

N-Ethylethanamine


 

Question 13.2(iii) What type of isomerism is exhibited by different pairs of amines?

Answer:

 

CH_{3}CH_{2}CH_{2}CH_{2}NH_{2}

Butanamine
(Chain isomerism + position isomerism)

Butan-2-amine
(chain isomerism + position isomerism)

2-Methylpropanamine
(chain isomerism)

N-Methylpropan-2-amine
(position isomerism + metamerism)

2-Methylpropan-2-amine
(chain isomerism)

N,N-Dimethylethanamine

CH_{3}CH_{2}CH_{2}CH_{2}NHCH_{3}

N-Methylpropanamine
(position isomerism)

N-Ethylethanamine
(no isomerism)

 

Question 13.3(i)      How will you convert

                  Benzene into aniline

Answer:

Nitration of benzene gives nitrobenzene. And now reduce the nitro group by catalytic hydrogenation.

 

Question 13.3(ii)     How will you convert

                  Benzene into N, N-dimethylaniline

Answer:

Nitration of benzene gives nitrobenzene and after catalytic hydrogenation of nitrobenzene, it gives aniline. Aniline on reacting with two moles of chloromethane to form N, N-dimethylaniline.

 

Question 13.3(iii)    How will you convert

                  Cl-(CH_{2})_{4}-Cl   into hexan-1 ,6-diamine?

Answer:

ON reacting the given reactant with ethanolic sodium cyanide, the CN molecules replace both chlorine atom. And after that catalytic hydrogenation, we get our desired product. (reduce the CN to -CH_{2}NH_{2} in both sides)

 

Question 13.4(i)     Arrange the following in increasing order of their basic strength:

                    C_{2}H_{5}NH_{2},  C_{6}H_{5}NH_{2},  NH_{3},  C_{6}H_{5}CH_{2}NH_{2}   and   (C_{2}H_{5})_{2}NH

Answer:

Considering the inductive effect, solvation effect and steric hindrance of the alkyl group which decides the basic strength of alkylamines. order of basic strength in ethyl substituted amine is-

NH_{3}<C_{2}H_{5}NH_{2}<(C_{2}H_{5})_{2}NH

and order in benzene substituted ring-

C_{6}H_{5}NH{2}<C_{6}H_{5}CH_{2}NH_{2}

Due to the -R effect of benzene C_{6}H_{5}NH{2} has less electron density than ammonia. So, the final order is -

C_{6}H_{5}NH{2}<NH_{3}<C_{6}H_{5}CH_{2}NH_{2}<C_{2}H_{5}NH_{2}<(C_{2}H_{5})_{2}NH

 

Question 13.4(ii)     Arrange the following in increasing order of their basic strength:

                C_{2}H_{5}NH_{2}, (C_{2}H_{5})_{2}NH, (C_{2}H_{5})_{3}N, C_{6}H_{5}NH_{2}

Answer:

C_{6}H_{5}NH_{2}<C_{2}H_{5}NH_{2}<(C_{2}H_{5})_{3}N<(C_{2}H_{5})_{2}NH

On considering the inductive effect, solvation effect and steric hindrance of the alkyl group the increasing order of basicity in ethyl as a substituted group is shown above.

 

Question 13.4(iii)     Arrange the following in increasing order of their basic strength:

                 (iii)  CH_{3}NH_{2}, (CH_{3})_{2}NH, (CH_{3})_{3}N, C_{6}H_{5}NH_{2}, C_{6}H_{5}CH_{2}NH_{2}.

Answer:

On considering the inductive effect, solvation effect and steric hindrance of the alkyl group which decides the basic strength of alkyl amines in the aqueous state. The order of basic strength in case of methyl substituted amines are - 

C_{6}H_{5}NH_{2}<C_{6}H_{5}CH_{2}NH_{2}<(CH_{3})_{3}N<CH_{3}NH_{2}<(CH_{3})_{2}NH

 

Question 13.5(i)      Complete the following acid-base reactions and name the products:

                  CH_{3}CH_{2}CH_{2}NH_{2}+HCl\rightarrow

Answer:

The above-given reaction is an acid-base reaction. so salt is formed.

CH_{3}CH_{2}CH_{2}NH_{2}+HCl\rightarrowCH_{3}(CH_{2})_{2}NH_{3}^{+}Cl^{-}

 

Question 13.5(ii)     Complete the following acid-base reactions and name the products:

                  (C_{2}H_{5})_{3}N+HCl\rightarrow

Answer:

(C_{2}H_{5})_{3}N+HCl\rightarrow(C_{2}H_{5})_{3}N^{+}HCl^{-}

It is an acid-base reaction, so the N will accept H from HCl and form a salt.

 

Question 13.6     Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.

Answer:

The methyl iodide reacts with aniline to give  N, N-dimethylaniline.


With the excess of methyl iodide in the presence of sodium carbonate solution (Na_{2}CO_{3}), N, N-dimethylaniline produces N, N, N-trimethyl anilinium carbonate.

 

Question 13.7     Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.

Answer:

When aniline reacts with benzoyl chloride HCl is produced as a by-product and N-phenyl benzamide is produced as a major product.
lone pair of N atom attacks the acidic carbon of benzoyl chloride.