Question 4: Arrange the following in increasing order of their basic strength:
(i) C2H5NH2, C H5NH2, NH3 , C6H5CH2NH2 and (C2H5)2NH
(ii) C2H5NH2 , (C2H5)2NH , (C2H5)3N , C6H5NH2
(iii) CH3NH2 , (CH3)2NH , (CH3)3N , C6H5NH2 , C6H5CH2NH2.
Answer: (i) Considering the inductive effect of alkyl groups, NH3, C2H5NH2, and (C2H5)2NH can be arranged in the increasing order of their basic strengths as:
Again, C6H5NH2 has proton acceptability less than NH3. Thus, we have:
Due to the −I effect of C6H5 group, the electron density on the N-atom in C6H5CH2NH2is lower than that on the N-atom in C2H5NH2, but more than that in NH3. Therefore, the given compounds can be arranged in the order of their basic strengths as:
(ii) Considering the inductive effect and the steric hindrance of the alkyl groups, C2H5NH2,(C2H5)3NH2 , and their basic strengths as follows:
Again, due to the −R effect of C6H5 group, the electron density on the N atom inC2H5NH2 is lower than that on the N atom in C2H5NH2. Therefore, the basicity of C2H5NH2 is lower than that of C2H5NH2. Hence, the given compounds can be arranged in the increasing order of their basic strengths as follows:
(iii) Considering the inductive effect and the steric hindrance of alkyl groups, CH3NH2, (CH3)2NH, and (CH3)3N can be arranged in the increasing order of their basic strengths as:
In C6H5CH2NH2, N is directly attached to the benzene ring. Thus, the lone pair of electrons on the N−atom is delocalized over the benzene ring. In C6H5CH2NH2, N is not directly attached to the benzene ring. Thus, its lone pair is not delocalized over the benzene ring. Therefore, the electrons on the N atom are more easily available for protonation in C6H5CH2NH2than in C6H5NH2 i.e.,C6H5CH2NH2 is more basic than C6H5NH2.
Again, due to the −I effect of C6H group, the electron density on the N−atom in C6H5CH2NH2 is lower than that on the N−atom in(CH3)3N. Therefore, (CH3)3N is more basic than C6H5CH2NH2 . Thus, the given compounds can be arranged in the increasing order of their basic strengths as follows.
Question 5: Complete the following acid-base reactions and name the products:
(i) CH3CH2CH2NH2 + HCl →
(ii) ( C2H5)3N + HCl →
Answer:
Question 6:
Write reactions of the final alkylation product of aniline with excess of methy liodide in the presence of sodium carbonate solution.
Answer:
Aniline reacts with methyl iodide to produce N, N-dimethylaniline.
With excess methyl iodide, in the presence of Na2CO3 solution, N, N-dimethylaniline produces N, N, N−trimethylanilinium carbonate.
Prepared By: Mr. MANISH TULI
mail to: [email protected]
