Arrange the following:
(i) In decreasing order of pKb values: .
C2H5NH2,C6H5NHCH3,(C2H5)2NH and C6H5NH2
(ii) In increasing order of basic strength:
C6H5NH2, C6H5N(CH3)2, (C2H5)2 NH and CH3NH2.
(iii) In increasing order of basic strength:
(а)Aniline,p-nitroaniline andp-toluidine
(b)C6H5NH2, C6H5NHCH3, C6H5CH2NH2
(iv) In decreasing order of basic strength in gas phase:
C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3
(v) In increasing order of boiling point:
C2H5OH, (CH3)2NH, C2H5NH2
(vi) In increasing order of solubility in water:
C6H5NH2,(C2H5)2NH,C2H5NH2
(i) Due to delocalisation of lone pair of electrons of the N-atom over the benzene ring,C6H5NH2 and C6H5NHCH3 are far less basic than C2H5NH2 and (C2H,)2NH. Due to +I-effect of the -CH3 group, C6H5NHCH3 is little more basic that C6H5NH2. Among C2H5NH2 and (C2H5)2NH, (C2H5)2NH is more basic than C2H5NH2 due to greater+I-effect of two -C2H5 groups. Therefore correct order of decreasing pKb values is:
(ii) Among CH3NH2 and (C2H5)2NH, primarily due to the greater +I-effect of the two -C2H5 groups over one -CH3 group, (C2H5)2NH is more basic than CH3NH2.In both C6H5NH2 and C6H5N(CH3)2 lone pair of electrons present on N-atom is delocalized over the benzene ring but C6H5N(CH3)2 is more basic due to +1 effect of two-CH3 groups.
(iii) (a) The presence of electron donating -CH3 group increases while the presence of electron withdrawing -NO2 group decreases the basic strength of amines.
(b) In C6H5NH2 and C6H5NHCH3, N is directly attached to the benzene ring. As a result, the lone pair of electrons on the N-atom is delocalised over the benzene ring. Therefore, both C6H5NH2 and C6H5NHCH3 are weaker base in comparison to C6H5CH2NH2. Among C6H5NH2 and C6H5NHCH3, due to +1 effect of-CH3 group C6H5NHCH3 is more basic.
(iv) In gas phase or in non-aqueous solvents such as chlorobenzene etc, the solvation effects i. e., the stabilization of the conjugate acid due to H-bonding are absent. Therefore, basic strength depends only upon the +I-effect of the alkyl groups. The +I-effect increases with increase in number of alkyl groups.Thus correct order of decreasing basic strength in gas phase is,
(v)Since the electronegativity of O is higher than thalof N, therefore, alcohols form stronger H-bonds than amines. Also, the extent of H-bonding depends upon flie number of H-atoms on the N-atom, thus the extent of H-bonding is greater in primary amine than secondary amine.
(vi)Solubility decreases with increase in molecular mass of amines due to increase in the size of the hydrophobic hydrocarbon part and with decrease iirthe number of H-atoms on the N-atom which undergo H-bonding.
Which of the following reactions belong to electrophilic aromatic substitution?
(a) Bromination of acetanilide
(b) Coupling reaction of aryldiazonium salts
(c) Diazotisation of aniline
(d) Acylation of aniline
Under which of the following reaction conditions, aniline gives p-nitro derivative as the major product?
(a) Acetyl chloride/pyridine followed by reaction with cone. H2SO4 + cone.
(b) Acetic anhydride/pyridine followed by cone. H2SO4 + cone. HNO3
(c) Dil. HCl followed by reaction with cone. H2SO4 + cone. HNO3
(d) Reaction with cone. HNO3 + cone. H2S04
The best reagent for converting-2-phenylpro-panamide into 1-phenylethana- mine is .
(a) excess H2/Pt (b) NaOH /Br2
(c) NaBH4/methanol (d) LiAlH4/ether
The best reagent for converting 2-phenylpro-panamide into-2-phenyl- propanamine is .
(a) excess H2
(b) Br2 in aqueous NaOH
(c) iodine in the presence of red phosphorus
(d) LiAlH4 in ether
Assertion (A): N, N-diethylbenzene sulphonamide is insoluble in alkalf. Reason (R): Sulphonyl group attached to nitrogen atom is strong electron withdrawing group.
Why is benzene diazonium chloride not stored and is used immediately after its preparation?
Which of the following amines can be prepared by Gabriel synthesis?
(a) Isobutyl amine (b) 2-Phenylethylamine
(c) N-Methylbenzylamine (d) Aniline
Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Under what reaction conditions (acidic/basic), the coupling reaction of aryl diazonium chloride with aniline is carried out?
Arrange the following in increasing order of their basic strength: ‘
(i) C2H5NH2,C6H5NH2,NH3,C6H5CH2NH2 and(C2H5)2 NH
(ii) C2H5NH2,(C2H5)2NH,(C2H5)3N,C6H5NH2
(iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2
Complete the following acid-base reactions and name the products:
(i) CH3CH2CH2NH2+HCl ——–>
(ii) (C2H5)3 N+HCl ——–>
Assertion (A): Hofmann's bromamide reaction is given by primary amines. Reason (R): Primary amines are more basic than secondary amines.
Which of the following cannot be prepared by Sandmeyer's reaction?
(a) Chlorobenzene (b) Bromobenzene
(c) Iodobenzene (d) Fluorobenzene
Arrange the following compounds in increasing order of dipole moment: CH3CH2CH3, CH3CH2NH2, CH3CH2OH.
The correct IUPAC name for CH2 = CHCH2NHCH3 is
(a) allylmethylamine (b) 2-amino-4-pentene
(c) 4-aminopent-l-ene . (d) N-methylprop-2-en-l-amine.
Arrange the following:
(i) In decreasing order of pKb values: .
C2H5NH2,C6H5NHCH3,(C2H5)2NH and C6H5NH2
(ii) In increasing order of basic strength:
C6H5NH2, C6H5N(CH3)2, (C2H5)2 NH and CH3NH2.
(iii) In increasing order of basic strength:
(а)Aniline,p-nitroaniline andp-toluidine
(b)C6H5NH2, C6H5NHCH3, C6H5CH2NH2
(iv) In decreasing order of basic strength in gas phase:
C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3
(v) In increasing order of boiling point:
C2H5OH, (CH3)2NH, C2H5NH2
(vi) In increasing order of solubility in water:
C6H5NH2,(C2H5)2NH,C2H5NH2
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3)2 CHNH2 (ii) CH3(CH2)2NH2 (iii) CH3NHCH(CH3)2
(iv) (CH3)3 CNH2 (v) C6H5NHCH3(vi) (CH3CH2)2NCH3
(vii)m-BrC6H4NH2
A compound Z with molecular formula C3H9N reacts with C6H5SO2Cl to give a solid, insoluble in alkali, identify Z.
How will you convert:
(i) Benzene into aniline
(ii) Benzene into N,N-dimethylaniline
(iii) Cl-(CH2)4-Cl into Hexane -1,6- diamine
Which of the following species are involved in the carbylamine test?
(a) R-NC (b) CHCl3 (c) COCl2 (d) NaN02 + HCl
Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
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.
Convert:
(i) 3-Methylanilineinto3-nitrotoluene
(ii) Aniline into 1,3,5- Tribromo benzene
The reagents that can be used to convert benzene diazonium chloride to benzene are
(a) SnCl2/HCl (b) CH3CH2OH
(c) H3PO2 (d) LiAlH4
A primary amine, RNH2 can be reacted with CH3-X to get secondary amine, R NHCH3 but the only disadvantage is that 3 ° amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where RNH2 forms only 2 ° amine?
Assertion (A): N-Ethylbenzene sulphonamide is soluble in alkali.
Reason (R): Hydrogen attached to nitrogen in sulphonamide is strongly acidic.
Assertion (A): Aromatic 1 ° amines can be prepared by Gabriel phthalimide synthesis.
Reason (R): Aryl halides undergo nucleophilic substitution with anion formed by phthalimide.
Assertion (A): Acetanilide is less basic than aniline.
Reason (R): Acetylation of aniline results in decrease of electron density on nitrogen.
