Sponsor Area
a) CH3-CH2-CH2-CH2-NH2
Butanamine (primary)
b) Buta-2-amine (sec.)
c) 2-methyl propamine (pri)
d) 2-methylpropan-2-amine (pri)
e) CH3-CH2-CH2-NH-CH3
N-methylpropanamine (sec.)
f) CH3-CH2-NH-CH2-CH3
N-ethyl ethanamine (sec.)
(g) N-Methyl propan-2-amine (sec.)
h) N,N Dimethylethanamine (tert)
a) CH3-CH2-CH2-CH2-NH2
Butanamine (primary)
b) Buta-2-amine (sec.)
c) 2-methyl propamine (pri)
d) 2-methylpropan-2-amine (pri)
e) CH3-CH2-CH2-NH-CH3
N-methylpropanamine (sec.)
f) CH3-CH2-NH-CH2-CH3
N-ethyl ethanamine (sec.)
(g) N-Methyl propan-2-amine (sec.)
h) N,N Dimethylethanamine (tert)
(iii) The pairs (a) and (b) and (e)and (g) exhibit position isomerism.
The pair (a) and (c); (a) and (d); (b) and (c); (b) and (d) exhibit chain isomerism.
The Pairs (e) and (f) and (f) and (g) exhibit metamerism.
All primary amines exhibit functional isomerism with secondary and tertiary amines and vice versa.
Sponsor Area
(a) CH3-CH2-CH2-NH2
Propan-1-amine (pri)
b) Propan -2-amine (pri)
(c) CH3-CH2-C2H5
N-methylethanamine (Sec)
(d) N,N-Dimethyl methanamine (tert)
primary amine, (a) Propan-1-amine, and (b) Propan-2-amine will liberate nitrogen gas on treatment with nitrous acid.
Convert 3-Methylaniline into 3-Nitrotoluene .
3-methyl aniline is converted to 3-nitrotoluene by nitration reaction.
Conversion of aniline to 1,3,5 -tribromobenzene. 
Classify the following amines as primary (1°), secondary (2°), or tertiary (3°).
Give the common names of the following compounds:
Give the common names of the following compounds:
Common name: o-toluidine
IUPAC name: 3-methyl benzenamine or 2- aminotoluene.
Write the major product of the reaction when amide is reduced with LiAlH4
Lithium aluminium hydride yield amines.
How is aniline obtained on a large scale?
Convert benzyl bromide to β-phenyl ethylamine.
Conversion of benzyl bromide to β-phenyl ethylamine.
Give the name of the product along with its structure when CH3NH2 reacts with an excess of CH3I.
Sponsor Area
How can a carboxylic acid be converted to an amine having one carbon atom less than the carboxyl acid used?
Arrange the following in decreasing order of their basic strength.
C6H5NH2, C2H5NH2, (C2H5)2NH, CH3NH2, C6H5CH2NH2, NH3
Why are amines basic in nature?
Why are amines soluble in dilute mineral acids?
What happens when silver chloride is treated with aqueous ethylamine?
What is diazotization?
Diazotization is a process of converting aromatic amine by nitrous acid below 5° to a diazonium salt.
NaNO2 + HCl → HNO3 + NaCl
C6H5NH2 + HNO2 → [C6H5N+ = N] Cl-
How acetanilide obtained from aniline?
Which amine in the following pair is stronger base?
because isopropyl group is electron releasing whereas -COOCH3 is electron withdrawing.
CH3—NH—CH2CH3 is stronger base because of greater inductive effect.
Which test is used to detect the presence of primary amine?
What happens when an alkylhalide reacts with AgNO2 and product is reduced?
Why aqueous ethylamine turns red litmus blue?
Why aniline dissolves in hydrochloric acid?
What happens when ethylamine is treated with methyl magnesium bromide?
What happens when aniline is treated with bromine?
Give one use of quaternary ammonium salt.
What would be the product of controlled oxidation of aniline with K2Cr2O7 + H2SO4 ?
when aniline oxidised in a controlled manner in the presence ofK2Cr2O7 + H2SO4 , p Benzoquinone is formed. 
Arrange the following in order of increasing basic strength.
Aniline, ethylamine, ethane, phenol.
Why cyclohexylamine is stronger base than aniline?
Benzene reacts with the nitric acid in presence sulphuric acid to form nitrobenzene and further reaction with S2/HCl to give aniline.
Arrange the following in the order of their increasing basicity; p-toluidine, N, N-dimethyl-p-toluidine, p-nitroaniline, aniline.
Why lower aliphatic amines are soluble in water?
Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecules.
Give the structure and IUPAC name of the product obtained in the following reaction:
The product is:
Arrange the following isomeric compounds in order of decreasing boiling points.
CH3CH2N(CH3)2 , CH3CH2CH2CH2NH2, (CH3CH2)2NH.
Sponsor Area
What is the correct order of decreasing boiling point of the following compounds?
CH3 CH2 CH2 OH, CH3 CH2 CH2 NH2, CH3 CH2OCH3, CH3 CH2 NHCH3
The correct order is:
CH3CH2CH2OH > CH3CH2CH2NH2 > CH3CH2NHCH3 > CH3CH2OCH3.
Describe the hybridization and shape of (CH3)3N.
Why do quaternary ammonium salts having four different group attached to nitrogen show optical activity?
When aniline reacts with benzaldehyde, it forms benzylidene aniline.
Give the structure and IUPAC name of the product obtained when formic acid reacts with dimethylamine .
and its IUPAC name is N, N—Dimethyl methana-mide.
The acid generated in the reaction can form a salt of the amine which thus will lose nucleophilic character and the reaction will not proceed to completion.
(CH3)2NH + HCl → (CH3)2NH2⊕ClΘ
So, a base is needed to facilitate the reaction.
Aniline cannot be prepared by Gabriel synthesis. Give reason.
Aniline cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Why are primary amines have higher boiling point than tertiary amines?
How is the basic strength of aromatic amines affected by the presence of an electron releasing group on the benzene ring ?
Which compound is formed when benzene diazonium chloride reacts with phenol in basic medium?
Name the product of the reaction of HNO2 with the CH3CH2NH2.
When a primary amine reacts with nitrous acid, it forms alcohol.
What happens when t-butylamine is treated with potassium permanganate?
Nitrobenzene on reaction with H2/pd gives aniline which on further reaction with benzaldehyde gives benzaniline.
insoluble in aqueous HCl?
Explain coupling reaction giving example.
When benzene diazonium chloride reacts with phenol in which the phenol molecules at its para position is coupled with the diazonium salt to form p-hydroxyazobenzene. This reaction is known as coupling reaction.
The Cl–, Br– and CN– nucleophiles can easily be introduced in the benzene ring in the presence of Cu(I) ion. This reaction is called Sandmeyer reaction.
Write the structure of the Zwitter ion of sulphanilic acid.
and the structure of Zwitter ion is 
.What happens when n-pentyl nitrite reacts with aniline in the presence of HCl in ice-cold solution?
Compare the reactions of MeN and Ph3N with BF3.
Why is it preferred to use methanol (and not methyl iodide HCHO or HCOOH) for preparing C6H5N(CH3)2 from aniline ?
Use of CH3I forms C6H5—N+ (CH3)3I- as a by-product. While the use of CH2O and HCOOH forms the intermediate electrophile, CH2OH which in turn can attack the activated benzene ring to give
p-NH2, C6H4, CH2OH.
Sulphanilic acid is insoluble in water and organic solvents. Explain.
Sponsor Area
The correct order of decreasing basicity is:
CH3CH2NH2 > HO(CH2)3NH2 > HO(CH2)2NH2
The electron withdrawing group -OH decreases the electron density on N, lowering its basicity. This effect decreases when the distance from amino group increases.
Why the order of boiling of isomeric amines are Primary > Secondary > Tertiary ?
Primary amines and secondary amines are engaged in intermolecular association due to hydrogen bonding between the nitrogen of one and hydrogen of another molecule. This intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it. Tertiary amines do not have intermolecular association due to the absence of hydrogen atom available for hydrogen bond formation. Therefore, the order of boiling points of isomeric amines is as.
Primary > Secondary > Tertiary
Giving a reason, arrange the following amines in order of decreasing basicity.
How will you purify aniline containing non-basic impurities ?
Write the structure of the isomeric aromatic amines with formula C7H9N and give their IUPAC names.
Give IUPAC names of the following compounds according to IUPAC system:
Give IUPAC names of the following compounds according to IUPAC system:
Draw structures for the following compound:
N-isopropylaniline.
Structure of N-isopropylaniline is;
Draw structures for the following compound:
p-toluidine
Structure of p-toluidine is given below.
Draw structures for the following compound:
p-amino acetophenone.
Structure of p-amino acetophenone is given as,
Write IUPAC names of the compounds given below:
p-amino acetophenone.
Write structure and IUPAC name of the amide which gives propanamine by Hoffmann bromamide reaction.
Write structure and IUPAC name of the amide produced by the Hoffmann degradation of benzamide.
How will you make the following conversions?
Ethanoic acid to Dimethylamine
Conversion of ethanoic acid to dimethylamine.
Explain what happens when aniline reacts with a mixture of sulphuric acid and nitric acid?
Account for the following:
Aqueous solution of methyl amine reacts with iron (III) chloride to precipitate iron (III) hydroxide.
Write one chemical reaction each to illustrate the following:
A coupling reaction of a diazonium salt.
Illustrate each of the following with an example of reaction:
Sandmeyer reaction.
Illustrate the following with an example of reaction:
Diazotization reaction.
Identify the substances A and B in each of the following sequences of reactions:
CH3CONH2 is a weaker base than CH3CH2NH2
ROH is a stronger acid than RNH2.
Show how p-amino azobenzene can be prepared from aniline?
Conversion of p-amino azobenzene from aniline.
Explain the following reactions by taking one suitable example in each case:
Hoffmann's bromamide reaction.
Explain the following reactions by taking one suitable example in each case:
Gattermann’s reaction.
How is ethylamine prepared from acetonitrile? Compare its basic character with that of ammonia.
How is aniline prepared commercially? Describe carbylamines test for primary amines.
Describe the method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
The three types of amines can be distinguished by Hinsberg method. The sample is treated with benzene sulphonyl chloride, C6H5SO2Cl (Hinsberg’s reagent) followed by treatment with aqueous KOH (5%) solution. Based upon the observations, the following conclusions may be drawn:
(i) If a clear solution is obtained , then it is a primary amine.
(ii) If the solution is turbid or ppt appears and remains unaffected by the addition of an acid, the given amine is a secondary amine.
(iii) If the sample remains insoluble in alkali and dissolves in an acid, then it is a tertiary amine.
The secondary amine forms dialkyl sulphonamide which is insoluble in alkali
Tertiary amine do not react with Hinsberg’s reagent. They do not dissolve in alkali, but dissolve in acid.
(CH3)2CHNH2 : 1-Methylethanamine (10 amines)
CH3(CH2)2NH2
1 - amino propane (primary)
CH3NHCH(CH3)2
Methyl isopropyl amine (secondary)
(CH3)3CNH2
Tert-butylamine (primary amine)
N-Methylbenzamine or N-methylaniline (20 amine)
(CH3CH2)2 NCH3
N-Ethyl-N-methylethanamine (30 amine)
m—Br C6H4 NH2.
3-Isobromo aniline (primary)
Give one chemical test to distinguish between the following pairs of compounds.
Methylamine and diethylamine.
Give one chemical test to distinguish between the following pairs of compounds.
Secondary and tertiary amines.
Shake the given amines separately with Hinsberg’s reagent (benzene sulphonyl chloride) in the presence of an excess of aqueous KOH solution.
A secondary amine forms N,N—dialkyl benzene sulphonamide which remains insoluble in aqueous KOH and even after acidification with dilute HCl.
A tertiary amine does not react with benzene sulphonyl chloride and remains insoluble in aqueous KOH.
Give one chemical test to distinguish between the following pairs of compounds.
Ethylamine and aniline
Ethylamine and aniline: To the ice cold solution of both these compounds prepared in excess of dilute HCl, add ice cold solution of sodium nitrite in water and of β—Naphthol (2—Naphthol) prepared in diluted in sodium hydroxide solution.
Further, cool the reaction mixture in both the cases. The mixture which forms a brilliant orange dye (azodye) contains aniline while the one in which no dye is formed has ethylamine present in it.
Give one chemical test to distinguish between the following pairs of compounds.
Aniline and benzylamine
Give one chemical test to distinguish between the following pairs of compounds.
Aniline and N — methlyaniline.
Account for the following:
pKb of aniline is more than that of methylamine.
Account for the following:
Ethylamine is soluble in water, whereas aniline is not.
Account for the following:
Methylamine in water reacts with ferric chloride to precipitate ferric hydroxide.
Due to the +I effect of -CH3 group, methylamine is more basic than water. therefore, in water methylamine produces OH- ions by accepting H+ ions form water.
Methyl amine is a base and dissolves in water to produce hydroxide ions.
a Friedel -crafts reaction is carried out in the presence of AlCl3. but AlCl3 is acidic in nature, while aniline is a strong base. Thus, aniline reacts with AlCl3 to form a salt. Due to the positive charge on the N-atom, electrophilic substitution in the benzene ring is deactivated.Hence, aniline does not under go the Friedel-crafts reactions.
Aniline does not under Friedel-Craft reaction (alkylation and acetylation) due to the salt formation with aluminium chloride, the Lewis acid which is used as a catalyst. Due to this, the nitrogen of aniline acquires a positive charge and hence acts as a strong deactivating group for further reaction.
Arrange the following:
In decreasing order of the pKb values C2H5NH2, C6H5 NHCH3, (C2H5)2NH and C6H5NH2.
Arrange the following:
In decreasing order of basic strength:
C6H5NH2, C6HsN(CH3)2, (C2H5)2NH and CH3NH2
Arrange the following:
In increasing order of basic strength:
(a) Aniline, p-nitroaniline and p-toluidine
(b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2
(i) p-nitro aniline < aniline < p-toluidine.
(ii) C6H5NHCH3 < C6H5NH2 < C6H5CH2NH2.
Arrange the following:
In decreasing order of basic strength in gas phase C2H5NH2, (C2H5)2NH, (C2H5)3 N and NH3
Arrange the following:
Increasing order of boiling point C2H5OH, (CH3)2NH, C2H5NH2
Arrange the following:
Increasing order of solubility in water:
C6H5NH2, (C2H5)2NH, C2H5NH2.
Write short notes on the following:
Carbylamine reaction.
Write short notes on the following:
Diazotization.
Write short notes on the following:
Hoffmann’s bromamide reaction
Write short notes on the following:
Coupling reaction
Coupling reaction: When benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with coupled with the diazonium salt to form p-hydroxyazobenzene. This is known as coupling reaction.
Write short notes on the following:
Ammonolysis
Ammonolysis: Alkyl halide reacts with ammonia to form primary amine. The reaction of ammonia with alkyl halide is known as ammonolysis.
CH3Cl + NH3 → CH3NH2.HCl
CH3NH2.HCl + NH3 → CH3NH2 + NH4Cl
Write short notes on the following:
Acetylation
is introduced, is called acety lation. It is done by reaction with acetyl chloride or acetic
Write short notes on the following:
Gabriel phthalimide synthesis
Conversion of Benzamide to toluene;
Why cannot aromatic primary amines be prepared by Gabriel pthalimide synthesis?
Give plausible explanation for each of the following:
Why are amines less acidic than alcohols of comparable molecular masses?
Oxygen in alcohol is more electro negative than nitrogen in amines. RC —H bond in alcohol is more polar with δ+ charge on
ROδ-Hδ+ as compared to RN—H bond in Alcohols can loose proton to some extent but are proton acceptors.
Give plausible explanation for each of the following:
Why are primary amines higher boiling than tertiary amines?
Give plausible explanation for each of the following:
Why are aliphatic amines stronger bases than aromatic amines?
Account for the fact that:
Sulphanilic acid is feebly basic.
Account for the fact that:
p-nitroaniline is a weak base.
Explain how does the presence or absence of hydrogen on N of amines affect the modes of their reactions with nitrous acid?
If‘N’ is attached with 2 atoms of hydrogen, we get alcohol and nitrogen gas.
If ‘N’ is attached to 1 atom of hydrogen, we get the yellow oily compound , nitroso amines by replacing one H by NO.
If nitrogen is not attached with hydrogen, this protonation of tertiary amines will take place to form salt soluble in water.
Why is an amide more acidic than an amine?
In structure (II), a positive charge is developed on nitrogen because of which it can liberate a proton and shows acidic character.
2CH3CONH2 + HgO → (CH3CONH)2 + HgO
mention the chief use of quaternary ammonium salts derived from long chain amines.
The quaternary ammonium salts derived from long chain aliphatic amines are used as detergents. i.e. CH3(CH2)15N+(CH3)2Cl-
Arrange the isomeric compounds:
(i) Ethyldimethyl amine
(ii) n-butyl amine
(iii) diethyl amine
in order of decreasing boiling point and give reason.
The correct order is II > III > I
(II) is a 1° amine (III) is 2° amine and can associate via intermolecular hydrogen bonding. But 1° amine has two H present at N-atom, therefore, more hydrogen bonding. Hence, (II) has a higher boiling point than (III)
(I) is a 3° amine, there is no H present on N atom, no H-bonding present. Hence, (I) has a lowest boiling point.
Explain the following general order of basicity in aqueous solution:
R2NH > RNH2 > R3N > NH3
Two effects: inductive and the solvation effect determines the basicity of alkyl amines.
As the number of alkyl group (electron releasing group with +I effect) increases, making the amine more basic. So, in terms of inductive effect, increasing the number of R's should increase the basicity.
Solvation is present only in the conjugate acid, more the number of H present in the conjugate acid more the H-bonding and more the stabilisation. Therefore the R' increases, the number of H-s in the conjugate acid decreases, solvation decreases. Hence, basicity should decrease.
The inductive and solvation effect both works in opposite direction. Inductive effect dominates to make all three amines stronger bases than NH3.
While in R3N solvation effect dominates over the inductive effect.
Explain the following general order of basicity in aqueous solution:
R3N ,R2NH , RNH2
What will be the basic strength order of these amines in gas phase?
In the gas phase there is no solvation effect, only the number of R's and their inductive effect determines the basicity.
More the number of R's, more the basicity. The basicity order will be their,
R3N > R2NH > RNH2
How can you find out whether a given amine is a primary amine? Write the chemical reaction involved in the test you perform.
How can you separate a mixture of primary, secondary and tertiary amines? Write chemical reactions involved in the process.
Hinsberg test is employed to separate primary, secondary and tertiary amines from a mixture. In this test the mixture of amines is treated with benzene sulphonyl chloride C6H5SO2Cl (Hinsberg’s reagent) followed by treatment with aqueous KOH (5%) solution and then shaken with ether in a separatory funnel.
(a) The primary amine reacts forming mono alkyl sulphonamide which is soluble in alkali forming a potassium salt which forms the lower aqueous layer.
(b) The secondary amine reacts forming dialkylsulphonamide which is insoluble in alkali.
The tertiary amines does not react at all.
Both dialkyl sulphonamide and tertiary amine go into the upper ether layer. While the monoalkyl sulphonamide remains in the lower aqueous layer. The two layers are separated, the aqueous layer containing the potassium salt of mono-alkyl sulphonamide is hydrolysed with concentrated hydrochloric acid when primary amine is regenerated as its hydrochloride.
Primary amine is recovered from the hydrochloride by distillation with alkali.
RNH2.HCl+ KOH → RNH2 + KCl + H2O
The ether layer distilled when tertiary amine distills over. The residue of dialkyl sulphonamide is hydrolysed with concentrated hydrochloric acid when secondary amine is regenerated as its hydrochloride.
C6H5SO2NR2 +H2O + HCI → C6H5SO2OH + R2NH.HCl
The hydrochloride on distillation with alkali gives the free secondary amine
R2NH.HCl + KOH → R2NH + KCl + H2O
Explain: In aqueous solution order is Et2NH,>Et3N > EtNH2.
In aqueous solution, the Kb order is Et2NH > Et3N > EtNH2
This may be explained by the following factors:
(a) Steric factor: The size of the alkyl group is more than that of hydrogen and, therefore, it hinders the attack of acid on the amine and therefore, basic strength decreases. Now crossing of alkyl group increases from primary to tertiary amines. As a result, their basic strength decreases. This is called steric hindrance. According to this effect, alkyl amines should be least basic.
(b) Solvation of ions: When amines are dissolved in water, they undergo hydration through hydrogen bonding. The protonated amines form hydrogen bonds with water molecules and release energy called hydration energy. Now, greater the extent of hydrogen bonding in protonated amine, more will be its stabilization and consequently, greater will be the tendency of amine to change into a cation and greater will be the strength of the corresponding amine. As can be seen below, the hydration of protonated amine due to hydrogen bonding decreases as:
Et NH3+ > Et2 NH2+ > Et3NH+
The tertiary ammonium ion is less hydrated than secondary ammonium ion, which is less hydrated than primary ammonium ion. Thus, tertiary amines have less tendency to form ammonium ion and consequently are less basic.
As a consequence of combining effects of inductive effect and solvation, the secondary amines are the strongest bases among amines and basic strength varies as:
Et2NH > Et3 N > Et NH2
As the compound (X) is giving a positive Hoffmann bromamide reaction, the compound (X) contains the CONH2 group. As only three carbons present, the structure of compound (X) is:
(X) → CH3CH2CONH2
Hoffmann bromamide reaction give 1° amine with one carbon less, therefore, the structure for compound (Y) is:
(Y) → CH3CH2NH2
The reaction involved are:
Why does bromination of aniline, even under very mild conditions give 2, 4, 6- tribromo aniline instantaneously?
An organic compound (A) with IMF C4H11N, capable of being resolved into optical isomers gives a base soluble product with benzene sulphonyl chloride. What is the structure and IUPAC name of (A)? What happens when this compound (A) is treated with ethanoyl chloride?
How is benzene diazonium chloride obtained?
are more stable than R-N2+ due to resonance. Aryl diazonium ions have more contributing structure than alkyl diazonium ions. Hence, aryl diazonium is more stable.
Explain what is meant by the 'protection of the amino group'.?
Show by means of equations how an amide may be converted into an amine containing the same number of carbon atoms.
Show by means of equations how an amide may be converted into an amine containing one carbon atom less than the amide.
The three classes of amines react differently with nitrous acid. Nitrous acid is very unstable and is generated only during the course of a reaction using NaNO2/HCl.
Primary amines have two hydrogens present on N-atoms.
Aromatic primary amine from diazonium salt with NaNO2/HCl
While Aliphatic primary amine form diazonium salt which is unstable and decompose to liberate N2 gas and form alcohol,
Secondary amines contain one hydrogen on N-atom. Both aliphatic and aromatic amines undergo electrophilic substitution reaction with HNO2 to form nitrosamines.
Tertiary amines have no H present on N-atom. Aliphatic tertiary amine form salt with HNO2 which is water soluble.
R3N + H—ONO → R3NH⊕®ONOΘ
Aromatic tertiary amine undergo electrophilic reaction at para position as the ring is highly activated by NR2 group.
Explain the following:
Aniline dissolves in aqueous HCl.
Explain the following:
Amines are stronger bases than ammonia.
Explain the following:
Aniline is a weak base than ethylamine.
Explain the following:
The amino group in ethylamine is basic whereas that in acetamide is not basic.
Explain the following:
Cyclo hexylamine is a stronger base than aniline.
Describe simple tests for distinguishing:
p-Toluidine and benzyl amine
Describe simple tests for distinguishing:
CH3ONHC6H5 and C6H5CONH2
Describe simple tests for distinguishing:
C6HsNH3 +Cl- and p-ClC6H4NH2
Describe simple tests for distinguishing:
(CH3)4N+OH and (CH3)2NCH2OH
Describe simple tests for distinguishing:
p-CH3COC6H4NH2 and CH3CONHC6H5
Arrange the following in the order of decreasing basicity:
NH3, CH3NH2, C6H5NH2 (C6H5)2NH.
Arrange the following in the order of decreasing basicity:
C6H3NH2, p-NO2C6H4NH2, m-NO2.C6H4NH2, p-CH3 OC6H4NH2.
p-CH3OC6H4NH2 > C6H5NH2 > m-NO2.C6H4. NH2 > p-NO2.C6H4.NH2-NO2 group in m-position weakens the base due to inductive effect while in o- and p- positions it weakens the base due to -I effect and extended 
bonding.
Arrange the following in the order of decreasing basicity:
p-Toluidine, N, N-dimethyl-p-toluidine, p-nitro aniline, aniline.
Arrange the following in the order of decreasing basicity:
Methyl amine, dimethyl amine, aniline, N-methyl aniline.
Arrange the following in the order of decreasing basicity:
C2H5NH2, iso-(C3H7)3N, CH3CONH2, CH3NH-Na+
CH3NH-Na > C2H5NH2 > (iso-C3H7)3N > CH3CONH2
CH3NH- is the conjugate base of the very weak acid CH3NH2 and therefore is the strongest base. The three bulky isopropyl groups on cause steric strain, but with an unshared electron pair on N, the strain is partially relieved by increasing the normal C—N—C bond angle (109°) to about 112°. If the unshared electron pair forms a bond to H, as in R3NH+, relief of strain by angle expansion is prevented. 3° Amines, therefore, resist forming the fourth bond and suffers a decrease in basicity. In CH3CONH2, the N atom is weakly basic because its electron pair can be delocalised to O of the CH3CO— group by extended 
bonding.
Potassium phthalimide can be used to prepare primary amines through a method known as the _________ synthesis.
Primary, secondary and tertiary amines can be separated from each other by means of ___________.
Sodium salt of benzene sulphonic acid when fused with sodium cyanide gives __________.
A.
Tertiary amines can not be acetylated.
B.
All types of amines give the isocyanide test.C.
Aniline becomes coloured in air owing to oxidation.D.
Cyclo hexylamine is a stronger base than aniline.E.
Triphenyl amine (C6H5)3N, forms an addition compound with BF3.A.
Benzamide, on treatment with Br2 and KOH gives benzyamine.B.
Benzene diazonium salts are usually stable.C.
p-methoxy aniline has a higher pka value than p-methylaniline.D.
Acetamide is a stronger acid than ethylamine.E.
Both p-amino benzoic acid and p-amino benzene sulphonic acid form Zwitter ion.A.
Bromination of aniline gives 2, 4, 6-tribromoaniline.B.
Trimethyl amine reacts with BF3 while triphenylamine does not.C.
Unpleasant smelling carbyl amine is formed by heating an alkali and chloroform with any primary amine.D.
Reduction of nitrobenzene in alkaline medium finally gives aniline.E.
N, N-dimethyl aniline gives positive carbylamine reaction.Acetamide react with NaOBr in alkaline medium to form
D.
CH3CH2NH2Which compound will liberate CO2 from NaHCO3 solution?
D.
CH3N+H3Cl-Amongst the following, the most basic compound is
A.
BenzylamineB.
2, 4-dimethyl anilineWhy are benzene diazonium salt soluble in water?
Benzene diazonium salt is soluble in water because diazonium salt is polar in nature. water is polar solvent, thus polar diazonium salt get dissolved in water.
An organic compound 'A' having molecular formula C2H7N on treatment with HNO2 gave an oily yellow substance- Identify A.
As the organic compounds has N and also it reacts with nitrous acid, it may be an amine.
CH3CH2NH2 +HNO2-->CH3CH2OH +N2+H2O
the compound A is ethylamine and the product is ethanol.
What happens when acetamide is reacted with bromine in the presence of potassium hydroxide?
Acetamide, when treated with bromine and concentrated potassium hydroxide solution methylamine, is formed. This reaction is known as Hofmann bromamide reaction.
CH3CONH2 + Br2+4KOH --> CH3NH2 +2KBr +K2CO3+2H2O
Aniline is more soluble in aq. HCl than in water. Explain.
Aniline does not undergo hydrogen bonding because of the presence of the benzene which is highly hydrophobic. therefore aniline is insoluble in water.
In the HCl the amine group becomes protonated (-NH3+) and the ionic hence soluble in HCl.
explain the sandmeyer's reaction?
The Cl-,Br- and CN- nucleophiles can easily be introduced in the benzene in the presence of Cu(I) ion. This reaction is called sand Meyer's reaction.
What happens when ethylamine is treated with silver chloride.
A soluble product is formed when ethylamine react with silver chloride.
Write the reaction of (i)aromatic and (ii) aliphatic primary amines with nitrous acid ?
Aromatic primary amines reacts with HNO2 at 273-278K to form aromatic diazonium salts.
aliphatic Primary amines also react with HNO2 at 273-278 K to form aliphatic diazonium salts. But these are unstable even at this low temperature and thus decompose readily to form a mixture of compounds consisting of alkyl chlorides, alkenes and alcohols,out of which alcohols generally predominates.
acylation of ethylamine?
ethylamine is readily acetylated by acetyl chloride and acetic anhydride in the cold to give ethyl acetamide.
How do you the following substances react?
Potassium phthalimide + ethyl bromide.
By treating potassium phthalimide with ethyl bromide and hydrolysing the resulting N-ethyl phthalimide by refluxing with 20% HCl under pressure .
Why are the secondary amines more basic than primary amines?
Amines are bases due to the lone pair of electrons on the nitrogen atom of amines. These react with water to form hydroxyl ions.
Bases are those species that donate OH- ions (hydroxyl ions). The more easily is the availability of the hydroxyl ions; more is the basicity of the amine. A secondary amine can donate its lone pair more easily, as the extra inductive effect stabilises the charge on the nitrogen atom
Mention two important uses of Sulphanilic acid.
sulphanilic acid is used in the manufacture of i) dyes ii) sulpha drugs.
Ethylamine is soluble in water because it forms a hydrogen bond with water. Whereas, aniline is hydrophobic to water. hence it is insoluble.
Name the reagents used in the following reactions ?
Name of the reagent: Sodium borohydride (NaBH4). It is a good reducing agent useful in the reduction of aldehyde and ketone to alcohol.
Name of the reagent: Alkaline potassium permanganate (KMnO4-KOH). Potassium permanganate is a strong oxidising agent.
Arrange the following in the increasing order of their boiling point: C2H5NH2, C2H5OH, (CH3)3N
Increasing order of boiling point: (CH3)3N < C2H5NH2 < C2H5OH
Alcohols have a higher boiling point as compared to that of amines because of oxygen, being more electronegative, forms strong hydrogen bond as compared to that of nitrogen. In tertiary amine, there is no hydrogen to form hydrogen bond and hence, has the lowest boiling point.Give a simple chemical test to distinguish between the following pair of compounds:
(CH3)2NH and (CH3)3N.
(CH3)2NH reacts with benzene sulphonyl chloride as follows:
(CH3)3N does not react with benzene sulphonyl chloride.
Arrange the following compounds in increasing order of solubility in water:
C6H5NH2, (C2H5)2NH, C2H5NH2
The increasing order of solubility in water of the given compounds is as follows:
C6H5NH2 < (C2H5)2NH < C2H5NH2
The more extensive the hydrogen bonding, the higher is the solubility of the compound in water. C2H5NH2 contains two H-atoms, whereas (C2H5)2NH contains only one H-atom. Therefore, C2H5NH2 is more soluble than (C2H5)2NH. The solubility of C6H5NH2 is least due to its high molecular mass.
How will you convert the following?
Ethanoic acid into methanamine
Ethanoic acid into methanamine
How will you convert the following?
Aniline into N-phenylethanamide
Aniline into N-phenylethanamide
Arrange the following compounds in an increasing order of their solubility in water:
C6H5NH2, (C2H5)2NH, C2H5NH2The solubility order of the given amines is as follows:
C6H5NH2 < (C2H5)2NH < C2H5NH2
The more extensive the Hydrogen bonding, the higher is the solubility. C2H5NH2 contains two H-atoms whereas (C2H5)2NH contains only one H-atom. Thus, C2H5NH2undergoes more extensive. Hydrogen bonding than (C2H5)2NH. Hence, the solubility in water of C2H5NH2 is more than that of (C2H5)2NH.
Further, the solubility of amines decreases with increase in the molecular mass. This is because the molecular mass of amines increases with an increase in the size of the hydrophobic part.
The molecular mass of C6H5NH2 is greater than that of C2H5NH2 and (C2H5)2NH.
Explain why an alkylamine is more basic than ammonia.
(a) The basicity of amines depends on the +I effect of the alkyl groups. With increase in alkyl group the +I effect will increase which will increase the ease of donation of lone pair electron. Amine will accept a proton and from cation will be stabilized in water by salvation (by hydrogen bonding).better the salvation by hydrogen bonding higher will be the basic strength.
Hence, alkyl amine is more basic than ammonia
CH3NH2 > NH3
Write the main products of the following reactions: 
(i) 
Products are Benzene, Nitrogen gas, Phosphorous acid and Hydrochloric acid.
(ii)
Products is 2,4,6-Tribromoaniline
Products are Methanamine, Sodium carbonate, Sodium bromide and Water.
Complete the following chemical reaction equations:
(i) C6H5N2Cl + H3PO2+ H2O --->
(ii) C6H5NH2+ Br2(aq.)--->
(i) C6H5N2Cl + H3PO2+ H2O ---> C6H6 + N2 + H3PO3+ HCl
(ii) 
Describe the following giving the relevant chemical equation in each case:
(i) Carbylamines reaction
(ii) Hofmann’s bromamide reaction.
(i) Carbylamines reaction
Carbylamines reaction is used as a test for the identification of primary amines. When aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide, carbylamines (or isocyanides) are formed. These carbylamines have very unpleasant odours.
Secondary and tertiary amines do not respond to this test.
For example,
(b) Hoffmann bromamide reaction
When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, a primary amine with one carbon atom less than the original amide is produced. This degradation reaction is known as Hoffmann bromamide reaction. This reaction involves the migration of an alkyl or aryl group from the carbonyl carbon atom of the amide to the nitrogen atom.
Write the IUPAC name of the given compound:
The IUPAC name of the compound is 2,4,6 tribromo aniline.
Give reasons for the following:
i) Aniline does not undergo Friedal-Crafts reaction.
ii) (CH3)3N is more basic than (CH3)3N in an aqueous solution.
iii) Primary amines have higher boiling point than tertiary amines.
(i) Aniline is a Lewis base while AlCl3 is lewis acid. They combine to form a salt.
(ii)Due to combined + I and solvation effects.
(iii)Due to the presence of H-bonding in primary amines.Give a chemical test to distinguish between ethylamine and aniline.
Ethylamine and aniline can be distinguished from each other by the azo-dye test.
A dye is obtained when aniline (an aromatic amine) reacts with HNO2 (NaNO2 + dil.HCl) at 0-5°C, followed by a reaction with the alkaline solution of 2-naphthol.
Ethylamine (an aliphatic amine) gives a brisk effervescence due (to the evolution of N2 gas) under similar conditions.
State reasons for the following:
(i) pKb value for aniline is more than that for methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not soluble in water.
(iii) Primary amines have higher boiling points than tertiary amines.(i) Aniline undergoes resonance and as a result, the electrons on the N-atom are delocalized over the benzene ring. Therefore, the electrons on the N-atom are less available to donate.
On the other hand, in case of methylamine (due to the +I effect of methyl group), the electron density on the N-atom is increased. As a result, aniline is less basic than methylamine. Thus, pKb of aniline is more than that of methylamine.
(ii) Ethylamine when added to water forms intermolecular H-bonds with water. Hence, it is soluble in water.
But aniline does not undergo H-bonding with water to a very large extent due to the presence of a large hydrophobic -C6H5 group. Hence, aniline is insoluble in water.
(iii) The boiling points of compounds depend on the extent of H-bonding present in that compound. The more extensive the H-bonding in the compound, the higher is the boiling point. Tertiary amines [(CH3)3N] does not contain H-atom whereas primary amine (CH3NH2) contains two H-atoms. Then, primary amines undergo more extensive H-bonding than tertiary amines. Hence, the boiling point of primary amines is higher than that of tertiary amines.
The conversion of primary aromatic amines into diazonium salts is known as __________ .
The conversion of primary aromatic amines into diazonium salts is known as diazotization reaction.
Account for the following:
(i) Primary amines (R-NH2) have higher boiling point than tertiary amines (R3N).
(ii) Aniline does not undergo Friedel - Crafts reactions:
(iii) (CH3)2NH is more basic than (CH3)3N in an aqueous solution.
OR
(i) The greater the extent of intermolecular hydrogen bonding, greater is the boiling point of the compound. Primary amines are involved in intermolecular hydrogen bonding, whereas tertiary amines do not have any available hydrogen atoms, and hence they are not involved in any intermolecular hydrogen bonding. Thus, primary amines (R-NH2) have higher boiling point than tertiary amines (R3N).
(ii) A Friedel-Crafts reaction is carried out in the presence of a lewis acid such as AlCl3. Aniline being basic in nature reacts with AlCl3 to form a salt (as shown in the following equation).
Due to the positive charge on the N-atom, electrophilic substitution in the benzene ring is deactivated. Hence, aniline does not undergo the Friedel-Crafts reaction.
(iii) The order of basic strength of aliphatic amines in the aqueous phase, is based on following factors:
Steric factor: Alkyl group is larger than hydrogen atom that causes steric hindrance to attack of acid. As the number of alkyl group increases from primary to tertiary. The steric hindrance also increases in the same direction.
Solvation of ions: when amines are dissolved in water, they form hydrogen bonds with a water molecule and release hydration energy; in the process converting themselves to the protonated amines to get stabilized. Greater the extent of hydrogen bonding, greater is the hydration energy released and more is the stability of the protonated amine; thus greater is the tendency of the amine to form cation leading to the greater basic strength of the amine. So, the actual order of basic strength in aqueous phase is found to be:
Secondary > Tertiary > Primary
Therefore, secondary amines are the strongest base in aqueous solution.
Give reasons:
Acetylation of aniline reduces its activation effect.
Acetylation of aniline reduces its activation effect Because it result in decrease of electron density on nitrogen. Due to the resonance, the electron pair of nitrogen atom gets delocalised towards carbonyl group therefore so activation effect reduces.
Give reasons :
CH3NH2 is more basic than C6H5NH2.
CH3NH2 is more basic than C6H5NH2 due to +I effect in methylamine, electron density in methylamine at nitrogen increases whereas in aniline resonance takes place and electron density on nitrogen
decreases.
Give reasons :
Although –NH2 is o/p directing group, yet aniline on nitration gives a significant amount of m-nitroaniline.
Nitration is carried out in an acidic medium ( a mixture of concentrated HNO3 and concentrated H2SO4). In an acidic medium, aniline is protonated to give anilinium ion (which is meta-directing).
Give reasons:
(CH3)2NH is more basic than (CH3)3N in an aqueous solution.
The order of basic strength of aliphatic amines in the aqueous phase is based on the following factors:
Steric factor: Alkyl group is larger than hydrogen atom that causes steric hindrance to attack of acid. As the number of alkyl group increases from primary to tertiary. The steric hindrance also increases in the same direction.
Solvation of ions: when amines are dissolved in water, they form hydrogen bonds with a water molecule and release hydration energy; in the process converting themselves to the protonated amines to get stabilized. Greater the extent of hydrogen bonding, greater is the hydration energy released and more is the stability of the protonated amine; thus greater is the tendency of the amine to form cation leading to the greater basic strength of the amine. So, the actual order of basic strength in the aqueous phase is found to be:
Secondary > Tertiary > Primary
Therefore, secondary amines are the strongest base in aqueous solution.
Give reasons:
Aromatic diazonium salts are more stable than aliphatic diazonium salts.
Aromatic diazonium salts are more stable than aliphatic diazonium salts. The stability of aromatic diazonium salts is due to resonance which is absent in aliphatic diazonium salts.
Give a simple chemical test to distinguish between Aniline and N, N-dimethylaniline.
Carbylamine test can be used to distinguish.
Arrange the following in the increasing order of their pKb values:
C6H5NH2, C2H5NH2, C6H5NHCH3
pKb tells the strength of a base. Lower is pKb value, stronger is the base. Now C6H5NH2 is a weak base because the lone pair is involved in resonance with the benzene and is thus not available for a donation while in C2H5NH2, C2H5 is an electron donating group, which increases the e- density on N thus felicitating the release of lone pair. Hence, C2H5NH2 is a strong base.
Thus, the correct order is:
C2H5NH2< C6H5NHCH3< C6H5NH2
In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine produced are:
Four moles of NaOH and two moles of Br2
Two moles of NaOH and two moles of Br2
Four moles of NaOH and one mole of Br2
One mole of NaOH and one mole of Br2
C.
Four moles of NaOH and one mole of Br2
Hofmann-bromamide degradation reaction is given as
RCONH2 + 4 NaOH + Br2 → RNH2 + Na2CO3 + 2NaBr + 2H2O
Hence four moles of NaOH and one mole of Br2 are used.
On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is
an alkanol
an alkaneodiol
an alkyl cyanide
an alkyl isocyanide
D.
an alkyl isocyanide
The reaction is an example of carbylamines reaction which includes conversion of amine to isocyanide.
Considering the basic strength of amines in aqueous solution, which one has the smallest pKb value?
(CH3)2NH
CH3NH2
(CH3)3N
C6H5NH2
A.
(CH3)2NH
Order of basic strength of aliphatic amine in aqueous solution is as follows (order of Kb)
So (CH3)2 NH will have smallest pKb value.
In the case of phenylamine, N is attached to sp2 hybridised carbon, hence it has highest pKb and least basic strength.
A compound with molecular mass 180 is acylated with CH3COCl to get a compound with molecular mass 390. The number of amino groups presents per molecule of the former compound is
2
5
4
6
B.
5
Ortho–Nitrophenol is less soluble in water than p– and m– Nitrophenols because
o–Nitrophenol is more volatile in steam than those of m – and p–isomers
o–Nitrophenol shows Intramolecular H–bonding
o–Nitrophenol shows Intermolecular H–bonding
Melting point of o–Nitrophenol is lower than those of m–and p–isomers
B.
o–Nitrophenol shows Intramolecular H–bonding
There is intramolecular H-bonding in o-nitrophenol and thus solubility in water is decreased.
In the chemical reactions,
the compounds 'A' and 'B' respectively are
nitrobenzene and fluorobenzene
phenol and benzene
benzene diazonium chloride and flurobenzene.
nitrobenzene and chlorobenzen
C.
benzene diazonium chloride and flurobenzene.
Which of the following compounds will the significant amount of meta product during mono-nitration reaction?
C.
(a) Nitration reactions take place in presence of concentrated HNO3+ concentrated H2SO4.
(b) Aniline acts as a base. In presence of H2SO4, its protonation takes place and anilinium ion is formed
(c)Anilinium ion is a strongly deactivating group and meta directing in nature so it gives meta nitration product in a significant amount.
The hydrocarbon which can react with sodium in liquid ammonia is
CH3CH2CH2C≡CCH2CH2CH3
CH3CH2C≡CH
CH3CH=CHCH3
CH3CH2C≡CCH2CH3
B.
CH3CH2C≡CH
In the chemical reaction, CH3CH2NH2 + CHCl3 + 3KOH → (A) + (B) + 3H2O, the compounds (A) and (B) are respectively
C2H5CN and 3KCl
CH3CH2CONH2 and 3KCl
C2H5NC and K2CO3
C2H5NC and 3KCl
D.
C2H5NC and 3KCl
It is example of carbylamine reaction. so, the product will be C2H5NC and 3KCl
Which one of the following is the strongest base in aqueous solution?
Trimethylamine
Aniline
Dimethylamine
Methylamine
C.
Dimethylamine
In aqueous solution basicity order of 1o, 2o and 3o amine with methyl group is 2o>1o>3o
In case of aniline lone pair of nitrogen is involved in resonance, so it is weaker base than aliphatic amines.
Amongst the following the most basic compound is
benzylamine
aniline
acetanilide
p-nitroaniline
A.
benzylamine
-NH2 group is not linked to the benzene ring.
Which one the following does not have sp2 hybridized carbon?
Acetone
Acetamide
Acetonitrile
Acetic acid
C.
Acetonitrile
Acetonitrile does not contain sp2 hybridised carbon.
Consider the acidity of the carboxylic acids
PhCOOH
p – NO2C6H4COOH
o – NO2C6H4COOH
m – NO2C6H4COOH
C.
o – NO2C6H4COOH
-NO2 group at any position shows electron withdrawing effect thus acid strength is increased. But o -nitro benzoate ion is stabilised by intramolecular H-bonding. Hence its acid strength is maximum. The effect is more at para position.
Which of the following is the strongest base?
B.
The basicity of amine group depends on the lone pair present on N atom. If this lone pair involved in the conjugation, the basic character of amine decreases. So, find the molecule in which lone pair of nitrogen does not involve in conjugation.
Which of the following compounds will be suitable for Kjeldah1’s method for nitrogen estimation?
D.
Kjeldahl's method is used for Aniline. This method is used for quantitative analysis of N compound in organic substance (NH3/NH4+).
This method is not used in the case of nitro, azo compounds and also to the compounds containing nitrogen in the ring e.g. Pyridine.
The increasing order of basicity of the following compounds is:
(d) < (b) < (a) < (c)
(a) < (b) < (c) < (d)
(b) < (a) < (c) < (d)
(b) < (a) < (d) < (c)
D.
(b) < (a) < (d) < (c)
Order of base nature depends on electron donation tendency.
It compounds (b) nitrogen is sp2 hybridized so least basic among all given compound
Compound (c) is a very strong nitrogeneous organic base as a lone pair of one nitrogen delocalize in resonance and make another nitrogen negatively charged and conjugate acid have two equivalent resonating structure.
Thus it is most basic in given compound.
(d) is secondary amin more than (a) which is a primary amine.
The correct statement regarding the basicity of arylamines is
Arylamines are generally more basic than alkylamines because the nitrogen lone pair electrons are not delocalized by interaction with the aromatic ring pi-electron system
Arylamines are generally more basic than alkylamines, because of aryl group
Arylamines are generally more basic than alkylamines because the nitrogen atom in arylamines is sp-hydridized.
Arylamines are generally less basic than alkylamines because the nitrogen lone pair electrons are delocalized by interaction with aromatic ring pi-electron system.
D.
Arylamines are generally less basic than alkylamines because the nitrogen lone pair electrons are delocalized by interaction with aromatic ring pi-electron system.
In the aryl amines, due to the delocalization of lone pair of electrons of N-atom to the benzene ring, it loses its basicity and becomes less basic than alkyl amine.
on the other hand, alkyl amines have +I alkyl effect of the alkyl group which increase electron density on N -atom. Hence the availability of free electron on amine as well as +I effects enhances its basic nature.
In the reaction,
X and Y are
X=2-butyne; Y= 3-hexyne
X=2-butyne;Y=2-hexyne
X=1-butyne ;Y = 2-hexyne
X=1-butyne; Y =3-hexyne
D.
X=1-butyne; Y =3-hexyne
NaNH2/liq.NH3 behaves as a base, so it abstracts a proton from acetylene to form acetylide anion followed by alkylation to give compound (X) i.e 1-butyne (X) further reacts with NaNH2/Liq NH3 followed by alkylation with ethyl bromide yields 3-hexyne (Y).
Consider the nitration of benzene using mixed conc. H2SO4 and HNO3. If a large amount of KHSO4 is added to the mixture, the rate of nitration will be
slower
uncharge
doubled
faster
A.
slower
IN the nitration of benzene in the presence of conc. H2SO4 and HNO3, benzene is formed.
HNO3 +H2SO4 
If a large amount of KHSO4 is added to this mixture more HSO4- ion furnishes and hence the concentration of electrophile decreases, rate of electrophilic aromatic reaction slows down.
The electrolytic reduction of nitrobenzene in strongly acidic medium produces
p-aminophenol
azoxybenzene
azobenzene
aniline
A.
p-aminophenol
Under weakly acidic conditions nitrobenzene on electrolytic reduction gives aniline but under strongly acidic conditions gives p-aminophenol.
In the following reaction, the product (A) is 
D.
Which of the following will most stable diazonium salt RN2+X-?
CH3N2+X-
C6H5N2+X-
CH3CH2N2+X-
C6H5CH2N2+X-
B.
C6H5N2+X-
Diazonium salt containing aryl group directly linked to the nitrogen atom is most stable due to resonance stabilization between the benzene nucleus and N-atom.
Nitrobenzene on reaction with conc. HNO3/H2SO4 at 80-1000C forms which one of the following products?
1,2-dinitrobenzene
1,3- dinitrobenzene
1,4-dinitrobenzene
1,2,4-trinitrobenzene
B.
1,3- dinitrobenzene
NO2 group being electron withdrawing reduces electron density at output positions. Hence, now the meta-position becomes electron rich on which the electrophile (nitronium ion) attacks during nitration.
HNO3 +H2SO4 ---> H2NO3+ +HSO-4 +H2O +NO2+
Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating?
-C 
N
-SO3H
-COOH
-NO2
D.
-NO2
The deactivation tendency of given groups follows the order
Thus,-NO2 is the most deactivation group.
Which of the following compounds will not undergo Friedel -Craft's reaction easily?
Cumene
Xylene
Nitrobenzene
Toluene
C.
Nitrobenzene
Nitro group being electron withdrawing, deactivates the benzene nucleus to such an extent so that it becomes incapable to give Friedel -Crafts reaction.
Nitrobenzene because of its unreactivity towards Friedel-Craft's reaction is used as a solvent for this reaction.
In the reaction,
HgSO4/H2SO4
Cu2Cl2
H3PO2 and H2O
H+/H2O
C.
H3PO2 and H2O
When diazonium salt is treated with H3PO2 followed by hydrolysis. It diazonium group is replaced by -H resulting to the formation of hydrocarbon. Thus, A must be H3PO2/H2O.
Among the following compounds, the one that is most reactive towards electrophilic nitration is
Benzoic acid
Nitrobenzene
toluene
benzene
C.
toluene
The presence of electron releasing group like -R, -OH etc., increases the electron density at o/p position and thus, makes the benzene ring more reactive (at(o/p position) towards electrophile. On the other hand, electron withdrawing group like-COOH,-NO2 etc. If present, reduces electron density and thus, reduces the activity benzene nucleus towards electrophile.Thus, the order of the given compounds towards electrophilic nitration is
Thus, toluene is most reactive towards electrophilic nitration.
Which of the following reagents will be able to distinguish between 1-butyne and 2-butyne?
NaNH2
HCl
O2
Br2
A.
NaNH2
NaNH2 is used to distinguish between 1-butyne and 2-butyne.
Consider the reaction,
RCHO +NH2NH2 →RCH =N-NH2
What sort of reaction is it?
Electrophilic addition elimination reaction
Free radical addition-elimination reaction
Electrophilic substitution elimination reaction
Nucleophilic addition-elimination reaction
D.
Nucleophilic addition-elimination reaction
acts as nucleophilic species which adds on carbon. Therefore this is the case of the nucleophilic addition reaction. Simultaneously, a mole of H2O also eliminated. Therefore, overall this reaction is a nucleophilic addition-elimination reaction.
An organic compound (C3H9N) (A), when treated with nitrous acid, gave an alcohol and N2 gas was evolved. (A) on warming with CHCl3 and Caustic potash gave (C) which on reduction gave iso-propyl methylamine predict the structure of (A)
CH3-CH2 -NH -CH3
CH3CH2CH2-NH2
A.
What is the product obtained in the following reaction?
D.
Reduction of nitrobenzene with Zn/NH4Cl (neutral medium) gives phenyl hydroxyamine.
Method by which aniline cannot be prepared is
hydrolysis phenyl isocyanide with the acidic solution
degradation of benzamide with bromine in alkaline solution
reduction of nitrobenzene with H2/Pd in ethanol
potassium salt of phthalimide treated with chlorobenzene followed by the hydrolysis with aqueous NaOH solution
D.
potassium salt of phthalimide treated with chlorobenzene followed by the hydrolysis with aqueous NaOH solution
Due to resonance in chlorobenzene C- Cl bond accquires double bond chracter hence, C -Cl bond is inert towards nucleophile (phthalimide ion)
Which of the following compound is most basic?
B.
The compound is most basic due to localised lone pair of electrons on nitrogen atom while in other compounds , as result of resonance, the lone pair of electrons on nitrogen atoms gets delocalised over benzene ring and thus is less easily available for donation.
The following reaction, 
is known by the name
Friedel -Crafts reaction
Perkins reaction
Acetylation reaction
Schotten - Baumann reaction
D.
Schotten - Baumann reaction
Schotten- Baumann reaction is a method to synthesise amides from amines and acid chlorides.
Which of the following statements about primary amines is false?
Alkyl amine are stronger bases than aryl amines
Alkyl amines reacts with nitrous acid to produce alcohols
Aryl amines react with nitrous acid to produce phenols
Alkyl amines are stronger base than ammonia
C.
Aryl amines react with nitrous acid to produce phenols
i) The presence of electron withdrawing substituent decrease the basicity while the presence of electron releasing substituent like, -CH3 - C2H5 etc increases the acidity.
ii) HNO2 converts - NH2 group of an aliphatic amine into - OH while that of aromatic amines into - N =NCl.
Since phneyl group is a electron withdrawing group, it decreases the basicity. Alkyl group, on the other hand, being electron releasing, increases the basicity. Thus, alkyl amines are more basic as compared to arylamines as well as ammonia.
Thus, HNO2 (nitrous acid ) converts alkyl amines to alcohols. But,
Thus, HNO2 does not convert aryl amines into phenol.
Predict the product,
A.
With nitric acid (NaNO2) + HCl), primary aromatic amines give azo compounds and secondary aromatic amines gives nitroso compound
Which one of the following on reduction with lithium aluminium hydride yield a secondary amine?
Nitroethane
Methylisocyanide
Acetamide
Methyl cyanide
B.
Methylisocyanide
On a catalytic reduction or with nascent hydrogen or with lithium aluminium hydride (LiAlH4) alkyl isocyanide yield secondary amine.
Which of the following is more basic than aniline?
Diphenylamine
Triphenylamine
p-nitroaniline
Benzylamine
D.
Benzylamine
Benzylamine C6H5 - NH2 is more basic than aniline because benzyl group C6H5CH2 is electron donating group due to +I effect. So, it is able to increase the electron density of N of -NH2 group. Thus due to higher electron density rate of donation of a free pair of electron is increased i.e, basic character is higher. Phenyl and nitro group are electron attractive groups. so they are able to decrease the electron density of N of NH2 group. Hence, they are less basic with aniline.
The correct increasing order of basic strength for the following compounds is 
II < III < I
III < I < II
III < II < I
II < I < III
D.
II < I < III
–NO2 has strong –R effect and –CH3 shows +R effect. Therefore, Order of basic strength is 
Which of the following compounds can form a zwitter ion?
Aniline
Acetanilide
Glycine
Benzoic acid
C.
Glycine
Amphoteric Nature of Glycine: In aqueous solution, glycine exists as an inner salt or a dipolar ion.
This doubly charged ion is known as a Zwitterion or an ampholyte. Due to formation of the ampholyte, glycine acts both as an acid or a base.
In aqueous solution, Zwitter- ion tends to lose a proton. Hence, if electrolyzed, the negative glycine ion will migrate to anode.
The pH value for the isoelectric point of glycine is 6.0.
Nitration of aniline in strong acidic medium also gives m-nitroaniline because
Inspite of substituents nitro group always goes to only m-position.
In electrophilic substitution reactions amino group is meta directive.
In acidic (strong) medium aniline is present as anilinium ion.
In absence of substituents nitro group always goes to m-position.
C.
In acidic (strong) medium aniline is present as anilinium ion.
In acidic medium, aniline is protonated to form anilinium ion which is meta-directing.
Hence besides para (51%) and ortho (2%), meta product (47%) is also formed in significant yield.
Which of the following carbocations is expected to be most stable?
C.
-NO2 group is meta-directing group, therefore,
Less stable due to more e- withdrawing effect of -NO2
More stable due to the less e-withdrawing effect of NO2
The main product formed in the following reaction is
A.
As -NH3 group is more reactive than -OH group, CH3COCl will attack -NH3 group and give as the product.
The correct order of the basic strength of the following are
I>II>III>IV
IV>II>III>I
III>IV>II>I
III>II>IV>I
D.
III>II>IV>I
Electron releasing groups increases the basic nature while electron withdrawing group decreases the basis strength of amines.
Therefore, the correct order is III> II> IV>I
Which of the following sodium compound/compound (s) are formed when an organic compound containing both nitrogen and sulphur is fused with sodium?
Cyanide and sulphide
Thiocyanate
Sulphite and Cyanide
Nitrate and sulphide
B.
Thiocyanate
When an organic compound containing both nitrogen and sulphur is fused with sodium, sodium-thiocyanate is formed as follows,
Na + S + C + N → NaSCN
Sponsor Area
Sponsor Area
into hexan - 1.6-diamine?
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C
