Class 12 Engineering Chemistry - Amines

$$pK_b$$ of aniline is more than that of methylamine.
If this is true enter 1, if false enter 0.

$$pK_b$$ of aniline is more than that of methylamine.

It is because in aniline or other arylamines, the$$ -NH_2$$ group is attached directly to the benzene ring. It results in the unshared electron pair on nitrogen atom to be in conjugation with the benzene ring and thus, making it less available for protonation whereas in methylamine, there is no such conjugation involved and so, lone pair of electrons are more freely available for protonation.

Ethylamine dissolves in water due to intermolecular H-bonding.
If this is true enter 1, if false enter 0.

Ethylamine when added to water forms intermolecular H−bonds with water. Therefore, it is soluble in water.

Discuss the basic strength of two nitrogens in benzimidazole.

N-3 can doante its lone pair of electrons more faster so more base than N-1
1157895_259133_ans_27a60f3433424a9c81edaa2982f2504e.PNG

What is the reason that aromatic diazonium salt is more stable than aliphatic diazonium salt?

Aliphatic primary amines react with nitrous acid in cold condition to give aliphatic diazonium salts which exist only as transient intermediates, quickly decomposing into a nitrogen molecule and carbonium ion. Aromatic amines react with nitrous acid to give aromatic diazonium salt which are stable for short time and slowly decompose even at low temperatures, hence are immediately used after preparation they are also stable due the resonance of its ion.

Primary amines have higher boiling points than tertiary amines
If this is true enter 1, if false enter 0.

Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between 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 amines is as follows.

Primary > Secondary > Tertiary

Ethylamine is not soluble in water whereas aniline is.
If this is true enter 1, if false enter 0.

Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecules. However, solubility decreases with increase in molar mass of amines due to increase in size of the hydrophobic alkyl part.

Aniline has very large size of hydrophobic part and hence, it is insoluble in water.

Diazonium salts of aliphatic amines are more stable than those of aromatic amines.
If this is true enter 1, if false enter 0.

Aliphatic amines form highly unstable alkyldiazonium salts. Primary aromatic amines form arenediazonium salts which are stable for a short time in solution at low temperatures (273-278 K). The stability of arenediazonium ion is explained on the basis of resonance.

It has four resonating structure which makes them very stable.

Predict W,X, Y and Z

(Refer to Image)

$$X=AgCN, Y=DIBAL-H$$ (Diisobutylaluminium Hydride)

$$W=HCN$$ $$Z$$=Reduction by $$Pd-C/H_2$$

1219118_1114213_ans_4c3f18a91c774ce1a29667fe62e3fafc.png

What are coupling reaction?

1376555_1157398_ans_0bc1fd8b364d4970a05bf11d77ec1abd.jpeg

Write IUPAC name of the given compound:
$$C_6H_5−NH−C_6H_5$$

In the given compound, $$2$$ phenyl groups are attached with $$N$$ atom of amine group. Hence the IUPAC name of the given compound is $$N,N$$-diphenylamine.

Write down IUPAC name of $$CH_3-\overset{CH_3}{\overset{|}{N}H}$$.

As the methyl group is attached to a Nitrogen atom in methylamine. Hence the IUPAC name of the given compound is N-methyl methanamine.

The correct order of basicities of the following compounds is

2)$$CH_{3}CH_{3}NH_{2}$$
3)$$(CH_{3})_{2}NH$$

The order of basicity for amines in aqueous solution is

Secondary amine> tertiary>primary

$$(CH_3)_2Nh>(CH_3)_3N>CH_3CH_2NH_2$$

And due to equivalent resonance,the basicity is highest is (refer to image 01).

The order is (Refer to image 02)

Phthalic acid $$\displaystyle + NH_{3}\rightarrow D\xrightarrow{\Delta }E$$

1379583_125733_ans_9d478c85a72540c199f24d35e6ba536c.jpg

What is B?

$$NaBH_4$$ known as reducing agent will reduce the $$CN \to CH_2-NH_2$$
1011929_205877_ans_598144fc5de5453189b5c457fc5983a4.PNG

2

Azo Compounds  "Derivatives of diazene (diimide), HN=NH, wherein both hydrogens are substituted by hydrocarbyl groups.
Aldehydes : An aldehyde containing a formyl group. The formyl group is a funtional group, with the structure R-CHO, consisting of a carbonyl center (a carbon double bonded to oxygen) bonded to hydrogen and R group.
An amide or acid amide, is a compound with the functional group $$R_nE(O)_xNR^{'2}$$ (R and R' refer to H or organic groups)or $$NH_2$$.
Amines are organic compounds and functional group that contain a basic nitrogen atom with a lone pair. Amines are derivatives of ammonia, wherein one or more hydrogen atoms have been replaced by a substituent  such as an alkyl or aryl group.
Thus, A = 2, B =  3, C=  4, D = 1 is correct match.

What are the possible products?

The possible products are shown in the figure:

Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

An example of aromatic amine is $$PhNH_2$$ and aliphatic amine is $$MeNH_2$$.

Give a 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 points than tertiary amines?
iii. Why are aliphatic amines stronger bases than aromatic amines?

(i) The loss of a proton from an amine gives an imide ion, while the loss of a proton from alcohol gives an alkoxide ion as shown below:
$$\displaystyle \underset { Amine }{ R } -{ NH }_{ 2 }\longrightarrow R-\overset { \ominus }{ \underset { Imide\quad ion }{ NH } } +{ H }^{ \oplus }$$
$$\displaystyle R-O-H\longrightarrow R-{ O }^{ \ominus }+{ H }^{ \oplus }$$
Since O is more electronegative than N, therefore, $$\displaystyle \left( \overset { \ominus }{ RO } \right) $$ can accomodate the negative charge more easily than $$\displaystyle R\overset { \ominus }{ NH }$$ can accomodate the negative charge.
In other words, $$\displaystyle { RO }^{ \ominus }$$ is more stable than $$\displaystyle R\overset { \ominus }{ NH }$$. Thus, alcohols are more acidic than amines. Conversely, amines are less acidic than alcohols.
(ii) $$\displaystyle { 1 }^{ o }$$ amines form stronger intermolecular H-bonding between N atom of one and H atom of another molecule.
(iii) Aliphatic amines are stronger bases due to the +I effect of the alkyl group, leading to high electron density on the N atom.

Convert $$\displaystyle { CH }_{ 3 }{ COOH }$$ into $$\displaystyle ({ CH }_{ 3 }{ ) }_{ 2 }NH$$.

Among $$(CH_3)_3N$$ and $$CH_3NH_2$$, amine with ________ methyl groups attached has the higher boiling point.

The boiling point of methyl amine is 267 K whereas the boiling point of trimethyl amine is 276 K.

Thus, the boiling point of trimethyl amine is higher than the boiling point of methyl amine.

This is due to higher molecular weight of trimethyl amine as compared to methylamine.

3 methyl group is attached to it.

How many of the following are soluble in water?
(i) Ethylamine (ii) Aniline

Ethylamine is a low molecular weight amine. It forms hydrogen bonds with water. Hence, it is soluble in water.
Aniline is a high molecular weight amine. It has large hydrocarbon part which interferes in the hydrogen bond formation.
Hence, it is insoluble in water.

In the given compound which site acts as an acid and which as a base?

$$-$$ve part of zwitterions acts as a base.

$$+$$ve part of zwitterions acts as an acid.

1174797_264397_ans_ed2c9cc910b64f76bde011104d7fc6c6.png

Write the compound $$(A)$$ and $$(B)$$ formed in this.

1155069_264400_ans_8dc231771e324497a637650ec054aed4.png

Arrange the following
(i) In decreasing order of the $$pK_b$$ values: $$C_2H_5NH_2, C_6H_5NHCH_3, (C_2H_5)_2NH$$ and $$C_6H_5NH_2$$
(ii) In increasing order of basic strength: $$C_6H_5NH_2, C_6H_5N(CH_3)_2, (C_2H_5)_2NH$$ and $$CH_3NH_2$$
(iii) In increasing order of basic strength:
(a) Aniline, p-nitroaniline and p-toluidine

(i) Stronger base has lower $$pK_b$$ value.
$$\displaystyle C_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5NH_2 > (C_2H_5)_2NH$$
(ii) Increasing order of basicity is as follows:
$$\displaystyle C_6H_5NH_2 < C_6H_5NHCH_3 < CH_3NH_2 < (C_2H_5)_2NH$$
(iii) Increasing order of basic strength.
(a) p-nitroaniline < aniline < p-toluidine
(b) $$\displaystyle PhNH_2 < PhNHCH_3 < PhCH_2NH_2$$
(iv) In decreasing order of basic strength in gas phase:
$$\displaystyle Et_3N > Et_2NH > Et NH_2 > NH_3$$
The stabilization of conjugate acids due to hydrogen bonding is absent.
(v) Increasing order of boiling point
$$\displaystyle Me_2NH < EtNH_2 < EtOH$$
(vi) Increasing order of solubility in water:
$$\displaystyle PhNH_2 < Et_2NH < EtNH_2$$

Amines are more acidic than alcohols of comparable molecular masses.
If this is true enter 1, if false enter 0.

Alcohols are more acidic than amines because of electronegativity of oxygen of alcohol, $${ H }^{ + }$$ can be easy removal thus acidic strength increases.

(a) Illustrate the following reaction giving suitable example(s) in each case:
(i) hoffmann bromamide degradation reaction
(ii) diazotisation
(iii) gabriel phthalimide synthesis
(b) Distinguish between the following pairs of compounds:
(i) aniline and N-methylaniline
(ii) $${ \left( C{ H }_{ 3 } \right) }_{ 2 }NH$$ and $${ \left( C{ H }_{ 3 } \right) }_{ 3 }N$$
OR
(a) Write the structures of main products when benzene diazonium chloride $$\left( { C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }^{ + }{ Cl }^{ - } \right) $$ reacts with the following reagents:
(i) $${ CuCN }/{ KCN }$$
(ii) $${ H }_{ 2 }O$$
(iii) $$C{ H }_{ 3 }C{ H }_{ 2 }OH$$
(b) Arrange the following:
(i) $${ C }_{ 2 }{ H }_{ 5 }N{ H }_{ 2 },{ C }_{ 2 }{ H }_{ 5 }OH,{ \left( C{ H }_{ 3 } \right) }_{ 3 }N$$ -in the increasing order of their boiling point
(ii) aniline, p-nitroaniline, p-methyl aniline - in the increasing order of their basic strength

(a)

(i) Hoffmann bromamide degradation reaction-

Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide.

In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. The amine so formed contains one carbon less than that present in the amide.

(ii) Diazotisation :-Benzenediazonium chloride is prepared by the reaction of aniline with nitrous acid at $$273-278K$$. Nitrous acid is produced in the reaction mixture by the reaction of sodium nitrite with hydrochloric acid. The

conversion of primary aromatic amines into diazonium salts is known

as diazotisation.

(iii) Gabriel phthalimide synthesis - Gabriel synthesis is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine. Aromatic primary amines cannot be prepared by this method

because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

(b)

(i) Given in image

(ii) $$(CH_3)_2NH$$ = Secondary amine

$$(CH_3)_3N$$ = tertiary amine

1016637_493155_ans_518e60b693cc4440b97e921011f68713.PNG

The conversion of primary aromatic amines into diazonium salts is known as ______.

The conversion of primary aromatic amines into diazonium salts is known as diazotisation.
Arylamines react with nitrous acid in cold condition to give arenediazonium salts.
The arenediazonium salts are stable for a short time and slowly decompose even at low temperature.
Hence, they are used immediately after preparation.
$$\displaystyle Ar-NH_2 + NaNO_2 + 2HCl \xrightarrow {273-278 K} Ar-N_2^+Cl^- + NaCl + 2H_2O$$

Arrange the following in the increasing order of their basic strength in aqueous solution: $$CH_3NH_2, (CH_3)_3N, (CH_3)_2NH$$

According to fact, when $$R=CH_3$$ group in aliphatic amine, the basic strength would be

$$(CH_3)_2NH>CH_3NH_2>(CH_3)_3N$$

This altered basicity is due to the solvation effect in aqueous solution

Account for the following:
(i) Primary amines $$\left( R-N{ H }_{ 2 } \right) $$ have higher boiling point than tertiary amines $$\left( { R }_{ 3 }N \right) $$.
(ii) Aniline does not undergo Friedel-Crafts reaction.
(iii) $${ \left( C{ H }_{ 3 } \right) }_{ 2 }NH$$ is more basic than $${ \left( C{ H }_{ 3 } \right) }_{ 3 }N$$ in an aqueous solution.

(i) Due to intermolecular hydrogen bonding in primary amines (presence of more number of H-atoms ).

(ii) Because aniline is basic in nature and reacts with $$AlCl_3$$ to form salt. The positive charge on the N-atom, electrophile substitution in the benzene ring is deactivated. (refer to given image)

(iii) Due to steric hindrance and solvation factor in case of ($$CH_3)_3N$$, it is less basic than ($$CH_3)_2NH$$ in aqueous solution.

Arrange the following in the decreasing order of their basic strength in aqueous solution:
$$CH_3NH_2, (CH_3)_2NH, (CH_3)_3 N$$ and $$NH_3$$

The order of basic strength is $$NH_3<(CH_3)_3N<CH_3NH_2<(CH_3)_2NH$$

Aliphatic amines are stronger bases than ammonia because of $$+I$$ (inductive) effect. The alkyl groups, which are electron releasing groups, increase the electron density around the nitrogen thereby, increasing the availability of the lone pair of electrons to proton or Lewis acids and making the amine more basic. But in tertiary amines crowding of alkyl groups cover nitrogen atom from all sides and thus makes it unable for protonation. Due to steric factors tertiary amines are less basic than primary amines.

An aromatic primary amine A with molecular formula $$C_{6}H_{7}N$$ undergoes diazotisation to give B. B when the aqueous solution of B is boiled it gives C. Identify A,B and C.

Compound A with molecular formula $$C_{6}H_{7}N$$ is Aniline $$\left ( C_{6} H_{5}NH_{2}\right )$$
Compound B is Benzene diazonium chloride $$\left ( C_{6} H_{5}-N=N-Cl\right )$$
Compound C is phenol $$\left ( C_{6}H_{5}OH \right )$$
Reaction:

Arrange the following compounds in increasing order of their basic strength. Explain reason
$$C_2H_5 NH_2, NH_3, C_6H_5 NH_2$$

The increasing order of their basic strength is $$\displaystyle C_6H_5 NH_2 < NH_3 < C_2H_5 NH_2 $$
In aniline, the lone pair of electrons on N atom is involved in resonance with benzene ring. Hence, it is less available for donation to suitable lewis acid. Hence, aniline is less basic than ammonia.
In ethyl amine, the ethyl group has +I effect. It increases electron density on N atom. Hence, N can easily donate its lone pair of electrons to suitable lewis acid. Hence, ethyl amine is more basic than ammonia.

Explain the following reactions:
(i) Diazotization
(ii) Carbylamine reaction

(i) Diazotization:
Aromatic primary amines reacts with nitrous acid in cold condition to give diazonium salts.
$$\displaystyle Ph -NH_2 + NaNO_2+2HCl \xrightarrow {273-278 K} PhN_2^+Cl^-+NaCl+2H_2O$$
Diazonium salts are stable for short time and decompose slowly even at low temperatures. They are used immediately after preparation. The diazonium group can be replaced by various nucleophiles such as halide anions, cyanide, thiols, water, and others.
(ii) Carbylamine reaction:
Aliphatic/aromatic primary amines are heated with chloroform/alcoholic KOH to form foul smelling alkyl isocyanides or carbylamines. This test is not given by secondary/tertiary amines.
$$\displaystyle R-NH_2+CHCl_3 + 3KOH \xrightarrow {\Delta} R-NC + 3KCl + 3H_2O$$
For example, ethylamine gives ethyl isocyanide.
$$\displaystyle C_2H_5-NH_2+CHCl_3 + 3KOH \xrightarrow {\Delta} C_2H_5-NC + 3KCl + 3H_2O$$

How can urea be detected by Biuret test?

Biuret is a condensation compound of urea equivalent to two molecules of urea less one of ammonia.
The parent compound can be prepared by heating urea above the melting point at which temperature of $$N{H}_{3}$$ is expected.
$$2CO{ \left( N{ H }_{ 2 } \right) }_{ 2 }\rightarrow { H }_{ 2 }N-CO-NH-CO-N{ H }_{ 2 }+N{ H }_{ 3 }\uparrow $$

What are functional groups? Name the class of compounds containing $$- NH_2$$ as the functional group.

The atom or group of atoms present in the molecule which determines characteristics property of organic compounds is called the functional group.
Functional groups are the sites where reactions occurs in organic molecules. $$-NH_2$$ is the functional group of amines. For example, methyl amine $$CH_3-NH_2$$ contains $$-NH_2$$ functional group.

Identify the reagents X, Y and Z.
$$C_2 H_2 Cl \xrightarrow{X} C_2H_5CN \xrightarrow{Y} C_2H_5CH_2 NH_2 \xrightarrow{Z} C_2H_5 CH_2 NHCOCH_3$$

The reagents X, Y and Z are alcoholic $$\displaystyle \,KCN, \, LiAlH_4,\, CH_3 COCl$$ respectively.
$$\displaystyle C_2 H_5 Cl \xrightarrow{alc \: KCN} C_2H_5CN \xrightarrow{LiAlH_4} C_2H_5CH_2 NH_2 \xrightarrow{CH_3COCl} C_2H_5 CH_2 NHCOCH_3$$
Alcoholic KCN converts ethyl chloride to ethyl cyanide. Chloride group is replaced with cyano group.
Lithium aluminium hydride reduces cyano group to amino group. The amino group is then acetylated with acetyl chloride.

Arrange the following in decreasing order of their basic strength:

$$C_6H_5NH_2, C_2H_5NH_2, (C_2H_5)_2 NH, NH_3$$

The decreasing order of basic strength is:
$$\displaystyle (C_2H_5)_2 NH > C_2H_5NH_2 > NH_3 > C_6H_5NH_2 $$
Ammonia is more basic than aniline as in aniline, the lone pair of electron on N atom is in resonance with benzene ring. Hence, it cannot be donated easily.
Ethyl amine is more basic than ammonia because in ethyl amine, the $$+I$$ effect of ethyl group increases the electron density on $$N$$.

Diethylamine is most basic due to $$+I$$ effect of two ethyl groups.

Give reasons of the following:
$$CH_{3}NH_{2}$$ is more basic than $$C_{6}H_{5}NH_{2}$$

In aniline lone pair present on nitrogen is in resonance with the benzene ring. Hence due to delocalisation of lone pair present on nitrogen basicity of aniline decreases.

Write a note on Hofmann bromamide degradation.

Hofmann bromamide degradation:
Treatment of an $$amide$$ with $$\displaystyle bromine $$ and aqueous / alcoholic $$\displaystyle sodium \: hydroxide$$ gives a $$\displaystyle primary \: amine$$
$$\displaystyle R-CONH_2+Br_2+4NaOH \rightarrow R-NH_2 + Na_2CO_3 +2NaBr + 2H_2O$$

The product $$\displaystyle amine$$ contains one carbon atom less than starting $$\displaystyle amide$$. This reaction involves migration of an alkyl or aryl group from the carbonyl carbon to the adjacent nitrogen atom.

For example, $$\displaystyle propanamide$$ on Hofmann bromamide degradation gives $$\displaystyle ethanamine$$
$$\displaystyle CH_3-CH_2-CONH_2+Br_2+4NaOH \rightarrow CH_3-CH_2-NH_2 + Na_2CO_3 +2NaBr + 2H_2O$$

What is Oil of mirbane.

Nitrobenzene is also used to mask unpleasant odors in shoe and floor polishes, leather dressings, paint solvents, and other materials. Redistilled, as oil of mirbane, nitrobenzene has been used as an inexpensive perfume for soaps.

The basic character of Amine is due to the presence of ________________ on Nitrogen atom.

The basic character of amine is due to the presence of lone pair of electron on nitrogen atom. Amines can easily donate this lone pair of electron on nitrogen atom to a suitable acid and form coordinate covalent bond.

$$\displaystyle CH_3-NH_2 + HCl \rightarrow CH_3-NH_3^++Cl^-$$

In the above reaction, $$\displaystyle methyl \: amine$$ donates a pair of electrons to proton to form
$$\displaystyle methyl \: ammonium \: ion$$. Thus, it acts as a base.

Methyl amine is __________ basic than ammonia.

Note that the strongest acid is ammonia. This means that ammonia is the weakest base of the four bases. Basicity order is

ammonia < primary amine < secondary amine < tertiary amine

Write the name of reaction:
$${ C }_{ 6 }{ H }_{ 5 }N{ H }_{ 2 }+CH{ Cl }_{ 3 }+3KOH\underset { \quad Solution\quad }{ \underrightarrow { \quad Alcoholic\quad } } { C }_{ 6 }{ H }_{ 5 }NC+3KCl+3{ H }_{ 2 }O$$

The reaction $$\displaystyle { C }_{ 6 }{ H }_{ 5 }N{ H }_{ 2

}+CH{ Cl }_{ 3 }+3KOH\underset { \quad Solution

\quad }{ \underrightarrow { \quad Alcoholic\quad }

} { C }_{ 6 }{ H }_{ 5 }NC+3KCl+3{ H }_{ 2 }O

$$ is Hoffmann's cabylamine test /isocyanide test. It is not given by secondary and tertiary amines. It is given by aliphatic / aromatic primary amines. The product is foul smelling isocyanide.

Write the name of reaction:
$${ C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }Cl\xrightarrow { \quad { CuCl }/{ HCl }\quad } { C }_{ 6 }{ H }_{ 5 }Cl+{ N }_{ 2 }$$

The name of reaction $$\displaystyle { C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }Cl\xrightarrow { \quad { CuCl }/{ HCl }\quad } { C }_{ 6 }{ H }_{ 5 }Cl+{ N }_{ 2 }$$ is Sandmeyer reaction. It converts aryl diazonium salts to aryl halides. The reagent used is $$\displaystyle CuCl/HCl$$, The byproduct is nitrogen gas.

Write the name of following compounds:
(A) $$C_{6}H_{5}N_{2}^{\oplus} HSO_{4}^{\ominus}$$
(B) Ref. image

The names of the compounds are given below.
Benzenediazonium hydrogen sulphate is similar to Benzenediazonium hydrogen chloride $$ \displaystyle Ph-N_2-Cl$$. 4-(phenyldiazenyl)phenol is obtained by azo coupling.

Write the reactions involved in the Diazotisation.

Diazotisation:
Action of nitrous acid on aniline at 273-278 K gives benzene diazonium chloride.

Nitrous acid is generated in-situ by the reaction between sodium nitrite and dilute hydrochloric acid.

$$C_6H_5NH_2+NaNO_2+2HCl \xrightarrow {273-278K} C_6H_5-N_2^+Cl^- + NaCl +2H_2O$$

Write the following reactions:
(i) Williamson's synthesis
(ii) Mendius reaction
(iii) Friedel Craft's Alkylation
(iv) Haloform reaction
(v) Gattermann reaction
(vi) Carbylamine reaction

(i) Williamson Synthesis:

$$\underbrace { RX }_{ (alyl\quad Halide) } +R^{\prime}ONa\longrightarrow \underbrace{ROR^{\prime}}_{Ether Formation}+NaX$$

(ii) Mendius Reaction:

$$RCN+\underbrace{[H]}_{nescent\quad Hydrogen}\longrightarrow R-CH_2NH_2$$

reducing of isonitrite.

(iii) Friedel Craft's Alkylation:

Refer to image 01

This is alkylation of aromatic compound.

(iv) Haloform Reaction:

Refer to Image 02.

This is specifically Iodoform Reaction.

(v) Gattermann Reaction

Refer to Image 03.

Diazonium Chloride on reaction with $$Cu$$ and $$HCl$$ gives.

1071495_771930_ans_1c4a0ef17947412cbb1644e080f109ff.png

Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

In this case, aromatic primary amine involves mesophilic substitution $$SN^2$$ of alkyl halides by the anion formed by phthalimide. But aryl halides do not undergo nucleophilic substitution with the anion form by the phthalimide. Hence aromatic primary animes cannot be prepared by this process.

1019368_849333_ans_3b1d48e28d3e4c41b2571d6220838921.png

How are ethylamine and ethyl methyl amine distinguished by using nitrous acid?

Ethylamine (a primary amine) reacts with nitrous acid in cold condition to give ethyl alcohol with liberation of nitrogen gas.

$$\displaystyle CH_3-CH_2-NH_2 + HNO_2 \xrightarrow [273-278 \ K]{NaNO_2+HCl} CH_3-CH_2-OH +H_2O+N_2 \uparrow$$

Ethyl methyl amine (a secondary amine) reacts with nitrous acid in cold condition to give yellow oily N-nitrosoethyl methyl amine.

$$\displaystyle CH_3-CH_2-NH-CH_3 + HNO_2 \xrightarrow [273-278 \ K]{NaNO_2+HCl} CH_3-CH_2-N(CH_3)-N=O +H_2O $$

Arrange the following in increasing order of their basic strength:
$$CH_3NH_2(I), (CH_3)_2NH (II), (CH_3)_3N (III), C_6H_5CH_2NH_2 (IV)$$

$${ CH }_{ 3 }{ NH }_{ 2 } $$, $$({ CH }_{ 3 })_{ 2 }{ NH } $$, $${ (CH }_{ 3 })_{ 3 }N$$, $${ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }{ NH }_{ 2 }$$

The basic strength $$(III)>(II)>(IV)>(I).$$

In case of compound $$(iv)$$ the inductive effect of phenyl is on the methyl, decreasing it's ability to donate electron density to nitrogen atom. As facility of $$1',2',3'$$ amines is determine by the tendency of donating electrons and $$3'$$ amines will have the height $$+I$$ effect due to presence of three methyl group attached to the $$N$$ atom , leading to greater electron density followed by $$2'$$ and $$1'$$ amine.

Which is more basic?

The structures are shown in the image$$.$$

The lone electrons of $$N$$ in pyrrole is delocalized $$.$$ It is part of the $$Pi$$ system$$,$$ thus involved in aromatization $$.$$ This $$L.P$$ is no longer available for a proton hence it is a weak base$$.$$

One the other hand$$,$$ aniline is already aromatic$$.$$ Through the lone pair of $${e}^{-}$$ of $$-N{H}_{2}$$ group of aniline is delocalized over the benzene ring$$,$$ it is not involved in aromatizetion $$.$$ This pair of lone electron still available for a proton$$,$$ hence aniline is stronger base than pyrrole$$.$$

1021054_878180_ans_c2a39626f52a497ebfec84db525f13be.png

What will be product of the reaction?

1443197_879332_ans_93767aa24a1f433f8f1f7f16075f6212.png

$$CH_3-NH_2$$ $$CH_3-NH-CH_3$$ $$CH_3-\underset{CH_3}{\underset{|}{N}}-CH_3$$
I II III

Compare boiling points of the given compounds:

$$CH_3-NH_2$$ $$CH_3-NH-CH_3$$ $$CH_3-\underset{CH_3}{\underset{|}{N}}-CH_3$$

1' amine 2' amine 3' amine

I II III

The first step refers to the image$$.$$

In case of compound $$(I)$$ and $$(II)$$ $$,$$ electron density in $$1'$$ amine is less than $$2'$$ amine and two methyl group attached to $$2'$$ amine having $$+I$$ effect$$.$$ i.e, why primary amine have more stable intermolecular $$H$$-bonding than secondary amine$$.$$ In $$1'$$ amine have $$2$$ $$H$$-atom and $$2'$$ amine have only one $$H$$-atom for intramolecular $$H$$-bonding$$.$$ So$$,$$ the boiling point of $$1'$$ amine is higher than the $$2'$$ and $$3'$$ amine$$.$$ Order$$\Rightarrow 1'>2'>3'.$$

Arrange the basic strength of the following compounds.

Cyclohexamine is more basic than aniline because$$,$$ In cyclohexamine$$,$$ the electron donating group is attached to $${sp}^{3}$$ hybridised cyclohexane ring while in case of aniline it attached with electron withdrawing benzene ring$$.$$ Since$$,$$ cyclohexamine is more basic than anilin and also $$Ph-NH-Ph.$$

The order is represented in the image$$.$$

1020975_875012_ans_5598a57aa6f743d0b0ad6432c39260b7.png

question

Solution:-

In the given compound, an aldehyde group, $$1$$ ethyl group, $$1$$ amine group and $$1$$ hydroxy group are attached to the benzene ring. Since the aldehyde group has highest priority, thus the numbering will be from aldehyde group.

Hence the IUPAC name of the given compound is $$2$$-hydroxy-$$4$$-ethyl-$$6$$-amino benzaldehyde.

Which reaction shows that amine is base?

$${ CH }_{ 3 }-{ CH }_{ 2 }-{ NH }_{ 2 }+{ H }^{ + }\longrightarrow { CH }_{ 3 }-{ CH }_{ 2 }-{ NH }_{ 3 }^{ \oplus }$$

Why diazonium salts do not show positive Lassaigne's test for nitrogen?

Diazonium salts are stable only at low temperatures $$(0-50^{o C})$$. At higher temperatures, they decompose to give nitrogen. If nitrogen is lost, then NaCN will not be formed and hence the compound will not give positive test for nitrogen. Thus, diazonium salts do not give Lassaigne's test

Rearrange the following in an increasing order of their basic strengths :
$$C_6H_5NH_2$$,$$C_6H_5N(CH_3)_2$$, $$(C_6H_5)_2NH$$ and $$CH_3NH_2$$.

$$({ C }_{ 6 }{ H }_{ 5 }{ ) }_{ 2 }NH<{ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }<{ C }_{ 6 }{ H }_{ 5 }N({ CH }_{ 3 }{ ) }_{ 2 }<{ CH }_{ 3 }{ NH }_{ 2 }$$

Due to the $$-I$$ effect of $${ C }_{ 6 }{ H }_{ 5 }$$ group, the electron density on N-atom $${ C }_{ 6 }{ H }_{ 5 }-{ NH }_{ 2 }$$ is lower than that on the N-atom $$({ C }_{ 6 }{ H }_{ 5 }{ ) }_{ 2 }NH$$ . But in case of $${ C }_{ 6 }{ H }_{ 5 }N({ CH }_{ 3 }{ ) }_{ 2 }$$, two $${ CH }_{ 3 }$$ group attached with N atom which increases the electron density on N-atom and in $${ CH }_{ 3 }-{ NH }_{ 3 }$$ only $$+I$$ effect of methyl group increases the basicity of the compound

Which is more basic and why?

Benzylamine $$\displaystyle (C_6H_5-CH_2-NH_2)$$ is more basic than aniline $$\displaystyle (C_6H_5-NH_2)$$. In benzylamine, the lone pair of electrons on N atom can be freely donated as it is not involved in resonance with the benzene ring. In aniline, the lone pair of electrons on N atom cannot be freely donated as it is involved in resonance with a benzene ring.

Explain.

Basic character in a compound involved easy donation of electrons. This becomes difficult in case of similar as the lone pair of electrons are not easily available where in case of $$Ph-CH_2-NH_2$$ does not have its lone pair of electrons for delocalized.

Classify the given amines a primary, secondary or tertiary:

Amines are organic compounds derived from ammonia ($$N{ H }_{ 3 }$$). They are derived by replacing its one or more hydrogen atoms by alkyl or aryl groups.

Amines are classified into 3 categories

(1)primary amines

In primary amines, one hydrogen atom of $$N{ H }_{ 3 }$$ is replaced by an alkyl or aryl group.

(2) secondary amines

In secondary amines, two hydrogen atoms of $$N{ H }_{ 3 }$$ are replaced by alkyl or aryl groups.

(3) In tertiary amine, all the three hydrogen atoms of $$N{ H }_{ 3 }$$ are replaced by alkyl or aryl groups.

Classification of the given amines are:

(iii) $$({ C }_{ 2 }{ H }_{ 5 })_{ 2 }CH{ NH }_{ 2 }$$ ------ Primary amine

(iv) $$({ C }_{ 2 }{ H }_{ 5 })_{ 2 }NH$$ ------- Secondary amine

1037474_1067973_ans_6d6a6631a0204e53abc1989389b48d1a.jpg

Arrange the given amines in the order of their decreasing basicity:

Basic character is due to the localized loan pair.

Steric inhibition of resonance is observed if bulky groups are attached due to which the loan pair gets localized since the nitrogen goes out of the plane.

Also, methyl group acts as electron donating group which increases the basicity.

1075015_1068128_ans_f081d0d44720471681ea58b0e6c1fcd5.jpg

Which among the following are the correct statements?
Statement 1: Methyl-amine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
Statement 2: Methylamine is more basic than ammonia.

Due to the +I effect of −CH₃ group, methylamine is more basic than ammonia. Therefore, in water, methylamine produces OH− ions by accepting H+ ions from water. Ferric chloride $$FeCl_3$$ dissociates in water to form $$Fe^{3+}$$ and $$Cl^{-}$$ ions. Then, $$OH^{-}$$ ion reacts with $$Fe^{3+}$$ ion to form a precipitate of hydrated ferric oxide.

Thus, both the statements are correct.

Justify the formation of given products and compare basic character of major organic product and aniline.

The factor effecting the basicity of amines are gradually for aromatic amines they are less basic than aliphatic amines due :-

(1) The lone pair of electrons on the nitrogen atom is declocalised into benzene rings as a result,the lone pair of electrons is less available for donation to an acid.

(2) Electron withdrawing group attached to the benzene ring decreases the basicity of amines. Eg:- refer to image

Due to more - I effect of halogen(i.e Bromine),the compound 2,4,6 tribromo aniline is less basic.

1013188_1069104_ans_8e1a1f9605904ec9929afae76070edfc.png

1) Butanenitrile to propanamine
2) propanenitrile to propanamine

$$s)$$ $$\displaystyle \underset{Butanenitrile}{CH_3 CH_2 CH_2CN} \xrightarrow[hydrolysis]{partial} \underset{\underset{(propanamine)}{\,\,\,\,CH_3CH_2CH_2-NH_2}}{\underset{\,\,\,\,\,\,\,\,\,\,\,\downarrow Br_2 \,KHO}{CH_3CH_2CH_2 - \overset{O}{\overset{||}{C}} - NH_2}}$$

$$t)$$ $$\underset{propanenitrile}{CH_3CH_2 CN} \xrightarrow{LiAl x_4} \underset{(propanamine)}{CH_3CH_2CH_2NH_2}$$

Arrange the following in the increasing order of their basic strength.
$${CH}_{3}{NH}_{2},{({CH}_{3})}_{2}NH,{C}_{6}{H}_{5}{NH}_{2},{C}_{6}{H}_{5}{CH}_{2}{NH}_{2}$$

Since the basic character depend on the inductive effect therefore correct order is

$$C_{6}H_{5}NH_{2}<C_{6}H_{5}CH_{2}NH_{2}<CH_{3}NH_{2}<\left ( CH_{3} \right )_{2}NH$$

Write the structures of the main products when benzene diazonium chloride reacts with the following reagents:
(a) $$CuCN$$ (b) $${CH}_{3}{CH}_{2}OH$$ (c) $$Cu/HCl$$

1153722_1071445_ans_88cd8ba00b784d8fa4f16fb539e1b2c5.png

In each of the following pair of compounds, which is more basic in aqueous solution? Give an explaination for your choice:

(i) Presence of an electron-withdrawing group decreases the lone pair availability on Nitrogen. Hence, $$CH_3NH_2$$ is more basic than $$CF_3NH_2$$.

(ii) $$2^{nd}$$ compound has more nitrogen atoms and hence more lone pair availability. Therefore, (Refer to Image) is more basic than $$CH_3CONH_2$$.

(iii) In $$CH_3CN$$, there is a triple bond between $$C$$ and $$N$$,. Its lone pair is less available. On the other hand, lone pair availability is more in $$PhNH_2$$. Therefore, $$PhNH_2$$ is more basic than $$CH_3CN$$.

1104064_1075933_ans_08bfa5d89bf84fd3857def54bc1813bb.png

Arrange the following compound according to their basic strength (with suitable explanation):
$${ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 },{ C }_{ 6 }{ H }_{ 5 }NHMe,{ C }_{ 6 }{ H }_{ 5 }NM{ e }_{ 2 },o-Me{ C }_{ 6 }{ H }_{ 4 }{ NH }_{ 2 }$$

$$\therefore$$$$O-M_cC_6H_4NH_2$$ is less than basic due to ortho effect decrease the basic character. Due to steric effect (equation refer to image ) is less basic.
1013282_1076964_ans_a4e73dd1da3f48f99bc325e8c376ab2e.png

1013282_1076964_ans_a4e73dd1da3f48f99bc325e8c376ab2e.png

Compare the basic strengths of the given compounds.

(a) As the number of $$-Ph$$ group increases, the availability of lone pair on nitrogen decreases. This is because the lone pair is attracted by the phenyl groups. Thus more the phenyl groups, less is the basic strength.

$$\therefore$$ Order of basic strength: $$PhNH_2 > Ph_2NH > Ph_3N$$

(b) In the $$3^{rd}$$compound, the lone pair is involved in making the ring aromatic. Thus it will be least basic.

The $$2^{nd}$$ compound has less availability of lone pair on the nitrogen due to the presence of phenyl group. The availability of lone pair is maximum in $$1^{st}$$ one. Hence it will be most basic.

$$\therefore$$ Order of Basic strength:

(Refer to Image)

(c) The $$2^{nd}$$ compound has the least availability of lone pair due to the presence of phenyl group directly attached to nitrogen. Hence it is least basic.

As the distance of phenyl group increases, the lone pair is more available. This is because the electron withdrawing effect of the phenyl group decreases.

Hence, the order of basic strength is:-

$$Ph-CH_2-CH_2-NH_2 > CH_3-\underset {|}{CH}-NH_2 > CH_3-CH_2-\underset {|}{NH}$$

$$Ph$$ $$Ph$$

1104049_1075210_ans_1406bc364f594b778930c5b93fee9724.png

Arrange the following in increasing order of their basic strength:
$$C_2H_5NH_2, C_6H_5NH_2, NH_3, C_6H_5CH_2NH_2$$ and $$(C_2H_5)_2NH$$

Basic strength:- Basic strength of an amine is determined by availability of electron from nitrogen.

Basic strength increasing order:-

$$(C_2H_5)_2NH > C_2H_5NH_2 > NH_3 >C_6H_5CH_2NH_2 > C_6H_5NH_2$$

Arrange the following in the increasing order of basic strength
$$C_6H_5NH_2, \; C_6H_5NHCH_3, \; C_6H_5CH_2NH_2$$

The increasing order of

basic strength is

$$\boxed{C_{6}H_{5}NH_{2}< C_{6}H_{5}NH CH_{3} < C_{6}H_{3}CH_{2}NH_{2}}$$

$$\rightarrow$$ This is because in $$C_{6}H_{5}NH_{2}$$

the lone pair of electrons in $$N$$ in aromatic amine is delocalized

due to resonance.

$$\rightarrow C_{6}H_{5}NH CH_{3}$$ is a secondary amine

having $$-C_{6}H_{5}$$ on one side & $$-CH_{3}$$
on another side $$-CH_{3}$$ is an

electron donating group & increases

the basicity.

$$\rightarrow C_{6}H_{5}CH_{2}NH_{2}$$ is a primary amine

& not directly attached to aromatic

ring. Thus it has done pair of

electrons on $$N$$ available for

donation.

Final product for this reaction is:

1406402_1100079_ans_d284a9acda184676a3d9d3eb1cf033cd.png

Arrange the following in increasing order of their basic strength:
$$C_2H_5NH_2, (C_2H_5)_2NH, (C_2H_5)_3N, C_6H_5NH_2$$

Basic strength:- Amine basic strength is determined by availability of electron [lone pair] from nitrogen.

Increasing order of basic strength:-

$$(C_2H_5)_2NH > (C_2H_5)_3N > C_2H_5NH_2 > C_6H_5NH_2$$

$$(C_2H_5)_3N$$ is less basic than $$(C_2H_5)_2NH$$ because of steric hindrance, electrons are less available from $$(C_2H_5)_3N$$ than $$(C_2H_5)_2N$$

Which is more basic? Explain in simple words.

Due to more resonating structure (b) shows more basic than structure (a).
1157821_1103306_ans_8b24917cb6a54ed190e0a9f20e24c5bd.png

Arrange the following amines in the decreasing order of their basic nature :

(a) Aniline, propane-1-amine and N-methyl ethanamine
(b) Benzene-1-diamine, ammonia and 4-aminobenzoic acid.
(c) N- Methylaniline, phenylmethylamine and N-Phenylaniline

N - methyl ethanamine > propane - $$1$$ - amine > aniline.

$$NH_3$$ > benzene-$$1$$,$$4$$-diamine > $$4$$-amino benzoic acid

N - methylaniline $$\approx$$ phenymethylamine > N - phenylalanine.

How does the formation of $$2^{\circ}$$ and $$3^{\circ}$$ amines can be avoided during the preparation of $$1^{\circ}$$ amines by alkylation?

The use of very large excess of $$NH_3$$ may partially avoid the formation of $$2^o$$ and $$3^o$$ amines during preparation of $$P$$ amines. But $$(a)$$ by treatment of with $$NaN_3$$ followed by reduction or $$(b)$$ Gabriel phthalimide synthesis or $$(c)$$ reduction of amide can give $$P$$ amines only.

(a) $$R-X \xrightarrow[]{NaN_3}R-N_3\xrightarrow[]{H_2/Pd}\underset {1^o amine}{R-NH_2+N_2}$$

(b) (Refer to Image)

$$R-\overset {||}{C}-NH_2 \xrightarrow []{LiAlH_4} R-CH_2NH_2$$

1184407_1111664_ans_e29790cfe6c44b4a83fbc6bd2a9f9290.JPG

Basic nature order of the following is:

Basic nature order -

c >a> b

Explanation for C- Lewis base is the entity that has the tendency to donate electrons, here in c , the lone pair of nitrogen remain at the nitrogen that's why it is easy to donate electron

Explanation for a- The lone pair of nitrogen is going in the resonance, that's why it is not stable at the nitrogen atom it is not very easy to donate electrons

Same reason is valid for b, but in b the lone pairs of nitrogen is in extended resonancex so its very much hard to donate electron in this case.

1380090_1119679_ans_6c19f927c59049a2a6a6c7430c943a19.jpg

Why are aliphatic amines stronger bases than aromatic amines?

In aliphatic amines, the $$-NH_2$$ group is attached to an alkyl group which is an electron donating group i.e, they have $$+I$$ effect. So, the lone pair of electrons on nitrogen are readily available.

In aromatic amines, the $$-NH_2$$ group is attached to a $$-C_6H_5$$ group, which is an electron withdrawing group. So, the availability of a lone pair of electrons on $$N$$ is decreased. Therefore aliphatic amines are more basic than aromatic amines.

Complete the following acid-base reactions and name the products:
(i)$$CH_3CH_2CH_2NH_2+HCl\rightarrow$$
(ii)$$(C_2H_5)N+HCl\rightarrow$$

Acid base reactions :

$$i) CH_3CH_2CH_2NH_2+HCl\longrightarrow \underset{\text{n-propyl ammonium chloride}}{{CH_3CH_2CH_2 N{H}_{3}}^{+} {Cl}^{-}}$$
$$ii) (C_2H_5)N+HCl\longrightarrow \underset{\text{tri-ethyl ammonium chloride}}{{C_2H_5N}^{+}{Cl}^{-}}$$

Assign reason:
a. There are two $$NH_2$$ groups in semi carbazide only one is involved in the formation of semicarbazide.

1372782_1148800_ans_7972928bdd574cedb5aa67282787cacc.jpg

The major product $$(X)$$ of the reaction is :

1372752_1149306_ans_f33bf40a5f324fa38c8c36b3e76b1215.jpg

What is A?

Benzylamine reacts with nitrous acid to form benzyl diazonium salt(A) which is highly unstable and hence ends up forming benzyl alcohol along with the evolution of Nitrogen gas

$$\underset{Benzylamine}{C_6H_5CH_2 - NH_2} + HNO_2 \xrightarrow{NaNO_2 + HCl} \underset{(Unstable)}{[C_6H_5CH_2 - N_2^+Cl^-]}$$

$$\downarrow H_2O$$

$$\underset{Benzyl \ alcohol}{N_2 \uparrow + C_6H_5CH_2 - OH + HCl}$$

Thus $$A$$ is benzy alcohol

What is the product of the reaction if the reactant is treated with ammonia ?

1038288_1172934_ans_a2e124193fab43da99e91ffe74163695.png

Arrange the following in the increasing order of their boiling point:

$${ C }_{ 2 }{ H }_{ 5 }OH,\ { (CH }_{ 3 })_{ 2 }NH,\ { C }_{ 2 }{ H }_{ 5 }{ NH }_{ 2 }$$

Boiling point depends upon $$H-$$bonding present in compound. Boiling point will be higher if extent of $$H-$$bonding is more in the compound.

$$(CH_{3})_{2}NH$$ has only one $$H-$$atom $$C_{2}H_{5}NH_{2}$$ has two $$H-$$atom then $$C_{2}H_{5}NH_{2}$$ has more extensive $$H-$$bonding than $$(CH_{3})_{2}NH$$.

Therefore the boiling point of $$C_{2}H_{5}NH_{2}$$ is more than $$(CH_{3})_{2}NH.$$

$$'O'$$ is more electromagnet iv than $$'N'$$. Thus, $$C_{2}H_{5}On$$ has more stronger $$n-$$bond than $$C_{2}H_{5}NH_{2}$$ and $$(CH_{3})_{2}NH$$

order $$(CH_{3})_{2} NH < C_{2} H_{5} NH_{2} < C_{2} H_{5} OH$$

Arrange the following compounds in decreasing order of basic strength ( In gaseous phase):
$${ NH }_{ 3 }$$, $$\left( { { C }_{ 2 }H }_{ 5 } \right) _{ 2 }{ NH }$$, $$\left( { { C }H }_{ 3 } \right) _{ 3 }{ N }$$.

The basicity order of the following compounds is:

$${ { (CH }_{ 3 }) }N_{ 3 }>{ ({ C }_{ 2 }{ H }_{ 5 }) }_{ 2 }-NH>{ NH }_{ 3 }$$

In case of $${ 3 }^{ 0 }$$ amine $$+I$$ effect of three methyl group attached to the nitrogen which increases the electron density on nitrogen atom.

On the other hand $$+I$$ effect of two ethyl group attached to the nitrogen atom which develops the electron density on nitrogen.

In case of ammonia,it is basic in nature but there is no effect $$+I$$ effect of alkyl group.

Give reason:
The $$pK_{b}$$ of aniline is greater than methylamine.

Higher the value of pKb lesser is the acidic character .

Aniline has higher pKb than methylamine because methylamine is more acidic than aniline as in methylamine delocalisation of lone pair of electrons belonging to nitrogen occurs as a result resonance structures are formed which increase the acidic character of methylamine as a result it has lesser pKb value.

Give reason:
Explain Gabriel Phtalimide reaction.

1326358_1319102_ans_ec64191ff16d41edacb74afd06d67d74.jpg

What is the action of sodium nitrite and hydrochloric acid on ethanamine, N-ethylethanamine and N, N-diethylethanamine?

Diethylamine reacts with NaNO2 +dilHCl gives Nitroso compound

Complete and balance the following reaction :
$$C_{2}H_{5}NH_{2}+HCl \longrightarrow$$

$${C}_{2}{H}_{5}N{H}_{2} + HCl \rightarrow {C}_{2}{H}_{5}{N}{H}_{3}^+Cl^-$$

When ethyl amine reacts with HCl, it forms ethyl ammonium chloride as a product.

The correct order of increasing basic nature for the following compound is

1318230_1347818_ans_1df15eb8f0924b46bda3b665f1855d99.jpg

Find the total no. of sigma and pi bonds in the given compound.

In Single bond , there is 1 $$\sigma$$ bond.

In Double bond , there is 1 $$\sigma$$ bond and 1 $$\pi$$ bond.

In this Compound there are 9 $$\sigma $$ bonds and 3 $$\pi$$ bonds.

How is methoxyethane prepared from
(i) Methyl lodide (ii) Diazomethane.

1314819_1352133_ans_c4919ef6dcea4d02a21286543ffd0f8e.jpg

What are amines?

An organic compound derived from ammonia by replacement of one or more hydragen atoms by organic group are called amines.

$$R-\ddot{N}H_2$$, $$R-\underset{H}{\underset{|}{\ddot{N}}}-R$$, $$R-\underset{R}{\underset{|}{\ddot{N}}}-R$$

Give reasons:
Aliphatic amines are stronger bases than ammonia.

Aliphatic amines are those in which one or more than one $$H$$ atoms of ammonia are replaced by alkyl groups.

$$R-NH_2$$ is a general representation of aliphatic amine.

In aliphatic amines, there is an alkyl group that is electron-donating and it increases electron density on $$N$$ and in ammonia, there are $$3H$$ and no electron-donating group.

That is why aliphatic amines are stronger bases than ammonia.

1499494_1700683_ans_dd2cf953a59c4846a158e1578d91e58d.png

Lower amines are soluble in water due to__________

The correct answer is hydrogen bonding

$${ (CH }_{ 3 })_{ 3 }{ { CNH }_{ 2 } }$$ is a______________ amine

this is primary amine

Ethylamine is __________ basic than ammonia.

Ethylamine is more basic than ammonia.

Basic nature of amine is due to the presence of ________ on nitrogen atom.

presence of electron pair

Give (a) conjugate acid (b) conjugate base of $$HO(CH_2)_3$$.$$NH_2$$.

(a) $$HO(CH_2)_3NH_2\xrightarrow {\displaystyle{ { H }^{ + } }} HO(CH_2)_3\overset { + }{ N } { H }_{ 3 }$$ (b) $$HO(C{ H }_{ 2 })_{ 3 }{ NH }_{ 2 }\xrightarrow {\displaystyle{ -H^{ + } }} { O }^{ - }-(C{ H }_{ 2 })_{ 3 }{ NH }_{ 2 }$$

Fill in the blanks:
Aniline when treated with sodium nitrite and $$HCl$$ at $$0^{\circ}C$$ forms benzene diazonium chloride. This reaction is termed is _______.

Diazotisation.

Write short note on Basic nature of amines.

Amines are basic in nature because of one the main facts the nitrogen atom which are present in amines they possess a lone pair that can be donated. According to a theory of "Lewis acid-base concept", amines can donate an electron pair, so they are Lewis bases

Name the final product of the following reaction:
Aniline reacts with sodium nitrite and hydrochloric acid at $$0^{\circ}$$.

This reaction is diazotisation of aniline, where benzene diazonium chloride is formed as yield.
1610327_1721626_ans_0404f5180f4c470b9e7493af7ba695a3.PNG

1610327_1721626_ans_0404f5180f4c470b9e7493af7ba695a3.PNG

Explain $$Me_4\overset {+ }{N }OH^-$$ is more basic than $$Me_3N$$.

$$Me_4\overset {+ }{N }OH^-$$ completely ionises while $$Me_3N$$ like $$NH_3$$ combines with water to produce $$OH^- ions. Me_3N+H_2O\rightleftharpoons [Me_3\overset { + }{ N } H\rightleftharpoons [Me_{ 3 }NH]^{ + }+[OH^{ - }]K_{ b }$$ of this reaction is very small. Thus, due to high concentration of $$OH^-$$ ions in $$Me_4\overset {+ }{N }OH^-$$, it is more basic.

Explain tertiary amine has lowest boiling point of a group of isomeric amines.

Primary and secondary amines can form hydrogen bonds whereas tritiary amines fail to do so.

Compare the basicities of (a) $$CH_3NH_2$$, (b) $$(CH_3)_2NH$$, (c)$$ NH_3$$, (d) $$C_6H_5NH_2.$$

$$(CH_3)_2NH>CH_3NH_2>NH_3>C_6H_5NH_2$$

Explain why are amines basic in nature?

Amines contains a lone pair of electrons on its nitrogen atom which is easily available for protonation. This is the cause of basic nature. Any other group which increases the electron density on nitrogen increases the basicity.

Explain the boiling point of methylamine is $$-6.3°C,$$ whereas ethane's boiling point is much lower at $$-88.6°C$$

The reason for the higher boiling points of the primary amines is that they can form hydrogen bonds with each other as well as van der Waals dispersion forces and dipole-dipole interactions. Hydrogen bonds can form between the lone pair on the very electronegative nitrogen atom and the slightly positive hydrogen atom in another molecule.

The boiling points of the primary amines increase as you increase chain length because of the greater amount of van der Waals dispersion forces between the bigger molecules.

1847373_1721693_ans_4fd68f0ff2a34f85b9707914bcc23a8f.gif

Explain Dimethylamine is a stronger base than trimethyl amine.

Two factors operate in deciding the basicity of alkyl amines, i.e., inductive effect and solvation effect.

Inductive effect: Alkyl group being electron releasing, increasing, increases the charge density on nitrogen. This in turn increases the basicity of amines. Thus, the expected order of basicity it: $$R_3N>R_2NH>RNH_2>NH_3$$ But the steric factor in $$R_3N$$ makes the availability of a lone pair of electrons on nitrogen poor than in $$R_2NH$$, predicting $$R_2NH$$ a stronger base than $$R_3N$$.

Explain there is decreasing order in basicity of: $$CH_3NH_2>CH_3N=CHCH_3>CH_3CN$$.

The hybrid atomic orbitals used by nitrogen in the compounds are $$sp^3, sp^2$$ and $$sp$$ respectively. $$CH_3\ddot { N } H_{ 2 }(sp^{ 3 }),CH_{ 3 }\ddot {N } =CHCH_3(sp^2), CH_3C\ddot {N } (sp),$$ The compound with least $$s$$-character is most basic, thus, more $$s$$-character.

Identify the unknown compounds:

(A) Aniline; (B) benzene diazonium chloride; (C) phenol.

Compare the basicities of:
$$C_2H_5NH_2$$, $$(C_3H_7)_3$$, $$CH_3CONH_2$$, $$CH_3N\overset { - }{ H } Na^+$$.

$$CH_3N\overset { - }{ H } Na^+$$ > $$C_2H_5NH_2$$>$$(C_3H_7)_3$$>$$CH_3CONH_2$$

Compare the basicities of:
Methylamine, Dimethylamine, Aniline, $$N$$-Methylaniline.

Dimethylamine>Methylamine>$$N$$-Methylaniline>Aniline

Compare the basicities of

IV< II < III < I

For an amine $$R{NH}_{2}$$, write the expression for $${K}_{b}$$ to indicate its strength.

$$R{ NH }_{ 2 }+{ H }_{ 2 }O\rightleftharpoons R\overset { \oplus }{ N } { H }_{ 3 }+{ OH }^{ - }\quad $$

$${ K }_{ b }=\cfrac { \left[ R\overset { \oplus }{ N } { H }_{ 3 } \right] \left[ { OH }^{ - } \right] }{ \left[ R{ NH }_{ 2 } \right] } $$

Give plausible explanation for: Why do primary amines have higher boiling points than tertiary amines?

Primary amines are engaged in intermolecular association due to hydrogen bonding between nitrogen of one and hydrogen of another molecule.

Tertiary amines do not have intermolecular associations due to the absence of hydrogen atom available for hydrogen bond formation.

Therefore, primary amines have higher boiling points than tertiary amines.

1691007_1770789_ans_1c7e5124699643ad83900cb5d92689ad.png

Account for: $$pK_{b}$$ of aniline is more than that of methylamine.

In aniline, the lone pair of electrons on the N atom is delocalized over the benzene ring. As a result, the electron density on nitrogen decreases.

In contrast, in $$CH_{3}NH_{2}\ ,+I$$ effect of $$CH_{3}$$ increases the electron density on the N atom. Therefore, aniline is a weaker base than methylamine, and hence its $$pK_{b}$$ value is higher than that of methylamine.

A lower $$pK_b$$ value indicates a stronger base.

Arrange the following in increasing order of base strength:
methylamine, dimethylamine, aniline, $$N-$$methylaniline.

Aliphatic amines are more basic than aromatic amine due to presence of alkyl groups which increase the electron density on nitrogen atom.

In aromatic amine, electron pair of nitrogen undergoes delocalization and hence less available to donate, so, less basic.

More is the alkyl group, more is the electron density and hence more basic. So, Dimethylamine is more basic than Methylamine.

In N− methyl aniline, there is one alkyl group is present. So more basic than aniline.

1691059_1771946_ans_abf449e4ffba438c88d3d60d5246497b.png

Give the decreasing order of basic strength of their following:
a.$$ NH_3 $$
b.$$ NH_2OH $$
c.$$ NH_2-NH_2 $$

Answer : c > a> b

(c) $$ \ddot {N}H_2 - \ddot {N}H_2 $$ two LP $$ \bar {e} > $$ (a) $$\ddot {N}H_3 $$ one LP $$ \bar {e} > (b) [ \ddot {N}H_2 \rightarrow OH , -I $$ effect of $$OH$$ group.]

Account for: Ethylamine is soluble in water, whereas aniline is not.

Ethylamine dissolves in water due to intermolecular H-bonding as in figure.

However, in aniline, due to the large hydrophobic part, i.e., hydrocarbon part, the extent of H-bonding decreases considerably, and hence aniline is insoluble in water.

1690968_1770740_ans_79f7317612f741468b806c95f7486144.png

Which of the following is the weakest base?
a. $$NH_{3}$$
b. $$C_{6}H_{5}NH_{2}$$
c. $$C_{6}H_{5}CH_{2}NH_{2}$$
d. $$CH_{3}NH_{2}$$

More is the availability of the lone pair on the nitrogen of amine for protonation, more basic in nature. As the number of alkyl group increases, then because of their $$+I$$ effect, the lone pair will become more available. Hence, the order of basicity of amine is:

$$Tertiary>Secondary>Primary>Ammonia$$

But in the case of aniline (aromatic amine), lone pair of $$N$$-atom is used in delocalization and is less available for protonation. Hence, aniline is less basic than an aliphatic amine.

However, in case of benzylamine $$(C_6H_5−CH_2−NH_2)$$, the N atom is attached to a saturated C atom. Hence, no resonance is possible and the lone pair of electron can be easily donated.

So, the order of basicity:

$$aliphatic \ amine>benzylamine>aniline$$

Hence, option $$'b'$$ is correct.

1687767_1769086_ans_858a7220154b4044ae0ca130229f9113.png

Give the decreasing order of acidic strength of the following :
a. $$ CH_4 $$
b.$$ H_2O $$
c.$$ HF $$
d.$$ NH_3 $$

Ans : $$ HF > H_2O > NH_3 > CH_4 ( c > b > d > a ) $$

Fluorine is the most electronegative, so $$F^−$$ (fluoride ion) is the least willing to donate electrons (the weakest base). Overall, the electronegativity order is $$C(2.5)<N(3.0)<O(3.5)<F(4.0)$$, the order of basicity is −$$CH_3$$ (strongest base) > −$$NH_2$$ > $$HO^−$$ > $$F^−$$

In the following case rearrange the compounds as directed:
In an increasing order of $$p{K}_{b}$$ values:
$${ C }_{ 2 }{ H }_{ 5 }{ NH }_{ 2 },{ C }_{ 6 }{ H }_{ 5 }NH{CH}_{3},{({ C }_{ 2 }{ H }_{ 5 })}_{2}NH$$ and $${ C }_{ 6 }{ H }_{ 5 }-{NH}_{2}$$

1. In aliphatic amines, the $$−NH_2$$ group is attached to an alkyl group which is an $$\textbf{electron donating group}$$ i.e, they have +I effect. So, the lone pair of electrons on nitrogen are readily available.Therefore aliphatic amines are more basic than aromatic amines. Because in aromatic amines the lone pair on $$N-$$atom is involved in resonance with the benzene ring.

Hence aniline is least basic.

2. $${ C }_{ 6 }{ H }_{ 5 }NH{CH}_{3}$$ is more basic than aniline due to presence of $$-CH_3$$ group directly on $$N-$$atom ,thereby showing $$+I$$ effect.

3. Diethylamine is the most basic due to presence of 2 ethyl groups showing $$+I$$ effect.

Hence order of basicity is:

$${({ C }_{ 2 }{ H }_{ 5 })}_{2}NH > { C }_{ 2 }{ H }_{ 5 }{NH}_{2} > { C }_{ 6 }{ H }_{ 5 }NH{CH}_{3} > { C }_{ 6 }{ H }_{ 5 }{NH}_{2}$$

Thus the order for $$pK_b$$ will be the exact opposite to it.

Answer:

$${({ C }_{ 2 }{ H }_{ 5 })}_{2}NH< { C }_{ 2 }{ H }_{ 5 }{NH}_{2}< { C }_{ 6 }{ H }_{ 5 }NH{CH}_{3}< { C }_{ 6 }{ H }_{ 5 }{NH}_{2}$$

Arrange the following:
In decreasing order of the $$pKb$$ values:
$$C_2H_5NH_2, C_6H_5NHCH_3, (C_2H_5)_2NH,$$ and $$CH_5NH_2$$

$$C_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5NH_2 > (C_2H_5)_2 NH$$
1739574_1828836_ans_f563ab2328c74fb18fb556e43c53d166.png

Classify as primary, secondary or tertiary amines.

$$\rightarrow$$ Primary amine
1739730_1829044_ans_19635407ae8a40518ffc7b740d5a9783.png

Arrange the following:
In increasing order of solubility in water:
$$C_6H_5NH_2, (C_2H_5)_2NH, C_2H_5NH_2.$$

$$C_6H_5NH_2 < (C_2H_5)_2NH < C_2H_5NH_2$$

as more H-bonding, more solubility due to the lone pair of Aniline is involved in conjugation with the ring, its intensity to form Hydrogen bonding decreases.

Classify as primary, secondary or tertiary amines.
$$(C_2H_5)_2 CHNH_2$$

$$(C_2H_5)_2 CHNH_2\Rightarrow $$ Primary

Because the nitrogen atom is attached to the only 1 carbon atom.

Note: The degree of amines depends on the no of atoms attached to the nitrogen atom of amine (except hydrogen).

In the given pairs which one is more basic and why?

(i) In $${CH}_{3}{NH}_{2}$$ the +I effect of $$-{CH}_{3}$$ group makes lone pair of electrons on $$N-$$atom more available for donation. On the other hand in $${C}_{6}{H}_{5}{NH}_{2}$$, the resonance effect causes delocalisation of lone pair of electrons over benzene ring and makes it less available for donation. Hence, $${CH}_{3}{NH}_{2}$$ is more basic than $${C}_{6}{H}_{5}{NH}_{2}$$

(ii) $${CH}_{3}{NH}_{2}$$ is more basic than $${NH}_{3}$$. The $$-{CH}_{3}$$ group due to its +I effect pushes electron towards nitrogen and this makes the unshared electron pair more available for sharing with the proton of the acid.

(iii) As $$-{CH}_{3}$$ group is electron releasing by +I effect ; hence p-toludine is more basic than aniline.

Write structural formulea for the following IUPAC name.
ethanamine

Write short notes on the following:
Diazotisation

Diazotization:

The process of conversion of a primary aromatic amino compound into a diazonium salt is known as diazotization. This process is carried out by adding an aqueous solution of sodium nitrite to a solution of primary aromatic amine (e.g., aniline) in excess of HCl at a temperature below 5°C.

1739647_1828889_ans_ae4fbd6428174864aed407c11c8303b9.png

Classify as primary, secondary or tertiary amines.
$$(C_2H_5)_2 NH$$

$$(C_2H_5)_2 NH\Rightarrow $$ secondary

Because the nitrogen atom is attached to the two carbon atom.

Note: The degree of amines depends on the no of atoms attached to the nitrogen atom of amine (except hydrogen).

For an amine $$RNH_2$$, write the expression for $$K_b$$ to indicate its base strength.
$$(DBS 2000)$$

Write the structures of $$(\mathrm{C}\mathrm{H}_3)_{3}\mathrm{N}$$ and $$(Me_3Si)_{3}\mathrm{N}$$. Are they isostructural? Justify your answer.

These are not isostructural because silicon uses its vacant dorbitals for back bonding with lone pair electrons of central $$N$$ atom so it is planar in shape while other has pyramidal shape.

Out of the given compounds $$A$$ and $$B$$, the number of $$N$$ atoms in the compound which is more basic are:

Compound B, i.e., guanidine is the strongest base amongst all nitrogenous compounds.

The increased basicity can be explained by drawing the resonance structures of the protonated guanidine.

It has 3 equivalent resonating structures and so is very stable.

The total number of basic groups in the following form of lysine is:

Explanation:

$$-COO^-$$ and $$-N{H_2}$$ groups can donate a lone pair of electrons. So these two groups can act as Lewis bases. Therefore these two groups are basic.

$$-NH_3^+$$ is acidic in nature as it can donate the $$H^+$$ ion.

Therefore, total number of basic groups in the above form of lysine is $$2.$$

An organic compound $$(A)$$ composed of $$C, H,$$ and $$O$$ gives a characteristic colour with ceric ammonium nitrate. Treatment of $$(A)$$ with $$PCl_5$$ gives $$(B)$$ which reacts with $$KCN$$ to form $$(C)$$. The reduction of $$(C)$$ with warm $$Na/C_2H_5OH$$ produces $$(D)$$ which on heating gives $$(E)$$ with the evolution of ammonia. Pyridine is obtained in treatment of $$(E)$$ with nitrobenzene.

1754937_247716_ans_827ac5c154dd4158a21635f34f456c31.png

Match.

Since $$pK_{a} $$values are given, for base the $$pK_{b}$$ values are required Lower the $$pK_{b}$$ value, stronger is the base,or higher the $$pK_{a}$$ value $$(pK_{b} = 14 - pK_{a})$$ stronger is the base.

$$p-Toluidine > m - Toluidine > Aniline > o - Toluidine$$

$$ pK_{a} = 5.12$$ $$ 4.69 $$ $$ 4.58 $$ $$ 4.39 $$

$$ pK_{b} = 8.88$$ $$ 9.31 $$ $$ 9.42 $$ $$ 9.61 $$

Compound (C) is___________.

Compound C is 3,5-dibromoaniline.

Compound (D) is :

Compound D is 1,3,5-tribromobenzene.

Match the reactions in Column I with appropriate options in Column II.

$$\displaystyle (a\rightarrow r,s,t;b\rightarrow t;c\rightarrow p,q;d\rightarrow r)$$
$$\displaystyle

(a\rightarrow r,s,t)$$ It is an example of electrophilic substitution reaction which results in a coupled product; hence it is coupling reaction also. It also involves carbocation intermediate.
($$\displaystyle b\rightarrow t$$) Pinacole-pinacolone rearrangement. In this reaction, the intermediate is carbocation.
($$\displaystyle

c\rightarrow p,q$$) It is an example of addition reaction by carbonyl compounds and both enantiomers will be formed. Hence, racemic mixture will be obtained. (see diag)
($$\displaystyle d\rightarrow r$$) It is an example of nucleophilic substitution.

Among methylamine and aniline, the number of $$C$$ atoms in compound with lower $$pK_b$$ is:

Lower is the value of $$pK_b$$, stronger is the base.
$$pK_b$$ of methylamine is less than that of aniline.
Thus methylamine is a stronger base than aniline.
This is because, in aniline, the lone pair on nitrogen is involved in resonance with aromatic nucleus. Hence, it is not available for donation.

Pyrroles are five-membered nitrogen-containing heterocyclic compounds present in heme which is a constituent of hemoglobin. Draw all the significantly contributing resonance structures of pyrrole (K).

How many resonance structures can be found?

Resonating structures of Pyrolle are shown above in the figure.

Hence, the no. of resonating structure of Pyrrole is 4.

1174484_305173_ans_dce6c1c7dd1e4410930e985e8868077b.PNG

Alkyl cyanides $$({CH}_{3}CN)$$ when treated with hydrogen in presence of $$Pt$$ or with $$LiAl{H}_{4}$$ produces same carbon number compound. What is the formula of that compound?

$${ H }_{ 3 }C-C{ H }_{ 2 }-C{ H }_{ 2 }-C\equiv N\quad +\quad 2{ H }_{ 2 }\xrightarrow { \quad Pd/c\quad } { H }_{ 3 }C-C{ H }_{ 2 }-C{ H }_{ 2 }-N{ H }_{ 2 }$$

This is reduction of nitrile to give amines

$${ H }_{ 3 }C-C\equiv N\quad +\quad 2{ H }_{ 2 }\xrightarrow [ Pt ]{ \quad LiAl{ H }_{ 4 }\quad } { H }_{ 3 }C-C{ H }_{ 2 }-N{ H }_{ 2 }$$

$${ { C }_{ 2 }H }_{ 3 }N\quad +\quad 2{ H }_{ 2 }\xrightarrow [ Pt ]{ \quad LiAl{ H }_{ 4 }\quad } { C }_{ 2 }{ H }_{ 7 }N$$

Thus the formula of the compound is $$CH_3-CH_2-NH_2$$

Write short notes on the following:
Carbylamine reaction

Carbylamine reaction

(i) Carbylamine reaction

When aliphatic/ aromatic primary amines are heated with chloroform and alc KOH, foul-smelling alkyl isocyanides or carbylamines are obtained. Secondary or tertiary amines do not give this test.

$$\displaystyle Ph-NH_2 + CHCl_3+3KOH \xrightarrow {\Delta} Ph-NC + 3KCl + 3H_2O$$

$$2^°$$and $$3^°$$ (aromatic/aliphatic) do not give this reaction.

Why can aromatic primary amines not be prepared by Gabriel phthalimide synthesis?

During Gabriel phthalimide synthesis, the reaction between phthalimide and ethanolic potassium hydroxide gives potassium salt of phthalimide.
The salt on heating with alkyl halide followed by alkaline hydrolysis gives corresponding primary amine.
Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis as aryl halides do not undergo nucleophilic substitution with the salt formed by phthalimide.

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?

(i) Amines lose a proton to form amide ion. Alcohols lose a proton to form alkoxide ion.
$$\displaystyle RNH_2 \rightarrow RNH^- + H^+ \\ R-OH \rightarrow R-O^- + H^+$$
$$O$$ is more electronegative than N, the negative charge is more easily accomodated in $$\displaystyle RO^-$$ than in $$\displaystyle R-NH^-$$. Hence, amines are less acidic than alcohols of comparable molecular masses.

(ii) In primary amines, $$N$$ atoms have two $$H$$ atoms which results in extensive intermolecular $$H$$ bonding. In tertiary amines, $$N$$ atoms do not have H atoms and hydrogen bonding is not possible.
Hence, primary amines have higher boiling point than tertiary amines.

(iii) Aliphatic amines stronger bases than aromatic amines due to following reasons:
(a) Aromatic amines have resonance due to which lone pair of electrons on $$N$$ atom is delocalized over benzene ring and is less available for protonation.
(b) The stability of aryl amine ions is lower than the stability of alkyl amines. Protonation of aromatic amines is not favoured.

An aromatic compound 'A' of molecules formula $$C_{7}H_{6}O_{2}$$ undergoes a series of reaction as shown below. Write the structures of A,B,C,D and E in the following reactions:
OR
(a) Write the structures of main products when benzene diazonium chloride reacts with the following regents:
(i) $$H_{3}PO_{2}+H_{2}O$$
(ii)$$CuCN/KCN$$
(iii)$$H_{2}O$$
(b) Arrange the following in the increasing order of their basic character in an aqueous solutions:
$$C_{2}H_{5}NH_{2},(C_{2}H_{5})_{2}NH,(C_{2}H_{5})_{3}N$$
(c) Give a simple chemical test to distinguish between the following pair of compounds:
$$C_{2}H_{5}-NH_{2}\;and\;C_{6}H_{5}-NH-CH_{3}$$

(b) Order of basic strenght

$$C_2H_5NH_2$$$$<(C_2H_5)_2NH$$$$<(C_2H_5)_3N$$

(c) $$C_6H_5-NH_2$$(Primary amines) and $$C_6H_5-NH-CH_3$$ (secondary amines) can be distinguished by Hinsberg's test. In this test amines are allowed to react with Hinsberg's reagent, benzeneesulphonyl chloride ($$C_6H_5SO_2Cl$$). Primary amines reacts with this reagent to form N-alkylbenzene sulphonyl amide which is soluble in alkali but secondary amines gives sulphonamide which is insoluble in alkali

Arrange the following in the decreasing order of their basic strength in aqueous solutions.
$$CH_3NH_2,\ (CH_3)_2NH,\ (CH_3)_3 N, \,NH_3$$

$$+I$$ groups, like $$R$$ groups, increases the electron density on nitrogen which increases the basicity (electron rich is basic) of amine. Of the given amines, more the number of R groups on nitrogen more is the basic nature of amine. Basicity also depends on the steric hindrance. More is the steric hindrance less is the basicity.
So, the correct order is $$(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$$.

Amines are classified as primary, secondary and tertiary.
(a) Write the IUPAC name of the following compound:
$$N{H}_{2} - {\left(C{H}_{2}\right)}_{6} - N{H}_{2}$$
(b) Which is stronger base - $$C{H}_{3}N{H}_{2}$$ or $${C}_{6}{H}_{5}N{H}_{2}$$? Why?

a) Hexa methylene diamine (6 methylene groups and two amino groups).

b) $$C_6H_5NH_2$$ isstrong base than $$CH_3NH_2$$ due to resonance effect.

Write a note a Gabriel phthalimide synthesis.

What are biodegradable polymers and non-biodegradable polymers? Write one example of each.

Explain cationic detergents.

Gabriel phthalimide synthesis: This synthesis is useful to get pure amino acids. The reaction involves the preparation of primary amines free from secondary and tertiary amines by using the reaction between alkyl halides with alkali metal phthalimide and subsequent hydrolysis.

Mechanism of Gabriel phthalimide ( Refer image).

For example, phthalimide is firstly treated with alcoholic $$KOH$$ to form potassium phthalimide which on further reaction with ethyl iodide gives N-ethyl phthalimide. The product N-ethylphthalimide is hydrolysed with dilute $$HCl$$ to form primary amine, i.e., ethylamine. This reaction is known as Gabriel phthalimide synthesis.

Biodegradable polymers: polymers that disintegrate by themselves after a certain period of time due to environmental degradation are called biodegradable polymers.

Non-biodegradable polymers: Polymers that do not disintegrate by themselves after a certain period are called non-biodegradable polymers.
Example of Biodegradable polymers: $$PHBV$$ i.e.

Example of Non-Biodegradable Polymers: Bakelite, Terylene, Nylon.

Cationic detergents: Cationic detergents are quaternary ammonium salts of amines with chlorides, acetates or bromides as anions. Cations are long-chain hydrocarbons and a positive charge on the nitrogen atom. Cetyltrimethylammonium chloride is used in hair conditioners.

Arrange the following in the increasing order of basicity:

$${CH}_{3}{NH}_{2}, {({CH}_{3})}_{2}NH, {({CH}_{3})}_{3}N$$

In the gas phase, the increasing order of basicity is
$$NH_3 <{CH}_{3}{NH}_{2} < {({CH}_{3})}_{2}NH < {({CH}_{3})}_{3}N$$
Note:
In the gas phase, the increasing order of basicity is

Ammonia < primary amine < secondary amine < tertiray amine.

The +I effect of alkyl groups increases electron density on N atom.
It increases the basicity of amines.

Account for the following.
p$$K_a$$ value of $$4$$-nitrobenzoic acid is lower than that of benzoic acid.

p$$K_a$$ value of $$4$$-nitrobenzoic acid (3.41) is lower than that of benzoic acid (4.19). Lower is the p$$K_a$$ value, greater is the acid strength. The electron withdrawing nitro $$(-NO_2)$$ group increases the acidity of benzoic acid.

Which of the following is most basic?
(Write 1 if $$I$$ is the answer)

The basicity order is $$III > II > I$$

Compoun (III) is attached with $$-I$$ effect of Electron withdrawing group but lone pair on N-atom is available and EWD group is far away from N-atom.

In case of (I), EWD group decrease the basicity through $$-I$$ effect from two side and compound (II) $$+I$$ effect of one methyl group attached to the N-atom.

So (II) is more basic than (I)

Arrange the following in the increasing order of their $$pK_b$$ values.
$$C_6H_5NH_2, C_2H_5NH_2, C_6H_5NHCH_3$$.

The increasing order for $$pK_b$$ values is $$C_2H_5NH_2 (3.29) <C_6H_5NHCH_3(9.30)< C_6H_5NH_2 (9.38) $$. The numbers in bracket indicates $$pK_b$$. Higher is the $$pK_b$$ value, lower is the basic strength of amine. Thus, the increasing order for $$pK_b$$ values indicates
decreasing order of the basic strength of amine. Aniline is least basic and ethanamine is most basic.
Aniline is least basic as unshared pair of electrons on N atom is involved in resonance with benzene ring and cannot be easily donated. When an alkyl group such as methyl or ethyl is present, the electron releasing inductive effect ($$+I$$ effect) of alkyl group increases the electron density on N atom and unshared pair of electrons on N atom can be easily donated.

Give a simple chemical test to distinguish between Aniline and N, N-dimethylaniline.

Aniline is aromatic primary amine.
Aliphatic or aromatic primary amines on heating with chloroform and alcoholic KOH give foul smelling alkyl isocyanides or carbylamines. Secondary or tertiary amines do not give this test.
$$C_6H_5-NH_2+ CHCl_3 + 3KOH \xrightarrow {\Delta} C_6H_5-NC+3KCl+3H_2O$$
N, N-dimethylaniline is a tertiary amine and do not give this test.

Give reasons $$(CH_3)_2NH$$ is more basic than $$(CH_3)_3N$$ in an aqueous solution.

$$(CH_3)_2NH$$ is more basic than $$(CH_3)_3N$$ in an aqueous solution.
$$(CH_3)_2NH_2^+$$ is hydrated to a greater extent than $$(CH_3)_3NH^+$$. As the number of methyl groups increases, the extent of hydration decreases due to steric hindrance. Greater is the extent of hydration, greater is the stability of ion and greater is the basic strength of amine.

Identify the compound $$'D'$$ in the following series of reactions.

$$CH_3-CH_2-Br \xrightarrow{alc. KOH} A \xrightarrow {Cl_2/ CCl_2}B \xrightarrow{NaNH_2} C \xrightarrow[1\% HgSO_4]{40\%H_2SO_4}D$$

1153737_1073574_ans_6e50a18cb640499a8973ac04346924fd.png

In the following pair which one is more basic and why? $${ CH }_{ 3 }{ NH }_{ 2 }$$ or $${ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$$

Methylamine is more basic than aniline. In aniline, ring structure delocalizes the electrons from individual atoms.

The source of basicity of amine group is the lone pair on the Nitrogen atom.

The lone pair is also delocalized in Aniline and resource structures are formed as the lone pair on Nitrogen is delocalized, it cannot donate an electron, therefore loses its basicity.

Methylamine, $${ CH }_{ 3 }{ NH }_{ 2 }$$, does not have resonance and thus it is more basic as the Nitrogen can donate electrons.

Therefore Methylamine is more basic than Aniline.

Arrange the following compound in decreasing order of basic strength $$NH_{3}, (C_{2}H_{s})_{2}NH, (CH_{3})_{3}N$$.

Decreasing order of basic strength:

$$ (CH_{3})_{3}N,(C_{2}H_{s})_{2}NH,NH_{3}$$.

Becuase presence of an electron releasing the basic strentght of amines.

In pyrrole
The electron density is maximum on

Pyrrole has maximum electron density on 2 and 5. It generally reacts with electrophiles at the C-2 or C-5 due to the highest degree of stability of the protonated intermediate.

Attack at position 3 or 4 yields a carbocation that is a hybrid of structures (I) and (II). Attack at position 2 or 5 yields a carbocation that is a hybrid not only of structures (III) and (IV) (analogous to I and II) but also of structure (V). The extra stabilization conferred by (V) makes this ion the more stable one. Also, attack at position 2 or 5 is faster because the developing positive charge is accommodated by three atoms of the ring instead of by only two.

1325419_1529351_ans_2efb11f0180e41abaef0cec12e40cd91.png

Account for the following:
(a) Gabriel phthalimide synthesis is not preferred for preparing aromatic primary amines.
(b) On reaction with benzene sulphonyl chloride, primary amine yields product soluble in alkali whereas secondary amine yields product insoluble in alkali.

(a) Gabriel phthalimide synthesis involves nucleophilic substitution $$\left( {S}_{{N}^{2}} \right)$$ of alkyl halides by the anion formed by the phthalimide but aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide and thus aromatic primary amines cannot be prepared by this process.
(b) The reaction of a primary amine with benzene sulphonyl chloride gives a corresponding $$N$$-alkyl benzene sulphonylamide. Owing to the presence of strong electron withdrawing sulphonyl group, attached to a nitrogen atom in $$N$$-alkyl benzene sulphonylamide is strongly acidic and dissolves in alkali.

On the other hand, a reaction of the secondary amine with benzene sulphonyl chloride gives corresponding $$N, N$$-dialkyl benzene sulphonylamide which do not contain acidic hydrogen and hence is insoluble in alkali.

Give reason:
Aromatic primary amines cannot be prepared by Gabriel's phthalimide synthesis.

In Gabriel's phthalimide synthesis the first reaction that happens is phthalimide with a base to form the salt. And that salt reacts with alkyl halide and forms product afterward.

The attack of salt on alkyl halide is $$S_N2$$ process. As we know, $$S_N2$$ process cannot happen in aromatic halides that is why we cannot prepare/ produce Aromatic primary amines.

1499528_1700678_ans_dea7767cf6cb4a29b2619e72e9cacaf1.png

Explain the following with at least one example :
Diazotisation.

Aromatic primary amines reacts with nitrous acid in cold condition to give diazonium salts at temperature 175 - 278K up to.

$$Ph-NH_2+NaNO_3+2HCl$$ $$\rightarrow$$ $$PhN_2^{+}Cl^{-}+NaCl+2H_2O$$

Diazonium salts are stable for short time and decompose slowly even at low temperatures. They are used immediately after preparation. The diazonium group can be replaced by various nucleophiles such as halide anions, cyanide, thiols, water, and others.

Write short note on Gabriel phthalimide synthesis.

The Gabriel synthesis is a chemical reaction that transforms primary alkyl halides into primary amines. Traditionally, the reaction uses potassium phthalimide. The reaction is named after the German chemist Siegmund Gabriel.

Compare the basicicties of (a) $$H_2C=CHCH_2NH_2$$. (b) $$CH_3CH_2CH_2NH_2$$, (c) $$HC=CCH_2NH_2$$.

[The significant difference among these tree bases is the kind of hybridization with $$\beta $$-carbon atom. The more $$s$$-character it has, the more electron withdrawing and base weakening it will be. Thus, decreasing order of basicity is (b)>(a)>(c).]

Compare the basicities of (a) $$Ho(CH_2)_2HH_2$$, (b) $$HO(CH_2)_3 NH_2$$, (c) $$CH_3CH_2NH_2$$.

[The electron withdrawing capacity of $$OH$$ decreases electron density on nitrogen atom thereby decreasing basicity. This effect decreases as the distance between $$OH$$ group and $$N$$ increases in the chain. Thus, dreceasing order of basicity is, (c)>(b)>(a)]

Primary amines are formed by the use of phthalimide. which is the name of the reaction?

Gabriel's phthalimide reaction.

Explain why amines are more basic than amides?

In simple amines, the unshared electron pair is usually localised on the nitrogen atom, hence available for protonation. In amides, the electron pair is delocalised to the carbonyl group through resonance.

Amongst the following, the total number of compounds soluble in aqueous $$HCl$$ is:

Hydrochloric acid reacts with amines to form hydrochloride salts.

In aqueous HCl contain a basic group such as $$RNH_2, R_2NH, or R_3N $$ are soluble.

1595832_1757933_ans_763165d9a465408aa5dc017550c8fb84.png

1595832_1757933_ans_763165d9a465408aa5dc017550c8fb84.png

Give the decreasing order of basic character at a, b, c, d in the following compounds:

I. $$ d > c > b > a $$ [ (d) $$ 1^0 $$ amine > (c) p-touldine > (b) ,-toluidine > o-toluidine]

II. $$ d > b > c > a $$ [ (d) $$ 2^0 $$ amine > (b) $$ 1^0 $$ amine > (c) $$ 3^0 $$ amine > (a) aniline (aromatic amine)]

III. $$ c> b > d > a $$ [ (c) $$ 3^0 (Et_3 \ddot {N}) > (b) 1^0 (EtNH_2) > (d) $$ p -nitroaniline > (a) o -Nitroaniline ]

IV. In gaseous phase, amines exhibit the basicities predicted from the electron-releasing effects of the organic substituents. Thus tertiary amines are more basic than secondary amines, which are more basic than primary amines

Hence c > b > a.

There are four amides with the formula $$C_{3}H_{7}NO$$.
a. Write their structures.
b. Identify the amide which has lower melting point and boiling point than the other three.

b. The compound (I) has the lowest boiling point and melting point because it does not have an $$H$$ atoms covalently bonded to an $$N$$ atom and, therefore, it cannot form $$H-bonds$$ with each other. On the other hand, the compounds (II), (III), and (IV) have $$H$$ atoms covalently bonded to $$N$$ atom and, therefore, form $$H-bonding$$.
1597010_1767959_ans_4612be8fc70544e48bfd725ef7a0f2fa.png

1597010_1767959_ans_4612be8fc70544e48bfd725ef7a0f2fa.png

Which of the following statements are correct?

a. In gas phase, the basic character of amine is $${3}^{0} > {2}^{0} > {1}^{0}$$, Due to the +I effect of (R-), the availability of LP $$\bar{e}$$'s on N increases.
b. In aqueous medium, the basic character of amines is $$Me_{2}NH>Me_{3}N>MeNH_{2}>NH_{3}$$.
c. In an aqueous medium, the addition of protons increases crowding and thus strains setup, which is the highest in $$3^{0}$$ amine decreases its basic character.
d. In an aqueous medium, the ammonium ions in solution are stabilised not only by alkyl groups but also by H-bond donation to the solvent.

The order is: $$Me_{2}NH>MeNH_{2}>Me_{3}N>NH_{3} ( 2^{0}>1^{0}>3^{0}>NH_{3})$$.

Statements (a), (c) and (d) are correct.

Gabriel synthesis is used for the preparation of:
a. $$1^{0}$$ amine
b. $$2^{0}$$ amine
c. $$3^{0}$$ amine
d. All can be prepared

Gabriel synthesis is a reaction that converts primary alkyl halides into primary amines. This reaction uses potassium phthalimide. This reaction is named after the German chemist Siegmund Gabriel.

Hence, option $$a$$ is correct.

Arrange the following compounds in an increasing order of basic strengths in their aqueous solutions:
$${ NH }_{ 3 },{ CH }_{ 3 }{ NH }_{ 2 },{ \left( { CH }_{ 3 } \right) }_{ 3 }N$$, $$\left(CH_{3}\right)_{2}NH$$

1. The introduction of an alkyl group into ammonia increases the basic strength.

2. The introduction of a second alkyl group further increases the basic strength, but the net effect of introducing the second alkyl group is very much less marked than compared to the first.

3. The introduction of a third alkyl group to yield a tertiary amine, however, actually decreases the basic strength. This is due to the fact that the basic strength of an amine in water is determined not only by electron-availability on the nitrogen atom but also by the extent to which the cation formed by uptake of a proton, can undergo solvation and thereby become stabilized.

Hence the correct order is:

$$NH_3 <3^o < 1^o < 2^o$$

Give reason for the following in one or two sentences.
'Dimethylamine is a stronger base than trimethylamine'.

Due to steric hindrance in trimethylamine, the availability for donation of lone pair of electron on N atom decreases, in spite of more (+I) inductive effect of $$(-CH_{3})$$ group. So, dimethylamine is a stronger base than trimethylamine.

Account for: Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

The diazonium salts of aromatic amines are more stable than those of aliphatic amines due to dispersal of the positive charge on the benzene ring as shown below:
1785261_1770748_ans_a8eac53705824fb9b0de7a06d22933d8.png

1785261_1770748_ans_a8eac53705824fb9b0de7a06d22933d8.png

Complete the following.

This is a Mannich reaction where an amino alkylation of an acidic proton placed next to a carbonyl functional group by formaldehyde and a primary or secondary amine or ammonia. The final product is a β-amino-carbonyl compound also known as a Mannich base.
1685059_1769784_ans_291caf895784440980f20d9f1c23ef5c.png

1685059_1769784_ans_291caf895784440980f20d9f1c23ef5c.png

Mention the chief use of quaternary ammonium salts derived from long chain amines.

Quaternary ammonium salts (QAS) are cationic compounds containing alkyl groups in a chain length of C8–C18, which are water soluble and these are chiefly used as detergents i.e. $$CH_3(CH_2)_{15}N^+(CH_3)_3Cl^-$$

Give a plausible explanation for: Why are aliphatic amines stronger bases than aromatic amines?

Aliphatic amines are stronger bases due to $$+I$$ effect of the alkyl group, leading to high $${e^-}$$ density on the N atom.

In aromatic amine (e.g.- aniline), lone pair of nitrogen atom involves in resonance. So, due to the electron delocalisation, electron density on nitrogen decreases and hence, less basic than aliphatic amine.

1691001_1770790_ans_d3409dd84eb146d99cfc85982e355a08.png

Give a plausible explanation for why are amines less acidic than alcohols of comparable molecular masses?

The loss of a proton from an amine gives an imide ion, while the loss of a proton from alcohol gives an alkoxide ion as shown below:

$$\underset{Amine}{R-NH_{2}}\rightarrow\underset{Imide\ ion}{R-\overset{\ominus}{N}H}+H^{\oplus}$$

$$R-O-H \rightarrow R-O^{\circleddash}+H^{\oplus}$$

Since O is more electronegative than N, therefore, ($$R\overset{\circleddash}{O}$$) can accommodate the negative charge more easily than $$R\overset{\circleddash}{N}H$$ can accommodate the negative charge.

In other words, $$R\overset{\circleddash}{O}$$ is more stable than $$R\overset{\circleddash}{N}H$$. Thus, alcohols are more acidic than amines. Conversely, amines are less acidic than alcohols.

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

Nitriles can be converted to 1° amines by reaction with LiAlH₄. During this reaction the hydride nucleophile attacks the electrophilic carbon in the nitrile to form an imine anion. Once stabilized by a Lewis acid-base complexation the imine salt can accept a second hydride to form a dianion. The dianion can then be converted to an amine by addition of water.
1647934_1781171_ans_0a7d2d18a8fd47c2a804087d3ad8b9e5.png

1647934_1781171_ans_0a7d2d18a8fd47c2a804087d3ad8b9e5.png

Complete the following:

1598438_1769781_ans_d45c7f1552924e1b9b0245200fbf8c6f.png

Convert: Hexanenitrile into 1-aminopentane

1785271_1770752_ans_c52ecba7df2e449dbfa154d2ca2ae254.png

Account for the following:
Although trimethylamine and n-propylamine have the same molecular weight, but the former boils at a lower temperature (276K) than the latter (322K). Explain why.

n-Propylamine has two H-atoms on the N-atom and hence undergoes intermolecular H-bonding, thereby raising its boiling point. Trimethylamine, $${({CH}_{3})}_{3}N$$, being a tertiary amine does not have any $$H-$$atom on the $$N-$$atom. As a result, it does not undergo H-bonding and hence its boiling point is low.

$$\left[ A \right] ,\left[ B \right] ,\left[ C \right] ,\left[ D \right] ,\left[ E \right] ,\left[ F \right]$$ are amines each of which forms a hydrochloride containing $$32.42$$% chlorine. $$\left[ A \right] ,\left[ B \right] ,\left[ C \right] ,\left[ D \right]$$ evolve $${N}_{2}$$ on reaction with $$H{NO}_{2}$$, but $$\left[ E \right] and \left[ F \right]$$ do not. Give structures of $$\left[ A \right] $$ to $$\left[ F \right] $$ with reasons.

Let the molecular formula of the six amines be $${C}_{n}{H}_{n+1}{NH}_{2}$$, where $$n=1,2,3,$$ etc. As all the six amines react with $$HCl$$ to form hydrochlorides, therefore, the molecular formula of their hydrochloride is

$${C}_{n}{H}_{2n+1}{NH}_{2}+HCl\rightarrow {C}_{n}{H}_{2n+1}\overset { + }{ N } { H }_{ 3 }{ Cl }^{ - }$$

As all these hydrochlorides contain $$32.4$$% of $$Cl$$, therefore molecular mass of amines

$$\cfrac{100\times 35.5}{32.42}=109.5$$

Now molecular mass of $$=109.5$$

$$12n+2n+1+14+3+35.5=109.5$$

$$\Rightarrow$$ $$14n=56$$

$$\Rightarrow n=4$$

Molecular formula of the six amines $$={C}_{4}{H}_{9}{NH}_{2}$$

As amines $$\left[ A \right] ,\left[ B \right] ,\left[ C \right] ,\left[ D \right] $$ evolve $${N}_{2}$$ on treatment wih $$H{NO}_{2}$$ they must be primary amines.

$${C}_{n}{H}_{2n+1}{NH}_{2}+HCl\rightarrow {C}_{n}{H}_{2n+1}\overset { + }{ N } { H }_{ 3 }{ Cl }^{ - }$$

The four primary amines having molecular formula $${C}_{4}{H}_{9}{NH}_{2}$$ are given above. Amines $$\left[ E \right] and \left[ F \right] $$ do not react with $$H{NO}_{2}$$ to evolve $${N}_{2}$$ therefore they must be either secondary or tertiary amines.

Hence the structures of compounds would be:

$$A$$ = $$CH_3-CH_2-CH_2-CH_2-NH_2$$
$$B$$ = $$(CH_3)_2CH-CH_2-NH_2$$
$$C$$ = $$CH_3-CH_2-CH(CH_3)-NH_2$$
$$D$$ = $$ (CH_3)_{3}C-NH_2$$
$$E$$ = $$CH_3-CH_2-NH-CH_2-CH_3$$
$$F$$ = $$CH_3-CH_2-CH_2-NH-CH_3$$

Why do primary amines have higher boiling points than the tertiary amines?

Tertiary amines have no hydrogen atoms bonded to the nitrogen atom and therefore are not hydrogen bond donors. Thus, tertiary amines cannot form intermolecular hydrogen bonds. As a result, they have lower boiling points than primary and secondary amines of comparable molecular weight.

Illustrate the following reactions by giving chemical equations in each case:
Gabriel phthalimide synthesis

Gabriel phthalimide synthesis is used for the preparation of $$\underline{aliphatic}$$ $$\underline{primary}$$ amines.In this reaction, phthalimide is first of all treated with ethanolic $$KOH$$ to form potassium phthalimide. Potassium phthalimide on treatment with alkyl halide gives N-alkyl phthalimide, which on hydrolysis with dilute hydrochloric acid gives a primary amine as the product.

1651870_1781570_ans_0fc25d89ce7d439c8cc405f490fcde7e.png

Give reasons for the following:
Primary amines have higher boiling point than tertiary amines.

In primary amines, two hydrogen atoms are present on $$N-$$atom and they undergo extensive intermolecular hydrogen bonding which results in the association of molecules while in tertiary amines, no hydrogen atom is present on N-atom. Hence, there is no hydrogen bonding in tertiary amines. As a result of this, primary amines have a higher boiling point than tertiary amines.

Give possible explanation for the following
Amides are more acidic than amines

In the case of amides $$(RCONH_2)$$ , the amine group which is bonded with a carbonyl group and due to the high electronegativity of oxygen, $$NH_2$$ group gets involved in resonance which in turn make it less basic or more acidic, while in amines $$(RNH_2)$$, nitrogen has a lone pair of electrons which accepts protons,thus making it more basic.

1652088_1781628_ans_4972b4bb6df54064b0674e47e157ea01.png

Give reasons for the following:
$${({CH}_{3})}_{2}NH$$ is more basic than $${({CH}_{3})}_{3}N$$ in an aqueous solution.

It is a combination of +I effect of the alkyl groups, H-bonding and steric factors which determine the stability of the ammonium cation.

In gaseous state : $$(CH_3)_3N$$ will be more basic.

In aqueous state : $$\left(CH_3\right)_2NH$$ will be more basic.

Illustrate the following reaction:
Coupling reaction:

1. The reaction of diazonium salts with phenols and aromatic amines to form azo compounds of the general formula, $$Ar-N=N-Ar$$ is called coupling reaction. The mechanism is basically that of electrophilic substitution where the diazonium ion is electrophile.In this reaction, the nitrogen atoms of the diazo group are retained in the product.

2. The coupling with phenols takes place in an alkaline medium while with amines, it occurs under faintly acidic conditions.

1651959_1781599_ans_bc099f2a5ee14e2d9b58d99541e3f868.png

Amino acids behave like salts rather than simple amines or carboxylic acids. Explain.

Amino acids behave like salts rather than simple amines or carboxylic acids. This behavior is due to the pressure of both acidic and basic groups in the same molecule. In aqueous solution, the carboxyl group can lose a proton, an amino group can accept a proton, giving rise to a dipolar ion known as a zwitter ion.

How will you covert:
Nitromethane into dimethlyamine

How will you covert:
Hexanenitrile into 1 - aminopentane

Arrange the following:
In decreasing order of basic strength in gas phase:
$$C_6H_5NH_2, (C_2H_5)_2NH, (C_2H_5)_3N$$ and $$NH_3$$

thus the correct order is

$$ (C_2H_5)_3N > (C_2H_5)_2NH > C_2H_5NH_2 > NH_3$$

Note: In GAS PHASE, 3 > 2 > 1 Ammonia [No Hydrogen bending and '+l' effect].

1740610_1828845_ans_cb41d2497df540a780f6f77016b20633.png

Which of the following amines has larges value of $$pK_b$$ ?
p-Toluidine, Aniline, p- Nitroaniline.

Higher the $$pK_b$$ value, weaker is the base

Presence of $$-CH_3$$ group in the ring will increase the basic character as it is an electron releasing group. Presence of $$-NO_2$$ group in the ring will decrease the basic character as it an electron withdrawing group. Thus p-nitro aniline will have largest value of $$pK_b$$.

1740652_1829099_ans_4572fdc0fc004584b96aea6f0ff1f195.png

Write short notes on the following:
Gabriel phthalimide synthesis

Garbriel phthalimide synethesis

Reaction between alkyl halide (R - X) and ammonia gives mixture, of $$1^{\circ },2^{\circ }$$. and $$3^{\circ }$$. amines with tetra alkyl ammonium halide. The mixture is not separable. To get pure 1 amine, Gabriel phthalimide synthesis is preferred.

1739691_1828904_ans_bb6f82f8139d4a59bbc0911a931caeb3.png

Account for the following:
Garbriel phthalimide synthesis is preferred for synthesising primary amines.

$$\text{Gabriel phthalimide synthesis}$$

The reaction between alkyl halide (R- X) and ammonia gives mixture, of 1, 2, and 3 amines with tetra alkyl ammonium halide. The mixture is not separable.

To get pure 1 amine, Gabriel phthalimide synthesis is preferred.

1739571_1828834_ans_a56c4181a3f34904abdb4137d0956166.png

Account for the following:
Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

In aromatic diazonium salts, due to resonance, there is dispersal of positive charge on benzene ring.

But in aliphatic diazonium salts, resonance is not possible, so aliphatic diazonium salts are less stable than aromatic diazonium salt.

Which is more soluble in water, $$NH_3$$ or methylamine?

$$NH_3$$ is more soluble in water. as ammonia has the ability to form hydrogen bonds. When the hydrogen bonds between water molecules are broken, they can be replaced by equivalent bonds between water and ammonia molecules.

Arrange the following:
In increasing order of boiling point:
$$C_2H_5OH, (CH_3)_2NH, C_2H_5NH_2$$

$$(CH_3)_2NH < C_2H_5NH_2 < C_2H_5OH$$

Note: Stronger H-bonding, more b.p more solubility
Alcohol > Amine In amines on H attached to N takes part in H-bond.

How is phenyl amionmethane obtained from phenyl nitrile. $$(AISB 2004)$$

1740726_1829122_ans_4d0054fe1668497a90af64902691dc20.png

Alkylamines are stronger base than ammonia. Explain.

Alkylamine is strongly basic than ammonia due to the electron releasing nature of the alkyl group.

Basicity $$\alpha$$ $$ +I \ effect$$

Alkylamine having more $$+ I$$ effect than $$NH_3$$.

Explain why silver chloride is soluble in aqueous solution of methyl amine.
$$(CBSE 1997, 1998, 2002, 2004)$$

Silver chloride dissolves in methylamine because of the formation of a coordination compound which is soluble in nature.

$$AgCl + 2CH_3NH_2 \rightarrow [Ag(CH_3NH_2)_2] + Cl$$

Why is an alkyl amine more basic than ammonia $$(CBSE Delhi 2009)$$

This is because of the presence of electron releasing alkyl group (positive l effect) which increases the electron density on nitrogen.

What does $$K_b$$ value for an amine stand for? $$(AISB 2000)$$

$$K_b$$ value for an amine stands for its basic strength. More the $$K_b$$ value of an amines more is its basic strength.

Why are aqueous solution of amines basic in nature ? $$(DSB 2006)$$

Aqueous solution of amines which are freely soluble in water (like $$CH_3NH_2, C_2H_5NH_2$$ etc) are basic in nature. It can be explained as follows,

For eg:-

$$CH_3NH_2 + H_2O \rightarrow CH_3 ^+NH_3OH.$$

$$\rightleftharpoons CH_3 NH_3 + OH$$

The aqueous solution of methyl amine will be basic in nature, due to the presence of OH ions.

1740665_1829102_ans_e631e3b55e5e4635862229cd8797f776.png

Write one chemical reaction each to illustrate the following
Hoffmann's bromamide reaction
Garbriel phthalimide synthesis (CBSE 2008)

Hofmann's bromamide reaction

Involves heating of an amide with bromine and caustic alkali to yield an amine with one carbon less than the original amide.

$$ R CONH_2 + Br_2 + 4 KOH \rightarrow RNH_2 + K_2 CO_3 + 2 KBr + 2H_2O$$

(1 AMINE)

Garbriel phthalimide synthesis

The reaction between alkyl halide (R - X) and ammonia gives mixture, of $$1, 2$$, and $$3$$ amines with tetraalkylammonium halide. The mixture is not separable.

To get pure $$1$$ amine, Gabriel phthalimide synthesis is preferred.

(a) Arrange the following in an increasing order of basic strength in water:
$$C_2H_5NH_2, (C_2H_5)_2NH, (C_2H_5)_3N$$ and $$NH_3$$
(b) Arrange the following in increasing order of basic strength in the gas phase
$$C_2H_5NH_2, (C_2H_5)_2NH, (C_2H_5)_3N$$ and $$CH_3NH_2.$$ (CBSE Delhi 2008)

(a) In aqueous solution, basic strength decreases in the order.

$$(C_2H_5)NH > (C_2H_5)_3N > C_2H_5NH_2 > NH_3$$

(b) In gaseous phase, (basic strength decreases in the order.

$$(C_2H_5)_3N > (C_2H_5)_2NH > C_2H_5NH_2 > NH_3$$.

The $$+I$$ effect increases with an increase in the alkyl group. Therefore the basic strength will be the highest in $$(C_2H_5)_3N$$ and least in $$NH_3.$$

1740826_1829169_ans_1ff94183a9fc4e6997c13048a59e5db0.png

Lassaigne's test is not shown by diazonium salts. Why?

Diazonium salts usually lose $$N_{2}$$ on heating much before they have a chance to react with fused sodium metal. Therefore, diazonium salts do not show positive Lassaigne's test for nitrogen.

How are the following conversions carried out ?
Aniline into bromobenzene

1935556_1877281_ans_6f873d1bd24f49e19dbff5b1f979390c.jpg

Amines - Class 12 Engineering Chemistry - Extra Questions

$$pK_b$$ of aniline is more than that of methylamine. If this is true enter 1, if false enter 0.

Ethylamine dissolves in water due to intermolecular H-bonding. If this is true enter 1, if false enter 0.