MCQ Questions for Class 12 Engineering Chemistry Amines Quiz 8

The basicity order of

$I, II$ and

$III$ is

0%
$III > I > II$
0%
$I > II > III$
0%
$III > II > I$
0%
$II > III > I$

Explanation

The basicity order will depend on the groups attached with 'N' other than benzene ring. In (III)

$-C_2H_5$ increases the basicity order due to its +I effect. Therefore (III) has highest basicity, (II) has lowest basicity due to -I effect of

$-COCH_3$ group.

Hence the correct order is

$(III)>(I)>(II)$ .

Option A.
1595458_1748593_ans_dfe933f20d194d51a21edb7cac31b903.png

Consider the following compounds, What is the correct order of basicity of the above compounds?

0%
$I > II > IIl$
0%
$III > I > II$
0%
$III > II > I$
0%
$I > III > II$

Which of the following is least basic?

0%
Aniline
0%
$p-Methylaniline$
0%
Diphenylamine
0%
Triphenylamine

The correct order of basicities of the following compounds is

0%
$2> 1> 3> 4$
0%
$1> 3> 2> 4$
0%
$3> 1> 2> 4$
0%
$1> 2> 3> 4$

Which of the following compound is the strongest base?

0%
Ammonia
0%
Aniline
0%
Methylamine
0%
N-methyl aniline

The order of basic strength among the following amines in benzene solution is

0%
$CH_{3}NH_{2}> (CH_{3})_{3} N> (CH_{3})_{2}NH$
0%
$(CH_{3})_{2}NH> CH_{3}NH_{2}> (CH_{3})_{3}N$
0%
$CH_{3}NH_{2}> (CH_{3})_{2}NH> (CH_{3})_{3}N$
0%
$(CH_{3})_{3}N> CH_{3}NH_{2}> (CH_{3})_{2}NH$

Explanation

$(CH_{3})_{2} NH \quad> CH_{3}NH_{2} \quad> (CH_{3})_{3}N$

$K_{b} = 5.4\times 10^{-4} \ \ 4.5\times 10^{-4} \quad 0.6\times 10^{-4}$

Order of basicity of ethyl amines in aqueous medium is

0%
$Secondry> primary> Tertiary$
0%
$primary>Secondry>Tertiary$
0%
$Secondry>Tertiary> primary$
0%
$Tertiary> primary> Secondry$

Which of the following is

$1^0$ amine

0%
Ethylene diamine
0%
Dimethyl amine
0%
Trimethyl amine
0%
N-methyl aniline

Explanation

in ethylene diamine here both amine are bond with different methyl group so this is

$1^0$ amine

A is correct answer

Gabriel's phthalimide synthesis is used for the preperation of

0%
primary aromatic amine
0%
Secondary acid
0%
Primary aliphatic amine
0%
Tertiary amine

Which of the following compound is expected to be most basic

0%
Aniline
0%
Methylamine
0%
Hydroxylamine
0%
Ethylamine

Explanation

Due to +ve I.E. of alkyl group, N-atom of amines acquires patrial -ve charge and thus electron pair is easily donated. Hence most basic compound is Ethylamine (

$C_2H_5NH_2$ ).
Option D is correct.

Compounds containing both amino and COOH groups are known as

0%
Diamines
0%
unknown
0%
Amino acids
0%
Enzymes

The strongest base among the following is

Correct the order of increasing basicity is

0%
$NH_3 < C_6H_5NH_2 < (C_2H_5)_2NH < C_2H_5NH_2 < (C_2H_5)_3N$
0%
$C_6H_5NH_2 < NH_3 < (C_2H_5)_3N < (C_2H_5)_2NH < C_2H_5NH_2$
0%
$C_{6}H_{5}NH_{2}< NH_{3}< C_{2}H_{5}NH_{2}< (C_{2}H_{5})_{3}N< (C_{2}H_{5})_{2}NH$
0%
$C_6H_5NH_2 < (C_2H_5)_3N < NH_3 < C_2H_5NH_2 < (C_2H_5)_2NH$

Explanation

Electron donors are bases. Electron withdrawing groups (eg, benzyl) decrease the basicity of amines and Electron donating groups (eg, alkyl) increase the acidity of amines.

so correct order is

$C_6H_5NH_2 < (C_2H_5)_3N < NH_3 < C_2H_5NH_2 < (C_2H_5)_2NH$

Hence D is correct answer

By the presence of a halogen atom in the ring, basic properties of aniline is

0%
Increased
0%
Decreased
0%
unchanged
0%
Doubled

Which of the following is most basic

0%
$C_{6}H_{5}NH_{2}$
0%
$(CH_{3})_{2}NH$
0%
$(CH_{3})_{3}N$
0%
$NH_{3}$

Explanation

In general, methylamine

$CH_3NH_2$ ) is a stronger base than ammonia (NH3) and dimethylamine

$((CH_3)_2NH$ is a stronger base than methylamine. However, in water, triethylamine

$((CH_3CH_2)_3N)$ which should be even stronger based on stronger inductive effect, is a weaker base than dimethylamine or methylamine.

Hence B is correct

Which one of the following is not a base

0%
$N_{2}H_{4}$
0%
$N_{2}OH$
0%
$(CH_{3})_{3}N$
0%
$NH_{3}$

Explanation

substance that accepts and H+ from water is considered a base Both

$NH_3$ and H2O are amphoteric

Which of the following amines would undergoes diazotisation

0%
Primary aliphatic amines
0%
Primary aromatics amines
0%
Both (a) and (b)
0%
None of these

If methyl is alkyl group, then which order of basicity is correct

0%
$R_{2}NH> RNH_{2}> R_{3}N> NH_{3}$
0%
$R_{2}NH> R_{3}N> RNH_{2}> NH_{3}$
0%
$RNH_{2}> NH_{3}> R_{2}NH> R_{3}N$
0%
$NH_{3}> RNH_{2}> R_{2}NH> R_{3}N$

Explanation

Greater is the stability of the substituted ammonium cation, stronger should be the corresponding amine as a base. Thus, the order of the basicity of aliphatic amines should be primary > secondary > tertiary, which is opposite to the inductive effect based order. Secondly, when the alkyl group is small, like

$CH_3$ group, there is no steric hindrance to H-bonding. In case the alkyl group is bigger than methyl group, there will be steric hindrance to H-bonding. Therefore, the change of nature of the alkyl group, e.g., from -

$CH_3$ to

$C_2H_5$ results in change of the order of basic strength. Thus, there is a subtle interplay of the inductive effect, solvation effect and steric hinderance of the alkyl group which decides the basic strength of alkyl amines in the aqueous state.

Which one less of alkaline

0%
0%
0%
0%
All of these

The decreasing order of the basic character of the three amines and ammonia is

0%
$NH_{3}> CH_{3}NH_{2}> C_{2}H_{5}NH_{2}> C_{6}H_{5}NH_{2}$
0%
$C_{2}H_{5}NH_{2}> CH_{3}NH_{2}> NH_{3}> C_{6}H_{5}NH_{2}$
0%
$C_{6}H_{5}NH_{2}> C_{2}H_{5}NH_{2}> CH_{3}NH_{2}> NH_{3}$
0%
$CH_{3}NH_{2}>C_{2}H_{5}NH_{2}>C_{6}H_{5}NH_{2}> NH_{3}$

Explanation

Electron donors are bases. Electron withdrawing groups (eg, benzyl) decrease the basicity of amines and Electron donating groups (eg, alkyl) increase the acidity of amines.

so The correct order of basicity of amines in

$C_{2}H_{5}NH_{2}> CH_{3}NH_{2}> NH_{3}> C_{6}H_{5}NH_{2}$

Hence B is correct answer

Among the following compounds nitrobenzene , benzene, aniline and phenol, the strongest basic behaviuor in acid medium is exhibited by :

0%
Phenol
0%
Aniline
0%
Nitrobenzene
0%
Benzene

Explanation

Because the N atom in aniline has a lone pair to donate and also due to +I effect of -

$NH_{2}$ group

Among the following the weakest base is

0%
$C_{6}H_{5}CH_{2}NH_{2}$
0%
$C_{6}H_{5}CH_{2}NHCH_{3}$
0%
$O_{2}N CH_{2} NH_{2}$
0%
$CH_{3} NH CHO$

Explanation

$CH_3CONH_2>CH_3CH_2NH_2>C_6H_5CH_2NH_2>C_6H_5CH_2NHCH_3$

So the weekest base is B

Arrange the following in increasing order of basicity

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

0%
$(CH_{3})_{3}N< (CH_{3})_{2}NH< CH_{3}NH_{2}< C_{6}H_{5}NH_{2}$
0%
$(CH_{3})_{3}N> (CH_{3})_{2}NH> CH_{3}NH_{2}> C_{6}H_{5}NH_{2}$
0%
$C_{6}H_{5}NH_{2}< (CH_{3})_{3}N< CH_{3}NH_{2}< (CH_{3})_{2}NH$
0%
$C_{6}H_{5}NH_{2}> (CH_{3})_{3}N> CH_{3}NH_{2}> (CH_{3})_{2}NH$

Explanation

The correct order is

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

Hence B is correct order

Decreasing order of basicity is
(1)

$CH_3CONH_2$
(2)

$CH_{3}CH_{2}NH_2$
(3)

$Ph-CH_{2}CONH_{2}$

0%
$1> 2> 3$
0%
$2> 1> 3$
0%
$3> 2>1$
0%
None of these

In the diazotisation of aniline with sodium nitrite and hdrochloric acid, an excess of hydrochloric acid is used primarily to

0%
Suppress the concentration of free aniline available for coupling
0%
Suppress hydrolysis of phenol
0%
Insure a stoichiometric amount of nitrous acid
0%
Neutralize the base liberated

Complete the following reaction

$R NH_{2}+H_{2}SO_{4}\rightarrow$

0%
$\left [ R NH_{3} \right ]^{+}HSO^{-}_{4}$
0%
$\left [ R NH_{3} \right ]^{+}_{2}SO^{2-}_{4}$
0%
$R NH_{2}.H_{2}SO_{4}$
0%
No reaction

Explanation

Bases react with acid to form salt.

∵

Amines are basic in nature

∵

It forms salt on reaction with

$H_2SO_4$

Azo dye is prepared by the coupling of pehnol and :

0%
Diazonium chloride
0%
o-nitro aniline
0%
Benzoic acid
0%
Chlorobenzene

Which one of the following compound is most basic

0%
(A)
0%
(B)
0%
(C)
0%
All are equally basic

The correct order of increasing basic nature for the bases

$NH_{3}$ ,

$CH_{3}NH_{2}$ and

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

0%
$CH_{3}NH_{2}< NH_{3}< (CH_{3})_{2}NH$
0%
$\left ( CH_{3} \right )_{2}NH< NH_{3}< CH_{3}NH_{2}$
0%
$NH_{3}< CH_{3}NH_{2}< (CH_{3})_{2}NH$
0%
$CH_{3}NH_{2}< (CH_{3})_{2}NH< NH_{3}$

The correct order of basicity of amines in water is

0%
$(CH_{3})_{2}NH> (CH_{3})_{3}N> CH_{3}NH_{2}$
0%
$CH_{3}NH_{2}> (CH_{3})_{2}NH> (CH_{3})_{3}N$
0%
$(CH_{3})_{3}N> (CH_{3})_{2}NH> CH_{3}NH_{2}$
0%
$(CH_{3})_{3}N> CH_{3}NH_{2}> (CH_{3})_{2}NH$

Explanation

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

As basicity of amines in water is determined by the tendency of donating electrons and 3degree amines will have the highest +I effect Leading to greater electron density followed by 2degree and primary amines

Which of the following is/are correct?

0%
$PhOH$ has a higher boiling point than $PhSH$ (benzenthiol).
0%
$p-Hydroquinone$ has a higher melting point than catechol.
0%
$o-Nitrophenol$ has a lower boiling point than $m-$ and $p-isomers$ .
0%
$o-Hydroxybenzaldehyde$ is more water soluble than $m-$ and $p-isomers$ .

Explanation

$a.$

$PhOH$ has higher boiling point due to

$H-bonding$ .

$b.$ Molecules of

$p-benzonoquinone$ can fit closer in solid state causing it to have higher melting point.

$c. and \,d.$ Intramolecular H-bonding (chelation) in the o-isomers inhibits the intermolecular attraction, lowers the boiling point, reduces H-bonding with

$H_2O$ , and decreases water solubilities. Intra-molecular chelation dots not occur in m- and

$p-isomers$ .

Hence, Options "A" ,"B" & "C" are correct answers.

The correct most basicity in amines is

0%
$C_{4}H_{5}NH_{2}$
0%
$CH_{3}NH_{2}$
0%
$(CH_{3})_{2}NH$
0%
$(CH_{3})_{3}N$

The correct IUPAC name for

$CH_{2}=CHCH_{2}NHCH_{3}$ is

0%
Allylmethylemine
0%
$2-$ amino $-4-$ pentene
0%
$4-$ aminopent $-1-$ ene
0%
$N-$ methylprop $-2-$ en $-1-$ amine

Explanation

$IUPAC$ name of the compound

$CH_2=CHCH_2NHCH_3$ is

$N-$ methyprop

$-2-$ en

$-1-$ amine.

The correct decreasing order of basic strength of the following species is:

$H_{2}O, NH_{3}, OH^{-}, NH_{2}^{-}$

0%
$NH_{2}^{-} > OH^{-} > NH_{3} > H_{2}O$
0%
$OH^{-} > NH_{2}^{-} > H_{2}O > NH_{3}$
0%
$NH_{3} > H_{2}O > NH_{2}^{-} > OH^{-}$
0%
$H_{2}O > NH_{3} > OH^{-} > NH_{2}^{-}$

Explanation

Basic strength depends upon the electron donating capacity of the central atom, here amide is most basic due to presence of negative charge and two lone pair of electrons on nitrogen atom.

$NH_{2}^{-} > OH^{-} > NH_{3} > H_{2}O$

Aniline dissolves in

$HCl$ due to the formation of:

0%
anilinium chloride
0%
o-chloroaniline
0%
azo dye
0%
diazonium chloride

Explanation

Aniline is basic in nature and

$HCl$ is an acid. So, acid-base reaction will take place.

In aniline, the lone pair of electrons is partially delocalized into the benzene ring and is thus, available for protonation by an acid.

Hence, aniline dissolves in acid like

$HCl$ forming anilinium chloride salt.

Therefore, the answer is option

$A$ .

Gabriel phthalimide reaction is used for the preparation of:

0%
primary aromatic amines
0%
secondary amines
0%
primary aliphatic amines
0%
tertiary amines

Explanation

Hint: Gabriel phthalimide is a type of chemical reaction where alkyl halide is converted to a primary amine.

Explanation:

In this reaction, phthalimide reacts with $KOH$ to form potassium salt of phthalimide. This salt reacts with an alkyl halide to form alkylphthalimide. This alkylphthalimide reacts with $NaOH$ to form primary aliphatic amine.

Write short notes on the following: (vii) Gabriel phthalimide synthesis.

Final Answer: Correct option is (C).

Diphenyl hydrazine is same as :

0%
azoxybenzene
0%
azobenzene
0%
hydrazobenzene
0%
aniline

The compound formed when aniline reacts with benzene diazonium chloride is:

0%
p-amino azobenzene
0%
anilinium chloride
0%
p-hydroxy azobenzene
0%
diphenyl amine

Out of the following compounds, which is the most basic in aqueous solution?

0%
${ CH }_{ 3 }{ NH }_{ 2 }$
0%
$\left( { CH }_{ 3 } \right) _{ 2 }NH$
0%
$\left( { CH }_{ 3 } \right) _{ 3 }N$
0%
${ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$

Explanation

$(CH_{3})_{2}NH$ is the most basic compound out of the compounds given.

The explanation is as follows :
It is more basic than

$CH_{3}NH_{2}$ because of one more methyl group that increases basicity by

$+I$ effect.

But a number of factors count into basicity.
One of the important ones being the solvation effect.

$N(CH_{3})_{3}$ , due to excess steric hinderance, the solvation of the resulting ion is reduced. So, the conjugate acid is not much stabilized. So, it is less basic.

Option B is correct.

Which of the following has the maximum value of

${ pk_b }$ ?

0%
$\left( { CH }_{ 3 } \right) _{ 2 }NH$
0%
$\left( { CH }_{ 3 }{ CH }_{ 2 } \right) _{ 2 }NH$
0%
$C_6H_5-NH-C_6H_5$
0%
$C_6H_5-NH-CH_3$

Explanation

In the compound mentioned in part C, the lone pair of nitrogen is in conjugation with both the phenyl rings.

So, they are greatly dispersed and are not available for donation.

As a result, the compound C is least basic.

Least basic means least

$K_{b}$ value and least

$K_{b}$ value means highest

$pK_{b}$ value.

Option C is correct.

Arrange the following in the correct order of their basic character in the aqueous phase.

$(i)$

${ NH }_{ 3 }$

$(ii)$

${ RNH }_{ 2 }$

$(iii)$

${ R }_{ 2 }NH$

$(iv)$

${ R }_{ 3 }N$

0%
$IV > III > II > I$
0%
$III > IV > II > I$
0%
$III > II > IV > I$
0%
$I > II > III > IV$

The colour of dye formed in the reaction between benzene diazonium chloride and phenol is:

0%
yellow
0%
orange
0%
purple
0%
blue

Which of the following when treated with aniline will give benzene diazonium chloride as the major product?

0%
${ NaNO }_{ 2 }$ and $HCl$ between $0-{ 5 }^{ 0 }C$
0%
${ HNO }_{ 3 }$ and $HCl$ between $0-{ 5 }^{ 0 }C$
0%
${ C }_{ 6 }{ H }_{ 5 }{ NO }_{ 2 }$ between $0-{ 5 }^{ 0 }C$
0%
${ NaNO }_{ 2 }$ between $0-{ 5 }^{ 0 }C$

Explanation

Benzene diazonium chloride is

$PhN_{2}^{+}Cl^{-}$ .

It is formed when aniline is treated with

$NaNO_{2}-HCl$ at a temperature between 0-5

$^{\circ} C$ .

$PhNH_{2} + NaNO_{2}+HCl \rightarrow PhN_{2}^{+}Cl^{-}$

Benzene diazonium chloride is an important synthetic intermediate and is used in a number of reactions.

Option A is correct.

Which of the following statements is not correct?

0%
Primary amines show intermolecular hydrogen bonding.
0%
Secondary amines show intermolecular hydrogen bonding.
0%
Tertiary amines show intermolecular hydrogen bonding.
0%
Amines have lower boiling points as compared to those of alcohols and carboxylic acids of comparable molar masses.

Explanation

For hydrogen bonding to occur, we need a polarized hydrogen atom (a hydrogen atom attached directly to

$F$ ,

$O$ or

$N$ is called as a polarized hydrogen).

In a tertiary amine, there is no polarized hydrogen because all the hydrogens are replaced by alkyl or aryl groups.

So, tertiary amines cannot show hydrogen bonding.

Which of the following orders is correct regarding the basic strength of subsituted aniline ?

0%
p - nitroaniline > p - aminobenzaldehyde > p- bromoaniline
0%
p - nitroaniline < p - bromoaniline < p- aminobenzaldehyde
0%
p - nitroaniline < p - aminobenzaldehyde < p- bromoaniline
0%
p - nitroaniline > p - aminobenzaldehyde < p-bromoaniline

Which of the following undergoes diazotization?

0%
${ CH }_{ 3 }{ NH }_{ 2 }$
0%
${ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$
0%
${ CH }_{ 3 }{ CONH }_{ 2 }$
0%
${ CH }_{ 3 }N\left( { CH }_{ 3 } \right) _{ 2 }$

Explanation

Only primary aromatic amines undergo diazotization to give diazonium salts.
Reaction is as follows :

$C_{6}H_{5}NH_{2} + NaNO_{2} \xrightarrow {HCl} C_{6}H_{5}N_{2}^{+}Cl^{-}$

So, B is the correct answer because it is a primary aromatic amine.

${ CaC }_{ 2 }\xrightarrow { Hydrolysis } A\xrightarrow [ Cu\:tube ]{ Red\:hot } B\xrightarrow [ { 50-60 }^{ 0 }C ]{ { HNO }_{ 3 }-{ H }_{ 2 }{ SO }_{ 4 } } C\xrightarrow { Fe+HCl } D \xrightarrow { { NaNO }_{ 2 }+HCl,\:{ 0}^{ 0 }C } E$

In the reaction sequence given above,

$E$ is:

0%
aniline
0%
benzene diazonium chloride
0%
phenyl osazone
0%
benzoyl chloride

Explanation

Hydrolysis of calcium carbide yields ethyne.

$CaC_{2} + 2H_{2}O \rightarrow C_{2}H_{2} + Ca(OH)_{2}$

Now, in the presence of red hot copper tube,

$3$ molecules of ethyne polymerize to give benzene.

$3C_{2}H_{2}$ + red hot

$Cu$ tube

$\rightarrow C_{6}H_{6}$ (benzene)

In the presence of

$HNO_{3}-H_{2}SO_{4}$ , nitration of benzene occurs (electrophilic aromatic substitution) and nitrobenzene is formed.

$C_{6}H_{6} + HNO_{3}-H_{2}SO_{4} \rightarrow C_{6}H_{5}NO_{2}$ (nitrobenzene)

In the presence of

$Fe-HCl$ , nitrobenzene reduces to aniline.

$C_{6}H_{5}NO_{2} + Fe-HCl \rightarrow C_{6}H_{5}NH_{2}$ (aniline)

Aniline on reaction with

$NaNO_{2}-HCl$ gives benzene diazonium chloride.

$C_{6}H_{5}NH_{2} + NaNO_{2}-HCl \rightarrow C_{6}H_{5}N_{2}^{+}Cl^{-}$

The complete reaction sequence is as follows:

${ CaC }_{ 2 }\xrightarrow { Hydrolysis } C_2H_2\xrightarrow [ Cu\:tube ]{ Red\:hot } C_6H_6\xrightarrow [ { 50-60 }^{ 0 }C ]{ { HNO }_{ 3 }-{ H }_{ 2 }{ SO }_{ 4 } } C_6H_5NO_2\xrightarrow { Fe+HCl } C_6H_5NH_2$

$\xrightarrow {NaNO_2 + HCl,\: 0^oC} C_6H_5N_2^+Cl^-$

Therefore,

$E$ is benzene diazonium chloride.

The correct order of basicity of the given compounds is :

0%
$CH_{3} CH_{2} NH_{2} < HO(CH_{2})_{3}NH_{2}< HO(CH_{2})_{2}NH_{2}$
0%
$CH_{3} CH_{2} NH_{2} < HO(CH_{2})_{3}NH_{2}> HO(CH_{2})_{2}NH_{2}$
0%
$CH_{3} CH_{2} NH_{2} > HO(CH_{2})_{3}NH_{2}> HO(CH_{2})_{2}NH_{2}$
0%
$CH_{3} CH_{2} NH_{2} > HO(CH_{2})_{3}NH_{2}< HO(CH_{2})_{2}NH_{2}$

Explanation

The correct order of basicity of the given compounds is as follows:

$\displaystyle CH_{3} CH_{2} NH_{2} > HO(CH_{2})_{3}NH_{2}> HO(CH_{2})_{2}NH_{2}$

$OH$ group is an electron withdrawing group and hence, it increases the partial positive charge on

$N$ atom. Therefore, lone pair of

$N$ cannot be easily donated and also, the basicity decreases.

As the number of carbon atoms between

$-OH$ group and amino group increases, the electron withdrawing capacity of

$-OH$ group on amine

$N$ decreases and basicity increases.

${ CH }_{ 3 }{ CH }_{ 2 }Cl\xrightarrow { NaCN } X\xrightarrow { Pt/{ H }_{ 2 } } Y\xrightarrow [ anhydride ]{ Acetic } Z$
In the above sequence,

$Z$ is:

0%
${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ NHCOCH }_{ 3 }$
0%
${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ NH }_{ 2 }$
0%
${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CONHCH }_{ 3 }$
0%
${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CONHCOCH }_{ 3 }$

Explanation

$CH_{3}CH_{2}Cl + NaCN \rightarrow CH_{3}CH_{2}CN$

So,

$X$ is

$CH_{3}CH_{2}CN$ and is formed by nucleophilic substitution reaction (

$S_N2$ )

Now,

$CH_{3}CH_{2}CN + H_{2}/Pt \rightarrow CH_{3}CH_{2}CH_{2}NH_{2}$

So,

$Y$ is

$CH_{3}CH_{2}CH_{2}NH_{2}$ and is formed by reduction reaction. The cyanide group on reduction gives amino group.

$CH_{3}CH_{2}CH_{2}NH_{2} + CH_{3}COOCOCH_{3} \rightarrow CH_{3}CH_{2}CH_{2}NHCOCH_{3}$

So,

$Z$ is

$CH_{3}CH_{2}CH_{2}NHCOCH_{3}$ and is formed by addition-elimination reaction.

Amongst the following, the most basic compound is :

0%
benzylamine
0%
aniline
0%
acetanilide
0%
p-nitroaniline

Explanation

Aliphatic amines are stronger bases as compared to aromatic amines. This is because in aromatic amines, the lone pair on nitrogen is involved in delocalization with aromatic ring.
Benzyl amine is an aliphatic amine, whereas other given amines are aromatic amines.
Hence, among the given amines, benzyl amine

$({ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }{ NH }_{ 2 })$ is the strongest base.

CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 8 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 8 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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