MCQ Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 9

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

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.

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

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.

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%
$I > II > III > IV$

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.

Reaction at

${ 0-5 }^{ 0 }C$ between aniline,

${ NaNO }_{ 2 }$ and HCl is known as____________

0%
Suphonation of benzene
0%
Acylation of aniline
0%
Alkylation of aniline
0%
Diazotisation

Explanation

This is the diazotization reaction.

The reaction goes as follows :

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

The benzendiazonium salt thus formed acts as a synthetic intermediate and can be used to synthesize a wide variety of substituted benzene compounds.

For example,

$PhN_{2}^{+} Cl^{-} + CuCN \rightarrow PhCN$ (cyanobenzene)

$PhN_{2}^{+} Cl^{-} + CuI (KI) \rightarrow PhI$ (iodobenzene)

Option D 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.

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.

Which of the following orders with regards to the basic strength of substituted aniline is correct?

0%
p - methylaniline > p - chloroaniline > p - aminoacetophenone
0%
p - methylaniline > p - aminoacetophenone > p - chloroaniline
0%
p - aminoacetophenone > p - methylaniline > p - chloroaniline
0%
p - aminoacetophenone > p - chloroaniline > p - methylaniline

${ CH }_{ 3 }{ CH }_{ 2 }Cl\xrightarrow { NaCN } X\xrightarrow { Pt/{ H }_{ 2 } } Y\xrightarrow [ anhydride ]{ Acetic } Z$
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.

Which of the following forms unstable diazonium ion when treated with

$NaNO_{2}$ in aqueous

$HCl$ ?

0%
p-nitrotoluene
0%
ethylamine
0%
N,N-dimethyl aniline
0%
All of the above

Explanation

Aromatic diazonium salts are stable if kept cold

$(0-5^oC)$ and in solution, but they often decompose violently when isolated. Their stability (compared to alkane diazonium salts) is due in part to resonance stabilization of the diazonium ion and in part to the very high energy of the aryl cation that would result from the loss of

$N_2$ (g).
Therefore except ethylamine all other given compounds forms stable diazonium ion when treated with

$NaNO_2$ in aqueous

$HCl$ .

Hence, the correct option is B.

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.

The correct order of increasing basic nature for the bases

$NH_{3} ,CH_{3} NH_{2} and (CH_{3})_{2}NH$ in gas phase is :

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

Which of the following is correct formula for BDC?

0%
${ C }_{ 6 }{ H }_{ 4 }{ N }_{ 2 }Cl$
0%
${ C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }$
0%
${ C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }Cl$
0%
${ C }_{ 6 }{ H }_{ 4 }{ N }_{ 2 }$

Explanation

${ C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }Cl$ or benzene diazonium chloride is stable for a short time only. It slowly decomposes even at low temperatures. It is used immediately after preparation.

Among the following which is more basic :

Explanation

Hint: If $N$ lone pair is more available to donation more will be the basicity.

Explanation:

According to lewis's theory, the species which donate, its lone pair of electrons are basic in nature.
The electron releasing groups increases the electron density in the ring and hence lone pair of electrons in $NH_2$ can be easily donated.
Electron withdrawing groups decrease the ring's electron density, and the $NH_2$ group doesn't donate its lone pairs easily.
Aniline is a basic compound. As the electron-donating group on aniline increases, basicity increases, and basicity decreases when electron-withdrawing groups increase.
From the given options, $NO_2$ and $COOH$ are electron-withdrawing groups, while $CH_3O$ is an electron releasing group. Therefore, the basicity of the compound is in order of $A>C>B>D$ .
Hence, option $A$ is the most basic.

Correct Option: $A$

The major product expected, when Phthalmide is treated with NaOH, is :

The product

$U$ is:

Explanation

Benzoic acid reacts with thionyl chloride to form benzoyl chloride. Reaction with dinitrogen pentoxide introduces nitro group. Reaction with lithium dimethyl cuprate converts acid chloride to methyl ketone. The carbonyl group is protected with ethylene glycol, nitro group is hydrogenated and the protecting group is removed to form

$1-(4-aminophenyl)$

$ethanone$ . This is followed by diazotization and coupling with phenol to obtain the end product

$U$ .

The starting reagents needed to make the azo compound shown below is:

The correct order of basicity of the above compounds is given.
Identify the correct statement(s).

0%
In III, lone pair of electrons is involved in delocalisation but not in II.
0%
In IV, lone pair of electrons is involved in delocalisation but not in III.
0%
I is more basic than II because six membered ring is more stable than five membered ring.
0%
In III, lone pair of electrons is present in sp orbital but in I, II and IV lone pair of electrons are present in $sp^{2}$ orbital.

Aniline when diazotized in cold and then treated with dimethylaniline gives a coloured product. Its structure would be:

Explanation

Aniline when diazotized in cold I.e 0-5

$^{0}$ C give benzene diazonium chloride . When this compound coupled with dimethyl aniline it give coloured compound I.e p-(N,N-dimerhyle) amino azobenzene I.e azo dye

The correct basicity order of various atoms is :

0%
1 < 2 < 3 < 4
0%
2 < 3 < 4 < 1
0%
3 < 2 < 1 < 4
0%
3 < 2 < 4 < 1

Explanation

Basicity order is: N

$sp^3$ (localised lone pair) > N

$sp^2$ (localised lone pair) > N

$sp^2$ (Delocalised lone pair).

What is the principle product when aniline is treated with

$NaNO_2$ and

$HCl$ at

$0-5^{\circ}C$ , and this mixture is added to p-toludine (p-methyl aniline)?

Predict the product (T) for the following reaction sequence.

$C_6H_5-CH=O \xrightarrow{KCN / OH^-} R \xrightarrow{CrO_2 / H^{\oplus}}S \xrightarrow[(ii)\, Acidification]{(i)\, conc\, KOH}T$ :

0%
Cinnamic acid
0%
Mandelic acid
0%
Benzilic acid
0%
Benzoic acid & Benzyl alcohol

Explanation

This is an example of Benzil-benzilic acid rearrangement. Benzaldehyde reacts with

$KCN$ in an alkaline medium to form benzaldehyde cyanohydrin.
Oxidation with acidified chromium trioxide gives benzil. Treatment with conc KOH followed by acidification gives rearranged product benzillic acid.

The increasing order of basic strength for the following compounds is:
(I)

$CH_{3}-CH_{2}-NH_{2}$
(II)

$CH_{3}-NH-CH_{3}$
(III)

$CH_{3}-NH-Ph$
(IV)

$CH_{3}-NH-CHO$

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

Explanation

Aliphatic amines are stronger bases than ammonia and aromatic amines are weaker bases than

$N{ H }_{ 3 }$ and

$3°>2°>1°\rightarrow$ basic strength in a gaseous phase.

Thus, the increasing order of basic strength from the options can be

$IV<III<II>I$

$II,III,IV$ are secondary amines.

$III$ is aromatic, hence is less basic than

$II, IV$ . The lone pair of amine resonates with the

$-CHO$ bond, hence it cannot be freely available, thus it is less basic.

Option C is correct.

The following product(s) will be obtained from the above reaction:

$\displaystyle -$ Ethyl pthalimide on hydrolysis gives:

0%
methyl alcohol
0%
ethyl amine
0%
dimethyl amine
0%
diethyl amine

Which of the following diazonium salt is relatively stable of

$0-5^\circ C$ - :

0%
$CH_{3}CH_2N\equiv N\left. \right \} ^\oplus Cl^{-}$
0%
$CH_{3}-C(CH_{3})_{2}-N\equiv N\left. \right \} ^\oplus Cl^{-}$
0%
$CH_{3}-N\equiv N\left. \right \} ^\oplus Cl^{-}$
0%
$(CH_{3})_{3}C-N\equiv N\left. \right \} ^\oplus Cl^{-}$

Explanation

The diazonium salt

$CH_{3}-N\equiv N\left. \right \} ^\oplus Cl^{-}$ obtained by treatment of methylamine with nitrous acid is relatively stable at

$0-5^\circ C$ .

Which compound does not give positive test in Lassaigne's test for nitrogen?

0%
Urea
0%
Hydrazine
0%
Azobenzene
0%
Phenyl hydrazine

Explanation

In Lassaigne's test, the organic compound is fused with sodium.

Carbon and nitrogen present in the organic compound gives sodium cyanide (

$NaCN$ ).

This is further used in the test for nitrogen.
Hydrazine

$(H_2N-NH_2)$ do not contain carbon. It cannot form

$NaCN$ on fusion with sodium.
Hence, hydrazine cannot give Lassaigne's test for nitrogen.

Which of the following is weakest base?

0%
0%
0%
$H_2N-CH_2-NO_2$
0%
$CH_3NH-CH=O$

Between any two of the following molecules, hydrogen bonding is not possible:

0%
two primary amine molecules
0%
two secondary amine molecules
0%
two tertiary amine molecules
0%
two ammonia molecules

Explanation

Between two tertiary amine molecules, hydrogen bonding is not possible.
The hydrogen bonding is possible when

$\displaystyle H$ atom is attached to electronegative

$\displaystyle N, O$ or

$\displaystyle F$ atom. In tertiary amines, no

$\displaystyle H$ atom is attached to

$\displaystyle N$ atom. Hence, hydrogen bonding is not possible.

Of the following statements :

$(P)\:C_{6}H_{5}N=CH-C_{6}H_{5}$ is a Schiff's base.

$(Q)\:$ A dye is obtained by the reaction of aniline and

$C_{6}H_{5}N^+ \equiv NCl$ .

$(R)\:C_{6}H_{5}CH_{2}NH_{2}$ on treatment with

$[NaNO_{2}+HCl]$ gives diazonium salt.

$(S)\:p-$ Toluidine on treatment with

$[HNO_{2}+HCl]$ gives diazonium salt.

0%
only $(P)\:and\:(Q)$ are correct
0%
only $(P)\:and\:(R)$ are correct
0%
only $(R)\:and\:(S)$ are correct
0%
$(P),\:(Q)\:$ and $\:(S)$ are correct

Explanation

A Schiff base is a substituted imine

$(R_2C=N-R')$ . It is used as antioxidant. Thus

$(P)\:C_{6}H_{5}N=CH-C_{6}H_{5}$ is a Schiff's base. Thus statment

$(P)$ is correct.

Aniline reacts with

$C_{6}H_{5}N^+\equiv NCl$ to form higly coloured azo dye. This reaction is known as azo coupling. Thus the statement

$(Q)$ is correct.

$\:C_{6}H_{5}CH_{2}NH_{2}$ is a primary aliphatic amine. When it is treated with

$[NaNO_{2}+HCl]$ , it gives unstable diazonium salt which decomposes to form benzyl alcohol. Hence, statement

$(R)$ is incorrect.

p-Toluidne is an aromatic primary amine. Hence, it on treatment with

$[NaNO_{2}+HCl]$ gives diazonium salt. Hence, statement

$(S)$ is correct.

Option D is correct.

Which compound is soluble in water :

0%
$\displaystyle \left [ \left ( CH_{3} \right )_{2}NH_{2}\right ]^{+}Cl^{-}$
0%
$\displaystyle \left [ CH_{3}NH_{3} \right ]^{+}Cl^{-}$
0%
$\displaystyle \left [ \left ( CH_{3} \right )_{3}NH\right ]^{+}Cl^{-}$
0%
$\displaystyle PhNO_{2}$

Explanation

Nitro benzene is insoluble in water.
Mono, di, tri and tetra alkyl ammonium halides are soluble in water.
Thus

$\displaystyle \left [ \left ( CH_{3} \right )_{2}NH_{2}\right ]^{+}Cl^{-}, \displaystyle \left [ CH_{3}NH_{3} \right ]^{+}Cl^{-} and \displaystyle \left [ \left ( CH_{3} \right )_{3}NH\right ]^{+}Cl^{-}$ are soluble in water.

Arrange the following in the increasing order of their basicities.
I. p - Toluidine
II. N, N - Dimethyl - p - toluidine
III. p - Nitroaniline
IV. Aniline

0%
III < IV < I < II
0%
II < IV < I < III
0%
III < II < I < IV
0%
I < IV < III < II

Explanation

N,N-Dimethyl p-toluidine is most basic whereas p-nitroaniline is least basic.
II +I effect of two Me groups of N atom and Me group at p position (+I and hyperconjugative. effects) are responsible for highest basicity.
I +I and H.C. effects of (Me) group at p position makes p-toluidine more basic than aniline.
IV (Standard) (Aniline)
III -I and -R effect of (-

$N{O}_{2}$ ) group at p position are responsible for least basicity.

Which one(s) is/are true?

0%
(I) and (III) are modest Bronsted bases whereas (II) is not
0%
In (III) ${N}^{a}$ is more basic than ${N}^{b}$
0%
When (II) is protonated in the presence of a strong acid, protonation occurs at $C - 2$
0%
All the nitrogen present in (I), (II), and (III) is $s{p}^{2}$ hybridised

Explanation

(I) and (III) are modest Bronsted bases whereas (II) is not
This is because, (II) is not protonated, hence the statement (A) is true.
In (II),

${N}^{b}$ is more basic due to the presence of a lone pair of electrons.
Hence, the statement (B) is wrong.
Statement (C): When (II) is protonated in the presence of a strong acid, protonation occurs at

$C - 2$ .
This is clear from the resonance structures as shown below.
Statement (C) is true.
Statement (D): Due to resonance all the

$N$ in I, II, and III is

$s{p}^{2}$ hybridized and hence the statement is true.

Which compound will liberate

$\displaystyle CO_{2}$ from

$\displaystyle NaHCO_{3}$ solution :

0%
$\displaystyle CH_{3}CO\:NH_{2}$
0%
$\displaystyle CH_{3}NH_{2}$
0%
$\displaystyle \left ( CH_{3} \right )_{4}N^{+}OH^{-}$
0%
$\displaystyle CH_{3}N^{+}H_{3}Cl^{-}$

Explanation

$\displaystyle CH_{3}N^{+}H_{3}Cl^{-}$ will react with

$\displaystyle NaHCO_{3}$ to liberate

$\displaystyle CO_{2}$ .

$\displaystyle CH_{3}N^{+}H_{3}Cl^{-} + \displaystyle NaHCO_{3} \rightarrow CH_{3}NH_2 + NaCl + H_2O + \displaystyle CO_{2}$

Arrange the following in their decreasing order of acidity.

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

Explanation

The decreasing order of acidity is

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

$+I$ effect of two

$Me$ groups makes it most basic
In (I) electron-withdrawing effect of the carbonyl group makes it less basic than ammonia.
In (II) electron-withdrawing effect of two (

$C = O$ ) groups makes it least basic.

Which of the following is the product of the reaction shown above?

0%
$(CH_3)_2CHCH_2CH_2NHNH_2$
0%
$[(CH_3)_2CHCH_2CH_2]_2NH$
0%
$(CH_3)_2CHCH_2CH_2NH_2$
0%
$(CH_3)_2CHCH_2CH_2CONH_2$

The decreasing order of basic characters of the following is :

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

Which of the following diazonium salt is relatively stable at

$\displaystyle 0-5^{\circ}$ C?

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$\displaystyle CH_{3}-N\equiv N^{\oplus }Cl^{-}$
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$\displaystyle CH_{3}CH\left ( CH_{3} \right )-N\equiv N ^{\oplus }Cl^{-}$
0%
$\displaystyle C_{6}H_{5}-N\equiv N ^{\oplus }Cl^{-}$
0%
$\displaystyle \left ( CH_{3} \right )_{3}C-N\equiv N ^{\oplus }Cl^{-}$

Explanation

Diazonium salts obtained from aliphatic primary amines are unstable. Those obtained from aromatic primary amines are relatively stable.
Benzene diazonium chloride

$(\displaystyle C_{6}H_{5}-N\equiv N ^{\oplus }Cl^{-})$ is obtained from aromatic primary amine. Hence, it is relatively stable.

Which of the following statement(s) is/are correct?

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(I) and (II) are aromatic and have equal basic strength
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(I) is aromatic and (II) is anti-aromatic, but (II) is stronger base than (I)
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The order of basicity of the above compounds is (IV) > (III) > (II) > (I)
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The conjugate acid of (IV) is more stabilised than the conjugate acid of (II)

Explanation

In IV, lone pair of two N makes it more basic and does not delocalize in the benzene ring.
In III, no delocalisation of lone pair of electrons on

$\ddot { N } { H }_{ 2 }$ .
In II, no delocalization of lone pair of electrons on N.
In I, delocalization of lone pair of electrons on N via resonance.
Hence, the order of basic character is: IV > III > II > I.
Also The conjugate acid of (IV) is more stabilized than the conjugate acid of (II).
This is due to resonance in II which is absent in IV.
(I) and (II) are aromatic (follow Huckel's Rule) and have different basic strengths.
Thus, the options A and B are incorrect whereas options C and D are correct.

Alkyl isocyanides (

$R N^{\bigoplus} \equiv C^{\ominus}$ ) are reduced to 2

$^{\circ}$ amines(

$R-NH-CH_3$ ) with:

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$LAH$
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$NaBH_4$
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$HI + P$
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$H_2 / Pt$

Explanation

Alkyl isocyanides

$(RN≡C)$ are reduced to

$2^o$ amines with sodium and alcohol. Alkyl isocyanides (RN≡C) are reduced to

$2 ^o$ amines with

$H_2 /Pt$ .

Compound

$(E)$ is:

Explanation

D.U in

$(A)=\cfrac { \left( 2{ n }_{ C }+2 \right) -{ n }_{ H } }{ 2 } =\cfrac { \left( 8\times 2+2 \right) -10 }{ 2 } ={ 4 }^{ o }$
4 D.U and

$C:H=1:1$ suggest that

$(A)$ contains benzene ring with two extra C atoms (i.e., two

$(Me)$ groups]. Since compound

$(A)$ is steam volatile and on nittration gives two nitro-derivatives, so

$(A)$ is ortho-xylene

Compound

$(A)$ is:

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Ethylbenzene
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$p-Xylene$
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$m-Xylene$
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$o-Xylene$

Explanation

D.U in

$(A)=\cfrac { \left( 2{ n }_{ C }+2 \right) -{ n }_{ H } }{ 2 } =\cfrac { \left( 8\times 2+2 \right) -10 }{ 2 } ={ 4 }^{ o }$
4 D.U and

$C:H=1:1$ suggest that

$(A)$ contains benzene ring with two extra C atoms (i.e., two

$(Me)$ groups]. Since compound

$(A)$ is steam volatile and on nittration gives two nitro-derivatives, so

$(A)$ is ortho-xylene

CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 9 - MCQExams.com

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 9 - MCQExams.com

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

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