$$\displaystyle { CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 3 }$$

$$\displaystyle CH_3-\underset {\displaystyle \underset {\displaystyle O}{ ||}}{C}-OH$$

$$\displaystyle { CH }_{ 3 }-O-{ C }_{ 3 }{ H }_{ 7 }$$

$$\displaystyle CH_3-\underset {\displaystyle \underset {\displaystyle O}{ ||}}{C}-CH_3$$

MCQ Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 3

Consider the Hofmann ammonolysis reaction:

For any value of

$R$ , the correct order of basic character of the above amines in gaseous phase of nonpolar solvent is :

0%
$3 > 2 > N{H}_{3} > 1$
0%
$3 > 2 > 1 > N{H}_{3}$
0%
$2 > 3 > N{H}_{3} > 1$
0%
$1 > 2 > 3 > N{H}_{3}$

Explanation

For any value of

$R$ , the correct order of basic character of the above amines in gaseous phase of nonpolar solvent is

$3 > 2 > 1 > N{H}_{3}$
This is because in gaseous or nonpolar solvent, the basic order is

$3 > 2 > 1 > N{H}_{3}$ .

The conjugate base of

$\displaystyle ({ CH }_{ 3 }{ ) }_{ 2 }{ NH }_{ 2 }^{ \oplus }$ is:

0%
$\displaystyle ({ CH }_{ 3 }{ ) }_{ 3 }N$
0%
$\displaystyle ({ CH }_{ 3 }{ ) }_{ 2 }NH$
0%
$\displaystyle ({ CH }_{ 3 }{ ) }_{ 2 }{ N }^{ \ominus }$
0%
$\displaystyle ({ CH }_{ 3 }{ ) }_{ 2 }{ N }^{ \oplus }$

Which one is more basic,

$CH_3NH_2 \,\, or \,\, CF_3NH_2$ ?

0%
$CH_3NH_2$
0%
$CF_3NH_2$
0%
Both have similar basicity
0%
None of the above

Explanation

Between

${ H }_{ 3 }C\ddot { N } H_{ 2 },{ F }_{ 3 }C\ddot { N } { H }_{ 2 },{ H }_{ 3 }\ddot { N } { H }_{ 2 }$ is more basic than

$F_3C\ddot NH_2$ because

$-CH_3$ group is an electron donating group so shows posetive inductive effect and donates electron density to nitrogen which makes lone pair on nitrogen more destabilished and are available for reaction .but

$CF_3$ group is an electron density from nitrogen as it shows negative inductive effect and so lone pairs on nitrogen are not available for reactions.

Which of the following statements is correct?

0%
Methyl amine is slightly acidic.
0%
Methyl amine is less basic than ammonia.
0%
Methyl amine is less basic than dimethyl amine.
0%
Methyl amine is less basic than aniline.

Explanation

Dimethyl amine (

$\displaystyle { Me }_{ 2 }NH$ ) is more basic than

$\displaystyle { MeNH }_{ 2 }$ , due to (+I) effect of two (Me) groups.

Diazo coupling is useful to prepare some:

0%
pesticides
0%
dyes
0%
proteins
0%
vitamins

Among the following, the strongest base is:

0%
$\displaystyle { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$
0%
$\displaystyle p-{ NO }_{ 2 }{ C }_{ 6 }{ H }_{ 4 }{ NH }_{ 2 }$
0%
$\displaystyle m-{ NO }_{ 2 }{ C }_{ 6 }{ H }_{ 4 }{ NH }_{ 2 }$
0%
$\displaystyle { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }{ NH }_{ 2 }$

Explanation

(D) is an aliphatic amine; so it is stronger base than aromatic amine.

Moreover, EWG

$( -{ NO }_{ 2 } )$ at o, p and m-positions decreases the basic character.

The basicity order is:

$\displaystyle (D)>(A)>(C)>(B)$ .

Option D is correct.

Aniline is soluble in which of the following organic reagents?

0%
Benzene
0%
Ether
0%
Alcohol
0%
All of above

Which one of the following is the strongest base in aqueous medium :

0%
$(C_2H_5)_3N$
0%
$C_2H_5NH_2$
0%
$NH_3$
0%
$(C_2H_5)_2NH$

Explanation

$+I$ effect of

$C_2H_5$ group increases electron density on

$N$ and make it stronger base. So the order of basic strength would be

$3^o>2^o>1^o>NH_3$ but in aqueous medium presence of hydrogen bonding and bulkier groups affect the basicity. In the

$(C_2H_5)_3N$ , alkyl groups hinder the attack of proton on

$N$ . Therefore it become less basic. And due the additive effects of steric hindrance and induction

$(C_2H_5)_2NH$ is strongest base in aquous medium.

Which of the following amines will form stable diazonium salt at 273-283 K?

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

Which of the above is more acidic?

0%
I
0%
II
0%
Both have similar acidity
0%
None of the above

Arrange the following in correct activating order towards EAS.

$-\underset{I}{NHCOCH_3} \quad - O - \underset{II}{\overset{O}{\overset{||}{C}}} - CH_3 \quad - \underset{III}{\overset{\!\!\!\!\!+}{NR_3}}\quad - \underset{IV}{NH_2}$

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

Explanation

In Electrophilic aromatic substitution reaction, the electrons donating group is favoured over electron withdrawing group.

$-NH_2 > \overset{+}{-NR_3} > - O - \overset{O}{\overset{||}{C}} - CH_3 > - NHCOCH_3$

This is the required order of electrophilic aromatic substitution.

Arrange the following in increasing order of pH value

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

Phenyl cyanide on reduction with

$Na/{C}_{2}{H}_{5}OH$ yields :

0%
${C}_{6}{H}_{5}{CH}_{2}{NH}_{2}$
0%
${C}_{6}{H}_{5}NH{CH}_{3}$
0%
0%
${C}_{6}{H}_{5}{NH}_{2}$

Explanation

Phenyl cyanide (benzonitrile) on reduction with sodium ethanol produces benzyl amine.

$\displaystyle -C \equiv N$ group is reduced to

$\displaystyle CH_2-NH_2$ group.

$\displaystyle Ph-C \equiv N \xrightarrow {Na/ ethanol} Ph-CH_2-NH_2$

Tertiary butyl amine is a:

0%
${1}^{o}$ amine
0%
${2}^{o}$ amine
0%
${3}^{o}$ amine
0%
quaternary salt

Explanation

Tertiary butyl amine is a primary amine as it contains

$R-NH_2$ group.

If only one hydrogen of ammonia is replaced by one alkyl group then it is primary amine.

Option A is correct.

Write the basicity order of the following.

0%
(II) > (I) > (III) > (IV)
0%
(I) > (III) > (II) > (IV)
0%
(III) > (I) > (II) > (IV)
0%
(I) > (II) > (III) > (IV)

Explanation

1,2,3,4 is more basic as it contains two electron donating

$CH_3$ groups which destabilise the lone pair on nitrogen by donating more electron density due to + I effect and so lone pair are available easily 2 is having only one electron donating

$H_3C$ group so is less basic than 3 and 4 is least basic as it contains an electron .co group which stabilizes lone pair on nitrogen and therefore are not available for donation

How many primary amines can be formulated by

${C}_{3}{H}_{9}N$ and how many

${1}^{o}$ hydrogen are associated with carbon atoms of each compound?

0%
Two primary amines $[3,6]$
0%
One primary amine $[3]$
0%
Three primary amines $[3,6,6]$
0%
Two primary amines $[5,6]$

Explanation

Two primary amines, n propyl amine and iso propyl amine can be formulated with molecular formula

$C_3H_9N$
In n propyl amine, five primary hydrogens are attached with carbon atoms. In isopropyl amine, six primary hydrogen atoms are attached with carbon atoms.

Methyl cyanide on reaction with sodium and

${C}_{2}{H}_{5}OH$ forms:

0%
ethyl amine
0%
ethanoic acid
0%
ethanimine
0%
none of these

Explanation

Methyl cyanide on reaction with sodium and ethanol forms ethylamine.

$CH_3-C \equiv N +4[H] \xrightarrow {Na/ ethanol} CH_3-CH_2-NH_2$
In this reaction, the nitrile group is hydrogenated to form a primary amine. This is a useful method to increase the length of a carbon chain in preparing a primary amine which contains one more carbon atom than the alkyl halide from which alkyl cyanide is obtained.

Pyridine is less basic than triethylamine because :

0%
Pyridine has aromatic character
0%
Nitrogen in pyridine is $sp^2$ hybridised
0%
Pyridine is a cyclic system
0%
In pyridine, lone pair of nitrogen is delocalised

Explanation

Pyridine, is much less basic than typical aliphatic amines, but for a very different reason: the unshared pair is in an

$sp^2$ , which is much lower in energy than the electron pair of aliphatic amines, which is in an

$sp^3$ . Therefore, pyridine is less easily protonated than typical aliphatic amines such as triethylamine .

Among the following substituted pyridines, the most basic compound is:

Which of the following method(s) does work for the preparation of secondary amine?

0%
$R_3N \xrightarrow[\begin{matrix} (ii) H_3O^+\\(iii) \Delta\end{matrix}]{(i) Br-CN}?$
0%
$R-\overset{O}{\overset{||}{C}}-NH - R' \xrightarrow{LiAlH_2}?$
0%
$\displaystyle RNC\xrightarrow[\Delta ]{Nl/H_{2}}?$
0%
All of these

Explanation

(A) Tertiary amines react with cyanogen bromide to form a dialkylcyanamide which, on hydrolysis, gives a secondary amine.

$R_3N + BrCN \rightarrow [R_3NCN]^+Br^- \rightarrow RBr + R_2NCN \overset {H^+}{\rightarrow} R_2NH$

(B) Secondary amides

$(RCONHR)$ react with lithium aluminium hydride (LiAlH4) to form secondary amines

$(RCH_2NHR')$

$R\overset{\overset{O}{||}}{C}NHR + H-Al^-H_3 \rightarrow R\overset {\overset{O^-}{|}}{C}=NR \rightarrow RCH_2NHR$

$RNC \overset {Ni / H_2}{\rightarrow} RNHCH_3$

Arrange the following in the decreasing order of boiling point:

${ C }_{ 2 }{ H }_{ 5}OH$

ii)

$(CH_3)_2NH$

iii)

$C_2H_5NH_2$

0%
i > ii > iii
0%
ii > i > iii
0%
iii > i > ii
0%
iii > ii > i

Explanation

All the given compounds have intermolecular hydrogen bonding due to which their boiling points are higher.

$C_2H_5OH$ possesses the strongest hydrogen bonding because the

$O$ is more electron negative than

$N$ in other compounds. Therefore it has the highest boiling point.

$(CH_3)_2NH$ and

$C_2H_5NH_2$ , the latter has a longer alkyl chain which makes hydrogen bonding weaker.

Hence the order of boiling points is

$i>ii>iii$

Which one of the amine have highest boiling point?

0%
Primary amine
0%
Secondary amine
0%
Tertiary amine
0%
all of these

Arrange the following in the increasing order of boiling point:

(i) $C_2H_5OH$

(ii) ${(CH_3)_2}NH$

(iii) $C_2H_5NH_2$

0%
i < ii < iii
0%
ii < i < iii
0%
iii < i < ii
0%
iii < ii < i

Explanation

All the given compounds have intermolecular hydrogen bonding due to which their boiling points are higher.

$C_2H_5OH$ possesses strongest hydrogen bonding because the

$O$ is more electron negative than

$N$ in other compounds. Therefore it has the highest boiling point. In

$(CH_3)_2NH$ and

$C_2H_5NH_2$ , the latter has longer alkyl chain which makes hydrogen bonding weaker. Hence the order of boiling points is

$iii<ii<i$

How many primary amines are possible for the formula

$C_4H_{11}N$ ?

0%
One
0%
Two
0%
Three
0%
Four

Arrange the following in the decreasing order of solubility in water :

$\displaystyle { C}_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$

ii)

$({ C }_{ 2 }{ H }_{5 }{ ) }_{ 2 }NH$

iii)

${ C }_{ 2 }{ H }_{ 5}{ NH }_{ 2 }$

0%
i > ii > iii
0%
ii > i > iii
0%
iii > i > ii
0%
iii > ii > i

Explanation

Ethylamine when added to water forms intermolecular H−bonds with water. And therefore it is most soluble in water.

Diethylamine having a bigger alkyl chain forms weaker intermolecular hydrogen bonds.

But aniline does not form H−bond with water to a very large extent due to the presence of a large hydrophobic

$−C_6H_5$ group. Hence, aniline has the least solubility in water.

Hence the order of solubility is

$iii>ii>i$

Which of the following is more basic than aniline?

0%
p-nitroaniline
0%
Benzyl amine
0%
Diphenyl amine
0%
Triphenyl amine

Explanation

Although both are primary amines, the nitrogen lone pair of electrons are involved in resonance with the aromatic benzene and are less available to accept a proton whereas benzylamine is an aliphatic amine with a larger

$K_b$ than aniline.

So, Option B is correct.

Arrange the following in the increasing order of solubility in water:

$\displaystyle { C}_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$

ii)

$({ C }_{ 2 }{ H }_{5 }{ ) }_{ 2 }NH$

iii)

${ C }_{ 2 }{ H }_{ 5}{ NH }_{ 2 }$

0%
i < ii < iii
0%
ii < i < iii
0%
iii < i < ii
0%
iii < ii < i

Explanation

The decreasing order of solubility in water; is iii)

${ C }_{ 2 }{ H }_{ 5}{ NH }_{ 2 }$ > ii)

$({ C }_{ 2 }{ H }_{5 }{ ) }_{ 2 }NH$ > i)

$\displaystyle { C}_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$

The extent of

$H$ bond formation in iii)

${ C }_{ 2 }{ H }_{ 5}{ NH }_{ 2 }$ is maximum as it has two

$H$ attached to

$N.$ Hence, it has highest solubility.

$\displaystyle { C}_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$ solubility is least due to large hydrophobic part.

So, the correct option is

$A.$

Gabriel phthalimide reaction is used for the preparation of :

0%
Primary aromatic amines
0%
Secondary amines
0%
Primary aliphatic amines
0%
Tertiary amines

Explanation

$Explanation:$

Gabriel phthalimide reaction is used to prepare primary amines from phthalimide the reaction steps are shown as below;

$Step\ 1:$ Potassium Phthalimide when treated with a primary alkyl halide gives N-alkylphthalimide.

$Step\ 2:$ N-alkylphthalimide with aqueous sodium hydroxide gives primary amine and sodium phthalate.

www.sarthaks.com (549×267)

Therefore, aliphatic primary amine is the correct answer.

$Final\ Answer:$ Option

$C$ .

Methyl amine reacts with methyl iodide. For completion of reaction, how many moles of methyl iodide are required?

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

Explanation

$\therefore$ 3 moles of

$C{H}_{3}I$ (methyl iodide) are required for reaction with methyl amine.

The correct answer is 3.

1056223_460299_ans_81226510599f4361933f1a5b351087c4.jpg

Acetonitrile

$\xrightarrow { Hydrolysis }$ ?

0%
Acetic acid
0%
Propionic acid
0%
Formic acid
0%
None of these

Explanation

Cyanides on hydrolysis give carboxylic acid and

$N{H}_{3}$ .

Option A is correct.

1142203_460301_ans_c2eb5a0dfea04663afae0fd8d4bc3d5b.jpg

Acetonitrile on reduction gives:

0%
propanamine
0%
methanamine
0%
ethanamine
0%
none of these

Explanation

Cyanides on reduction give

$1$ amine.
1142090_460287_ans_0411db8b614f4750a94a4bfaca9714cc.jpg

Amine that cannot be prepared by Gabriel phthalimide synthesis is :

0%
aniline
0%
benzylamine
0%
methylamine
0%
iso-butylamine

Explanation

$Explanation:$

In Gabriel phthalimide synthesis is a two-step process in which the first step, potassium phthalimide treated with a primary alkyl halide to produce N-Alkyl phthalimide.

In the second step, N-Alkyl phthalimide. is treated with aqueous sodium hydroxide or with hydrazine (

$NH_2NH_2$ ) which gives primary amine.

In the given options, Aniline, Benzyl amine, and methylamine are primary amines therefore we can synthesis these three amines by Gabriel's phthalimide synthesis reaction.

In the synthesis of iso-Butylamine, we have to take iso-butyl halide as a reactant in the first step where iso-butyl cation will undergo rearrangement to form a tert. butyl cation and will gives N-tert-butyl phthalimide instead of N-Isobutyl phthalimide.

Solved: The isobutyl cation spontaneously rearranges to the tert-b... | Chegg.com

Therefore, in the second step, N-tert-butyl phthalimide will produce tert-butyl amine instead of iso-butyl amine.

Therefore, iso-butyl amine cannot be prepared by Gabriel phthalimide synthesis.

$Final\ Answer:$ Option

$D$ .

Primary amine

$(RNH_2)$ reacts with nitrous acid to give :

0%
$RNH_3^+NO_2$
0%
ROH
0%
ROR
0%
None of these

Explanation

$1^o$ aliphatic amines on reaction with

$HNO_2$ from alcohol and evolve

$N_2$

$RNH_2+HONO \xrightarrow [ ]{NaNO_2-HCl} ROH + N_2+H_2$

The decreasing order of 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

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

$\therefore$ The correct order of basicity of amines in

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

How many primary amines are possible with the formula

$\displaystyle { C }_{ 4 }{ H }_{ 11 }N$ ?

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

Explanation

The possible primary amine with the formula

$\displaystyle { C }_{ 4 }{ H }_{ 11 }N$ are

(i)

$\displaystyle { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CH }_{ 2 }{ NH }_{ 2 }$

(ii)

$\displaystyle { CH }_{ 3 }\overset { \overset { { CH }_{ 3 } }{ | } }{ CH } -{ CH }_{ 2 }-{ NH }_{ 2 }$

(iii)

$\displaystyle { CH }_{ 3 }-\overset { \overset { { CH }_{ 3 } }{ | } }{ \underset { \underset { { NH }_{ 2 } }{ | } }{ C } } -{ CH }_{ 3 }$

(iv)

$\displaystyle { CH }_{ 3 }-{ CH }_{ 2 }-\underset { \underset { { CH }_{ 3 } }{ | } }{ CH } -{ NH }_{ 2 }$

Ethylamine is ?

0%
$\displaystyle { CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 3 }$
0%
$\displaystyle CH_3-\underset {\displaystyle \underset {\displaystyle O}{ ||}}{C}-OH$
0%
$\displaystyle { CH }_{ 3 }-O-{ C }_{ 3 }{ H }_{ 7 }$
0%
$\displaystyle CH_3-\underset {\displaystyle \underset {\displaystyle O}{ ||}}{C}-CH_3$
0%
$\displaystyle CH_3 - CH_2- NH_2$

Explanation

Ethylamine is

$\displaystyle CH_3CH_2NH_3$ . It is organic derivative of ammonia in which one H atom attached to nitrogen is replaced by ethyl group.

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

0%
decarboxylation
0%
diazotization
0%
ammonolysis
0%
none of these

Explanation

$3 ksw + fe^{3+} \rightarrow fe(sw)_{3}+3k+$

red colour

1155134_492481_ans_8ff751679655480da681e1891ee666b9.png

Structure containing an amine group is :

0%
$\displaystyle { CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 3 }$
0%
$\displaystyle CH_3-\underset {\displaystyle \underset {\displaystyle O}{ ||}}{C}-OH$
0%
$\displaystyle { CH }_{ 3 }-O-{ C }_{ 3 }{ H }_{ 7 }$
0%
$\displaystyle CH_3-\underset {\displaystyle \underset {\displaystyle O}{ ||}}{C}-CH_3$
0%
$\displaystyle CH_3 - CH_2- NH_2$

Explanation

Structure containing an amine group is

$\displaystyle CH_3CH_2NH_2$ . It is the structure of ethylamine. It is organic derivative of ammonia in which one H atom attached to nitrogen is replaced by one ethyl group.

$CH_3CH_2CH_2CN+LiAlH_4\rightarrow$
Which of the following amine is formed in the above reaction :

0%
$CH_3CH_2CH_2CH_2NH_2$
0%
$CH_3CH_2CH_2NH_2$
0%
$CH_3CH_2NH_2$
0%
$CH_3CH_2CH_2CH(NH_2)_2$

Explanation

$CH_3CH_2CH_2CN+LiAlH_4\rightarrow CH_3CH_2CH_2CH_2NH_2$
Hence option B is correct.

Which compound is different from the others?

0%
Methyl ethyl ketone
0%
Pentan-2-one
0%
2-Pentanone
0%
Methyl propyl ketone

Explanation

Methyl ethyl ketone (

$\displaystyle CH_3 - \underset { \underset {\displaystyle O}{||} }{C} - CH_2-CH_3$ ) is different from the others.

Methyl ethyl ketone (also called butanone) contains

$4 C$ atoms.

Pentan-2-one, 2-pentanone and methyl propyl ketone are different names of same compound

$\displaystyle CH_3 - \underset { \underset {\displaystyle O}{||} }{C} - CH_2-CH_2-CH_3$ which contains

$5 C$ atoms.

$R-CN+2H_2\overset{Pd}{\rightarrow}R-CH_2NH_2$ is :

0%
Oxidation Reaction
0%
Reduction reaction
0%
Combustion
0%
None

Which compound is a secondary alcohol?

0%
Butan-1-ol
0%
Butan-2-ol
0%
Isobutyl alcohol
0%
2-Methylpropan-2-ol

Explanation

Butan-2-ol is a secondary alcohol. The

$-OH$ group is attached to a

$C$ atom which is attached to 2 other

$C$ atoms and 1

$H$ atom.

Which of the folllowing catalysts are used in the preparation amines by the reduction of isonitriles:

0%
Platinum
0%
Palladium
0%
Nickel
0%
All of the above

Alkyl and aryl amines are prepared using which of the following reducing agent.

0%
$LiAlH_4$
0%
$H_2$
0%
$NaBH_4$
0%
Both A and B

Explanation

Alkyl and aryl cyanides (nitriles) can be reduced to their corresponding primary amines using lithium aluminium hydride ( $LiAlH_4$ ) or catalytic hydrogenation.

Hence option A is correct.

Reduction of alkyl nitriles in presence of

$LiAlH_4$ Gives :

0%
Alkyl Amines
0%
Alkyl Amides
0%
Alkyl Nitro compounds
0%
Alkyl Iso cyanides

Explanation

Reduction of alkyl nitriles in presence of

$LiAlH_4$ Gives Alkyl Amines. Alkyl and aryl cyanides (nitriles) can be reduce by

$LiAlH_4$ ) or catalytic hydrogenation.

Hence option A is correct.

Which of the following is the product of the reaction?

0%
$C_6H_5CH_2NH_2$
0%
$C_5H_5CH_2NH_2$
0%
$C_6H_4CH_2NH_2$
0%
$C_4H_5CH_2NH_2$

Explanation

Cyano

$C_6H_5CH_2NH_2$ benzene when reacts with

$LiAlH_4$ it forms benzyl amine(

$C_6H_5CH_2NH_2$ ).
Hence option A is correct.

Amines are almost colourless but develop colour on keeping in the air for a long time, because:

0%
they are readily oxidised in air to form coloured oxidation product.
0%
they react with oxygen present in air in very less amount.
0%
they regain color in presence of air.
0%
none of these

Reduction of aryl nitriles in presence of

$LiAlH_4$ Gives :

0%
Aryl nitro compounds
0%
Aryl amines
0%
Aryl iso cyanides
0%
Alky amines

Explanation

Reduction of aryl nitriles in presence of $LiAlH_4$ Gives Aryl Amines. Alkyl and aryl cyanides (nitriles) can be reduce by $LiAlH_4$ ) or catalytic hydrogenation

Hence option B is correct.

Which of the following is true?

0%
Aliphatic amines are stronger bases than ammonia.
0%
Aliphatic amines are weaker bases than ammonia.
0%
Aliphatic amines ammonia have approximately same basic strength.
0%
Low molecular weight aliphatic amines are stronger bases than ammonia.

Explanation

Aliphatic amines are stronger bases than ammonia due

$+I$ (electron releasing) effect of alkyl groups. The electron density on N increases and lone pair of electrons can be easily donated.

Aromatic amines are generally:

0%
pleasant in smell.
0%
toxic in nature.
0%
low melting solids.
0%
none of these.

CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 3 - 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 3 - MCQExams.com

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

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