MCQ Questions for Class 12 Engineering Chemistry Amines Quiz 12

A 55- kDa protien was acid hydrolysed to obtain a misture of amino acids. How many amino acids could be present in the solution?

0%
550
0%
500
0%
1000
0%
1100

Phenyl acetonitrile reacts with sodium and alcohol to give

0%
$Ph-CH_2\ CH_2NH_2$
0%
$Ph\ CH_2-NH_2$
0%
$CH_3-CH_2-NH-Ph$
0%
$Ph-NH_2$

The most basic compound among the following is

0%
Benzylamine
0%
Aniline
0%
Acetanilide
0%
p-nitroaniline

Which of the following amines cannot be prepared by Gabriel Phthalimide reaction?

0%
0%
0%
0%
$CH_3-\underset{NH_2}{\underset{|\,\,\,\,\,\,\,}{NH_2}}-CH_3$

Which of the following will be most stable diazonium salt

R N_{2}^{+} X^{-}

$RN_2^{+}X^{-}$ ?

0%
$CH_3N_2^{+}X^{-}$
0%
$C_6H_5N_2^{+}X^{-}$
0%
$CH_3 CH_2 N_2^{+}X^{-}$
0%
$C_6H_5 CH_2 N_2^{+}X^{-}$

Which of the following is obtained by reducing methyl cyanide with

N a + C_{2} H_{5} O H

$Na+ C_2H_5OH$ ?

0%
Methyl alcohol
0%
Acetic acid
0%
Ethyl amine
0%
Methane

Explanation

Sodium in alcohol is a strong reducing agent and reduces nitriles to amines.

C H_{3} - C N + N a + C_{2} H_{5} O H \to C_{2} H_{5} - N H_{2}

$CH_3-CN+Na+C_2H_5OH\rightarrow C_2H_5-NH_2$

This reaction is also known as Mendius Reduction.

Factors responsible for base strength comparison of amines in aqueous solution ?

0%
Availability of lone pair on nitrogen
0%
Lone pair donating power of nitrogen
0%
Stability of conjugate acid formed by solvation through H-bonding.
0%
All of the above

Explanation

The basicity of amines depend upon its ability to donate the lone pair of electrons on nitrogen.

Order of basicity of amines in aqueous solution is,

N H_{3} < P r i m a r y a m i n e s (R N H_{2}) < T e r t i a r y a m i n e s (R_{3} N) < S e c o n d a r y a m i n e s (R_{2} N H)

$NH_3<Primary \ amines(RNH_2)<Tertiary \ amines(R_3N)<Secondary \ amines(R_2NH)$

This is due to the combined effect of +I effect of alkyl group, steric hindrance caused by alkyl groups and solvation of amines through H-bonding.

The power to donate lone pair of electrons increases as the number of alkyl groups increase. Therefore, making

$R_3N$ most basic, followed by

$R_2NH$ and

$RNH_2$ .

But as the number of alkyl groups increase, the steric hindrance caused by them also increases. Thus, decreasing the electron donating power of amines. Hence, making

$RNH_2$ most basic followed by

$R_2NH$ and

$R_3N$ .

Solvation is the formation of protonated amines when they are dissolved in water. As the number of H-atoms on nitrogen increases the possibility of H-bonding also increases, providing greater stability to the amine. Thus, primary amine

$(RNH_2)$ is more stable than secondary amine

$(R_2NH)$ , which is more stable than tertiary amine

$(R_3N)$ .

The combined effect of these three factors give the observed order of basicity of amines in aqueous solution.

Strongest base is

0%
$CH_2$ = $CH -CH_2-NH_2$
0%
$CH_2$ = $CH- CH_2 - NH_2$
0%
$CH_3CH_2CH_2NH_2$
0%

Explanation

Hint: Basic strength is directly proportional to the electron pair intensity on the nitrogen atom.

Correct Answer: Option C

Explanation of correct option:

Amines are basic in nature as they have lone pair of electrons on nitrogen. Therefore, they have a strong tendency to donate this lone pair of electrons to electron acceptors.

Here, in the case of option C

It is a primary amine and the presence of the alkyl group in alkyl amines increases the electron density on the nitrogen atom by virtue of the +I effect. Hence it is most basic.

Explanation of incorrect options:

Options A and B: The double-bonded carbon is more electronegative, so it shows -I effect. Hence electron richness on

$N$ decreases. Hence Both A andB are less basic.

Option D: As we know Aryl amines are less basic than alkyl-substituted amines because some electron density provided by the nitrogen atom is distributed throughout the aromatic ring. Hence it is also less basic.

Which of the following order is correct ?

0%
0%
0%
0%
All of the above

Minimum C-atom required for

$2^\circ$ amine to show chain isomerism are:

0%
$4$
0%
$6$
0%
$3$
0%
$5$

$CH_3NH_2+CHCl_3+KOH \rightarrow$ Nitrogen containing compound +

$KCl +H_2O$ . Nitrogen containing compound is:

0%
$CH_3CN$
0%
$CH_3NHCH_3$
0%
$CH_3-\bar{N}\equiv\overset{+}{C}$
0%
$CH_3-\overset{+}{{N}}\equiv {C}:^-$

Explanation

The reaction of primary amine (

$R-NH_2$ ) with chloroform

$(CHCl_3)$ and a base

$(KOH)$ is known as carbylamine reaction. The reaction yields isocyanide as the product, along with salt and water.

$CH_3NH_2+CHCl_3+KOH\rightarrow CH_3N^+\overline{=}C^-+KCl+H_2O$

$-NH_2$ group in aniline is:

0%
m- directing and deactivating
0%
o,p- directing and deactivating
0%
o,p- directing and activating
0%
m- directing and activating

Which of the following compound has highest boiling point ?

Which of the following are soluble in aq.

$NaOH$ solution ?

$H_3C-CH_2-NH_2$
3)

$(CH_3)_2NH$
Write basicity order of the above

0%
1>2>3
0%
3>2>1
0%
1>3>2
0%
2>3>1

$H_3C-CH_2-NH_2$
3)

$(CH_3)_2NH$
Write basicity order of the above

0%
1>2>3
0%
3>2>1
0%
1>3>2
0%
2>3>1

Which nitrogen atom is most basic in the above given compound?

0%
$3$
0%
$4$
0%
$5$
0%
$2$

What will be the final product in the following reaction sequence?

$CH_3CH_2CN\xrightarrow {H^{+}/H_2O } A \xrightarrow [ \Delta ]{ { NH }_{ 3 } } B \xrightarrow { NaOBr }$

0%
$CH_3CH_2CONH_2$
0%
$CH_3CH_2COBr$
0%
$CH_3CH_2NH_2$
0%
$CH_3CH_2CH_2NH_2$

Find the B in the following reaction

Which of the following is correct about Guanidine base?

0%
Resonance possible
0%
Odd number of p- electrons
0%
Localized lone pair of $e^{-}$
0%
Delocalized lone pair of $e^{-}$

The correct order of increasing basic nature for the bases

$NH_{3},CH_{3}NH_{2}$ and

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

0%
$NH_{3}<CH_{3}NH_{2}$ < $(CH_{3})_{2}NH$
0%
$(CH_3)_2NH<CH_3NH_2<NH_3$
0%
$CH_3NH_2<NH_3<(CH_3)_2NH$
0%
$NH_3 < (CH_3)_2NH<CH_3NH_2$

What is the decreasing order of basicity of pri-, sec-, and tert-, ethyl amines and

$NH_3$ ?

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

$[Cr(H_2O)_6]^{3+}$ typically absorbs at around 574 nm. It is allowed to react with ammonia to form a new complex

$[Cr(NH_3)_6]^{3+}$ that should have absorption at

0%
800 nm
0%
462 nm
0%
620 nm
0%
574 nm

Which of the following compounds has the most basic nitrogen?

0%
$a$
0%
$b$
0%
$c$
0%
$d$

Compare basic strength:

0%
$a > b > c$
0%
$b > a > c$
0%
$c > a > b$
0%
$c > b > a$

Arrange the following in the decreasing order of their

$pK_{a}$ value:
I .

$CH_{2} = CHCH_{2}NH_{2}$
II .

$CH_{3}CH_{2}CH_{2}NH_{2}$
III .

$CH \equiv CCH_{2}NH_{2}$

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

Which of the following will not react with sodiumnitrite and

$HCl$ ?

Among the following which amine can be prepared by Gabriel phthalimide method?

0%
$C_6H_5NH_2$
0%
$CH_3 - NH - CH_2CH_3$
0%
0%

Which of the following is an Incorrect statement about benzenediazonium chloride?

0%
It is prepared by the reaction of aniline with nitrous acid at 273-278 K
0%
It is a colourless crystalline solid
0%
Alkyl diazonium salts are more stable than arenediazonium salts
0%
Benzene diazonium fluoroborate is water insoluble & stable at room temperature

$1^0, 2^0$ and

$3^0$ amine can be distinguished by?

0%
Hinsberg reaction
0%
Hoffmann reaction
0%
Reaction with Nitrous acid
0%
All of these

Arrange the following compounds in decreasing order of basicity:

0%
P > Q > R > S
0%
Q > P > R > S
0%
R > S > Q > P
0%
Q > P > S > R

Among the following compounds, the most basic compound is:

Which of the following is most basic:

0%
$CH_{3}CH_{2}NH_{2}$
0%
$CH_{3}CH_{2}NHCH_{3}$
0%
0%
$CH_{3}CONH_{2}$

Most basic nitrogen among the following

Which of the following given compounds is the strongest base?

A mixture containing the following four compounds is extracted with $1$ M $HCl$ . The compound that goes to aqueous layer is :

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

The correct order of basicities of the following compounds is-

0%
B > A > C > D
0%
A > C > B > D
0%
A > A > B > D
0%
A > B > C > D

Which of the following is most basic in nature?

Arrange basicity of given compounds in decreasing order :
(1)

$CH_3-CH_2-NH_2$
(2)

$CH_2= CH-NH_2$
(3)

$CH\equiv C-NH_2$
(4)

$C_6H_5-NH_2$

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

The given imine can be prepared from which of the following

0%
0%
0%
0%
all of these

Arrange the compounds in decreasing order of basicity

Ammonia, Triethylamine, Aniline, Ethylamine, Diethylamine

0%
Diethylamine > Triethylamine > Ethylamine > Ammonia > Aniline
0%
Diethylamine > Ethylamine >Triethylamine> Ammonia > Aniline
0%
Triethylamine > Diethylamine > Ethylamine> Ammonia > Aniline
0%
Aniline > Ethylamine> Diethylamine >Triethylamine> Ammonia

Highest basic strength of the following compounds is of:

Arrange in decreasing order of basicity:

0%
R>S>P>Q
0%
Q>P>R>S
0%
S>R>P>Q
0%
R>S>Q>P

For the successful diazotization of arylamines, how many mole of mineral acid

$(HCl \ or\ H_2SO_4 )$ are required for each mole of the amine?

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

Which of the following order regarding basicity of amines is correct?

Amines are basic in nature because

0%
They produce $OH^-$ ions when treated with $H_2O$
0%
They have polar H atom attached to nitrogen atom
0%
They have lone pair of electrons on nitrogen atom
0%
They have alkyl group attached with nitrogen atom

The most stable diazonium salt is ........

0%
${ CH }_{ 3 }N_{ 2 }X$
0%
${ C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }X$
0%
${ C }_{ 2 }{ H }_{ 5 }{ N }_{ 2 }X$
0%
${ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }N_2X$

Among the following amines, the strongest Bronsted base is

Which of the following shows he correct order of decreasing the basic strength in benzene medium?

0%
$CH_3NH_2 > (CH_3)_3N > (CH_3)_2NH > NH_3$
0%
$(CH_3)_2NH > NH_3 > (CH_3)_3N > CH_3NH_2$
0%
$(CH_3)_2NH > (CH_3)_3N > NH_3 > CH_3NH_2$
0%
$(CH_3)_3N > (CH_3)_2NH > CH_3NH_2 > NH_3$

$HCONR\xrightarrow[pyridine ]{POCl_3}(a)+H_2O$ ; (a) in the above reaction is:

0%
$RCH=NOH$
0%
$R-N=C=O$
0%
$R-C=N$
0%
$R-N=C$

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

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.