Download App

Chemistry M.C.Q (1 Marks)

Amines · Bihar Board (BSEB) STD 12 Science (BSEB - English Medium). 165 questions.

Questions · Page 2 of 4

M.C.Q (1 Marks)

MCQ 511 Mark
Which of these alkyl halides can be used to prepare amines using Gabriel phthalimide synthesis?
  • A
    Vinyl bromide
  • ✓
    $1 -$ bromo $- 3 -$ methylpentane
  • C
    Bromobenzene
  • D
    $2 -$ bromo $- 2, 3 -$ dimethylbutane
Answer
Correct option: B.
$1 -$ bromo $- 3 -$ methylpentane
View full question & answer→
MCQ 521 Mark
Which of the following is most basic?
  • A
    Diphenylamine
  • B
    Triphenylamine
  • C
    $p -$ Nitroaniline
  • ✓
    Benzylamine
Answer
Correct option: D.
Benzylamine
Benzylamine $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2$​ is more basic because benzyl group is electron donating group due to $+I$ effect.
So it is able to increase electron density of $N$ of $-NH_2$​ group.
Thus due to higher electron density rate of donation of a free pair of electron is increased $I.e$ basic character is higher While phenyl and nitro group are electron withdrawing group so they are able to decrease the electron density of $N$ of $−NH_2$​ group. Hence they are less basic.
View full question & answer→
MCQ 531 Mark
The correct statement among the following is:
  • A
    Boiling point of amine is more than that of alcohol of nearly same molecular mass.
  • B
    The solubility of amines in water increases with increase in molecular mass.
  • ✓
    The order of boiling point of isomeric amines is primary $>$ secondary $>$ tertiary.
  • D
    None of these
Answer
Correct option: C.
The order of boiling point of isomeric amines is primary $>$ secondary $>$ tertiary.
  1. Boiling point of amines have high boiling point due to presence of hydrogen bonding between amine molecules.
  2. But the boiling point decreases with increase in alkyl groups attached to the $N$ atom, because the number of $H$ atoms present for hydrogen bonding decreases.
View full question & answer→
MCQ 541 Mark
Aniline and methyl amine can be differentiated by:
  • A
    Reaction with chloroform and aqueous solution of $\text{KOH}.$
  • ✓
    Diazotisation followed by coupling with phenol.
  • C
    Reaction with $\ce{HNO_2​.}$
  • D
    None of these.
Answer
Correct option: B.
Diazotisation followed by coupling with phenol.
Phenol react with aniline to give diazonium salt by coupling but Methyl amine not react with phenolss.
View full question & answer→
MCQ 551 Mark
In order to prepare a $1^\circ $ amine from an alkyl halide with simultaneous addition of one $CH_2$ group in the carbon chain, the reagent used as source of nitrogen is $........$
  • A
    Sodium amide$, \ce{NaNH_2}$.
  • B
    Sodium azide$, \ce{NaN_3}$
  • ✓
    Potassium cyanide$, \text{KCN}.$
  • D
    Potassium phthalimide, $\mathrm{C}_6 \mathrm{H}_4(\mathrm{CO})_2 \mathrm{\sim N}^-K^{+}$.
Answer
Correct option: C.
Potassium cyanide$, \text{KCN}.$
$\text{KCN}$ is used to increase number of carbon atoms.
$\text{RX + KCN}\xrightarrow{\ \ \ \ \ \ }\text{RCN}+\text{KX}$
$\text{R}-\text{CN}+4\text{H}\xrightarrow[]{\text{H}_2/\text{Raney Ni}}\text{RCH}_2\text{NH}_2$
View full question & answer→
MCQ 561 Mark
$\ce{C_3​H_9N}​$ cannot represent:
  • A
    $10$ amine
  • B
    $20$ amine
  • C
    $30$ amine
  • ✓
    Quaternary salt
Answer
Correct option: D.
Quaternary salt
Quaternary salts of ammonia contain atleast $4$ carbon atoms
Here we are given only $3$ carbon atoms
So, it cannot represent a quaternary salt
View full question & answer→
MCQ 571 Mark
In the diazotization of aryl amines with sodium nitrite and hydrochloric acid, an excess of hydrochloric acid is used primarily to :
  • ✓
    Supress the concentration of free aniline available for coupling.
  • B
    Supress hydrolysis of phenol.
  • C
    Ensure a stoichiometric amount of nitrous acid.
  • D
    Neutralise the base liberated.
Answer
Correct option: A.
Supress the concentration of free aniline available for coupling.
If excess $\text{HCl}$ is not used, the arene diazonium salt will react with free aniline to form azo compound.
Excess $\text{HCl}$ forms hydrochloride salt of aniline. This prevents azo coupling.
View full question & answer→
MCQ 581 Mark
By heating ammonium chloride with two equivalents of formaldehyde it forms:
  • A
    Dimethylamine
  • B
    Ethylamine
  • ✓
    Methylamine
  • D
    Ammonium formate
Answer
Correct option: C.
Methylamine
$6 \mathrm{HCHO}+4 \mathrm{NH}_4 \mathrm{Cl} \rightarrow\left(\mathrm{CH}_{26}\right) \mathrm{N}_4+4 \mathrm{HCl}+6 \mathrm{H}_2 \mathrm{O}$
So Methylamine is formed
View full question & answer→
MCQ 591 Mark
Identify the most basic compound from the following.
  • A
    Primary amines
  • ✓
    Secondary amines
  • C
    Aromatic amines
  • D
    Aliphatic amines
Answer
Correct option: B.
Secondary amines
View full question & answer→
MCQ 601 Mark
What is the decreasing order of basicity?
  • A
    $ \mathrm{NH}_3 > \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 > \left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} > \left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N} $
  • B
    $ \left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N} > \left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} > \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 > \mathrm{NH}_3 $
    ​
  • ✓
    $ \left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} > \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 > \left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{N} > \mathrm{NH}_3 $
    ​
  • D
    $ \left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} > \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 > \mathrm{NH}_3 > \left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{N}$
Answer
Correct option: C.
$ \left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} > \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 > \left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{N} > \mathrm{NH}_3 $
​
View full question & answer→
MCQ 611 Mark
Basic strength of different alkyl amines depends upon:
  • A
    $+I$ effect
  • B
    Steric effect
  • C
    Solvation effect
  • ✓
    All of these
Answer
Correct option: D.
All of these
The basic strength of different alkyl amines depends on many factors like:
  1. $+I$ effect is the polarization of a sigma bond due to electron donating effect of adjacent groups or atoms.
  2. In water, the ammonium salts of primary and secondary amines undergo solvation effects due to hydrogen bonding to a much greater degree than ammonium salts of tertiary amines.
  3. Steric effect is the effect faced by incoming group or hydrogen by the already present bulky $−R$ groups on the Nitrogen atom.
View full question & answer→
MCQ 621 Mark
Benzylamine is a stronger base than $......$
  • A
    Aniline
  • B
    Acetamide
  • C
    $O -$ methylaniline
  • ✓
    All of these
Answer
Correct option: D.
All of these
View full question & answer→
MCQ 631 Mark
Which compound is a secondary alcohol?
  • A
    Butan $- 1- ol$
  • ✓
    Butan $- 2 - ol$
  • C
    Isobutyl alcohol
  • D
    $2 -$ Methylpropan $- 2 - ol$
Answer
Correct option: B.
Butan $- 2 - ol$
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.
View full question & answer→
MCQ 641 Mark
Under which of the following reaction conditions, aniline gives $p-$nitro derivative as the major product?
$a.$ Acetyl chloride/ pyridine followed by reaction with conc$. \ce{H_2SO_4} +$ conc$. \ce{HNO_3}$
$b.$ Acetic anyhdride/ pyridine followed by conc$. \ce{H_2SO_4} +$ conc$. \ce{HNO_3}.$
$c.$ Dil$. \text{HCl}$ followed by reaction with conc$. \ce{H_2SO_4} +$ conc$. \ce{HNO_3}.$
$d.$ Reaction with conc$. \ce{HNO_3} +$ con$c.\ce{H_2SO_4}$
  • ✓
    $a$ and $b$
  • B
    $a$ and $c$
  • C
    $b$ and $c$
  • D
    $b$ and $d$
Answer
Correct option: A.
$a$ and $b$
Aniline or reaction with acetyl chloride or acetic anhydride in the presence of pyridine produces $N-$acetyl aniline which is a ortho, para directing group which on further reaction with nitrating mixture $($cone$.\ce{HNO_3}+$ conc$. \ce{H_2SO_4})$ produces $p-$nitroaniline preferentially as shown below:
View full question & answer→
MCQ 651 Mark
The correct $\text{IUPAC}$ name for $\text{CH}_2=\text{CHCH}_2\text{NHCH}_3$ is:
  • A
    Allylmethylamine.
  • B
    $2-$amino$-4-$pentene.
  • C
    $4-$aminopent$-1-$ene.
  • ✓
    $N-$methylprop$-2-en-1-$amine.
Answer
Correct option: D.
$N-$methylprop$-2-en-1-$amine.
View full question & answer→
MCQ 661 Mark
The first organic compound synthesized in the laboratory from an inorganic compound is:
  • A
    $\mathrm{NH}_4 \mathrm{NCO}$
  • ✓
    $\mathrm{NH}_2-\mathrm{CO}-\mathrm{NH}_2$
  • C
    $\mathrm{CH}_3 \mathrm{COOH}$
  • D
    $\mathrm{CH}_4$
Answer
Correct option: B.
$\mathrm{NH}_2-\mathrm{CO}-\mathrm{NH}_2$
The first organic compound synthesized in the laboratory from an inorganic compound is urea which is $\mathrm{NH}_2\mathrm{CO}\mathrm{NH}_2$​.
View full question & answer→
MCQ 671 Mark
The diazonium salts are the reaction product in presence of excess of mineral acid with nitrous acid and:
  • A
    Primary aliphatic amine
  • B
    Secondary aromatic amine
  • ✓
    Primary aromatic amine
  • D
    Tertiary alipathic amine
Answer
Correct option: C.
Primary aromatic amine
The diazonium salts are the reaction product in presence of excess of mineral acid with nitrous acid and primary aromatic amine.
View full question & answer→
MCQ 681 Mark
Which of the following amines can be prepared by Gabriel synthesis.
$a.$ Isobutyl amine.
$b. 2-$ Phenylethylamine.
$c. N-$methylbenzylamine.
$d.$ Aniline.
  • ✓
    $a$ and $b$
  • B
    $a$ and $c$
  • C
    $b$ and $c$
  • D
    $a$ and $d$
Answer
Correct option: A.
$a$ and $b$
Only primary aliphatic amines such as $(a) \left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{CH}_2 \mathrm{NH}_2$ and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2 (b)$ can be prepared by Gabriel synthesis. $2^{\circ}$ amines, i.e., $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NHCH}_3$ (C) and $1^{\circ}$ amine, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 (d),$ however, cannot be prepared.
View full question & answer→
MCQ 691 Mark
The correct most basicity in amines is:
  • A
    $\mathrm{C}_4 \mathrm{H}_5 \mathrm{NH}_2$
  • B
    $\mathrm{CH}_3 \mathrm{NH}_2$
  • C
    $\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$
  • ✓
    $\left(\mathrm{CH}_3\right)_3 \mathrm{~N}$
Answer
Correct option: D.
$\left(\mathrm{CH}_3\right)_3 \mathrm{~N}$
The observed order in the case of lower members is found to be as secondary $>$ primary $>$ tertiary.
This anomalous behavior of tertiary amines is due to steric factors i.e. crowding of alkyl groups cover nitrogen atom from all sides and thus makes it unable for protonation.
Thus the relative strength is in order $\mathrm{CH}_3 \mathrm{NH}_2<\left(\mathrm{CH}_3\right)_2 \mathrm{NH}>\left(\mathrm{CH}_3\right)_3 \mathrm{N}$
View full question & answer→
MCQ 701 Mark
The correct decreasing order of basic strength of the following species is $........ \mathrm{H}_2 \mathrm{O}, \mathrm{NH}_3, \mathrm{OH}^{-}, \mathrm{NH_2}^{-}$
  • ✓
    $\mathrm{NH_2}^{-} > \mathrm{OH}^{-} > \mathrm{NH}_3 > \mathrm{H}_2 \mathrm{O}$
  • B
    $\mathrm{OH}^{-} > \mathrm{NH_2}^{-} > \mathrm{H}_2 \mathrm{O} > \mathrm{NH}_3$
  • C
    $\mathrm{NH}_3 > \mathrm{H}_2 \mathrm{O} > \mathrm{NH_2}^{-} > \mathrm{OH}^{-}$
  • D
    $\mathrm{H}_2 \mathrm{O} > \mathrm{NH}_3 > \mathrm{OH}^{-} > \mathrm{NH}_2$
Answer
Correct option: A.
$\mathrm{NH_2}^{-} > \mathrm{OH}^{-} > \mathrm{NH}_3 > \mathrm{H}_2 \mathrm{O}$
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.
View full question & answer→
MCQ 711 Mark
A compound $A$ has molecular formula $\ce{C_7​H_7​NO}.$ On treatment with $Br_2$ and $\text{KOH},$ A gives an amine $B$ which gives carbylamine test$. B$ upon diazotization and coupling with phenol gives an azo dye$. A$ can be:
  • A
    $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CONHCOCH}_3$
  • ✓
    $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CONH}_2$
    ​
  • C
    $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2$
    ​
  • D
    $\mathrm{O}, \mathrm{m}$ or $\mathrm{p}-\mathrm{C}_6 \mathrm{H}_4\left(\mathrm{NH}_2\right) \mathrm{CHO}$
Answer
Correct option: B.
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CONH}_2$
​
According to the reactions given for compound $B, B$ should be aniline as it gives carbylamine test and also couples with phenol to form dye.
The action of $Br_2$​ and $\text{KOH}$ is Hoffmann bromamide reaction in which aniline is formed.
Therefore$, A$ should be an amide.
Thus$, A$ is $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CONH}_2$.
View full question & answer→
MCQ 721 Mark
Which of the following is an arylalkyl amine?
  • A
    $ \mathrm{C}_6 \mathrm{H}_5 \mathrm{NHC}_6 \mathrm{H}_5 $
  • B
    $ \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 $
  • ✓
    $ \left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2\right) \mathrm{NH} $
  • D
    $ \left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \mathrm{\sim N} $
Answer
Correct option: C.
$ \left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2\right) \mathrm{NH} $
Arylalkyl amines are aromatic amines which are side chain substituted, i.e., the nitrogen is not directly attached to the phenyl group but to a side chain of the benzene ring.
In $ \left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2\right) \mathrm{NH} ,$ there are two same groups attached to the nitrogen where it is attached to a benzyl carbon, instead of an aryl carbon.
View full question & answer→
MCQ 731 Mark
In the diazotisation of aniline with sodium nitrite and hdrochloric acid, an excess of hydrochloric acid is used primarily to:
  • ✓
    Suppress the concentration of free aniline available for coupling.
  • B
    Suppress hydrolysis of phenol.
  • C
    Insure a stoichiometric amount of nitrous acid.
  • D
    Neutralize the base liberated.
Answer
Correct option: A.
Suppress the concentration of free aniline available for coupling.
Excess of $\text{HCl}$ is used to convert free aniline to aniline hydrochloride. Other wise free aniline would undergo coupling reaction with benzene diazonium chloride
View full question & answer→
MCQ 741 Mark
Gabriel phthalimide synthesisis used in the preparation of:
  • ✓
    $1^\circ $ amine
  • B
    $2^\circ $ amine
  • C
    $3^\circ $ amine
  • D
    $4^\circ $ amine
Answer
Correct option: A.
$1^\circ $ amine
The chemical reaction used for transforming primary alkyl halides into primary amines is called Gabriel phthalimide synthesis. It uses potassium phthalimide.
Therefore, Gabriel phthalimide synthesis is used in the preparation of $1^\circ$ amide.
View full question & answer→
MCQ 751 Mark
Amine that can't be prepared by Gabriel phthalimide synthesis is:
  • ✓
    Aniline
  • B
    Benzylamine
  • C
    Methyl amine
  • D
    Iso $-$ butyl amine
Answer
Correct option: A.
Aniline
Gabriel phthalamide cannot be used to prepare aromatic amines since aromatic halides donot undergo nucleophilic subsitution from the salt formed by phthalamide.
View full question & answer→
MCQ 761 Mark
In the nitration of benzene using a mixture of conc$. \ce{H_2SO_4}$ and conc$. \ce{HNO_3},$ the species which initiates the reaction is $.......$
  • A
    $\text{NO}_2$
  • B
    $\text{NO}^+$
  • ✓
    $\text{NO}_2^+$
  • D
    $\text{NO}_2^-$
Answer
Correct option: C.
$\text{NO}_2^+$
$NO_2\ ($Nitronium ion$)$ electrophile initiates the process of nitration. It is obtained as:
$\text{H}_2\text{SO}_4(\text{conc.})\xrightarrow{\ \ \ \ \ \ \ \ }\text{H}^++\text{HSO}_4^-$
$\text{H}^++\text{HNO}_3\xrightarrow{\ \ \ \ \ \ \ }\text{H}_2\text{NO}_3^+$
$\text{H}_2\text{NO}_3^+\xrightarrow{\ \ \ \ \ \ \ }\text{NO}_2^++\text{H}_2\text{O}$
View full question & answer→
MCQ 771 Mark
Amines are more basic than:
  • A
    Alcohols
  • B
    Ethers
  • C
    Esters
  • ✓
    All of these
Answer
Correct option: D.
All of these
$\ce{-OH, -COOR, -COC}$ group
View full question & answer→
MCQ 781 Mark
Which of the following orders is correct regarding the basic strength of subsituted aniline ?
  • A
    $P -$ nitroaniline $> p -$ aminobenzaldehyde $> p-$ bromoaniline.
  • B
    $P -$ nitroaniline $< p -$ bromoaniline $< p -$ aminobenzaldehyde.
  • ✓
    $P -$ nitroaniline $< p -$ aminobenzaldehyde $< p -$ bromoaniline.
  • D
    $P -$ nitroaniline $> p -$ aminobenzaldehyde $< p -$ bromoaniline.
Answer
Correct option: C.
$P -$ nitroaniline $< p -$ aminobenzaldehyde $< p -$ bromoaniline.
Nitro group is strong electron withdrawing group.
Hence$, p -$ nitroaniline is least basic.
Halogens and aldehydes are weakly deactivating or weak electron withdrawing groups.
The order of the ability to withdraw electron density is nitro $>$ aldehyde $>$ bromine.
Higher is the ability of the substituent to withdraw electrons, lower is the basicity of amine.
Hence, the order of basicity is
$P -$ nitroaniline $< p -$ amino benzaldehyde $< p -$ bromoaniline.
View full question & answer→
MCQ 791 Mark
Which is most basic among the following?
  • A
    $ \mathrm{CH}_3 \mathrm{NH}_2 $
    ​
  • B
    $ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2$
    ​
  • C
    $\mathrm{NH}_3$
    ​
  • ✓
    $\left(\mathrm{CH}_3\right)_2 \mathrm{CHNH}_2$
Answer
Correct option: D.
$\left(\mathrm{CH}_3\right)_2 \mathrm{CHNH}_2$
The basic character of a compound is determined by the stability of the conjugate acid.
Among the following alternatives$, D$ will be the most basic due to stabilization of the conjugate acid by the positive inductive as well as the hyperconjugation effect of the two methyl groups
View full question & answer→
MCQ 801 Mark
Which of the following has the maximum value of $\text{pkb}​?$
  • A
    $ \left(\mathrm{CH}_3\right)_2 \mathrm{NH} $
  • B
    $ \left(\mathrm{CH}_3 \mathrm{CH}_2\right)_2 \mathrm{NH} $
  • ✓
    $ \mathrm{C}_6 \mathrm{H}_5-\mathrm{NH}-\mathrm{C}_6 \mathrm{H}_5 $
    ​
  • D
    $ \mathrm{C}_6 \mathrm{H}_5-\mathrm{NH}-\mathrm{CH}_3 $​
Answer
Correct option: C.
$ \mathrm{C}_6 \mathrm{H}_5-\mathrm{NH}-\mathrm{C}_6 \mathrm{H}_5 $
​
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 $Kb$​ value and least $Kb​$ value means highest $\text{pKb}$​ value.
View full question & answer→
MCQ 811 Mark
Methylamine reacts with $H\ce{NO_2}$ to form $........$
  • A
    $\ce{CH_3-O-N=O}$
  • B
    $\ce{CH_3-O-CH_3}$
  • ✓
    $\ce{CH_3OH}$
  • D
    $\ce{CH_3CHO}$
Answer
Correct option: C.
$\ce{CH_3OH}$
$\text{R}-\text{NH}_2+\text{HNO}_2\xrightarrow{\ \ \ \ \text{NaNO}_2+\text{HCI}\ \ \ }[\text{R}-\stackrel{+}{\hbox{N}}_2\text{C}\stackrel{-}{\hbox{l}}]\xrightarrow{\ \ \text{H}_2\text{O}\ \ }\text{ROH}+\text{N}_2+\text{HCl}$
View full question & answer→
MCQ 821 Mark
Which of the following names of amines belong strictly to the common system?
  • ✓
    Ethylmethylamine
  • B
    Aniline
  • C
    Benzenamine
  • D
    Propan$-1-$amine
Answer
Correct option: A.
Ethylmethylamine
In the common system, aliphatic amines are named by prefixing the alkyl group to amine, i.e., alkylamine. In case of two different alkyl groups, it is named by listing the alkyl groups in alphabetical order before the word amine, just like in ethylmethylamine. Aniline is a common name but is also accepted by $\text{IUPAC}.$
View full question & answer→
MCQ 831 Mark
Which of the following is not an amine?
  • ✓
    $NH_3$
  • B
    $\ce{CH_3NH_2}$
  • C
    $\ce{C_6H_5NH_2}$
  • D
    $\ce{CH_3NHCH_3}$
Answer
Correct option: A.
$NH_3$
Amines are clearly defined as ‘derivatives’ of ammonia obtained by replacement of one or more $H$ atoms by an alkyl/aryl group.
Therefore$, NH_3$ itself is not an amine, but the base compound for all amines.
View full question & answer→
MCQ 841 Mark
Which of the following is the weakest Brönsted base?
 
  • ✓
  • B
  • C
  • D
    $\mathrm{CH_3NH_2}$
Answer
Correct option: A.
Due to delocalization of lone pair of electrons on the $N-$atom into the benzene ring$, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$ is the weakest base.
Resonating Structure of Aniline.
View full question & answer→
MCQ 851 Mark
Benzylamine may be alkylated as shown in the following equation: $\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2+\text{R}-\text{X}\xrightarrow{\ \ \ \ \ \ \ \ }\text{C}_6\text{H}_5\text{CH}_2\text{NHR}$ Which of the following alkylhalides is best suited for this reaction through $\mathrm{S}_{\mathrm{N}}{_1}$ mechanism?
  • A
    $\mathrm{CH}_3 \mathrm{Br}$
  • B
    $\mathrm{C}_6 \mathrm{H}_5 \mathrm{Br}$
  • ✓
    $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Br}$
  • D
    $\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}$
Answer
Correct option: C.
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Br}$
$\mathrm{S}_{\mathrm{N}}{_1}$ reaction proceeds through the formation of carbocation since in $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Br}$ benzyl carbocation is formed which is stabilized by resonance.
View full question & answer→
MCQ 861 Mark
How many primary amines can be formulated by $\ce{C_3​H_9​N}$ and how many $1^\circ $ hydrogen are associated with carbon atoms of each compound?
  • A
    Two primary amines $[3, 6]$
  • B
    One primary amine $[3]$
  • C
    Three primary amines $[3, 6, 6]$
  • ✓
    Two primary amines $[5, 6]$
Answer
Correct option: D.
Two primary amines $[5, 6]$
Two primary amines$, n$ propyl amine and iso propyl amine can be formulated with molecular formula $\ce{C_3​H_9​N}$
In n propyl amine, five primary hydrogens are attached with carbon atoms.
In isopropyl amine, six primary hydrogen atoms are attached with carbon atoms.
View full question & answer→
MCQ 871 Mark
Which of the following statements is correct?
  • A
    Methyl amine is slightly acidic.
  • B
    Methyl amine is less basic than ammonia.
  • ✓
    Methyl amine is less basic than dimethyl amine.
  • D
    Methyl amine is less basic than aniline.
Answer
Correct option: C.
Methyl amine is less basic than dimethyl amine.
Dimethyl amine $\ce{(Me_2NH)}$ is more basic than $\ce{MeNH_2},$ due to $(+I)$ effect of two $(Me)$ groups.
View full question & answer→
MCQ 881 Mark
Based on the concept of the hydration of protonated amines, which of the following represents correct order of decreasing basic strength of amines?
  • ✓
    Primary amine $>$ secondary amine $>$ tertiary amine.
  • B
    Tertiary amine $>$ primary amine $>$ secondary amine.
  • C
    Tertiary amine $>$ secondary amine $>$ primary amine.
  • D
    Primary amine $>$ tertiary amine $>$ secondary amine.
Answer
Correct option: A.
Primary amine $>$ secondary amine $>$ tertiary amine.
Based on the concept of the hydration of protonated amines, the following represents the correct order of decreasing basic strength of amines
primary amine $>$ secondary amine $>$ tertiary amine
This is the same order in which the degree of hydrogen bond formation, the extent of hydration and the stabilisation of protonated amines decrease.
View full question & answer→
MCQ 891 Mark
Identify the statement about the basic nature of amines.
  • A
    Alkylamines are weaker bases than ammonia.
  • B
    Arylamines are stronger bases than alkyl amines.
  • ✓
    Secondary aliphatic amines are stronger bases than primary aliphatic amines.
  • D
    Tertiary aliphatic amines are weaker bases than arylamines.
Answer
Correct option: C.
Secondary aliphatic amines are stronger bases than primary aliphatic amines.
The lone pair of the electron on nitrogen atom present is arylamines is less available for protonation partly due to $-I$ effect of a phenyl group and mostly due to resonance.
Hence aryl amines are weaker bases than alkyl amines.
Also, the secondary amines are a stronger base than the primary amines as the number of alkyl groups attached to the nitrogen atoms are more in secondary alkyl amines.
View full question & answer→
MCQ 901 Mark
The reaction $\text{Ar}\stackrel{+}{\hbox{N}_2}\text{Cl}^-\xrightarrow{\ \ \text{Cu/HCl}\ \ \ }\text{ArCl}+\text{N}_2+\text{CuCl}$ is named as:
  • A
    Sandmeyer reaction.
  • ✓
    Gatterman reaction.
  • C
    Claisen reaction.
  • D
    Carbylamine reaction.
Answer
Correct option: B.
Gatterman reaction.

is named Gatterman reaction.
View full question & answer→
MCQ 911 Mark
The compound shown is a $.......$ amine.
  • ✓
    $1^\circ $ aryl
  • B
    $1^\circ $ arylalkyl
  • C
    $2^\circ $ aryl
  • D
    $2^\circ $ arylalkyl
Answer
Correct option: A.
$1^\circ $ aryl
Only one hydrogen atom is replaced by an aromatic group, in which the nitrogen is directly linked to the $s p^2$ hybridised carbon of benzene ring. Hence, it is a primary aromatic aryl amine.
View full question & answer→
MCQ 921 Mark
Diazo coupling is useful to prepare some:
  • A
    Pesticides
  • ✓
    Dyes
  • C
    Proteins
  • D
    Vitamins
Answer
Correct option: B.
Dyes
The first use of diazonium salts was to produce water$-$fast dyed fabrics by immersing the fabric in an aqueous solution of the diazonium compound, followed by immersion in a solution of the coupler $($the electron $-$ rich ring that undergoes electrophilic substitution$).$
The major applications of diazonium compounds remains in the dye and pigment industry.
View full question & answer→
MCQ 931 Mark
What is the bond angle in ammonia molecule?
  • A
    $106.5^\circ $
  • ✓
    $107^\circ $
  • C
    $108^\circ $
  • D
    $109.5^\circ $
Answer
Correct option: B.
$107^\circ $
The bond angle for a standard tetrahedral geometry is $109.5^\circ .$ But the ammonia molecule contains a lone electron pair on nitrogen.
As it is closer to the $N$ atom than the other orbitals, it creates a repulsive effect and pushes the Hydrogen atoms close to each other, thus reducing the bond angle to $107^\circ .$
View full question & answer→
MCQ 941 Mark
Which of the following compound will not undergo azo coupling reaction with benzene diazonium chloride.
  • A
    Aniline.
  • B
    Phenol.
  • C
    Anisole.
  • ✓
    Nitrobenzene.
Answer
Correct option: D.
Nitrobenzene.
Diazonium cation is a weak electrophile and hence reacts with electron rich compounds containing electron donating groups such as $-\mathrm{OH},-\mathrm{NH}_2$ and $-\mathrm{OCH}_3$ groups and not with compounds containing electron withdrawing groups such as $-\mathrm{NO}_2,$ etc.
View full question & answer→
MCQ 951 Mark
The correct increasing order of basic strength for the following compounds is $.......$
  • A
    $\text{II < III < I.}$
  • B
    $\text{III < I < II.}$
  • C
    $\text{III < II < I.}$
  • ✓
    $\text{II < I < III.}$
Answer
Correct option: D.
$\text{II < I < III.}$

Electron withdrawing group decreases the basic strength while electron releasing groups increases the basic strength of aniline.
View full question & answer→
MCQ 961 Mark
The best reagent for converting $2-$phenylpropanamide into $2-$phenylpropanamine is $........$
  • A
    Excess $\mathrm{H}_2$.
  • B
    $\mathrm{Br}_2$ in aqueous $\text{NaOH}.$
  • C
    Iodine in the presence of red phosphorus.
  • ✓
    $\mathrm{LiAlH}_4$ in ether.
Answer
Correct option: D.
$\mathrm{LiAlH}_4$ in ether.
The best reagent for converting $2-$phenylpropanamide into $2-$phenylpropanamine is $\mathrm{LiAlH}_4$ in ether. Reaction is as given below:
​​​​​​​
View full question & answer→
MCQ 971 Mark
Which of the following explode on heating?
  • ✓
    Azides
  • B
    Sulphates
  • C
    Chlorides
  • D
    Phosphides
Answer
Correct option: A.
Azides
some experiments describe on calcium and barium azid and tetrazene, which explode while they are solid.
The explosion of the molten azides is due to self heating of the liquid.
Explosion is facilitated by the presence of an inert gas above the decomposing liquid self heating of the liquid azides is due to the retention of heat products of reaction near the surface.
View full question & answer→
MCQ 981 Mark
$\text{A 55- kDa}$ protien was acid hydrolysed to obtain a misture of amino acids. How many amino acids could be present in the solution?
  • A
    $550$
  • ✓
    $500$
  • C
    $1000$
  • D
    $1100$
Answer
Correct option: B.
$500$
View full question & answer→
MCQ 991 Mark
Nitrobenzene when reduced with tin and hydrochloric acid, i.e., in acidic medium, the product formes is:
  • ✓
    Aniline
  • B
    Benzene
  • C
    Both $a$ and $b$
  • D
    None of these
Answer
Correct option: A.
Aniline
Nitrobenzene is reduced to phenylammonium ions using a mixture of tin and concentrated hydrochloric acid.
The mixture is heated under reflux in a boiling water bath for about half an hour.
Under the acidic conditions, rather than getting phenylamine directly, you instead get phenylammonium ions formed.
View full question & answer→
MCQ 1001 Mark
How many primary amines are possible for the formula $\mathrm{C}_4 \mathrm{H}_{11} \mathrm{N}\ ?$
  • A
    One
  • B
    Two
  • C
    Three
  • ✓
    Four
Answer
Correct option: D.
Four
Four primary amines are possible for this structure.
View full question & answer→
Generate Paper FreeAll Amines topics
M.C.Q (1 Marks) - Page 2 - Chemistry STD 12 Science Questions - Vidyadip