For carbylamine reaction, we need hot alc. KOH and
1
any primary amine and chloroform
2
chloroform and silver powder
3
a primary amine and an alkyl halide
4
a mono alkyl amine and trichloromethane
Official Solution
Correct Option: (1)
Aliphatic and aromatic primary amines when warmed with chloroform and an alcoholic solution of KOH, form isocyanide or carbylamine which has very unpleasant smell.
02
PYQ 1994
medium
chemistryID: neet-ug-
What is the decreasing order of basicity of 1 , 2 and 3 ethyl amines and ammonia?
1
2
3
4
Official Solution
Correct Option: (4)
As the number of alkyl groups increases, the electron density on nitrogen atom also increases, so the basic character increases but 3 amines are less basic than 2 amines due to steric hindrance of 3 amines, so the correct order of basicity is Ammonia < 1 < 3 < 2
03
PYQ 1998
medium
chemistryID: neet-ug-
Aniline is reacted with bromine water and the resulting product is treated with an aqueous solution of sodium nitrite in presence of dilute hydrochloric acid. The compound so formed is converted into a tetrafluoroborate which is subsequently heated. The final product is
1
1,3, 5-tribromobenzene
2
p-bromofluorobenzene
3
p-bromoaniline
4
2, 4, 6-tribromofluorobenzene
Official Solution
Correct Option: (4)
The correct answer is D:2, 4, 6-tribromofluorobenzene - group is greatly activating group. Hence, reaction takes place rapidly.
04
PYQ 1999
medium
chemistryID: neet-ug-
Amides can be converted into amines by a reaction named after
1
Perkin
2
Claisen
3
Hofmann
4
Kekule
Official Solution
Correct Option: (3)
Amides can be converted into amines by Hofmann's bromamide reaction. The reaction is -
05
PYQ 2000
medium
chemistryID: neet-ug-
Then A is:
1
2
3
4
Official Solution
Correct Option: (2)
'C' must be an isocyanide, which is derived from a primary amine through the Carbylamine reaction . Additionally, the primary amine must be produced through the reduction of a nitro hydrocarbon. Therefore, 'A' is nitrobenzene.
So, the correct option is (B).
06
PYQ 2001
medium
chemistryID: neet-ug-
Intermediates formed during reaction of RCNH2||O with Br2 and KOH are
1
RCONHBr and RNCO
2
RNHCOBr and RNCO
3
RNH-Br and RCONHBr
4
RCONBr2
Official Solution
Correct Option: (1)
The correct option is(A): RCONHBr and RNCO.
The process underlying the Hoffmann hypobromite reaction's mechanism.the mechanism involves the conversion of an amide (RCONH2) into an isocyanate (RNCO) through the reaction with bromine and sodium hydroxide (NaOH). This process results in the formation of RCONHBr (an intermediate) and eventually leads to the desired isocyanate product, RNCO.
07
PYQ 2005
easy
chemistryID: neet-ug-
Aniline in a set of reactions yielded a product D.
The structure of the product D would be:
1
C6H5CH2NH2
2
C6H5NHCH2CH3
3
C6H5NHOH
4
C6H5CH2OH
Official Solution
Correct Option: (4)
Therefore, the correct option is (D): C6H5CH2OH
08
PYQ 2006
medium
chemistryID: neet-ug-
Which of the following is more basic than aniline ?
1
Diphenylamine
2
Triphenylamine
3
p-nitroaniline
4
Benzylamine
Official Solution
Correct Option: (4)
The lone pair of electrons are not involved in conjugation and is readily available for donation
09
PYQ 2010
medium
chemistryID: neet-ug-
Which of the following statements about primary amines is false ?
1
Alkyl amines are stronger bases than aryl amines
2
Alkyl amines react with nitrous acid to produce alcohols
3
Aryl amines react with nitrous acid to produce phenols
4
Alkyl amines are stronger bases than ammonia
Official Solution
Correct Option: (3)
Aryl amines react with nitrous acid to produce diazonium salts.
10
PYQ 2013
medium
chemistryID: neet-ug-
Nitrobenzene on reaction with conc. at forms which one of the following products ?
1
1, 2, 4-Trinitrobenzene
2
1, 2-Dinitrobenzene
3
1, 3-Dinitrobenzene
4
1, 4-Dinitrobenzene
Official Solution
Correct Option: (3)
group being electron withdrawing reduces electron density at ortho and para-positions. Hence, now the meta-position becomes
electron rich on which the electrophile (nitronium ion) attacks during nitration.
11
PYQ 2014
medium
chemistryID: neet-ug-
Which of the following will be most stable diazoniumsalt RN2+X- ?
1
2
3
4
Official Solution
Correct Option: (1)
The correct option (A): .
12
PYQ 2014
medium
chemistryID: neet-ug-
Which of the following will not be soluble in sodiumhydrogen carbonate?
1
Benzoic acid
2
o-Nitrophenol
3
Benzenesulphonic acid
4
2,4,6-trinitrophenol
Official Solution
Correct Option: (2)
The correct option (B): o-Nitrophenol.
13
PYQ 2014
medium
chemistryID: neet-ug-
In the following reaction, the product (A) is :
1
2
3
4
Official Solution
Correct Option: (3)
The correct option (C): .
14
PYQ 2014
medium
chemistryID: neet-ug-
Which of the following will be most stable diazonium salt ?
1
2
3
4
Official Solution
Correct Option: (2)
Diazonium salt containing aryl group directly linked to the nitrogen atom is most stable due to resonance stabilisation between the benzene nucleus and N-atom. Diazonium ion act as a electrophile. [Resonance structure of benzene diazonium ion]
15
PYQ 2016
medium
chemistryID: neet-ug-
The correct statement regarding the basicity of arylamines is:
1
Arylamines are generally more basis than alkylamines because the nitrogen lone-pair electrons are not delocalized by interaction with the aromatic ring electron system.
2
Arylamines are generally more basic than alkylamines because of aryl group.
3
Arylamines are generally more basic than alkylamines, because the nitrogen atom in arylamines is sp-hybridized
4
Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring electron system.
Official Solution
Correct Option: (4)
The correct statement regarding the basicity of arylamines is:
Select the correct option:
Option 1: Arylamines are generally more basic than alkylamines because the nitrogen lone-pair electrons are not delocalized by interaction with the aromatic ring ππ electron system.
Option 2: Arylamines are generally more basic than alkylamines because of the aryl group.
Option 3: Arylamines are generally more basic than alkylamines, because the nitrogen atom in arylamines is sp-hybridized.
Option 4: Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring ππ electron system.
Answer: Option 4 - Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring ππ electron system.
16
PYQ 2017
medium
chemistryID: neet-ug-
The correct increasing order of basic strength for the following compounds is
1
II < III < I
2
III < I < II
3
III < II < I
4
II < I < III
Official Solution
Correct Option: (4)
The correct option is (D): II < I lt; III.
17
PYQ 2017
medium
chemistryID: neet-ug-
Which of the following reactions is appropriate for converting acetamide to methanamine ?
1
Hoffmann hypobromamide reaction
2
Stephen's reaction
3
Gabriel phthalimide synthesis
4
Carbylamine reaction
Official Solution
Correct Option: (1)
It is called Hoffman bromamide reaction.
18
PYQ 2017
medium
chemistryID: neet-ug-
The correct increasing order of basic strength for the following compounds is :
1
III < I < II
2
III < II < I
3
II < I < III
4
II < III < I
Official Solution
Correct Option: (3)
has strong effect and shows effect.
Order of basic strength is
19
PYQ 2018
medium
chemistryID: neet-ug-
Nitration of aniline in strong acidic medium also gives m-nitroaniline because
1
In spite of substituents nitro group always goes to only m-position.
2
In absence of substituents nitro group always goes to m-position.
3
In electrophilic substitution reactions amino group is meta directive.
4
In acidic (strong) medium aniline is present as anilinium ion.
Official Solution
Correct Option: (4)
To understand why the nitration of aniline in a strong acidic medium results in m-nitroaniline, we need to consider the behavior of aniline in such a medium, especially during electrophilic substitution reactions.
Aniline ( ) is a primary amine with an amino group (-NH2) attached to the benzene ring. It is a reactive compound and typically undergoes electrophilic substitution reactions. The NH2 group is an activating group and is ortho/para directing under normal conditions. However, the scenario changes in a strong acidic medium.
In a strong acidic medium, such as concentrated sulfuric and nitric acid used for nitrations, aniline gets protonated to form an anilinium ion ( ). The structure of anilinium ion is different compared to unprotonated aniline.
The positive charge on the nitrogen atom reduces the electron density on the benzene ring significantly. As a result, the directing effect of the NH2 group changes from ortho/para to meta. Therefore, the nitro group, being an electron-withdrawing group, tends to go into the meta position because the ortho and para positions are less favorable due to electronic factors.
Hence, the correct answer to the question is that in acidic (strong) medium aniline is present as anilinium ion.
Other options can be ruled out because:
The nitro group does not always go to the meta position without influence; it depends on the directing groups present.
The amino group in unprotonated form is not meta-directing; it is ortho/para directing.
This detailed explanation shows why under these conditions, anilinium ion promotes meta substitution during nitration.
20
PYQ 2019
easy
chemistryID: neet-ug-
The correct order of the basic strength of methyl substituted amines in aqueous solution is :
1
2
3
4
Official Solution
Correct Option: (3)
In aqueous solution, electron donating inductive effect, solvation effect (H-bonding) and steric hindrance all together affect basic strength of substituted amines Basic character : \underset{{ 2^{\circ}}}{{ (CH_3)_2NH }} \underset{{ 1^{\circ}}}{{CH_3NH_2 }} \underset{{ 3^{\circ}}}{{ (CH_3)_3N}}
21
PYQ 2020
medium
chemistryID: neet-ug-
Which of the following is not correct about carbonmonoxide ?
1
It forms carboxyhaemoglobin.
2
) It reduces oxygen carrying ability of blood.
3
The carboxyhaemoglobin (haemoglobinbound to CO) is less stable thanoxyhaemoglobin.
4
It is produced due to incomplete combustion.
Official Solution
Correct Option: (3)
The correct option is (C) :The carboxyhaemoglobin (haemoglobin bound to CO) is less stable than oxyhaemoglobin.
22
PYQ 2021
medium
chemistryID: neet-ug-
Identify the compound that will react with Hinsberg’s reagent to give a solid which dissolves in alkali
1
2
3
4
Official Solution
Correct Option: (1)
To solve this question, we need to understand the chemical reaction involved with Hinsberg's reagent. Hinsberg's reagent is typically benzene sulfonyl chloride. This reagent reacts with primary and secondary amines differently. Primary amines form a sulfonamide that is soluble in an alkali, whereas secondary amines form an insoluble sulfonamide. Tertiary amines do not react with Hinsberg's reagent.
The question asks which compound will react with Hinsberg's reagent to give a solid that dissolves in alkali. This means we need to identify a primary amine since the sulfonamide from primary amines dissolves in alkali to form a soluble salt.
Let's evaluate the given options:
The correct compound that will react with Hinsberg's reagent to give a solid which dissolves in alkali is a primary amine.
The image associated with the correct answer is:
This compound is a primary amine. When it reacts with Hinsberg's reagent, it forms a sulfonamide that is soluble in alkali. Therefore, this compound meets the criteria specified in the question.
The other options might involve secondary or tertiary amines, which would not give a solid that dissolves in alkali. Hence, they can be ruled out as potential correct answers for this question.
23
PYQ 2022
medium
chemistryID: neet-ug-
The Kjeldahl' s method for the estimation of nitrogen can be used to estimate the amount of nitrogen in which one of the following compounds?
1
2
3
4
Official Solution
Correct Option: (3)
The Kjeldahl's method is a widely used technique for estimating the amount of nitrogen present in organic compounds. It is particularly effective for compounds containing nitrogen in the form of amines or ammonia, typically bound within organic structures. Let's analyze the given options to determine which compound is suitable for nitrogen estimation using the Kjeldahl method.
1. In the Kjeldahl procedure, the nitrogen in the organic compound is converted into ammonium sulfate by digestion with concentrated sulfuric acid.
2. The Kjeldahl method is effective for compounds where nitrogen does not exist as nitro (-NO2), azo (-N=N-), nitroso (-NO), or within any other complex derivative of nitrogen.
Options
Conclusion
1.
Nitroglycerin contains nitrogen as nitro groups, unsuitable for Kjeldahl.
2.
Thiourea has nitrogen in a form suitable for Kjeldahl method.
3.
The compound contains nitrogen in a non-complex form suitable for Kjeldahl.
4.
Azobenzene has nitrogen as azo groups, unsuitable for Kjeldahl.
Based on the evaluation of these compounds, the compound represented by can be estimated using the Kjeldahl method as it contains nitrogen in a simple form. This validates the selection of the third option as the compound suitable for nitrogen estimation via the Kjeldahl method.
24
PYQ 2023
easy
chemistryID: neet-ug-
Which of the following reactions will NOT give primary amine as the product
1
2
3
4
Official Solution
Correct Option: (2)
The question asks which reaction will NOT produce a primary amine. To determine this, we must analyze each reaction path to understand the type of product formed. Primary amines have the general formula R-NH2.
Let's evaluate each option:
Option 1: This reaction involves the reduction of a nitrile (RCN) using hydrogen in the presence of a catalyst, typically producing a primary amine (RNH2).
Option 2: This process involves a reaction of an alkyl halide with ammonia (NH3). Typically, this method results in a mixture of primary, secondary, and tertiary amines due to multiple alkylation. Therefore, it does not exclusively produce a primary amine.
Option 3: The reduction of an amide (RCONH2) using a reducing agent like LiAlH4 results in a primary amine (RNH2).
Option 4: The reduction of an azide (RN3) usually results in a primary amine (RNH2).
Based on the analysis, Option 2 is the reaction which will NOT exclusively produce a primary amine due to the possibility of over-alkylation.
