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JEE • CHEMISTRY • ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES

Organic chemistry-some basic principles and techniques practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 345 questions
1 Medium
1 of 345
Analyzing +4 | -1
Keto-enol tautomerism is not observe in
View answer and explanation
Correct option: A
C6H5−C∣∣O−H{C_6}{H_5} - \mathop C\limits_{\mathop {||}\limits_O } - H
Option A (C6H5−C∣∣O−H{C_6}{H_5} - \mathop C\limits_{\mathop {||}\limits_O } - H) is correct because it lacks an alpha-hydrogen, which is necessary for keto-enol tautomerism.
2 Medium
2 of 345
Analyzing +4 | -1
How many isomers of C5H11OH{C_5}{H_{11}}OH will be primary alcohols?
View answer and explanation
Correct option: B
4
Option 4 is correct because there are four primary alcohols that are isomers of C5H11OH.
3 Hard
3 of 345
Analyzing +4 | -1
Which statement is true?
View answer and explanation
Correct option: D
By breaking two bonds on the chiral centre configuration changes
Option D (By breaking two bonds on the chiral centre configuration changes) is correct because breaking and reforming two bonds at a chiral center can lead to inversion of configuration.
4 Easy
4 of 345
Remembering +4 | -1
Common reactions of benzene and its derivatives are:
View answer and explanation
Correct option: B
Electrophilic substitution reactions
Option B (Electrophilic substitution reactions) is correct because benzene and its derivatives primarily undergo electrophilic substitution reactions due to the stability of the aromatic ring.
5 Hard
5 of 345
Analyzing +4 | -1
How many structural formulae are possible for C5H11Cl?{\mathrm{C}_5}{\mathrm{H}_{11}}\mathrm{Cl}?
View answer and explanation
Correct option: B
8
Option B (8) is correct. There are eight structural isomers possible for C5H11Cl{\mathrm{C}_5}{\mathrm{H}_{11}}\mathrm{Cl}, considering different positions of the chlorine atom and branching of the carbon chain.
6 Medium
6 of 345
Understanding +4 | -1
Assertion (A): The compounds that have the same constitution and sequence of covalent bonds but differ in relative positions of their atoms or groups in space are called stereoisomers.

Reason (R): Stereoisomerism is a special type of isomerism which can be classified as geometrical and optical isomerism
View answer and explanation
Correct option: B
Both A\mathrm{A} and R\mathrm{R} are correct; R\mathrm{R} is not the correct explanation of A\mathrm{A}.
Option B (Both A and R are correct; R is not the correct explanation of A) is correct because both the definition of stereoisomers and the classification of stereoisomerism are correct statements, but the classification doesn't directly explain the definition.
7 Easy
7 of 345
Remembering +4 | -1
Isomerism among compounds due to the migration of a proton is known as :
View answer and explanation
Correct option: C
Tautomerism
Option C (Tautomerism) is correct because tautomerism is a type of isomerism involving the migration of a proton, resulting in two or more interconvertible structures.
8 Medium
8 of 345
Remembering +4 | -1
Which of the following is used as an adsorbent in adsorption chromatography ?
View answer and explanation
Correct option: D
Both (1) and (2)
Option D (Both (1) and (2)) is correct because both silica gel and alumina are commonly used as adsorbents in adsorption chromatography.
9 Easy
9 of 345
Understanding +4 | -1
Which of the following is electrophile
View answer and explanation
Correct option: B
:CH2:C{H_2}
The correct answer is B. Carbene (:CH2CH_2) is an electrophile because it is electron deficient and seeks electrons.
10 Medium
10 of 345
Analyzing +4 | -1
Base strength of,
H3C⊖CH2{H_3}\mathop {\mathrm{ }C}\limits^ \ominus C{H_2}
H2C=C⊖H{H_2}C = \mathop {\mathrm{ }C}\limits^ \ominus H
H−C=C⊖H - C = \mathop C\limits^ \ominus is in the order of
View answer and explanation
Correct option: C
(1) > (2) > (3)
Option A is correct because sulfite ions react with sulfuric acid to produce sulfur dioxide, which has an irritating smell and can reduce dichromate to a green solution of chromium(III) sulfate.
11 Easy
11 of 345
Remembering +4 | -1
Ammonia molecule is :
View answer and explanation
Correct option: D
An acid
Option A (A nucleophile) is correct because ammonia has a lone pair of electrons on the nitrogen atom, making it a good nucleophile.
12 Medium
12 of 345
Analyzing +4 | -1
Indicate the wrongly named compound.
View answer and explanation
Correct option: D
CH3CH=CH−CH2−CO∣∣−CH3(3−hexen−2−one)\mathop {C{H_3}CH = CH - C{H_2} - \mathop C\limits^{\begin{array}{cccccccccccccccccccc} O \\ {||} \end{array}} - C{H_3}}\limits_{\left( {3 - {\text{hexen}} - 2 - {\text{one}}} \right)}
Option D (CH3CH=CH−CH2−CO∣∣−CH3(3−hexen−2−one)\mathop {C{H_3}CH = CH - C{H_2} - \mathop C\limits^{\begin{array}{cccccccccccccccccccc} O \\ {||} \end{array}} - C{H_3}}\limits_{\left( {3 - {\text{hexen}} - 2 - {\text{one}}} \right)}) is wrongly named because the correct IUPAC name should be hex-4-en-2-one, not 3-hexen-2-one.
13 Easy
13 of 345
Understanding +4 | -1
The basic strength of
CH≡C_I,CH2=C_HII  ,CH3C_H2III\mathop {\mathrm{CH} \equiv \mathop \mathrm{C}\limits^\_ }\limits_I \mathrm{,}\mathop {\mathrm{C}{\mathrm{H}_2} = \mathop \mathrm{C}\limits^\_ \mathrm{H}}\limits_{II\;} ,\mathop {\mathrm{C}{\mathrm{H}_3}\mathop \mathrm{C}\limits^\_ {\mathrm{H}_2}}\limits_{III}
Will be in order
View answer and explanation
Correct option: A
I<II<IIII\mathrm{ } < II < III
Option A (I<II<IIII < II < III) is correct because the basic strength is inversely proportional to the stability of the conjugate acid, and the stability increases in the order: alkyne < alkene < alkane.
14 Medium
14 of 345
Analyzing +4 | -1
Assertion: Pyridine is more basic than Pyrrole.

Reason: In Pyrrole, nitrogen has sp2sp^2 hybridisation
and the lone pair of nitrogen is not available for
donation.
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and
Reason is the correct explanation of
Assertion
Option A is correct. Pyridine is more basic than pyrrole because the lone pair of electrons on nitrogen in pyridine is not involved in resonance and is readily available for donation, while in pyrrole, the lone pair is part of the aromatic system and less available.
15 Medium
15 of 345
Application +4 | -1
In Kjeldahl’s method, ammonia from 5g of food neutralizes 30 cm3c{m^3} of 0.1 N acid. The percentage of nitrogen in the food is
View answer and explanation
Correct option: A
0.84
Option A (0.84) is correct because the percentage of nitrogen is calculated using the formula involving the normality of the acid, its volume, and the weight of the food sample.

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16 Medium
16 of 345
Remembering +4 | -1
Aniline is purified by:
View answer and explanation
Correct option: A
Steam distillation
Option A (Steam distillation) is correct because aniline is steam volatile and can be separated from non-volatile impurities by steam distillation.
17 Medium
17 of 345
Understanding +4 | -1
The presence of carbon in an organic compound can be shown by
View answer and explanation
Correct option: C
Heating it with copper oxide to convert it into CO2\mathrm{C}{\mathrm{O}_2}
Option C (Heating it with copper oxide to convert it into CO2\mathrm{C}{\mathrm{O}_2}) is correct because the carbon in the organic compound reacts with copper oxide to form carbon dioxide, which can be detected.
18 Easy
18 of 345
Understanding +4 | -1
The ratio of σ−\sigma - and π−\pi -bonds in benzene is:
View answer and explanation
Correct option: B
4
Option B (4) is correct because the ratio of sigma to pi bonds in benzene is 12:3, which simplifies to 4:1.
19 Easy
19 of 345
Remembering +4 | -1
Naphthalene can be easily purified by
View answer and explanation
Correct option: A
Sublimation
Option A (Sublimation) is correct because naphthalene is a volatile solid that can be easily purified by sublimation, where it transitions directly from solid to gas.
20 Medium
20 of 345
Analyzing +4 | -1
Among the following the strongest acid is;
View answer and explanation
Correct option: B
CH3CH(NO2)COOH\mathrm{CH}_3 \mathrm{CH}\left(\mathrm{NO}_2\right) \mathrm{COOH}
Option B (CH3CH(NO2)COOH\mathrm{CH}_3 \mathrm{CH}(\mathrm{NO}_2) \mathrm{COOH}) is the strongest acid because the electron-withdrawing nitro group has the strongest inductive effect, stabilizing the conjugate base.
21 Easy
21 of 345
Remembering +4 | -1
In organic compounds sulphur is estimated as:
View answer and explanation
Correct option: B
BaSO4\mathrm{BaS}{\mathrm{O}_4}
Option B (BaSO4\mathrm{BaSO_4}) is correct because in organic compounds, sulfur is estimated as barium sulfate (BaSO4\mathrm{BaSO_4}) using Carius method.
22 Medium
22 of 345
Analyzing +4 | -1
Which technique among the following, is most appropriate in separation of a mixture of 100mg100 \mathrm{mg} of pp-nitrophenol and picric acid?
View answer and explanation
Correct option: D
Preparative TLC (Thin Layer Chromatography)
Option D (Preparative TLC (Thin Layer Chromatography)) is correct because TLC is well-suited for separating small quantities of non-volatile mixtures like p-nitrophenol and picric acid.
23 Medium
23 of 345
Understanding +4 | -1
Most stable carbanion is
View answer and explanation
Correct option: C
HC≡C⊖{HC} \equiv \stackrel{\ominus}{{C}}
Option C (HC≡C⊖{HC} \equiv \stackrel{\ominus}{{C}}) is correct because the stability of a carbanion increases with the s-character of the carbon bearing the negative charge.
24 Medium
24 of 345
Understanding +4 | -1
Consider the following statements :

I. The presence of functional group disables systematisation of organic compound into different classes.

II. In polyfunctional compounds, one of the functional group is chosen as the principal functional group.

III. The order of priority for functional group is

−COCl>−COOR>−SO3H>−COOH-\mathrm{COCl} > -\mathrm{COO}R > -\mathrm{SO}_3 \mathrm{H} > -\mathrm{COOH}

IV. CH2(OH)CH2(OH)\mathrm{CH}_2(\mathrm{OH}) \mathrm{CH}_2(\mathrm{OH}) is named as ethane-1, 2-diol.
Read the given statements and choose the correct option.
View answer and explanation
Correct option: B
Both II and IV
Option B (Both II and IV) is correct because statements II and IV are accurate descriptions of organic compound nomenclature.
25 Easy
25 of 345
Understanding +4 | -1
Among the following compounds nitrobenzene, benzene, aniline and phenol, the strongest basic behaviour in acid medium is exhibited by :
View answer and explanation
Correct option: B
Aniline
Option B (Aniline) is correct because aniline is the most basic among the given compounds due to the availability of the lone pair of electrons on the nitrogen atom for protonation.
26 Medium
26 of 345
Analyzing +4 | -1
The correct order of decreasing acid strength
of following acids is

(I) Trichloro acetic acid

(II) Trifluoro acetic acid

(III) Acetic acid

(IV) Formic acid
View answer and explanation
Correct option: C
II >> I >> IV >> III
Option C is correct. The correct order of decreasing acid strength is Trifluoroacetic acid > Trichloroacetic acid > Formic acid > Acetic acid, based on the -I effect of the halogens.
27 Medium
27 of 345
Understanding +4 | -1
Which of the following process is used to separate volatile liquid from non-volatile impurities?
View answer and explanation
Correct option: A
Distillation
Option A (Distillation) is correct because distillation separates liquids based on differences in boiling points, making it suitable for separating volatile liquids from non-volatile impurities.
28 Medium
28 of 345
Understanding +4 | -1
Which of the following element(s) is/are confirmed by appearance of blood red colour with FeCl3\mathrm{FeCl}_3 in Lassaigne's test?
View answer and explanation
Correct option: B
Presence of N & S\mathrm{N} \,\&\, \mathrm{S}
Option B is correct because the appearance of a blood-red color with FeCl3 in Lassaigne's test indicates the presence of both nitrogen and sulfur, forming [Fe(SCN)]2+.
29 Easy
29 of 345
Application +4 | -1
Haemoglobin is a chromoprotein having four atoms of Fe\mathrm{Fe} in each molecule. Analysis showed 0.35%0.35 \% Fe.\text{Fe}. What is the molecular weight of haemoglobin?
View answer and explanation
Correct option: A
64000 amu{6 4 0 0 0} \mathrm{~a m u}
Option C (9.5) is correct because it is the pH calculated from the dissociation constants and concentrations of the acid and base.
30 Hard
30 of 345
Analyzing +4 | -1
Which of the following statements is correct?
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the C-Cl bond in chlorobenzene is shorter and has some double bond character due to resonance, making it more difficult to replace than in benzoyl chloride.
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Key Takeaways

  • ✓ This set contains 345 single-correct-answer MCQs on ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES.
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