Skip to main content
Back to ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES

Your Progress

0 / 32 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES

Organic chemistry-some basic principles and techniques practice · Hard

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

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.

Take a chapter quiz →
Showing 1-30 of 32 questions
1 Hard
1 of 32
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.
2 Hard
2 of 32
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.
3 Hard
3 of 32
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.
4 Hard
4 of 32
Application +4 | -1
0.5 g of an organic compound containing nitrogen on Kjeldahlising required 29 mL of N/5H2SO4N/5{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} for complete neutralization of ammonia. The percentage of nitrogen in the compound is:
View answer and explanation
Correct option: B
16.216.2
Option B (16.216.2) is correct because the percentage of nitrogen is calculated using the stoichiometry of the reaction and the given data.
5 Hard
5 of 32
Analyzing +4 | -1
Among the acids given in the options, identify the acid which has maximum tendency to ionize in aqueous solution
View answer and explanation
Correct option: C
FCH2COOH\mathrm{FCH}_2 \mathrm{COOH}
The correct answer is FCH2COOH\mathrm{FCH}_2 \mathrm{COOH} because the presence of the highly electronegative fluorine atom increases the acidity of the carboxylic acid due to the -I effect.
6 Hard
6 of 32
Application +4 | -1
Caffeine has a molecular weight of 194. If it contains 28.9%28.9 \% by mass of nitrogen, number of atoms of nitrogen in one molecule of caffeine is
View answer and explanation
Correct option: A
4
Option A (4) is correct because the calculation of nitrogen atoms based on the given percentage and molecular weight results in approximately 4 atoms.
7 Hard
7 of 32
Analyzing +4 | -1
Correct statement among the following is
View answer and explanation
Correct option: C
(2R,3S)(2 R, 3 S) - Tartaric acid and (2R, 3R) - Tartaric acid are diastereomers
Option C is correct because (2R, 3S) - Tartaric acid and (2R, 3R) - Tartaric acid are diastereomers.
8 Hard
8 of 32
Application +4 | -1
29.5 mg of organic compound containing nitrogen was digested according to Kjeldahl’s method and the evolved ammonia was absorbed in 20mL of 0.1 M HCl solution. The excess of the acid required 15mL of 0.1M NaOH solution for complete neutralisation. The percentage of nitrogen in the compound is
View answer and explanation
Correct option: C
23.7
Option C (23.7) is correct because it is calculated by finding the millimoles of ammonia evolved and then determining the percentage of nitrogen in the organic compound.
9 Hard
9 of 32
Application +4 | -1
In the estimation of nitrogen by Duma’s method 1.18 g of an organic compound gave 224 mL of N2{\mathrm{N}_2} at NTP. The percentage of nitrogen in the compound is about:
View answer and explanation
Correct option: D
23.723.7
Option D (23.723.7) is correct because it is the percentage of nitrogen calculated using Duma\'s method formula with the given values.
10 Hard
10 of 32
Analyzing +4 | -1
Which of the following resonance structure is least stable?
View answer and explanation
Correct option: B
CH2=CH−C + H−C−H−OCH3{\text{C}}{{\text{H}}_{\text{2}}} = {\text{CH}} - \mathop {\text{C}}\limits^{\text{ + }} {\text{H}} - \mathop {\text{C}}\limits^ - {\text{H}} - {\text{OC}}{{\text{H}}_{\text{3}}}
Option B (CH2=CH−C + H−C−H−OCHext3{\text{C}}{{\text{H}}_{\text{2}}} = {\text{CH}} - \mathop {\text{C}}\limits^{\text{ + }} {\text{H}} - \mathop {\text{C}}\limits^ - {\text{H}} - {\text{OC}}{{\text{H}}_{ ext{3}}}) is the least stable because it has adjacent positive and negative charges on carbon atoms.
11 Hard
11 of 32
Analyzing +4 | -1
Arrange the following in decreasing order of acidic strength.

(i) Benzoic acid

(ii) 4-nitrobenzoic acid

(iii) 3,4-dinitrobenzoic acid

(iv) 4-methoxy benzoic acid
View answer and explanation
Correct option: A
III >> II >> I >> IV
Option A (III >> II >> I >> IV) is correct because electron-withdrawing groups increase acidity, and electron-donating groups decrease acidity. More nitro groups lead to higher acidity.
12 Hard
12 of 32
Application +4 | -1
0.58 g of hydrocarbon on combustion gave 0.9 g water. The percentage of carbon is about :
View answer and explanation
Correct option: B
82.782.7
Option B (82.782.7) is correct because it is the percentage of carbon calculated from the mass of water produced during combustion.
13 Hard
13 of 32
Analyzing +4 | -1
An organic compound on heating with CuO produces CO2\mathrm{C}{\mathrm{O}_2} but no water. The organic compound may be
View answer and explanation
Correct option: A
Carbon tetrachloride
The correct answer is A (Carbon tetrachloride) because it contains carbon but no hydrogen, thus producing CO2CO_2 but no water upon heating with CuO.
14 Hard
14 of 32
Understanding +4 | -1
Assertion (A): Components of a mixture of red and blue inks can be separated by distributing the components between stationary and mobile phases in paper chromatography.

Reason (R): The coloured components of inks migrate at different rates because paper selectively retains different components according to the difference in their partition between two phases.
View answer and explanation
Correct option: A
Both A\mathrm{A} and R\mathrm{R} are correct ; R\mathrm{R} is the correct explanation of AA.
Option A is correct because both the assertion and reason are true, and the reason correctly explains why paper chromatography can separate ink components based on their differential partitioning.
15 Hard
15 of 32
Analyzing +4 | -1
Increasing order of expected keto content
View answer and explanation
Correct option: C
CH3CHO>CH3COC2H5>CH3COCH3>CH3COCH2COCH3\mathrm{C}{\mathrm{H}_3}\mathrm{CHO} > C{\mathrm{H}_3}\mathrm{CO}{\mathrm{C}_2}{\mathrm{H}_5} > C{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3} > C{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_2}\mathrm{COC}{\mathrm{H}_3}
Option C is correct because it accurately reflects the increasing order of keto content based on the stability of the enol form.

Explore Other Chapters in CHEMISTRY

View All →
P-BLOCK ELEMENTS 736 MCQs Equilibrium 696 MCQs STRUCTURE OF ATOM 634 MCQs THERMODYNAMICS 536 MCQs CHEMICAL BONDING & MOLECULAR STRUCTURE 514 MCQs SOLUTIONS 480 MCQs
16 Hard
16 of 32
Analyzing +4 | -1
Increasing order of acid strength is

a) Phenol \quad \quad b) 3-Nitrophenol

c) 3,5-dinitro phenol \quad 4) 2,4,6-trinitro phenol
View answer and explanation
Correct option: C
a < b < c < d
The correct answer is C (a < b < c < d) because the acidic strength increases with the number of electron-withdrawing nitro groups attached to the phenol ring.
17 Hard
17 of 32
Remembering +4 | -1
An indicator ‘XX’ is used for studying the effect of variation in concentration of iodide on the rate of reaction of iodide ion with H2O2{H_2}{O_2} at room temp. The indicator ‘XX’ forms blue colored complex with
compound ‘AA’ present in the solution. The indicator ‘XX’ and compound ‘AA’ respectively are
View answer and explanation
Correct option: B
Starch and iodine
Option B (Starch and iodine) is correct because starch is a common indicator that forms a blue complex specifically with iodine.
18 Hard
18 of 32
Application +4 | -1
1.216 g1.216 \mathrm{~g} of an organic compound was reacted under Kjeldahl's method and the ammonia evolved was absorbed in 100 mL NH2SO4.100 \mathrm{~mL} \mathrm{~N} \mathrm{H}_2 \mathrm{SO}_4. The remaining acid solution was made up to 500 mL500 \mathrm{~mL} by addition of water. 20 mL20 \mathrm{~mL} of this dilute solution required 32 mL32 \mathrm{~mL} of N/10\mathrm{N} / 10 caustic soda solution for complete neutralisation. Calculate the percentage of nitrogen in the compound.
View answer and explanation
Correct option: A
23.03%{2 3 . 0 3} \%
Option A (23.03%) is correct because it is the percentage of nitrogen calculated using Kjeldahl's method based on the amount of acid neutralized by the evolved ammonia.
19 Hard
19 of 32
Analyzing +4 | -1
Indicate the complex/complex ion which did not show any geometrical isomerism.
View answer and explanation
Correct option: B
[Co(CN)5(NC)]3−{\left[ {Co{{\left( {CN} \right)}_5}\left( {NC} \right)} \right]^{3 - }}
Option B is correct because [Co(CN)5(NC)]3- is of the type MA5B, which does not exhibit geometrical isomerism.
20 Hard
20 of 32
Remembering +4 | -1
In phosphorus detection of an organic compound, the oxidised phosphorus compound solution is boiled with nitric acid and then treated with ammonium molybdate. Yellow colour is due to the formation of
View answer and explanation
Correct option: B
(NH4)3PO4⋅12MoO3\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3
Option B is correct because the yellow color in phosphorus detection is due to the formation of ammonium phosphomolybdate, which has the formula (NH₄)₃PO₄ ⋅ 12MoO₃.
21 Hard
21 of 32
Analyzing +4 | -1
The correct decreasing order of acidic strength
of (A) trichloroacetic acid, (B) trifluoroacetic acid, (C) acetic acid, (D) formic acid is
View answer and explanation
Correct option: C
B >> A >> D >> C
The correct answer is C because the electron-withdrawing effect of fluorine is greater than chlorine, and both are greater than hydrogen, leading to the acidity order B > A > D > C.
22 Hard
22 of 32
Application +4 | -1
In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBrAgBr. The percentage of bromine in the compound is (at. Mass AgAg = 108, BrBr = 80)
View answer and explanation
Correct option: A
24
Option A (24) is correct because the percentage of bromine is calculated correctly using Carius method formula.
23 Hard
23 of 32
Analyzing +4 | -1
The correct decreasing order of nucleo-philicity of the following species are -

(i) (CH3)3COΘ\left(\mathrm{CH}_3\right)_3 \mathrm{CO}^{\Theta}

(ii) CH3OΘ\mathrm{CH}_3 \mathrm{O}^{\Theta}

(iii) (CH3)2CHOΘ\left(\mathrm{CH}_3\right)_2 \mathrm{CHO}^{\Theta}

(iv) CH3CH2OΘ\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{O}^{\Theta}
View answer and explanation
Correct option: B
(i) >> (iii) >> (iv) >> (ii)
Option C (ii) >> (iv) >> (iii) >> (i) is correct because it reflects the correct decreasing order of nucleophilicity based on steric hindrance and inductive effects.
24 Hard
24 of 32
Analyzing +4 | -1
Among the compounds

(i) H−C≡C−COOH,\text{(i) }\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{COOH}, (ii) CH2=CH−COOH,\text{(ii) } \mathrm{CH}_2=\mathrm{CH}-\mathrm{COOH}, (iii) CH3−CH2COOH,\text{(iii) }\mathrm{CH}_3-\mathrm{CH}_2 \mathrm{COOH}, (iv) CH3−CH2−OH\text{(iv) }\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OH}

The correct of acid strength is
View answer and explanation
Correct option: A
(i) >> (ii) >> (iii) >> (iv)
Option A is correct because the acidity increases with the s-character of the carbon atom bonded to the carboxylic acid group.
25 Hard
25 of 32
Application +4 | -1
An organic compound is found to have the formula C5H10ONCl\mathrm{C}_5 \mathrm{H}_{10} \mathrm{ONCl}. The percentage of nitrogen present in it is
View answer and explanation
Correct option: B
10.3%10.3 \%
Option B (10.3 %) is correct because it is calculated by dividing the mass of nitrogen by the molecular mass of the compound and multiplying by 100.
26 Hard
26 of 32
Analyzing +4 | -1
How many structural isomers are possible for C6H14{C_6}{H_{14}}
View answer and explanation
Correct option: A
5
Option A is correct because there are 5 structural isomers possible for the molecular formula C6H14{C_6}{H_{14}}.
27 Hard
27 of 32
Analyzing +4 | -1
The order of decreasing stability of the carbanions

(CH3)3C−(CH3)2CH−CH3CH2−C6H5CH2−(I)(II)(III)(IV)\begin{array}{cccccccccccccccccccc} {{{\left( {{\text{C}}{{\text{H}}_3}} \right)}_3}{{\text{C}}^ - }}&{{{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}{\text{C}}{{\text{H}}^ - }}&{{\text{C}}{{\text{H}}_3}{\text{CH}}_2^ - }&{{{\text{C}}_6}{{\text{H}}_5}{\text{CH}}_2^ - } \\ {\left( {\text{I}} \right)}&{\left( {{\text{II}}} \right)}&{\left( {{\text{III}}} \right)}&{\left( {{\text{IV}}} \right)} \end{array}
View answer and explanation
Correct option: B
IV >> III >> II >> I
Option B (IV > III > II > I) is correct because carbanion stability increases with delocalization and decreases with +I effect.
28 Hard
28 of 32
Analyzing +4 | -1
Optical isomerism is not shown by:
View answer and explanation
Correct option: C
C2H5CH(Br)CH2CH3{C_2}{H_5}CH\left( {Br} \right)C{H_2}C{H_3}
The correct answer is C (C2H5CH(Br)CH2CH3{C_2}{H_5}CH\left( {Br} \right)C{H_2}C{H_3}) because it does not have a chiral center, which is necessary for optical isomerism.
29 Hard
29 of 32
Remembering +4 | -1
In kjeldahl’s method, nitrogen present is estimated as:
View answer and explanation
Correct option: B
NH3\mathrm{N}{\mathrm{H}_3}
Option B ( NH3\mathrm{N}{\mathrm{H}_3} ) is correct because in Kjeldahl's method, nitrogen is estimated as ammonia (NH3\mathrm{N}{\mathrm{H}_3}).
30 Hard
30 of 32
Understanding +4 | -1
Identify the correct statements from the
following.

(i) Electromeric effect is a permanent effect

(ii) Hyper conjugation is a temporary effect

(iii) Fractional distillation is used to separate
two liquids from a mixture if the difference
in their boiling points is less

(iv) Different compounds are adsorbed on an
adsorbent different extents
View answer and explanation
Correct option: D
iii, iv
Option D (iii, iv) is correct because fractional distillation separates liquids with boiling point differences, and different compounds adsorb differently.
Revision notes and key takeaways

Review ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

  • ✓ This set contains 32 single-correct-answer MCQs on ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
Showing 1-30 of 32 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING CHEMISTRY - ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES MCQs (All) 345 MCQs → ENGINEERING CHEMISTRY - ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES MCQs (Easy) 118 MCQs → ENGINEERING CHEMISTRY - ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES MCQs (Medium) 195 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ P-BLOCK ELEMENTS 736 MCQs → Equilibrium 696 MCQs → STRUCTURE OF ATOM 634 MCQs → THERMODYNAMICS 536 MCQs
📚 Switch Set (HARD)
Chapter Sets
ALL 345 MCQs EASY 118 MCQs HARD 32 MCQs MEDIUM 195 MCQs
0 /32
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →