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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 211-240 of 345 questions

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211 Easy
211 of 345
Understanding +4 | -1
The −I- I effect is shown by :
View answer and explanation
Correct option: A
−COOH- \mathrm{COOH}
Option A (−COOH- \mathrm{COOH}) is correct because carboxylic acid groups are electron-withdrawing and exhibit a -I effect.
212 Easy
212 of 345
Understanding +4 | -1
Which of the following molecules contain asymmetric carbon atom?
View answer and explanation
Correct option: A
CH3CHClCOOH\mathrm{C}{\mathrm{H}_3}\mathrm{CHClCOOH}
Option A is correct because the second carbon in CH3CHClCOOH is bonded to four different groups (H, Cl, CH3, COOH), making it a chiral center.
213 Medium
213 of 345
Remembering +4 | -1
Anthracene is purified by:
View answer and explanation
Correct option: D
Sublimation
Option D (Sublimation) is correct because Anthracene can be purified by sublimation due to its ability to transition directly from solid to gas phase.
214 Medium
214 of 345
Analyzing +4 | -1
Assertion (A): Pyrrole, is aromatic and undergoes electrophilic aromatic substitution extremely readily and predominant by at position adjacent to nitrogen.

Reason (R): Nitrogen in the ring bearing a lone pair in conjugation with π\pi-electrons brings aromaticity to the pyrrole.
View answer and explanation
Correct option: B
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
The correct answer is B because both the assertion and reason are true, but the reason doesn't fully explain why electrophilic substitution occurs predominantly at the position adjacent to nitrogen.
215 Easy
215 of 345
Understanding +4 | -1
Example of chain isomerism are
View answer and explanation
Correct option: C
pentanoic acid and 2 – methyl butanoic acid
Option C (pentanoic acid and 2 – methyl butanoic acid) is correct because they have the same molecular formula but differ in the arrangement of the carbon chain.
216 Easy
216 of 345
Understanding +4 | -1
The maximum bond energy is present
View answer and explanation
Correct option: A
C−H\mathrm{C} - \mathrm{H}
Option A (C−H\mathrm{C} - \mathrm{H}) is correct because the C-H bond generally has the highest bond energy among the given options due to the smaller size and higher electronegativity of hydrogen.
217 Easy
217 of 345
Remembering +4 | -1
The IUPAC suffix name of −COX- COX is:
View answer and explanation
Correct option: A
Oyl halide
Option A (Oyl halide) is correct because the IUPAC suffix for acyl halides (-COX) is 'oyl halide'.
218 Easy
218 of 345
Application +4 | -1
Which of the following molecules represents the order of hybridization SP2,  SP2,  SP,   SPS{P^2},\,\,S{P^2},\,\,SP,\,\,\,SP from left to right atoms.
View answer and explanation
Correct option: B
CH2=CH−C≡CHC{H_2} = CH - C \equiv CH
Option B (CH2=CH−C≡CHC{H_2} = CH - C \equiv CH) is correct because it correctly assigns SP2SP^2, SP2SP^2, SPSP, and SPSP hybridization to the carbon atoms from left to right.
219 Easy
219 of 345
Remembering +4 | -1
The energy difference between the chair and boat form of cyclohexane is :
View answer and explanation
Correct option: A
44 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}}
Option A (44 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}}) is correct because the energy difference between the chair and boat form of cyclohexane is 44 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}}.
220 Medium
220 of 345
Analyzing +4 | -1
The number of asymmetric C\mathrm{C}-atom created and number of possible stereoisomers when benzil (PhCOCOPh)(\mathrm{Ph} \mathrm{COCO} \mathrm{Ph}) is reduced with LiAlH4\mathrm{LiAlH}_4.
View answer and explanation
Correct option: A
2,3
Option A (2,3) is correct because the reduction of benzil with LiAlH4 creates two chiral carbon atoms, leading to three stereoisomers (two enantiomers and one meso compound).
221 Medium
221 of 345
Understanding +4 | -1
Assertion (A): Simple distillation can help in separating a mixture of propan-1-ol (boiling point 97∘C)97^{\circ} \mathrm{C}) and propanone (boiling point 56∘C).56^{\circ} \mathrm{C}).
Reason (R): Liquids with a difference of more than 20∘C20^{\circ} \mathrm{C} in their boiling points can be separated by simple distillation.
View answer and explanation
Correct option: A
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
The correct answer is A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) because simple distillation is effective for separating liquids with a significant difference in boiling points.
222 Hard
222 of 345
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.
223 Medium
223 of 345
Analyzing +4 | -1
Identify, which of the below does not possess any element of symmetry?
View answer and explanation
Correct option: A
(+)(-) tartaric acid
Option A ((+)(-) tartaric acid) is correct because a racemic mixture of tartaric acid does not possess any element of symmetry.
224 Easy
224 of 345
Understanding +4 | -1
Out of following which is not property of homologous series
View answer and explanation
Correct option: C
same physical property
Option C (same physical property) is correct because while homologous series share the same functional group, chemical properties, and general formula, their physical properties vary due to differences in molecular weight.
225 Easy
225 of 345
Remembering +4 | -1
Detection of sulphur in sodium extract is done by
View answer and explanation
Correct option: C
Both (a) and (b)
Option C (Both (a) and (b)) is correct because both lead acetate and sodium nitroprusside can be used to detect the presence of sulfur in a sodium extract.
226 Medium
226 of 345
Understanding +4 | -1
Compound that will give positive Lassaigne's test for both nitrogen and halogen is
View answer and explanation
Correct option: A
CH3NH2⋅HCl\mathrm{CH}_{3} \mathrm{NH}_{2} \cdot \mathrm{HCl}
Option A (CH3NH2⋅HCl\mathrm{CH}_{3} \mathrm{NH}_{2} \cdot \mathrm{HCl}) is correct because it contains both nitrogen and halogen, which will be detected in the Lassaigne's test.
227 Medium
227 of 345
Analyzing +4 | -1
But-2-yne is reacted separately with one mole of Hydrogen as shown below
B‾←liq NH3NaCH3−C+H2≡C−CH3→ΔPd/CA‾\underline B \xleftarrow[{{\text{liq }}N{H_3}}]{{Na}}C{H_3} - \mathop C\limits_{ + {H_2}} \equiv C - C{H_3}\xrightarrow[\Delta ]{{Pd/C}}\underline A
(A) A is more than soluble than B.
(B) The boiling point & melting point of A are higher and lower than B respectively.
(C) A is more polar than B because dipole moment of A is zero.
(D) Br2Br_2 adds easily to B than A.
Identify the incorrect statements from the option given below
View answer and explanation
Correct option: B
B, C & D only
The correct answer is B (B, C & D only) because the boiling point of A (cis-But-2-ene) is higher than B (trans-But-2-ene), A is more polar than B, and Br2Br_2 adds more easily to A than B.
228 Easy
228 of 345
Remembering +4 | -1
Which of the following is not a preliminary test used to detect ions?
View answer and explanation
Correct option: C
Brown ring test
Option C (Brown ring test) is correct because the brown ring test is a confirmatory test for nitrates, not a preliminary test.
229 Medium
229 of 345
Understanding +4 | -1
Statement I : A tertiary carbocation is more stable than a secondary carbocation which is more stable than a primary carbocation
Statement II : The inductive effects operate through sigma bonds and decrease rapidly with increase in distance between the substituent and the reaction site
View answer and explanation
Correct option: C
Both statements are correct
Option C (Both statements are correct) is correct because tertiary carbocations are indeed more stable due to the inductive effect and hyperconjugation, and the inductive effect diminishes with distance.
230 Medium
230 of 345
Understanding +4 | -1
The production of an optically active compound from a symmetric molecule without resolution in termed as :
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Correct option: C
Asymmetric synthesis
Option C (Asymmetric synthesis) is correct because asymmetric synthesis is the production of an optically active compound from a symmetric molecule without resolution.
231 Medium
231 of 345
Understanding +4 | -1
Arrange the Anions (p).CH3,  (q).NH2,  (r).OH,  (s).F−(\mathrm{p}). {CH}_3, \, \, (\mathrm{q}). {NH}_2, \, \, (\mathrm{r}). {OH}, \, \, (\mathrm{s}). {F}^{-} in decreasing order of their basic strength
View answer and explanation
Correct option: A
p >> q >> r >> s
Option A (p >> q >> r >> s) is correct because the basic strength of anions is inversely proportional to the acidic strength of their conjugate acids. The conjugate acids are CH4CH_4, NH3NH_3, H2OH_2O, and HFHF, with increasing acidity, thus the basicity decreases in the order given.
232 Easy
232 of 345
Understanding +4 | -1
Which of the following sodium compound/compounds are formed when an organic compound containing both nitrogen and sulphur is fused with sodium?
View answer and explanation
Correct option: B
Thiocyanate
Option B (Thiocyanate) is correct because when an organic compound containing both nitrogen and sulfur is fused with sodium, it forms sodium thiocyanate (NaSCN).
233 Medium
233 of 345
Analyzing +4 | -1
Incorrect statement for the use of indicators in acid-base titration is :
View answer and explanation
Correct option: C
Methyl orange may be used for a weak acid vs weak base titration.
The correct answer is C (Methyl orange may be used for a weak acid vs weak base titration) because there is no suitable indicator for weak acid vs weak base titrations.
234 Medium
234 of 345
Analyzing +4 | -1
The number of sp2s{p^2} hybridized carbon atoms present in ‘Aspartame’ are
View answer and explanation
Correct option: C
9
Option C (9) is correct because the structure of Aspartame contains 9 carbon atoms with sp² hybridization.
235 Medium
235 of 345
Understanding +4 | -1
Which of the following would be the best (most reactive) nucleophile in the polar medium?
View answer and explanation
Correct option: A
I−I^{-}
Option A (I−I^{-}) is the best nucleophile in a polar protic medium because larger ions are more polarizable and less solvated, making them more reactive.
236 Medium
236 of 345
Analyzing +4 | -1
Which of the following statement(s) is/are correct ?

(i) A carbon atom having an sps p hybrid orbital is less electronegative than carbon atoms possessing sp2s p^2 or sp3s p^3 hybridised orbitals.

(ii) π\pi-bonds provide the most reactive centres in the molecules containing multiple bonds

(iii) The number of σ\sigma and π\pi bonds in compound CH2=C=CHCH3{CH}_2={C}={CHCH}_3 are 7 and 2 respectively.
View answer and explanation
Correct option: C
(ii) only
The correct answer is C ((ii) only) because π-bonds are indeed reactive centers, while the other statements are incorrect regarding electronegativity and bond counts.
237 Medium
237 of 345
Understanding +4 | -1
Arrange the following free radicals in order of decreasing stability.

Methyl (I), Vinyl (II), Allyl (III), Benzyl (IV)
View answer and explanation
Correct option: D
IV >> III >> I >> II
Option D (IV >> III >> I >> II) is correct because benzyl is most stable due to resonance, followed by allyl, methyl, and then vinyl due to resonance and hyperconjugation.
238 Easy
238 of 345
Understanding +4 | -1
Isomerism exhibited by acetic acid and methyl formate is :
View answer and explanation
Correct option: A
Functional
Option A (Functional) is correct because acetic acid and methyl formate have different functional groups (carboxylic acid and ester, respectively).
239 Easy
239 of 345
Analyzing +4 | -1
Which one of the following does not have sp2s p^2-hybridized carbon?
View answer and explanation
Correct option: C
CH3−C≡N\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{N}
The correct answer is C (CH3−C≡N\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{N}) because the carbon atom bonded to the nitrogen atom is sp hybridized, not sp2.
240 Easy
240 of 345
Remembering +4 | -1
Which method is used to separate sugars?
View answer and explanation
Correct option: C
Chromatography
Option C (Chromatography) is correct because chromatography is a technique used to separate sugars based on their different affinities for a stationary and mobile phase.
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