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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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Showing 301-330 of 345 questions

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301 Easy
301 of 345
Understanding +4 | -1
Select the correct statements
View answer and explanation
Correct option: B
−NH2- N{H_2} group activates benzene nucleus for attack of electrophile at ortho and parasites.
Option B (−NH2- N{H_2} group activates benzene nucleus for attack of electrophile at ortho and parasites.) is correct because the amino group is an activating and ortho/para-directing group.
302 Hard
302 of 345
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.
303 Hard
303 of 345
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.
304 Medium
304 of 345
Analyzing +4 | -1
Statement I: trans-2-Butane on reaction with Br2{Br}_2 gives meso-2,3-dibromobutane.

Statement II: The reaction involves syn-addition of bromine.
View answer and explanation
Correct option: C
Statement I is correct but statement II is incorrect
Option C is correct because trans-2-butene reacts with Br2 to give meso-2,3-dibromobutane, which involves anti-addition, not syn-addition.
305 Medium
305 of 345
Analyzing +4 | -1
What is the correct order of decreasing stability of the following cations

I) CH3−C⊕H−CH3C{H_3} - \mathop C\limits^ \oplus H - C{H_3}

II) CH3−C⊕H−OCH3C{H_3} - \mathop C\limits^ \oplus H - OC{H_3}

III) CH3−C⊕H−OCH3C{H_3} - \mathop C\limits^ \oplus H - OC{H_3}
View answer and explanation
Correct option: A
II > I > III
The correct answer is B (II > III > I) because the carbocation is stabilized by the +R effect of the methoxy group in II and III, with II being more stable due to less steric hindrance.
306 Hard
306 of 345
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}).
307 Easy
307 of 345
Understanding +4 | -1
Which of the following is a primary halide?
View answer and explanation
Correct option: D
Neo hexyl chloride
Option D (Neo hexyl chloride) is correct because the chlorine atom is attached to a primary carbon atom.
308 Medium
308 of 345
Analyzing +4 | -1
The number of isomers in C4H10O{\mathrm{C}_4}{\mathrm{H}_{10}}\mathrm{O} are
View answer and explanation
Correct option: A
7
Option A (7) is correct because there are 7 structural isomers of C4H10O{\mathrm{C}_4}{\mathrm{H}_{10}}\mathrm{O}, including 4 alcohols and 3 ethers.
309 Easy
309 of 345
Understanding +4 | -1
Ethanoyl chloride is the IUPAC name of
View answer and explanation
Correct option: B
CH3COCl\mathrm{CH}_3 \mathrm{COCl}
The correct answer is B (CH3COCl\mathrm{CH}_3 \mathrm{COCl}) because ethanoyl chloride is an acyl chloride with the formula CH3COClCH_3COCl.
310 Medium
310 of 345
Analyzing +4 | -1
Which is the correct order of stability of the following three carbonium ions?
CH2=CHC+H3I,  CH=CCH3∣∣−CH2II,  CH3CH=CHC+H2III\mathop {{\text{C}}{{\text{H}}_2} = {\text{CH}}\mathop {\text{C}}\limits^ + {{\text{H}}_3}}\limits_{\text{I}} ,\,\,\mathop {{\text{CH}} = \mathop {\text{C}}\limits^{\begin{array}{cccccccccccccccccccc} {{\text{C}}{{\text{H}}_3}} \\ {||} \end{array}} - {\text{C}}{{\text{H}}_2}}\limits_{{\text{II}}} ,\,\,\mathop {{\text{C}}{{\text{H}}_3}{\text{CH}} = {\text{CH}}\mathop {\text{C}}\limits^ + {{\text{H}}_2}}\limits_{{\text{III}}}
View answer and explanation
Correct option: C
I ≈\approx III >> II
Option C is correct. Carbocations are stabilized by +I effect and delocalization. I and III are more stable than II due to more delocalization.
311 Easy
311 of 345
Remembering +4 | -1
Blood red colour solution is produced when ferric chloride solution is treated with:
View answer and explanation
Correct option: B
KSCN
Option B (KSCN) is correct because ferric chloride reacts with potassium thiocyanate (KSCN) to form a blood-red colored solution of ferric thiocyanate.
312 Easy
312 of 345
Remembering +4 | -1
The intermediate formed during the homolytic cleavage is
View answer and explanation
Correct option: A
free radical
Option A (free radical) is correct because homolytic cleavage involves the breaking of a covalent bond in such a way that each atom gets one electron, forming free radicals.
313 Medium
313 of 345
Understanding +4 | -1
Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl due halides to:
View answer and explanation
Correct option: B
Resonance stabilization
Option B (Resonance stabilization) is correct because the aryl halides are less reactive towards nucleophilic substitution reactions due to the resonance stabilization of the C-X bond, which makes it stronger and harder to break.
314 Easy
314 of 345
Understanding +4 | -1
Nitrogen detection in an organic compound is carried out by Lassaigne’s test. The blue colour formed corresponds to which of the following formulae
View answer and explanation
Correct option: B
Fe4[Fe(CN)6]3F{e_4}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_3}
Option B (Fe4[Fe(CN)6]3F{e_4}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_3}) is correct because the Prussian blue color in Lassaigne's test is due to the formation of ferric ferrocyanide.
315 Medium
315 of 345
Analyzing +4 | -1
The sum of geometrical isomers in [Pt(H2 N−CH(CH3)COO−)2]\left[\mathrm{Pt}\left(\mathrm{H}_2 \mathrm{~N}-\mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{COO}^{-}\right)_2\right] complex and stereoisomers of [Pt( gly )3]+\left[\mathrm{Pt}(\text { gly })_3\right]^{+}complex is
View answer and explanation
Correct option: A
8
Option A (If both assertion and reason are true and reason is the correct explanation of assertion) is correct because alcoholic silver nitrite (AgNO2AgNO_2) is covalent, leading to the formation of nitroethane instead of ethyl nitrite.
316 Medium
316 of 345
Analyzing +4 | -1
Assertion (A): Carboxylic acid is more acidic than carbolic acid.

Reason (R): Carboxylic acid have equivalent resonating structure.
View answer and explanation
Correct option: C
(A) is true but (R) is false
Option C is correct. Carboxylic acids are more acidic than phenols, but the reason given about equivalent resonating structures is false. The resonance structures in carboxylate ions are not equivalent.
317 Hard
317 of 345
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.
318 Medium
318 of 345
Analyzing +4 | -1
Assertion : When inductive and electromeric effects operate in opposite directions, the inductive effect predominates.

Reason : Inductive effect is the complete transfer of shared pair of π\pi electrons to one of the atoms.
View answer and explanation
Correct option: D
Both assertion and reason are false.
The correct answer is C (If Assertion is True but the Reason is False) because while Eu²⁺ and Yb²⁺ are indeed reducing agents, the reason is due to their tendency to achieve stable f⁷ and f¹⁴ configurations, respectively, not a half-filled configuration.
319 Easy
319 of 345
Remembering +4 | -1
First Nobel Prize winner in chemistry is:
View answer and explanation
Correct option: A
Van’t Hoff
Option A (Van’t Hoff) is correct because Jacobus Henricus van 't Hoff was the first Nobel Prize winner in Chemistry in 1901.
320 Easy
320 of 345
Remembering +4 | -1
Aniline is separated by:
View answer and explanation
Correct option: C
Steam distillation
Option C (Steam distillation) is correct because aniline is steam volatile and can be separated from non-volatile impurities using steam distillation.
321 Easy
321 of 345
Understanding +4 | -1
The Lassaigne's extract is boiled with dil HNO3\mathrm{HN}{\mathrm{O}_3} before testing for halogens because
View answer and explanation
Correct option: C
Na2S\mathrm{N}{\mathrm{a}_2}\mathrm{S} and NaCN\mathrm{NaCN} are decomposed by HNO3\mathrm{HN}{\mathrm{O}_3}
Option C (Na2S\mathrm{N}{\mathrm{a}_2}\mathrm{S} and NaCN\mathrm{NaCN} are decomposed by HNO3\mathrm{HN}{\mathrm{O}_3}) is correct because Na2SNa_2S and NaCNNaCN interfere with the halogen test by forming Ag2SAg_2S and AgCNAgCN precipitates, respectively, which are not silver halides.
322 Easy
322 of 345
Remembering +4 | -1
In case of a homologous series each member differs from the preceding or the succeeding member by :
View answer and explanation
Correct option: A
a CH2\mathrm{C}{\mathrm{H}_2} group
Option A (a CH2\mathrm{C}{\mathrm{H}_2} group) is correct because successive members in a homologous series differ by a methylene group (CH2\mathrm{C}{\mathrm{H}_2}).
323 Medium
323 of 345
Application +4 | -1
Compound (A) is formed on treating a benzilwith NaBH4\mathrm{NaBH}_4. The number of possible meso isomers for (A) are
View answer and explanation
Correct option: A
1
The element X could have an atomic weight of 7 and its oxide is X2O{\mathrm{X}_2}\mathrm{O} because the ratio of their masses is consistent with this.
324 Hard
324 of 345
Analyzing +4 | -1
Assertion A: Hydrolysis of alkyl chloride is a slow reaction but in the presence of NaI,NaI, the rate of the hydrolysis increases
Reason: I−I^- is a good nucleophile as well as good leaving group
Correct answer from the options given below
View answer and explanation
Correct option: C
Both A and R are true and R is the correct explanation of A
Option C (Both A and R are true and R is the correct explanation of A) is correct because the rate of hydrolysis increases due to the formation of alkyl iodide, where iodide is a good nucleophile and leaving group.
325 Medium
325 of 345
Analyzing +4 | -1
The number of optical isomers of pent-3-en-2-ol is :
View answer and explanation
Correct option: A
2
Option B (KCl3+F2→KF3+Cl2KCl_3 + F_2 \rightarrow KF_3 + Cl_2) is incorrect because fluorine does not form F3−F_3^− due to the absence of available d-orbitals for expansion of its octet.
326 Easy
326 of 345
Remembering +4 | -1
Fractional distillation is useful in distillation of
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because fractional distillation is useful in the distillation of Petroleum, Coal-tar, and Crude alcohol.
327 Medium
327 of 345
Remembering +4 | -1
Which is purified by steam distillation
View answer and explanation
Correct option: A
Aniline
Option A (Aniline) is correct because aniline, being an organic compound immiscible with water and having a high boiling point, can be effectively purified by steam distillation.
328 Medium
328 of 345
Analyzing +4 | -1
The number of asymmetric carbon atoms and the number of optical isomers in CH3(CHOH)2COOH\mathrm{C}{\mathrm{H}_3}{\left( {\mathrm{CHOH}} \right)_2}\mathrm{COOH} are respectively :
View answer and explanation
Correct option: C
2 and 4
Option C (2 and 4) is correct because the compound has two asymmetric carbon atoms, leading to 22=42^2 = 4 stereoisomers.
329 Medium
329 of 345
Understanding +4 | -1
In chlorobenzene solution, the basic strength of amines increases in the order
View answer and explanation
Correct option: B
C2H5NH2<(C2H5)2NH<(C2H5)3N\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}
Option B is correct because in chlorobenzene solution, the basic strength of amines increases with the number of ethyl groups due to the +I effect, making the order 1∘<2∘<3∘1^{\circ} < 2^{\circ} < 3^{\circ}.
330 Medium
330 of 345
Analyzing +4 | -1
How many stereoisomers does this molecule have?
CH3CH=CHCH2CHBrCH3\mathrm{C}{\mathrm{H}_3}\mathrm{CH} = \mathrm{CHC}{\mathrm{H}_2}\mathrm{CHBrC}{\mathrm{H}_3}
View answer and explanation
Correct option: A
4
The correct answer is A (4) because the molecule has one chiral center, leading to two enantiomers, and exhibits geometrical isomerism (cis/trans), resulting in a total of 4 stereoisomers.
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