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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 181-210 of 345 questions

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181 Medium
181 of 345
Analyzing +4 | -1
Which structure can be explained by taking ground state configuration of atom?
View answer and explanation
Correct option: D
H2O{\mathrm{H}_2}\mathrm{O}
Option D (H2O{\mathrm{H}_2}\mathrm{O}) is correct because oxygen has two unpaired electrons in its ground state configuration, allowing it to form two bonds with hydrogen without requiring hybridization.
182 Medium
182 of 345
Application +4 | -1
At low temperature, the slow addition of molecular bromine to H2C=CH−CH2−C≡CH{\mathrm{H}_2}\mathrm{C} = \mathrm{CH}-\mathrm{C}{\mathrm{H}_2}-\mathrm{C} \equiv \mathrm{CH} gives:
View answer and explanation
Correct option: C
H2C=CH−CH2−CH2−CBr3{\mathrm{H}_2}\mathrm{C} = \mathrm{CH}-\mathrm{C}{\mathrm{H}_2}-\mathrm{C}{\mathrm{H}_2}-\mathrm{CB}{\mathrm{r}_3}
Option B (BrCH2−CHBr−CH2−C≡CH\mathrm{BrC}{\mathrm{H}_2}-\mathrm{CHBr}-\mathrm{C}{\mathrm{H}_2}-\mathrm{C} \equiv \mathrm{CH}) is correct because at low temperatures, the addition of bromine favors the more reactive double bond over the triple bond.
183 Medium
183 of 345
Analyzing +4 | -1
In Carius tube, an organic compound ' X\mathrm{X} ' is treated with sodium peroxide to form a mineral acid ' YY '.

The solution of BaCl2\mathrm{BaCl}_2 is added to ' Y\mathrm{Y} ' to form a precipitate ' ZZ '. ' ZZ ' is used for the quantitative estimation of an extra element. ' X\mathrm{X} ' could be
View answer and explanation
Correct option: C
Methionine
Option C (Methionine) is correct because the formation of BaSO4 precipitate indicates the presence of sulfur in the organic compound, and Methionine is the only option containing sulfur.
184 Hard
184 of 345
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.
185 Medium
185 of 345
Understanding +4 | -1
Which purification technique is used for high boiling organic liquid compound (decomposes near its boiling point)?
View answer and explanation
Correct option: D
Reduced pressure distillation
Option D (Reduced pressure distillation) is correct because it lowers the boiling point, preventing decomposition of the high-boiling organic liquid.
186 Medium
186 of 345
Remembering +4 | -1
LiAlH4LiAl{H_4} is used as:
View answer and explanation
Correct option: B
Reducing agent
Option B (Reducing agent) is correct because LiAlH4LiAl{H_4} is a strong reducing agent, donating hydride ions to reduce other compounds.
187 Hard
187 of 345
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.
188 Medium
188 of 345
Analyzing +4 | -1
Which of the following statements is/are incorrect?
View answer and explanation
Correct option: B
Tautomeric structures are the resonating structures of a molecule
The correct answer is "Tautomeric structures are the resonating structures of a molecule" because tautomers are constitutional isomers that interconvert, while resonance structures are different representations of the same molecule.
189 Easy
189 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 the hybridization of the carbon atoms from left to right is SP2,SP2,SP,SPS{P^2},S{P^2},SP,SP.
190 Easy
190 of 345
Analyzing +4 | -1
Among the following, which one has more than one kind of hybridization?

(i) CH3CH2CH2CH3\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_3}

(ii) CH3CH=CHCH3\mathrm{C}{\mathrm{H}_3}\mathrm{CH} = \mathrm{CHC}{\mathrm{H}_3}

(iii)CH2=CH−CH≡CH\mathrm{C}{\mathrm{H}_2} = \mathrm{CH} - \mathrm{CH} \equiv \mathrm{CH}

(iv) CH ≡ CH
View answer and explanation
Correct option: A
(ii) and (iii)
Option A ((ii) and (iii)) is correct because these molecules contain carbon atoms with both sp2 and sp hybridization.
191 Medium
191 of 345
Remembering +4 | -1
The structure of tert-Butyl group in an organic compound is
View answer and explanation
Correct option: D
CH3−C∣CH3CH3∣−C{H_3} - \mathop {\mathop C\limits_{\begin{array}{cccccccccccccccccccc} | \\ {C{H_3}} \end{array}} }\limits^{\begin{array}{cccccccccccccccccccc} {C{H_3}} \\ | \end{array}} -
Option D (CH3−C∣CH3CH3∣−C{H_3} - \mathop {\mathop C\limits_{\begin{array}{cccccccccccccccccccc} | \\ {C{H_3}} \\ \end{array}} }\limits^{\begin{array}{cccccccccccccccccccc} {C{H_3}} \\ | \\ \end{array}} -) is correct because it represents a carbon atom bonded to three methyl groups and one other group.
192 Medium
192 of 345
Understanding +4 | -1
During elimination reactions, the hybrid state of carbon atoms involved in change shows:
View answer and explanation
Correct option: D
Either of the above
Option A is correct. During elimination reactions, a sigma bond is broken and a pi bond is formed, leading to a change in hybridization from sp3 to sp2 or sp2 to sp depending on the specific reaction.
193 Medium
193 of 345
Application +4 | -1
Total number of geometrical isomers in the following compound.
CH3−CH=CH−CH=CH−CH=CH−C6H5C{H_3} - CH = CH - CH = CH - CH = CH - {C_6}{H_5}
View answer and explanation
Correct option: D
8
Option D (8) is correct because with three double bonds, each capable of cis/trans isomerism, the total number of geometrical isomers is 23 = 8.
194 Easy
194 of 345
Understanding +4 | -1
IUPAC name of C6H5COCl{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{COCl} is :
View answer and explanation
Correct option: C
Benzene carbonyl chloride
Option C is correct because according to IUPAC nomenclature, C6H5COCl{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{COCl} is named as Benzene carbonyl chloride.
195 Medium
195 of 345
Remembering +4 | -1
Rearrangement is a property of_____
View answer and explanation
Correct option: B
Carbocation
Option B (Carbocation) is correct because carbocations are known to undergo rearrangement reactions.
196 Medium
196 of 345
Application +4 | -1
1.6 g1.6 \mathrm{~g} of an organic compound gave 2.6 g2.6 \mathrm{~g} of magnesium pyrophosphate. The percentage of phosphorus in the compound is
View answer and explanation
Correct option: A
45.38%45.38 \%
Option A (45.38 %) is correct because it is calculated by determining the mass of phosphorus in magnesium pyrophosphate and then finding its percentage in the original organic compound.
197 Medium
197 of 345
Analyzing +4 | -1
The species which cannot exhibit optical isomerism is
View answer and explanation
Correct option: D
(CH3CH2)4C\left({CH}_3 {CH}_2\right)_4 {C}
Option D is correct because (CH3CH2)4C\left({CH}_3 {CH}_2\right)_4 {C} has four identical groups attached to the central carbon, making it achiral and unable to exhibit optical isomerism.
198 Medium
198 of 345
Understanding +4 | -1
Which of the following is prepared by heating its hydrated salt?
View answer and explanation
Correct option: D
None of the above
Option D (None of the above) is correct because anhydrous AlCl3, SnCl2, and SnCl4 are not prepared by simply heating their hydrated salts due to hydrolysis.
199 Hard
199 of 345
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.
200 Medium
200 of 345
Analyzing +4 | -1
In which of the following options chlorine will act as the best leaving group
View answer and explanation
Correct option: D
{\text{C}}{{\text{H}}_{\text{3}}} - {\text{C}}{{\text{H}}_{\text{2}}} - \mathop {\text{C}}\limits_{\begin{array}{cccccccccccccccccccc} | \\ {{\text{C}}{{\text{H}}_{\text{3}}}} \end{array}}^{\begin{array}{cccccccccccccccccccc} {{\text{C}}{{\text{H}}_{\text{3}}}} \\ |{\text{C}}{{\text{H}}_{\text{3}}} - {\text{C}}{{\text{H}}_{\text{2}}} - \mathop {\text{C}}\limits_{\begin{array}{cccccccccccccccccccc} | \\ {{\text{C}}{{\text{H}}_{\text{3}}}} \end{array}}^{\begin{array}{cccccccccccccccccccc} {{\text{C}}{{\text{H}}_{\text{3}}}} \\ |
Option D is correct because the stability of the carbocation formed after chlorine leaves determines its leaving group ability, with more substituted carbocations being more stable.
201 Medium
201 of 345
Remembering +4 | -1
Which of the following compound is synthesised by Kolbe and Berthelot respectively that shows that organic compounds could be synthesised from inorganic sources in a laboratory?
View answer and explanation
Correct option: D
Acetic acid, methane
Option D (Acetic acid, methane) is correct because Kolbe synthesized acetic acid and Berthelot synthesized methane from inorganic sources.
202 Medium
202 of 345
Understanding +4 | -1
The reaction of propene with HBr in the presence of Benzoyl peroxide is an example of
View answer and explanation
Correct option: D
Free radical addition reaction
Option D (Free radical addition reaction) is correct because the presence of benzoyl peroxide initiates a free radical mechanism for the addition of HBr to propene.
203 Easy
203 of 345
Understanding +4 | -1
Which is optically active?
View answer and explanation
Correct option: D
α-chloropropionic acid
Option D (α-chloropropionic acid) is correct because it contains a chiral carbon atom, which is a carbon atom bonded to four different groups, making it optically active.
204 Medium
204 of 345
Remembering +4 | -1
Among the following four compounds:
View answer and explanation
Correct option: A
Phenol
Option D (parapara-nitrophenol) is the most acidic because the nitro group (-NO₂) is an electron-withdrawing group that stabilizes the phenoxide ion through resonance, and the para position provides the greatest stabilization.
205 Easy
205 of 345
Understanding +4 | -1
Statement I : Stereoisomers have the same molecular formula and same structural formual but they differ in configuration
Statement II : cis but-2-ene and trans but-2-ene are stereosisomers
View answer and explanation
Correct option: C
Both statements are correct
Option C is correct because stereoisomers have the same molecular and structural formula but differ in spatial arrangement, and cis/trans but-2-ene exemplify this.
206 Easy
206 of 345
Analyzing +4 | -1
How many isomers will C3H6{\mathrm{C}_3}{\mathrm{H}_6} have?
View answer and explanation
Correct option: B
2
Option B (2) is correct because C3H6 has two structural isomers: propene and cyclopropane.
207 Hard
207 of 345
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.
208 Hard
208 of 345
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.
209 Medium
209 of 345
Remembering +4 | -1
Which chromatography is based on the fact that different compounds are adsorbed on an adsorbent to different degrees?
View answer and explanation
Correct option: B
Adsorption chromatography
Option B (Adsorption chromatography) is correct because it separates compounds based on their different affinities for an adsorbent material.
210 Medium
210 of 345
Remembering +4 | -1
Inductive effect involves
View answer and explanation
Correct option: A
Displacement of σ\sigma - Electrons
Option A (Displacement of σ\sigma - Electrons) is correct because the inductive effect involves the displacement of sigma electrons along a chain of atoms.
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