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

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

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Showing 241-270 of 392 questions

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241 Medium
241 of 392
Remembering +4 | -1
Least stable conformer of cyclohexane is
View answer and explanation
Correct option: D
Planar hexagon
Option D is correct because the planar hexagon is the least stable conformer of cyclohexane due to significant angle strain and torsional strain. The chair form is the most stable.
242 Medium
242 of 392
Application +4 | -1
The maximum number of possible optical isomers in 1-bromo-2-methyl cyclobutane is
View answer and explanation
Correct option: A
4
Option A (4) is correct because 1-bromo-2-methyl cyclobutane has two chiral centers, leading to a maximum of 2^n = 22 = 4 stereoisomers.
243 Medium
243 of 392
Analyzing +4 | -1
Statement-I: In group III metal ion analysis we should not use only NH4OH{NH}_4{OH} without saturating by NH4Cl{NH}_4 {Cl}.

Statement-II: In the highly alkaline solution Mn2+{Mn}^{2+} and Ca2+{Ca}^{2+} both precipitated together.
View answer and explanation
Correct option: C
If Statement-I is true but Statement-II is false
Option C (If Statement-I is true but Statement-II is false) is correct because Statement-I is true, but Statement-II is false.
244 Medium
244 of 392
Remembering +4 | -1
In Duma’s method for determining the nitrogen content of an organic compound, the nitrogen content is determined in the form of:
View answer and explanation
Correct option: C
Gaseous N2{\mathrm{N}_2}
Option C (Gaseous N2{\mathrm{N}_2}) is correct because in Duma's method, the nitrogen content is determined by collecting and measuring the volume of nitrogen gas produced.
245 Medium
245 of 392
Understanding +4 | -1
Carbanion can undergo :
View answer and explanation
Correct option: D
All of the above are correct
Option D (All of the above are correct) is correct because carbanions are capable of rearrangement, combination with cations, and addition to carbonyl groups due to their negative charge and electron-rich nature.
246 Easy
246 of 392
Remembering +4 | -1
Rotation of plane polarized light can be measured by :
View answer and explanation
Correct option: C
Polarimeter
Option C is correct because the rotation of plane-polarized light can be measured by a polarimeter. A polarimeter is specifically designed for this purpose.
247 Easy
247 of 392
Remembering +4 | -1
The tautomeric form which is less stable is called :
View answer and explanation
Correct option: C
Labile form
Option C (Labile form) is correct because the labile form in tautomerism refers to the less stable tautomer.
248 Medium
248 of 392
Understanding +4 | -1
Unpaired electron in C.H3\mathop C\limits^. {H_3} occupies:
View answer and explanation
Correct option: B
sp3s{p^3} - hybrid orbital
Option D (sp3s{p^3} - hybrid orbital) is correct because the methyl radical has three sigma bonds and one unpaired electron, resulting in sp3sp^3 hybridization.
249 Medium
249 of 392
Understanding +4 | -1
Which of the organic compounds will give red colour in Lassaigne test?
View answer and explanation
Correct option: B
S∣∣NH2−C−NH2\mathop {\mathop \mathrm{S}\limits_{||} }\limits_{N{H_2} - C - N{H_2}}
Option A (NaCNS\mathrm{NaCNS}) is correct because in the Lassaigne's test, if both nitrogen and sulfur are present, they combine to form NaCNS, which reacts with Fe3+Fe^{3+} to form a blood-red color.
250 Hard
250 of 392
Understanding +4 | -1
Geometrical isomerism is not shown by
View answer and explanation
Correct option: A
1, 1-dichloro-1-pentene
Option A (1, 1-dichloro-1-pentene) is correct because geometrical isomerism requires different groups on each carbon of the double bond, and this compound has two identical chlorine atoms on one carbon.
251 Medium
251 of 392
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 temperature, bromine adds to the alkene first, as alkynes are less reactive than alkenes towards electrophilic addition.
252 Medium
252 of 392
Understanding +4 | -1
The stability of 2,3-dimethyl but-2-ene is more than 2-butene. This can be explained in terms of :
View answer and explanation
Correct option: B
Hyperconjugation
Option B (Hyperconjugation) is correct because 2,3-dimethylbut-2-ene has more alpha hydrogens than 2-butene, leading to greater stabilization through hyperconjugation.
253 Medium
253 of 392
Understanding +4 | -1
A molecule is R3C−H{R_3}\mathrm{C}-\mathrm{H}. If H is replaced by Z(R3C−Z)\mathrm{Z}\left( {{R_3}\mathrm{C}-\mathrm{Z}} \right) and on doing so electron density on R3−C{R_3}-\mathrm{C} part increases, then Z is :
View answer and explanation
Correct option: C
Electron repelling group
Option C (Electron repelling group) is correct because if the electron density on R3CR_3C increases when H is replaced by Z, then Z must be donating electrons, thus repelling them.
254 Medium
254 of 392
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 it is suitable for separating small amounts of compounds with different polarities.
255 Medium
255 of 392
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.
256 Easy
256 of 392
Remembering +4 | -1
The first organic compound urea was synthesized in the laboratory by:
View answer and explanation
Correct option: D
Wöhler
Option D (Wöhler) is correct because Friedrich Wöhler synthesized urea in the laboratory, disproving the vitalism theory.
257 Easy
257 of 392
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 in (ii) there are both sp3sp^3 and sp2sp^2 hybridized carbons, and in (iii) there are sp3sp^3, sp2sp^2 and spsp hybridized carbons.
258 Easy
258 of 392
Understanding +4 | -1
Identify which of the following group act as +M+ M as well as −m- m?
View answer and explanation
Correct option: B
−Cl- Cl
Option B (−Cl- Cl) is correct because chlorine can act as both a +M and -I group due to its lone pairs and electronegativity, but its +M effect is weaker than its -I effect.
259 Medium
259 of 392
Remembering +4 | -1
A Dewar flask is usually used to:
View answer and explanation
Correct option: D
Store liquid air
The correct answer is Store liquid air because Dewar flasks are designed to minimize heat transfer, making them ideal for storing cryogenic liquids like liquid air.
260 Hard
260 of 392
Analyzing +4 | -1
Total number of isomeric aldehydes and ketones that can exist with the molecular formula C5H10O{\mathrm{C}_5}{\mathrm{H}_{10}}\mathrm{O} :
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Correct option: D
7
Option D (7) is correct because there are 7 isomeric aldehydes and ketones with the molecular formula C5H10O{\mathrm{C}_5}{\mathrm{H}_{10}}\mathrm{O}.
261 Easy
261 of 392
Understanding +4 | -1
The isomerism which exists between CH3CHCl2\mathrm{C}{\mathrm{H}_3}\mathrm{CHC}{\mathrm{l}_2} and CH2ClCH2Cl\mathrm{C}{\mathrm{H}_2}\mathrm{ClC}{\mathrm{H}_2}\mathrm{Cl} is :
View answer and explanation
Correct option: C
Positional
Option C (Positional) is correct because the compounds have the same molecular formula but differ in the position of the chlorine atoms on the carbon chain.
262 Hard
262 of 392
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 is correct because breaking two bonds on a chiral center will change the spatial arrangement of the remaining groups, thus altering the configuration.
263 Hard
263 of 392
Understanding +4 | -1
Which of the following would yield a carbon radical on heating?
View answer and explanation
Correct option: B
Et4Pb\mathrm{Et}_4 \mathrm{Pb}
Option B (Et4Pb) is correct because tetraethyllead (Et4Pb) is known to decompose upon heating, generating ethyl radicals.
264 Easy
264 of 392
Remembering +4 | -1
The relative adsorption of each component of the mixture is expressed in terms of
View answer and explanation
Correct option: B
retention factor
The correct answer is retention factor because it quantifies the relative migration of a substance in chromatography.
265 Medium
265 of 392
Understanding +4 | -1
Which of the following statements is correct?
View answer and explanation
Correct option: A
Allyl carbonium ion (CH2=CH−C+H2)\left( {C{H_2} = CH - \mathop {\mathrm{ }C}\limits^ + {H_2}} \right)  is more stable then propyl carbonium ion
Option B (7σ and 3π7\sigma \,and\,3\pi) is correct because but-1-ene-3-yne has 7 sigma bonds and 3 pi bonds when you draw out its structure and count them.
266 Medium
266 of 392
Understanding +4 | -1
The IUPAC name of CH3−CH2−CHO\mathrm{C}{\mathrm{H}_3} - \mathrm{C}{\mathrm{H}_2} - \mathrm{CHO}
View answer and explanation
Correct option: A
Propanal-1
Option A (Propanal-1) is correct because the aldehyde group is always at the end of the chain (carbon 1), and the parent chain is propane.
267 Medium
267 of 392
Application +4 | -1
Total isomers with the molecular formula, C4H11N\mathrm{C}_4 \mathrm{H}_{11} \mathrm{N} is
View answer and explanation
Correct option: C
8
Option C (8) is correct because there are eight structural isomers of C4H11N, considering different arrangements of the carbon chain and the amine group.
268 Medium
268 of 392
Understanding +4 | -1
A molecule having three different chiral carbon atoms, how many stereoisomers it will have?
View answer and explanation
Correct option: A
8
Option A (8) is correct because the number of stereoisomers is given by 2n2^n, where n is the number of chiral centers. Here, n=3, so 23=82^3 = 8.
269 Hard
269 of 392
Understanding +4 | -1
The halogen compound which most readily undergoes nucleophilic substitutions is
View answer and explanation
Correct option: D
CH2=CHCH2Cl\mathrm{C}{\mathrm{H}_2} = \mathrm{CHC}{\mathrm{H}_2}\mathrm{Cl}
Option D (CH2=CHCH2Cl\mathrm{C}{\mathrm{H}_2} = \mathrm{CHC}{\mathrm{H}_2}\mathrm{Cl}) is correct because the allylic halide is more reactive towards nucleophilic substitution due to the stability of the resulting carbocation by resonance.
270 Medium
270 of 392
Understanding +4 | -1
Given below are two statements.
Statement I: The presence of weaker π - \pi {\text{ - }} bonds make alkenes less stable than alkanes.
Statement II: The strength of the double bond is greater than that of carbon-carbon single bond.
In the light of the above statements, choose the correct answer from the options given below.
View answer and explanation
Correct option: A
Both Statement I and Statement II are correct.
Option A (Both Statement I and Statement II are correct) is correct because alkenes are indeed less stable due to weaker pi bonds, and the overall double bond strength is greater than a single bond.
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