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

Organic chemistry-some basic principles and techniques practice · Medium

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121 Medium
121 of 173
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.
122 Medium
122 of 173
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 CH3(CHOH)2COOH\mathrm{C}{\mathrm{H}_3}{\left( {\mathrm{CHOH}} \right)_2}\mathrm{COOH} has two asymmetric carbon atoms, leading to 22 = 4 stereoisomers.
123 Medium
123 of 173
Understanding +4 | -1
Arrange the following in increasing order of stability
(A) (CH3)2C+−CH2CH3{\left( {C{H_3}} \right)_2}\mathop C\limits^ + - C{H_2}C{H_3} (B) (CH3)3C+{\left( {C{H_3}} \right)_3}\mathop C\limits^ +
(C) (CH3)2CH+{\left( {C{H_3}} \right)_2}\mathop {CH}\limits^ +
(D) CH3−CH2+C{H_3} - {\mathop {CH}\limits^ + _2} (E) CH3+{\mathop {CH}\limits^ + _3}
View answer and explanation
Correct option: B
E<D<C<A<BE < D < C < A < B
Option B (E<D<C<A<BE < D < C < A < B) is correct because carbocation stability increases with the number of alkyl groups attached to the positively charged carbon due to hyperconjugation and inductive effects.
124 Medium
124 of 173
Application +4 | -1
An organic compound contains 29.27% carbon, 5.69 % hydrogen and 65.04% bromine. Its empirical formula is
View answer and explanation
Correct option: D
C3H7Br{\mathrm{C}_3}{\mathrm{H}_7}\mathrm{Br}
Option D (C3H7Br{\mathrm{C}_3}{\mathrm{H}_7}\mathrm{Br}) is correct because the mole ratio of C:H:Br is approximately 3:7:1, which corresponds to the empirical formula C3H7Br{\mathrm{C}_3}{\mathrm{H}_7}\mathrm{Br}.
125 Medium
125 of 173
Remembering +4 | -1
In Kjeldahl’s method, the Nitrogen present in the organic compound is quantitatively converted into
View answer and explanation
Correct option: D
NH3N{H_3}
Option D (NH3N{H_3}) is correct because in Kjeldahl's method, the nitrogen present in the organic compound is quantitatively converted into ammonia.
126 Medium
126 of 173
Remembering +4 | -1
Anthraquinone is prepared by the oxidation of anthracene by chromic acid. The crude reaction product is usually purified by
View answer and explanation
Correct option: D
Sublimation
The correct answer is Sublimation because anthraquinone can be purified by sublimation due to its ability to transition directly from solid to gas phase upon heating, leaving impurities behind.
127 Medium
127 of 173
Analyzing +4 | -1
Which of the following alkenes is most reactive for electrophilic addition reactions?
View answer and explanation
Correct option: A
CH3−CH=CH−CH3\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3
Option A (CH3−CH=CH−CH3\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3) is the most reactive because alkyl groups are electron-donating, increasing the electron density of the alkene, which favors electrophilic attack.
128 Medium
128 of 173
Understanding +4 | -1
In Lassaigne’s test, a black colour is obtained if the organic compound contains sulphur. The black colour is due to
View answer and explanation
Correct option: B
PbSPbS
Option B (PbSPbS) is correct because in Lassaigne's test, the black color indicates the presence of sulfur, which reacts with lead acetate to form lead sulfide (PbSPbS).
129 Medium
129 of 173
Remembering +4 | -1
Highest priority functional group according to IUPAC is &ndash;
View answer and explanation
Correct option: B
−COCl- COCl
Option B (-COCl) is correct because acyl halides (-COCl) have higher priority than nitriles (-CN), aldehydes (-CHO), and amines (-NH₂) according to IUPAC nomenclature rules.
130 Medium
130 of 173
Analyzing +4 | -1
The isomeric ciscis-2-butene and transtrans-2-butene can be distinguished on the basis of :
View answer and explanation
Correct option: D
The products they give on addition to bromine
Option D (The products they give on addition to bromine) is correct because the stereochemistry of the addition of bromine to ciscis and transtrans isomers differs, leading to different products.
131 Medium
131 of 173
Analyzing +4 | -1
Tests on an aqueous solution of sodium salt having an anion Xn−\mathrm{X}^{\mathrm{n}-} gives the following results.

Anion Xn−⟶K2Cr2O7\mathrm{X}^{\mathrm{n}-} \stackrel{\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7}{\longrightarrow} Green solution

Anion Xn−⟶Pb(OAc)2\mathrm{X}^{\mathrm{n}-} \stackrel{\mathrm{Pb}(\mathrm{OAc})_2}{\longrightarrow} Black precipitate

Which of the following could be Xn−\mathrm{X}^{\mathrm{n}-} ?
View answer and explanation
Correct option: C
S2−\mathrm{S}^{2-}
Option C (S2−\mathrm{S}^{2-}) is correct because sulfide ions react with dichromate to form a green solution and with lead acetate to form a black precipitate.
132 Medium
132 of 173
Remembering +4 | -1
The change in optical rotation with time of freshly prepared solution of sugar is known as:
View answer and explanation
Correct option: B
Inversion
Option D (Mutarotation) is correct because mutarotation is the change in optical rotation observed when freshly prepared solutions of certain sugars are allowed to stand.
133 Medium
133 of 173
Understanding +4 | -1
In E2 elimination, some compounds follow Hofmann’s rule which means:
View answer and explanation
Correct option: D
The double bond goes mainly towards the least substituted carbon
Option D (The double bond goes mainly towards the least substituted carbon) is correct because Hofmann's rule states that in E2 elimination, the major product is the least substituted alkene.
134 Medium
134 of 173
Understanding +4 | -1
The number of optical enantiomorphs of tartaric acid :
View answer and explanation
Correct option: B
2
Option B (2) is the correct answer because tartaric acid has two chiral carbons, leading to three stereoisomers, but only two are optically active enantiomers.
135 Medium
135 of 173
Understanding +4 | -1
The isomers which can be converted into another forms by rotation of the molecules around single bond are:
View answer and explanation
Correct option: B
Conformers
Option B (Conformers) is correct because conformers are isomers that interconvert through rotation around single bonds.
136 Medium
136 of 173
Understanding +4 | -1
The best method for the separation of naphthalene and benzoic acid from their mixture is
View answer and explanation
Correct option: B
Crystallisation
Option B is correct because naphthalene and benzoic acid have different solubilities in suitable solvents at different temperatures, allowing separation by crystallization.
137 Medium
137 of 173
Understanding +4 | -1
Which of the following compounds will show metamerism?
View answer and explanation
Correct option: B
C2H5−S−C2H5{\mathrm{C}_2}{\mathrm{H}_5} - \mathrm{S} - {\mathrm{C}_2}{\mathrm{H}_5}
Option B (C2H5−S−C2H5{\mathrm{C}_2}{\mathrm{H}_5} - \mathrm{S} - {\mathrm{C}_2}{\mathrm{H}_5}) is correct because metamerism requires different alkyl groups attached to the same functional group, which is possible with sulfur.
138 Medium
138 of 173
Application +4 | -1
If a compound on analysis was found to contain C=18.5%,  H=1.55%,  Cl=55.04%\mathrm{C} = 18.5\mathrm{\% },\mathrm{\;H} = 1.55\mathrm{\% },\mathrm{\;Cl} = 55.04\mathrm{\% } and O=24.81%\mathrm{O} = 24.81\mathrm{\% } then its empirical formula is
View answer and explanation
Correct option: D
CHClO\mathrm{CHClO}
Option D (CHClO\mathrm{CHClO}) is correct because the mole ratio of C:H:Cl:O is approximately 1:1:1:1, which corresponds to the empirical formula CHClO\mathrm{CHClO}.
139 Medium
139 of 173
Application +4 | -1
The number of optical isomers of 2, 3-dihydroxy butanal is
View answer and explanation
Correct option: B
4
Option B (4) is correct because 2,3-dihydroxybutanal has two chiral centers, and since it's not a meso compound, the number of stereoisomers is 2^n = 22 = 4.
140 Medium
140 of 173
Understanding +4 | -1
A strong base can abstract an α-hydrogen from :
View answer and explanation
Correct option: B
Ketone
Option B (Ketone) is correct because ketones have alpha hydrogens that are relatively acidic due to the adjacent carbonyl group, making them susceptible to abstraction by a strong base.
141 Medium
141 of 173
Analyzing +4 | -1
Among the following the most stable compound is:
View answer and explanation
Correct option: D
trans-1,3-cyclohexanediol
Option D is correct because trans-1,3-cyclohexanediol is the most stable due to the hydroxyl groups being in equatorial positions, minimizing steric hindrance.
142 Medium
142 of 173
Remembering +4 | -1
The weakest +M group of the given species is:
View answer and explanation
Correct option: C
−I-\text{I}
Option C (−I-\text{I}) is correct because Iodine is the least electron-donating group among the given options due to its lower electronegativity compared to the others.
143 Medium
143 of 173
Understanding +4 | -1
Rotational angle requires to get maximum stable conformer from minimum stable conformer in nn-butane is
View answer and explanation
Correct option: B
1800{180^0}
Option B (1800{180^0}) is correct because the most stable conformer (anti) and the least stable conformer (eclipsed) in n-butane are 180 degrees apart in terms of rotational angle.
144 Medium
144 of 173
Remembering +4 | -1
The permanent displacement of electron through a chain involving only σ\sigma-bonds is called
View answer and explanation
Correct option: A
inductive effect
Option A (inductive effect) is correct because the permanent displacement of electrons through a chain involving only sigma bonds is the definition of the inductive effect.
145 Medium
145 of 173
Analyzing +4 | -1
Which of the following organic compounds has same hybridization as its combustion product (CO2)?\left(\mathrm{CO}_2\right)?
View answer and explanation
Correct option: B
Ethyne
Option B (Ethyne) is correct because ethyne (C₂H₂) has sp hybridization, and its combustion product, CO₂, also has sp hybridization.
146 Medium
146 of 173
Analyzing +4 | -1
In which of the following p−p -electrons of the halogens are not involved in delocalisation?
View answer and explanation
Correct option: C
Allyl chloride
Option C (Allyl chloride) is correct because the p-electrons of the halogen are not directly conjugated with the pi system of the double bond. The chlorine is attached to an sp3 carbon adjacent to the double bond.
147 Medium
147 of 173
Application +4 | -1
Which one of the following will show optical isomerism?
View answer and explanation
Correct option: B
H3C−C∣∣OHH−CO2H{\mathrm{H}_3}\mathrm{C} - \mathop {\mathop \mathrm{C}\limits_|^| }\limits_{OH}^\mathrm{H} - \mathrm{C}{\mathrm{O}_2}\mathrm{H}
Option B (H3C - CH(OH) - CO2H) is correct because the central carbon atom is bonded to four different groups (H, CH3, OH, and CO2H), making it a chiral center.
148 Medium
148 of 173
Analyzing +4 | -1
How many isomers of C5H11OH{C_5}{H_{11}}OH will be primary alcohols?
View answer and explanation
Correct option: B
4
Option A (4) is correct because there are four isomers of C5H11OHC_5H_{11}OH that are primary alcohols.
149 Medium
149 of 173
Remembering +4 | -1
Which is most commonly used to dry organic liquids?
View answer and explanation
Correct option: B
Sodium
The correct answer is Sodium because it reacts with water present in the organic liquid to form hydrogen gas and sodium hydroxide, effectively drying the liquid.
150 Medium
150 of 173
Understanding +4 | -1
Which of the following has maximum resonance energy?
View answer and explanation
Correct option: D
Phenanthrene
Option D (Phenanthrene) is correct because it has the most stable resonance structures among the given options due to its angular structure, which reduces steric hindrance and allows for better electron delocalization.
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