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

Organic chemistry-some basic principles and techniques practice · Medium

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Showing 1-30 of 195 questions
1 Medium
1 of 195
Analyzing +4 | -1
Keto-enol tautomerism is not observe in
View answer and explanation
Correct option: A
C6H5−C∣∣O−H{C_6}{H_5} - \mathop C\limits_{\mathop {||}\limits_O } - H
Option A (C6H5−C∣∣O−H{C_6}{H_5} - \mathop C\limits_{\mathop {||}\limits_O } - H) is correct because it lacks an alpha-hydrogen, which is necessary for keto-enol tautomerism.
2 Medium
2 of 195
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 4 is correct because there are four primary alcohols that are isomers of C5H11OH.
3 Medium
3 of 195
Understanding +4 | -1
Assertion (A): The compounds that have the same constitution and sequence of covalent bonds but differ in relative positions of their atoms or groups in space are called stereoisomers.

Reason (R): Stereoisomerism is a special type of isomerism which can be classified as geometrical and optical isomerism
View answer and explanation
Correct option: B
Both A\mathrm{A} and R\mathrm{R} are correct; R\mathrm{R} is not the correct explanation of A\mathrm{A}.
Option B (Both A and R are correct; R is not the correct explanation of A) is correct because both the definition of stereoisomers and the classification of stereoisomerism are correct statements, but the classification doesn't directly explain the definition.
4 Medium
4 of 195
Remembering +4 | -1
Which of the following is used as an adsorbent in adsorption chromatography ?
View answer and explanation
Correct option: D
Both (1) and (2)
Option D (Both (1) and (2)) is correct because both silica gel and alumina are commonly used as adsorbents in adsorption chromatography.
5 Medium
5 of 195
Analyzing +4 | -1
Base strength of,
H3C⊖CH2{H_3}\mathop {\mathrm{ }C}\limits^ \ominus C{H_2}
H2C=C⊖H{H_2}C = \mathop {\mathrm{ }C}\limits^ \ominus H
H−C=C⊖H - C = \mathop C\limits^ \ominus is in the order of
View answer and explanation
Correct option: C
(1) > (2) > (3)
Option A is correct because sulfite ions react with sulfuric acid to produce sulfur dioxide, which has an irritating smell and can reduce dichromate to a green solution of chromium(III) sulfate.
6 Medium
6 of 195
Analyzing +4 | -1
Indicate the wrongly named compound.
View answer and explanation
Correct option: D
CH3CH=CH−CH2−CO∣∣−CH3(3−hexen−2−one)\mathop {C{H_3}CH = CH - C{H_2} - \mathop C\limits^{\begin{array}{cccccccccccccccccccc} O \\ {||} \end{array}} - C{H_3}}\limits_{\left( {3 - {\text{hexen}} - 2 - {\text{one}}} \right)}
Option D (CH3CH=CH−CH2−CO∣∣−CH3(3−hexen−2−one)\mathop {C{H_3}CH = CH - C{H_2} - \mathop C\limits^{\begin{array}{cccccccccccccccccccc} O \\ {||} \end{array}} - C{H_3}}\limits_{\left( {3 - {\text{hexen}} - 2 - {\text{one}}} \right)}) is wrongly named because the correct IUPAC name should be hex-4-en-2-one, not 3-hexen-2-one.
7 Medium
7 of 195
Analyzing +4 | -1
Assertion: Pyridine is more basic than Pyrrole.

Reason: In Pyrrole, nitrogen has sp2sp^2 hybridisation
and the lone pair of nitrogen is not available for
donation.
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and
Reason is the correct explanation of
Assertion
Option A is correct. Pyridine is more basic than pyrrole because the lone pair of electrons on nitrogen in pyridine is not involved in resonance and is readily available for donation, while in pyrrole, the lone pair is part of the aromatic system and less available.
8 Medium
8 of 195
Application +4 | -1
In Kjeldahl’s method, ammonia from 5g of food neutralizes 30 cm3c{m^3} of 0.1 N acid. The percentage of nitrogen in the food is
View answer and explanation
Correct option: A
0.84
Option A (0.84) is correct because the percentage of nitrogen is calculated using the formula involving the normality of the acid, its volume, and the weight of the food sample.
9 Medium
9 of 195
Remembering +4 | -1
Aniline is purified by:
View answer and explanation
Correct option: A
Steam distillation
Option A (Steam distillation) is correct because aniline is steam volatile and can be separated from non-volatile impurities by steam distillation.
10 Medium
10 of 195
Understanding +4 | -1
The presence of carbon in an organic compound can be shown by
View answer and explanation
Correct option: C
Heating it with copper oxide to convert it into CO2\mathrm{C}{\mathrm{O}_2}
Option C (Heating it with copper oxide to convert it into CO2\mathrm{C}{\mathrm{O}_2}) is correct because the carbon in the organic compound reacts with copper oxide to form carbon dioxide, which can be detected.
11 Medium
11 of 195
Analyzing +4 | -1
Among the following the strongest acid is;
View answer and explanation
Correct option: B
CH3CH(NO2)COOH\mathrm{CH}_3 \mathrm{CH}\left(\mathrm{NO}_2\right) \mathrm{COOH}
Option B (CH3CH(NO2)COOH\mathrm{CH}_3 \mathrm{CH}(\mathrm{NO}_2) \mathrm{COOH}) is the strongest acid because the electron-withdrawing nitro group has the strongest inductive effect, stabilizing the conjugate base.
12 Medium
12 of 195
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 TLC is well-suited for separating small quantities of non-volatile mixtures like p-nitrophenol and picric acid.
13 Medium
13 of 195
Understanding +4 | -1
Most stable carbanion is
View answer and explanation
Correct option: C
HC≡C⊖{HC} \equiv \stackrel{\ominus}{{C}}
Option C (HC≡C⊖{HC} \equiv \stackrel{\ominus}{{C}}) is correct because the stability of a carbanion increases with the s-character of the carbon bearing the negative charge.
14 Medium
14 of 195
Understanding +4 | -1
Consider the following statements :

I. The presence of functional group disables systematisation of organic compound into different classes.

II. In polyfunctional compounds, one of the functional group is chosen as the principal functional group.

III. The order of priority for functional group is

−COCl>−COOR>−SO3H>−COOH-\mathrm{COCl} > -\mathrm{COO}R > -\mathrm{SO}_3 \mathrm{H} > -\mathrm{COOH}

IV. CH2(OH)CH2(OH)\mathrm{CH}_2(\mathrm{OH}) \mathrm{CH}_2(\mathrm{OH}) is named as ethane-1, 2-diol.
Read the given statements and choose the correct option.
View answer and explanation
Correct option: B
Both II and IV
Option B (Both II and IV) is correct because statements II and IV are accurate descriptions of organic compound nomenclature.
15 Medium
15 of 195
Analyzing +4 | -1
The correct order of decreasing acid strength
of following acids is

(I) Trichloro acetic acid

(II) Trifluoro acetic acid

(III) Acetic acid

(IV) Formic acid
View answer and explanation
Correct option: C
II >> I >> IV >> III
Option C is correct. The correct order of decreasing acid strength is Trifluoroacetic acid > Trichloroacetic acid > Formic acid > Acetic acid, based on the -I effect of the halogens.

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16 Medium
16 of 195
Understanding +4 | -1
Which of the following process is used to separate volatile liquid from non-volatile impurities?
View answer and explanation
Correct option: A
Distillation
Option A (Distillation) is correct because distillation separates liquids based on differences in boiling points, making it suitable for separating volatile liquids from non-volatile impurities.
17 Medium
17 of 195
Understanding +4 | -1
Which of the following element(s) is/are confirmed by appearance of blood red colour with FeCl3\mathrm{FeCl}_3 in Lassaigne's test?
View answer and explanation
Correct option: B
Presence of N & S\mathrm{N} \,\&\, \mathrm{S}
Option B is correct because the appearance of a blood-red color with FeCl3 in Lassaigne's test indicates the presence of both nitrogen and sulfur, forming [Fe(SCN)]2+.
18 Medium
18 of 195
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 nucleophilic and can undergo rearrangements, combine with cations, and add to carbonyl groups.
19 Medium
19 of 195
Understanding +4 | -1
Which one of the following is the stablest structure of cyclohexatriene?
View answer and explanation
Correct option: D
Planar form
Option D (Planar form) is correct because cyclohexatriene, also known as benzene, has sp² hybridization for all carbon atoms, resulting in a planar structure due to resonance and stability.
20 Medium
20 of 195
Understanding +4 | -1
Boiling point of a compound does not depend on:
View answer and explanation
Correct option: B
Solubility of the compound in water
Option B (Solubility of the compound in water) is correct because boiling point depends on intermolecular forces, size, and polarity, but not directly on water solubility.
21 Medium
21 of 195
Remembering +4 | -1
Coal-tar is obtained as by product during:
View answer and explanation
Correct option: B
Destructive distillation of coal
Option B (Destructive distillation of coal) is correct because coal-tar is a byproduct obtained during the destructive distillation of coal.
22 Medium
22 of 195
Remembering +4 | -1
When HgCl2{HgCl}_2 reacts with CO32−{CO}_3^{2-} anion, then:
View answer and explanation
Correct option: A
Reddish brown precipitate of HgCO3.2HgO{HgCO}_3 .2 {HgO} is formed
Option A (Reddish brown precipitate of HgCO3.2HgO{HgCO}_3 .2 {HgO} is formed) is correct because when HgCl2{HgCl}_2 reacts with CO32−{CO}_3^{2-} anion, a reddish-brown precipitate of HgCO3.2HgO{HgCO}_3 .2 {HgO} is formed.
23 Medium
23 of 195
Understanding +4 | -1
Aniline is weaker base than ethylamine. This is due to
View answer and explanation
Correct option: D
+R+\mathrm{R} effect of NH2\mathrm{NH}_2 in aniline
Option D (+R+\mathrm{R} effect of NH2\mathrm{NH}_2 in aniline) is correct because the lone pair on the nitrogen in aniline is delocalized into the benzene ring through resonance, reducing its availability for protonation.
24 Medium
24 of 195
Analyzing +4 | -1
The correct statements about the compounds a,b\mathrm{a}, \mathrm{b} and c\mathrm{c} is/are -
View answer and explanation
Correct option: D
a\mathrm{a} and b\mathrm{b} are enantiomers
Option D (a and b are enantiomers) is correct because compounds a and b are non-superimposable mirror images of each other.
25 Medium
25 of 195
Analyzing +4 | -1
When CH3COOCH3+HCl{CH}_3 {COOCH}_3+{HCl} is titrated with NaOH{NaOH} then at netural point the colour of phenolphthalein becomes colourless from pink
View answer and explanation
Correct option: B
Due to formation of CH3COOH{CH}_3 {COOH} which acts as a weak acid.
The correct answer is B (Due to formation of CH₃COOH which acts as a weak acid) because the hydrolysis of methyl acetate produces acetic acid, which neutralizes the NaOH, causing the phenolphthalein to become colorless at the endpoint.
26 Medium
26 of 195
Analyzing +4 | -1
Which of the following has maximum PKa{P^{{K_a}}}
View answer and explanation
Correct option: C
CH3COOHC{H_3}COOH
Option C (CH3COOHC{H_3}COOH) is correct because it is the weakest acid among the given options, and pKapK_a is inversely proportional to acid strength.
27 Medium
27 of 195
Analyzing +4 | -1
Compound with molecular formula C3H6O{C_3}{H_6}O can show
View answer and explanation
Correct option: D
functional group isomerism
Option D (functional group isomerism) is correct because C3H6O{C_3}{H_6}O can exist as an aldehyde or a ketone, which are different functional groups.
28 Medium
28 of 195
Analyzing +4 | -1
An organic compound (molar mass = 60) is found to containC= 20%,  H= 6.67% and N=  46.67%.C = {\text{ }}20\% ,\,\,H = {\text{ }}6.67\% {\text{ and}}\,N = \;46.67\% . While rest is oxygen. On heating it gives NH3N{H_3} along with a solid residue. The solid residue gives violet with alkaline CuSO4.CuS{O_4}. The organic compound is:
View answer and explanation
Correct option: C
NH2CONH2N{H_2}CON{H_2}
Option C (NH2CONH2N{H_2}CON{H_2}) is correct because the compound's composition, molar mass, and reaction products (NH3 and a biuret-positive residue) match urea.
29 Medium
29 of 195
Understanding +4 | -1
In which of the following compounds delocalisation is not possible
View answer and explanation
Correct option: A
1, 4-pentadiene
Option A (1, 4-pentadiene) is correct because the double bonds are not conjugated, preventing delocalization.
30 Medium
30 of 195
Analyzing +4 | -1
The correct decreasing order of acidity is represented by

(a) CH3−NO2\mathrm{CH}_3-\mathrm{NO}_2

(b) NO2−CH2−NO2\mathrm{NO}_2-\mathrm{CH}_2-\mathrm{NO}_2

(c) CH3−CH2−NO2\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{NO}_2

(d) NO2−CH∣NO2−NO2{\text{N}}{{\text{O}}_{\text{2}}} - \mathop {{\text{CH}}}\limits_{\begin{array}{cccccccccccccccccccc} | \\ {{\text{N}}{{\text{O}}_{\text{2}}}} \end{array}} - {\text{N}}{{\text{O}}_{\text{2}}}
View answer and explanation
Correct option: A
(d) >> (b) >> (a) >> ( c)
Option A ((d) >> (b) >> (a) >> ( c)) is correct because the acidity increases with the number of electron-withdrawing nitro groups attached to the carbon chain.
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  • ✓ This set contains 195 single-correct-answer MCQs on ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES.
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