Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

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Q.1. A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO3 to give fraction A. The left-over organic phase was extracted with dilute NaOH solution to give fraction B. The final organic layer was labeled as fraction C. Fractions A, B and C, contain respectively    (2020)
(1) m-chlorobenzoic acid, m-chloroaniline and m-chlorophenol.
(2) m-chlorobenzoic acid, m-chlorophenol and m-chloroaniline.
(3) m-chlorophenol, m-chlorobenzoic acid and m-chloroaniline.
(4) m-chlorobenzoic acid and m-chlorophenol.
Ans.
(2)
For the given extraction of the given organic compounds the fractions left over are:

Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

Fraction A: m-Chlorobenzoic acid
Fraction B: m-Chlorophenol
Fraction C: m-Chloroaniline

Q.2. A chromatography column, packed with silica gel as stationary phase, was used to separate a mixture of compounds consisting of (I) benzanilide (II) aniline and (III) acetophenone. When the column is eluted with a mixture of solvents, hexane: ethyl acetate (20:80), the sequence of obtained compounds is    (2020)
(1) (II), (III) and (I)
(2) (II), (I) and (III)
(3) (III), (I) and (II)
(4) (I), (II)  and (III)
Ans.
(3)
The separation of the substances in chromatography column is depend upon the polarity of substance. In polar solvent hexane : ethyl acetate (20:80), more polar compound will react faster with solvent and separate first.
Compound (I): Benzanilide

Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

Compound (II): Aniline

Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

The lone pairs of electrons delocalized into the ring, thus, the polarity will decrease.
Compound (III): Acetophenone

Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

The vectors of dipole moments get add so the dipole moment increases.
Hence, compound (III) will be separated out first followed by compound (I) than compound (II).

Q.3. A flask contains a mixture of isohexane and 3-methylpentane. One of the liquids boils at 63°C while the other boils at 60°C. What is the best way to separate the two liquids and which one will be distilled out first?    (2020)
(1) Fractional distillation, isohexane
(2) Simple distillation, 3-methylpentane
(3) Simple distillation, isohexane
(4) Fractional distillation, 3-methylpentane
Ans.
(1)
Liquids with difference in boiling point less than 5 K are separated by fractional distillation. Iso-hexane have lower boiling point than 3-methylpentane, thus, iso-hexane will be distilled out first.

Q.4. Kjeldahl’s method cannot be used to estimate nitrogen for which of the following compounds?    (2020)
(1) C6H5NH2
(2) CH3CH2 - C ≡ N
(3) C6H5NO2
(4) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Ans.
(3)
In Kjeldahl’s method, the nitrogen is estimated as NH3. But on reaction with dil. H2SO4, nitrobenzene does not give ammonia, so nitrogen in nitrobenzene cannot be estimated by Kjeldahl’s method.

Q.5. A chemist has 4 samples of artificial sweetener A, B, C and D. To identify these samples, he performed certain experiments and noted the following observations:
(I) A and D both form blue-violet color with ninhydrin.
(II) Lassaigne extract of C gives positive AgNO3 test and negative Fe4[Fe(CN)6]3 test.
(III) Lassaigne extract of B and D gives positive sodium nitroprusside test.
Based on these observations which option is correct?    (2020)
(1) A: Aspartame; B: Saccharin: C: Sucralose; D: Alitame
(2) A: Alitame; B: Saccharin; C: Aspartame; D: Sucralose
(3) A: Saccharin; B: Alitame; C: Sucralose; D: Aspartame
(4) A: Aspartame; B: Alitame; C: Saccharin; D: Sucralose
Ans.
(1)
The ninhydrin test is given by amino acids; thus, the artificial sugar A and D are amino acids. Lassaingne test of C give positive test for AgNO3, that is, Cl is present and negative test for ferri ferrocyanide, that is, absence of nitrogen in it.
Lassaingne test of B and D give positive test sodium nitroprusside, that is, sulphur is present.
From the above observations, we conclude that,
A: Aspartame
  Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
B: Saccharin
 Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
C: Sucralose
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
D: Alitame
 Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

Q.6. The increasing order of basicity for the following intermediates is (from weak to strong)    (2020)
(i) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(ii) CH2 = CH - CH2-
(iii) CH ≡ C-
(iv) CH3-
(v) CN-
(1) (III) < (I) < (II) < (IV) < (V)
(2) (V) < (I) < (IV) < (II) < (III)
(3) (V) < (III) < (II) < (IV) < (I)
(4) (III) < (IV) < (II) < (I) < (V)
Ans. 
(3)
The basicity of the given carbanions depends upon the stability of the carbanion and also it is inversely proportional to the electronegativity of the carbon bearing the negative charge.
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Thus, (I) is the most basic carbanion due to hyperconjugation effect followed by methyl anion (IV) and allylic carbanion (II). Out of acetylide anion (III) and cyanide anion (V), due to the electronegative nitrogen atom in cyanide anion, it is less basic than acetylide anion (III). Hence, the correct order of basicity is: (V) < (III) < (II) < (IV) < (I).

Q.7. Which of the following has the shortest C – Cl bond?    (2020)
(1) Cl – CH = CH2
(2) Cl – CH = CH – NO2
(3) Cl – CH = CH – CH3
(4) Cl – CH = CH – OCH3
Ans.
(2)
Due to the presence of nitro group, the lone pair of electrons on chlorine is involved in long resonance, thus, the bond between C–Cl becomes partial double bond and bond length become shortest.
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev 

Q.8. The correct decreasing order for acidic strength is:    (2019)
(1) NO2CH2COOH > FCH2COOH > CNCH2COOH > CICH2COOH
(2) FCH2COOH > NCCH2COOH > NO2CH2COOH > CICH2COOH
(3) CNCH2COOH > O2NCH2COOH > FCH2COOH > CICH2COOH
(4) NO2CH2COOH > NCCH2COOH > FCH2COOH > CICH2COOH
Ans. (4)

The acidic strength of a compound or an acid depends on the inductive effect (-I). Higher the (-I) effect of a substituent higher will be acidic strength. Now, the decreasing order of (-I) effect of the given substituents is NO2 > CN > F > Cl.
∴ The correct decreasing order of acidic strength amongst the given carboxylic acids is:
O2NCH2COOH > NCCH2COOH > FCH2COOH > ClCH2COOH

Q.9. Which of the following compounds is not aromatic?    (2019)
(1) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(2) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(3) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(4) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Ans.
(1)
Compounds (2), (3) and (4) are containing 6πe- in complete conjugation and are aromatic.
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRevCompound (1) is anti-aromatic as it has 4πe- in complete conjugation and doesn’t obey Huckel rule.

Q.10. The increasing order of the pKa values of the following compounds is:    (2019)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) C < B < A < D
(2) B < C < D < A
(3) D < A< C < B
(4) B < C < A < D
Ans.
(4)
Electron withdrawing substituents increase the acidic strength while electron releasing groups decrease the acidic strength.
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
pKa : B < C < A < D

Q.11. What is the IUPAC name of the following compound?    (2019)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) 3-Bromo-1, 2-dimethylbut-1 -ene
(2) 3-Bromo-3-methyl-1, 2-dimethylprop-1-ene
(3) 2-Bromo-3-methylpent-3-ene
(4) 4-Bromo-3-methylpent-2-ene
Ans.
(4)
IUPAC name: 4-Bromo-3-methylpent-2-ene
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

Q.12. Which compound (s) out of the following is/are not aromatic?    (2019)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) (B), (C) and (D)
(2) (C) and (D)
(3) (B)
(4) (A) and (C)
Ans.
(1)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev is non aromatic due to the presence of sp3 carbon.

Q.13. In the following compound, Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev  the favourable site/s for protonation is/are:    (2019)
(1) (a) and (e)
(2) (b), (c) and (d)
(3) (a) and (d)
(4) (a)
Ans.
(b), (c) and (d) are the favourable sites for protonation because these are localised lone pair electron.
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

Q.14. The correct order for acidic strength of compounds CH ≡ CH, CH3-C ≡ CH and CH≡ CH2 is as follows:    (2019)
(1) CH ≡ CH > CH2 ≡ CH2 > CH- C ≡ CH
(2) CH- C ≡ CH > CH ≡ CH > CH2 ≡ CH2
(3) CH3 - C ≡ CH > CH2 ≡ CH> HC ≡ CH
(4) HC = CH > CH3 - C ≡ CH > CH2 ≡ CH2
Ans.
(4)
Order of acidic strength is
CH ≡ CH > CH3 - C ≡ CH > CH2 = CH2

Q.15. The IUPAC name of the following compound is:    (2019)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) 4, 4-Dimethyl-3-hydroxybutanoic acid
(2) 2-Methyl-3-hydroxypentan-5-oic acid
(3) 3-Hydroxy-4-methylpentanoic acid
(4) 4-Methyl-3-hydroxypentanoic acid
Ans.
(3)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
3-Hydroxy-4-methyl pentanoic acid

Q.16. Which of the following compounds will show the maximum 'enol' content?    (2019)
(1) CH3COCH2COOC2H5 
(2) CH3COCH2COCH3 
(3) CH3COCH3    
(4) CH3COCH2CONH2
Ans. 
(2)
Enol content ∝ Acidity of active methylene hydrogens.
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Maximum enol content

Q.17. The correct IUPAC name of the following compound is:    (2019)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) 5-chloro-4-methyl-1 -nitrobenzene
(2) 2-chloro-1 -methy 1-4-nitrobenzene
(3) 3-chloro-4-methyl-l-nitrobenzene
(4) 2-methyl-5-nitro-1 -chlorobenzene
Ans.
(2)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
2-Chloro-1 -methyl-4-nitrobenzene

Q.18. The increasing order of nucleophilicity of the following nucleophiles is:    (2019)
(a) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(b) H2O
(c) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(d) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) (a) < (d) < (c) < (b)
(2) (b) < (c) < (d) < (a)
(3) (d) < (a) < (c) < (b)
(4) (b) < (c) < (a) < (d)
Ans.
(4)
(i) Negatively charge molecule is more nucleophilic than neutral molecule.
(ii) Concentrated negatively charged molecule is more nucleophilic than delocalised negatively charge molecule.

Q.19. Which of these factors does not govern the stability of a conformation in acyclic compounds?    (2019)
(1) Steric interactions
(2) Torsional strain
(3) Electrostatic forces of interaction
(4) Angle strain
Ans.
(4)
Angle strain is present in cyclic compounds.

Q.20. The increasing order of the pKb of the following compounds is:    (2019)
(a)Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(b) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(c) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(d) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) (A) < (C) < (D) < (B)
(2) (C) < (A) < (D) < (B)
(3) (B) < (D) < (A) < (C)
(4) (B) < (D) < (C) < (A)
Ans.
(3)
Electron withdrawing group attached to benzene ring will reduce the basic strength and increase pKb, while electron donating group decreases pKb.
Correct increasing order of pKb.
(B) < (D) < (A) < (C)

Q.21. The decreasing order of electrical conductivity of the following aqueous solutions is:    (2019)
0.1 M Formic acid (A),
0.1 M Acetic acid (B),
0.1 M Benzoic acid (C).
(1) A > C > B
(2) C > B > A
(3) A > B > C
(4) C > A > B
Ans.
(1)
Order of acidic strength is HCOOH > C6H5COOH > CH3COOH
More the acidic strength more will be the dissociation of acid into ions and more will be the conductivity. Thus, order of conductivity will be,
HCOOH > C6H5COOH > CH3COOH
(A)    >    (C)    >    (B)

Q.22. The IUPAC name for the following compound is:    (2019)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) 3-methyl-4-(3 -methylprop-1-enyl)-1-heptyne
(2) 3,5-dimethyl-4-propylhept-6-en-1-yne
(3) 3-methyl-4-(1-methylprop-2-ynyl)-1-heptene
(4) 3,5-dimethyl-4-propylhept-1 -en-6-yne
Ans. 
(4)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
3, 5-dimethyl-4-propylhept-1-en-6-yne

Q.23. Number of stereo centers present in linear and cyclic structures of glucose are respectively:    (2019)
(1) 5 & 4
(2) 4 & 4
(3) 5 & 5
(4) 4 & 5
Ans.
(4)
Linear structure of glucose,
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Cyclic structure of glucose.
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Here, * represents stereocenters.

Q.24. In the following skew conformation of ethane, H' - C - C - H" dihedral angle is:    (2019)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) 58°
(2) 149°
(3) 151°
(4) 120°
Ans.
(2)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
∴ Angle between H' and H" = 120° + 29° = 149°

Q.25. The correct match between Item I and Item II is:    (2019)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) (A) → (Q); (B) → (P); (C) → (R)
(2) (A) → (R); (B) → (Q); (C) → (P)
(3) (A) → (Q); (B) → (R); (C) → (P)
(4) (A) → (P); (B) → (R); (C) → (Q)
Ans.
(2)
Benzaldehyde is an absorbate, alumina is an adsorbent (stationary phase) and acetonitrile is in mobile phase.

Q.26. If dichloromcthane (DCM) and water (H20) are used for differential extraction, which one of the following statements is correct?    (2019)
(1) DCM and H2O would stay as lower and upper layer respectively in the S.F.
(2) DCM and H2O will make turbid/colloidal mixture
(3) DCM and H2O would stay as upper and lower layer respectively in the separating funnel (S.F.)
(4) DCM and H2O will be miscible clearly
Ans.
(1)
Due to higher density of dichloromethane than water DCM would be the lower layer and water will form the upper layer in the separating funnel.

Q.27. The correct match between items I and II is :    (2019)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) (A) → (S); (B) → (R); (C) → (P)
(2) (A) → (Q); (B) → (R); (C) → (S)
(3) (A) → (R); (B) → (P); (C) → (S)
(4) (A) → (Q); (B) → (R); (C) → (P)
Ans.
(2)
H2O : Sugar-Recrystallisation
H2O : Aniline - Separation by steam distillation
H2O : Toluene - Differential extraction

Q.28. An organic compound is estimated through Dumas method and was found to evolve 6 moles of CO2, 4 moles of H2O and 1 mole of nitrogen gas. The formula of the compound is:    (2019)
(1) C12H8N
(2) C12H8N2
(3) C6H8N2
(4) C6H8N
Ans.
(3)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

Q.29. An organic compound ‘X’ showing the following solubility profile is:    (2019)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) o-Toluidine
(2) Oleic acid
(3) m-Cresol
(4) Benzamide
Ans. 
(3)
Phenols (e.g. m-cresol), being weak acid, are soluble in dil NaOH, but insoluble in NaHCO3.
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

Q.30. The principle of column chromatography is:    (2019)
(1) Gravitational force.
(2) Capillary action.
(3) Differential absorption of the substances on the solid phase.
(4) Differential adsorption of the substances on the solid phase.
Ans.
(4)
In column chromatograph; a solid adsorbent is packed on a column and a solution containing number of solute particles is allowed to flow down the column. The solute molecules get adsorbed on the surface of adsorbent and move through column at different rates based on differential adsorption of the substances on the solid phase.

Q.31. In chromatography, which of the following statement is INCORRECT for Rf?    (2019)
(1) Rf value depends on the type of chromatography.
(2) The value of Rf can not be more than one.
(3) Higher Rf value means higher adsorption.
(4) Rvalue is dependent on the mobile phase.
Ans. 
(3)
In chromatography, Rf represents retardation factor.
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
∴ Higher Rf value means lower adsorption.

Q.32. An organic compound ‘A’ is oxidized with Na2O2, followed by boiling with HNO3. The resultant solution is then treated with ammonium molybdate to yield a yellow precipitate.
Based on above observation, the element present in the given compound is:    (2019)
(1) Nitrogen 
(2) Phosphorus
(3) Fluorine
(4) Sulphur
Ans.
(2)
Phosphorus is detected in the form of yellow ppt of ammonium phosphate molybdate on reaction with ammonium molybdate.

Q.33. 25 g of an unknown hydrocarbon upon burning produces 88 g of CO2 and 9 g of H2O. This unknown hydrocarbon contains:    (2019)
(1) 20 g of carbon and 5 g of hydrogen
(2) 22 g of carbon and 3 g of hydrogen
(3) 24 g of carbon and 1 g of hydrogen
(4) 18 g of carbon and 7 g of hydrogen
Ans.
(3)
Let the hydrocarbon be CxHy.
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
x = 2 y = 1
∵ x = 2 and y = 1 the hydrocarbon will be C2H. 2 mol carbon contains 24 g and 1 mol hydrogen contains 1g.

Q.34. The ratio of mass present of C and H of an organic compound (CXHYOZ) is 6 : 1. If one molecule of the above compound (CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of compound CXHYOZ is:    (2018)
(1) C3H6O3 
(2) C2H4
(3) C3H4O2 
(4) C2H4O3 

Ans. (4)
Ratio of mass % of C and H in CxHyOz is 6 : 1.
Therefore,
Ratio of mole % of C and H in CxHyOz will be 1 : 2.
Therefore x : y = 1 : 2, which is possible in options 1, 2 and 3.
Now oxygen required to burn CxHy 
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Now z is half of oxygen atoms required to burn CxHy.
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Now putting values of x and y from the given options:
Option (1), x = 2, y = 4
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Therefore correct option is 1 (C2H4O3)

Q.35. Which of the following compounds will be suitable for Kjeldahl’s method for nitrogen estimation?    (2018)
(1) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(2) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(3) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(4) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Ans.
(3)
Kjeldahl’s method is not used in the case of nitro, azo compounds and also to the compounds containing nitrogen in the ring e.g. Pyridine.

Q.36. The correct match between List-I and List-II is:    (2018)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) (A)-(R), (B)-(P), (C)-(Q), (D)-(S) 
(2) (A)-(R), (B)-(S), (C)-(P), (D)-(Q) 
(3) (A)-(P), (B)-(S), (C)-(R), (D)-(Q) 
(4) (A)-(R), (B)-(P), (C)-(S), (D)-(Q) 
Ans. (1)
O – P ⇒ difference in Boiling point  ⇒ Steam distillation 
Coloured impurity → Chromatography 

Q.37. The IUPAC name of the following compound is:    (2018)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) 4, 4-diethyl-3-methylbut-2-ene 
(2) 4-ethyl-3-methylhex-2-ene 
(3) 3-ethyl-4-methylhex-4-ene 
(4) 4-methyl-3-ethylhex-4-ene 
Ans. (2)
Basic Nomenclature. 
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

Q.38. Which of the following molecules is least resonance stabilized?    (2017)
(1) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(2) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(3) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(4) Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Ans. (4)
However, all molecules given in options are stabilised by resonance but compound given in option (4) is least resonance stabilised (other three are aromatic)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

Q.39. Which of the following statements is not true about partition chromatography?    (2017)
(1) Stationary phase is a finely divided solid adsorbent
(2) Separation depends upon equilibration of solute between a mobile and a stationary phase
(3) Paper chromatography is an example of partition chromatography
(4) Mobile phase can be a gas
Ans. (4)

Q.40. The IUPAC name of the following compound is    (2017)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) 2-Ethyl-1, 1-dimethylcyclohexane
(2) 1, 1-Dimethyl 1-2-ethylcyclohexane
(3) 2, 2-Dimethyl-1-1-ethylcyclohexane
(4) 1-Ethyl-2,2-dimethylcyclohexane
Ans. (1)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

Q.41. In the following structure, the double bonds are marked as I, II, III and IV. Geometrical isomerism is not possible at site (s):    (2017)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) I and III
(2) III
(3) I
(4) III and IV
Ans.
(3)
Different group should be attached to each sp2 hybridised c-atom.

Q.42. Which of the following compounds will show highest dipole moment?    (2017)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) (II)
(2) (IV)
(3) (III)
(4) (I)
Ans.
(1)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
A dipolar resonances structure has aromatic character in the ring and would be expected to make a major contribution to the overall structure.

Q.43. The distillation technique most suited for separating glycerol from spent-lye in the soap industry is:    (2016)
(1) Simple distillation 
(2) Fractional distillation
(3) Steam distillation
(4) Distillation under reduced pressure
Ans.
(4)
Glycerol and spent-lye can be separated by distillation under reduced pressure.

Q.44. The absolute configuration of    (2016)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev is:  
(1) (2R, 3S)
(2) (2S, 3R)
(3) (2S, 3S)
(4) (2R, 3R)
Ans.
(2)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev

Q.45. The hydrocarbon with seven carbon atoms containing a neopentyl and a vinyl group is:    (2016)
(1) 2, 2-dimethyl - 4 - pentene
(2) 4, 4-dimethyl pentene
(3) Isopropyl-2-butene
(4) 2, 2-dimethyl-3-pentene
Ans.
(2)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev has seven carbon atoms containing a neopentyl and a vinyl group.

Q.46. An organic compound contains C, H and S. The minimum molecular weight of the compound containing 8% sulphur is:    (2016)
(atomic weight of S = 32 amu)
(1) 300 g mol–1 
(2) 400 g mol–1 
(3) 200 g mol–1 
(4) 600 g mol–1 
Ans.
(2)
8 g sulphur present in = 100 g of organic compound.
32 g sulphur present in = 100/8 x 32 = 400 g of organic compound.
Hence, minimum molecular weight of compound = 400 g/mol

Q.47. The "N" which does not contribute to the basicity for the compound is:    (2016)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
(1) N 7
(2) N 1
(3) N 9
(4) N 3
Ans.
(3)
Lone pair of N9 is involve in aromaticity.

Q.48. Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is:    (2016)
(1) 6.72 L
(2) 2.24 L
(3) 4.48 L
(4) 8.96 L
Ans.
(4)
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Total moles of gases Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev
Previous year Questions (2016-20) - Organic Chemistry — Some Basic Principles & Technique Notes | EduRev 

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