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Basics of Organic Chemistry (GOC)

JEE Chemistry · 87 questions · Page 5 of 9 · Click an option or "Show Solution" to reveal answer

Q41
Compound with molecular formula C3H6O can show :
A Positional isomerism
B Both positional isomerism and metamerism
C Metamerism
D Functional group isomerism
Correct Answer
Option D
Solution

C3H6O (degree of unsaturation = 1) Functional group isomerism.

Q42
The interaction between π\pi bond and lone pair of electrons present on an adjacent atom is responsible for
A Resonance effect
B Electromeric effect
C Hyperconjugation
D Inductive effect
Correct Answer
Option A
Solution

It is a type of conjugation responsible for resonance.

Q43
The correct order in aqueous medium of basic strength in case of methyl substituted amines is :
A NH3>Me3N>MeNH2>Me2NH\mathrm{NH_3 > Me_3N > MeNH_2 > Me_2NH}
B Me3N>Me2NH>MeNH2>NH3\mathrm{Me_3N > Me_2NH > MeNH_2 > NH_3}
C Me2NH>MeNH2>Me3N>NH3\mathrm{Me_2NH > MeNH_2 > Me_3N > NH_3}
D Me2NH>Me3N>MeNH2>NH3\mathrm{Me_2NH > Me_3N > MeNH_2 > NH_3}
Correct Answer
Option C
Solution

In aqueous medium basic strength is dependent on electron density on nitrogen as well as solvation of cation formed after accepting H+\mathrm{H}^{+}.

After considering all these factors overall basic strength order is

Me2NH>MeNH2>Me3N>NH3\mathrm{Me}_2 \mathrm{NH}>\mathrm{MeNH}_2>\mathrm{Me}_3 \mathrm{N}>\mathrm{NH}_3
Q44
Which of the following compound is not chiral?
A 1- chloropentane
B 3-chloro-2- methyl pentane
C 1-chloro -2- methyl pentane
D 2- chloropentane
Correct Answer
Option A
Solution
11

- chloropentane is not chiral while others are chiral in nature

Q45
Which one of the following techniques is not used to spot components of a mixture separated on thin layer chromatographic plate?
A I2 (Solid)
B U.V. Light
C Visualisation agent as a component of mobile phase
D Spraying of an appropriate reagent
Correct Answer
Option C
Solution

TLC is a technique used to separate mixture of compounds based on differences in polarity.

In TLC a glass plate coated with a stationary phase is spotted with the mixture to be separated.

Q46
Which of the following will have meso-isomer also?
A 2- chlorobutane
B 2- hydroxyopanoic acid
C 2,3 – dichloropentane
D 2-3- dichlorobutane
Correct Answer
Option D
Solution

NOTE : The compounds containing two similar assymmetric

CC

-atoms have plane of symmetry and exist in Meso form.

Q47
Consider the acidity of the carboxylic acids: (a) PhCOOH (b) o – NO2C6H4COOH (c) p – NO2C6H4COOH (d) m – NO2C6H4COOH Which of the following order is correct?
A (b) > (d) > (a) > (c)
B (b) > (d) > (c) > (a)
C (a) > (b) > (c) > (d)
D (b) > (c) > (d) > (a)
Correct Answer
Option D
Solution

In aromatic acids presence of electron withdrawing substituent e.g.

−NO2- N{O_2}

disperses the negative charge of the anion and stablises it and hence increases the acidity of the parent benzoic acid.

Further

OO

- isomer will have higher acidity than corresponding

mm

and

pp

isomers. Since nitro group at

pp

-position have more pronounced electron withdrawing than

−NO2- N{O_2}

group at

mm

-position hence the correct order is the one given above.

Q48
A similarity between optical and geometrical isomerism is that
A each forms equal number of isomers for a given compound
B if in a compound one is present then so is the other
C both are included in stereoisomerism
D they have no similarity
Correct Answer
Option C
Solution

Stereoisomerism involve those isomers which contain same ligands in their co-ordination spheres but differ in the arrangement of those ligands in space.

Stereo-isomerism is of two type geometrical isomerism ligands occupy different positions around the central metal atom or ion.

NOTE : In optical isomerism isomers have same formula but differ in their ability to rotate directions of the plane of polarized light.

Q49
Arrangement of (CH3)3C-, (CH3)2CH-, CH3-CH2- when attached to benzyl or an unsaturated group in increasing order of inductive effect is
A (CH3)3C- < (CH3)2CH- < CH3-CH2
B CH3-CH2- < (CH3)2CH- < (CH3)3C-
C (CH3)2CH- < (CH3)3C- < CH3-CH2
D (CH3)3C- < CH3-CH2- < CH3)2CH-
Correct Answer
Option B
Solution
−CH3- C{H_3}

group has

+I+I

effect, as number of

−CH3- C{H_3}

group increases, the inductive effect increases. Therefore the correct order is

CH3−CH2−<(CH3)2CH−<(CH3)3C−C{H_3} - C{H_2} - < {\left( {C{H_3}} \right)_2}CH - < {\left( {C{H_3}} \right)_3}C -
Q50
The increasing order of stability of the following free radicals is
A (C6H5)2C∙H{({C_6}{H_5})_2}\mathop C\limits^ \bullet H < (C6H5)3C∙{({C_6}{H_5})_3}\mathop C\limits^ \bullet < (CH3)3C∙{(C{H_3})_3}\mathop C\limits^ \bullet < (CH3)2C∙H{(C{H_3})_2}\mathop C\limits^ \bullet H
B (CH3)2C∙H{(C{H_3})_2}\mathop C\limits^ \bullet H < (CH3)3C∙{(C{H_3})_3}\mathop C\limits^ \bullet < (C6H5)2C∙H{({C_6}{H_5})_2}\mathop C\limits^ \bullet H < (C6H5)3C∙{({C_6}{H_5})_3}\mathop C\limits^ \bullet
C (CH3)2C∙H{(C{H_3})_2}\mathop C\limits^ \bullet H < (CH3)3C∙{(C{H_3})_3}\mathop C\limits^ \bullet < (C6H5)3C∙{({C_6}{H_5})_3}\mathop C\limits^ \bullet < (C6H5)2C∙H{({C_6}{H_5})_2}\mathop C\limits^ \bullet H
D (C6H5)3C∙{({C_6}{H_5})_3}\mathop C\limits^ \bullet < (C6H5)2C∙H{({C_6}{H_5})_2}\mathop C\limits^ \bullet H < (CH3)3C∙{(C{H_3})_3}\mathop C\limits^ \bullet < (CH3)2C∙H{(C{H_3})_2}\mathop C\limits^ \bullet H
Correct Answer
Option B
Solution

The order of stability of free radicals

(C6H5)3 C∙>(C6H5)2C∙H>(CH3)3C∙>(CH3)2C∙H{\left( {{C_6}{H_5}} \right)_3}\,\mathop C\limits^ \bullet > {\left( {{C_6}{H_5}} \right)_2}\mathop C\limits^ \bullet H > {\left( {C{H_3}} \right)_3}\mathop C\limits^ \bullet > {\left( {C{H_3}} \right)_2}\mathop C\limits^ \bullet H

The stabilisation of first two is due to resonance and last two is due to inductive effect.

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