Basics of Organic Chemistry (GOC) — Page 6 | JEE Chemistry

Q51

Increasing order of stability among the three main conformations (i.e. Eclipse, Anti, Gauche) of 2-fluoroethanol is

A Eclipse, Gauche, Anti

B Gauche, Eclipse, Anti

C Eclipse, Anti, Gauche

D Anti, Gauche, Eclipse

Correct Answer

Option C

Solution

Due to hydrogen bonding between

H

&

F

gauche conformation is most stable hence the correct order is Eclipse, Anti, Gauche

Q52

CH3Br + Nu-

\to

CH3 - Nu + Br- The decreasing order of the rate of the above reaction with nucleophiles (Nu–) A to D is [Nu– = (A) PhO–, (B) AcO–, (C) HO–, (D) CH3O–]

A A > B > C > D

B B > D > C > A

C D > C > A > B

D D > C > B > A

Correct Answer

Option C

Solution

TIPS/Formulae : The stronger the acid, the weaker the conjugate base formed. The acid character follows the order :

C{H_3}COOH > {C_6}{H_5}OH > {H_2}O > C{H_3}OH

The basic character will follow the order

C{H_3}CO{O^ - } < {C_6}{H_5}{O^ - } < {O^ - }H < C{H_3}{O^ - }

Q53

Presence of a nitro group in a benzene ring

A activates the ring towards electrophilic substitution

B renders the ring basic

C deactivates the ring towards nucleophilic substitution

D deactivates the ring towards electrophilic substitution

Correct Answer

Option D

Solution

Nitro group is electron withdrawing group, so it deactivities the ring towards electrophilic substitution.

Q54

The number of stereoisomers possible for a compound of the molecular formula CH3 - CH = CH - CH(OH) - Me is :

A 2

B 4

C 6

D 3

Correct Answer

Option B

Solution

exhibits both geometrical as well as optical isomerism.

cis\,\,

-

R

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,cis\,\,

S

trans\,\,

-

R

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,trans\,\,

-

S

Q55

The correct order of increasing basicity of the given conjugate bases (R = CH3) is

A

RCO\overline O

< HC =

\overline C

<

\overline R

<

\overline NH_2

B

\overline R

< HC =

\overline C

<

RCO\overline O

<

\overline NH_2

C

RCO\overline O

<

\overline NH_2

< HC =

\overline C

<

\overline R

D

RCO\overline O

< HC =

\overline C

<

\overline NH_2

<

\overline R

Correct Answer

Option D

Solution

The correct order of basicity is

Q56

Out of the following, the alkene that exhibits optical isomerism is

A 3–methyl–2–pentene

B 4–methyl–1–pentene

C 3–methyl–1–pentene

D 2–methyl–2–pentene

Correct Answer

Option C

Solution

For a compound to show optical isomerism, presence of chiral carbon atom is a necessary condition.

Q57

For the estimation of nitrogen, 1.4 g of organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M/10 sulphuric acid. The unreacted acid required 20 ml of M/10 sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is:

A 3%

B 5%

C 6%

D 10%

Correct Answer

Option D

Solution

\% \,\,

of

\,\,N = {{1.4\,\, \times \,\,meq.of\,\,acid} \over {mass\,\,of\,\,organic\,\,compound}}

meq. of

\,\,{H_2}S{O_4} = 60 \times {M \over {10}} \times 2 = 12

meq. of

\,\,NaOH = 20 \times {M \over {10}} = 2

$\therefore$ meq. of acid consumed

= 12 - 2 = 10

$\therefore$

\,\,\,\% \,\,\,

N = {{1.4 \times 10} \over {1.4}} = 10\%

Q58

Which of the following compounds will exhibit geometrical isomerism?

A 3-Phenyl-1-butene

B 2-Phenyl-1-butene

C 1,1-Diphenyl-1-propane

D 1-Phenyl-2-butene

Correct Answer

Option D

Solution

1

- Phenyl-

2

-butene the two groups around each of the doubly bonded carbon Because, all are different. This compound can show

cis

-and

trans

- isomerism.

Q59

In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is: (at. Mass Ag = 108; Br = 80)

A 48

B 60

C 36

D 24

Correct Answer

Option D

Solution

Mass of substance

=250

mg

=0.250

g

Mass of

AgBr

=141

mg=0.141

g

1

mole of

AgBr

=1

g

atom of

Br

188

g

of

AgBr

=80

g

of

Br

188

g

of

AgBr

contain bromine

=80

g

0.141

g

of

AgBr

contain bromine

= {{80} \over {188}} \times 0.141

This much amount of bromine present in

0.250

g

of organic compound $\therefore$

\%

of bromine

= {{80} \over {188}} \times {{0.414} \over {0.250}} \times 100 = 24\%

Q60

The distillation technique most suited for separating glycerol from spent-lye in the soap industry is :

A Simple distillation

B Fractional distillation

C Steam distillation

D Distillation under reduced pressure

Correct Answer

Option D

Solution

The best technique for separating glycerol from spent-lye in the soap industry is Distillation under Reduced Pressure, also known as Vacuum Distillation (Option D).

The reason for this is that glycerol has a high boiling point (~

{290^ \circ }

C

).

If we were to use simple or fractional distillation, which involves boiling at standard atmospheric pressure, the high temperature could degrade or decompose the glycerol.

In contrast, vacuum distillation allows the separation to occur at a much lower temperature ( around

{180^ \circ }C

) by reducing the pressure.

This helps to prevent degradation of the glycerol, making it a better method for this particular separation.

Steam distillation isn't suitable here, as it's typically used for temperature-sensitive, volatile substances, which isn't the case for glycerol.