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In the following structure, the double bonds are marked as I, II, III and IV Geometrical isomerism is not possible at site (s) :
Option: 1  III
Option: 2  I
Option: 3  I and II
Option: 4  III and IV  
 

Geometrical isomerism is not possible at Site I as two identical methyl groups are attached to the same carbon bearing the double bond.

Hence, the answer is Option (2)

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Posted by

vishal kumar

The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight (in kg) which a 75 kg person would gain if all ^{1}H atoms are replaced by ^{2}H atoms is :  
Option: 1 7.5
Option: 2 10
Option: 3 15
Option: 4 37.5
 

Given that

Mass of the person = 75 kg

Mass of 1H1 present in person = 10% of 75 kg = 7.5 kg

Since Mass of 1H2 is double the Mass of 1H1

So, Mass of 1H2 will be in person = 2 X 7.5 kg =15 kg

Thus, increase in weight = 15 - 7.5 = 7.5 kg

Therefore, Option (1) is correct

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Posted by

vishal kumar

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Which of the following compounds will show the maximum 'enol' content?
Option: 1 CH_{3}COCH_{2}COOC_{2}H_{5}
Option: 2 CH_{3}COCH_{2}COCH_{3}
Option: 3 CH_{3}COCH_{3}
Option: 4 CH_{3}COCH_{2}CONH_{2}
 

Difinition of Tautomerism -

Tautomers are isomers of a compound that differ only in the position of the protons and electrons. The carbon skeleton of the compound is unchanged. A reaction that involves simple proton transfer in an intramolecular fashion is called tautomerism.

The stability of enol depends on the factors

(1) Resonance

(2) Hydrogen-bonding

(3) Hyperconjugation

(4) Hydrogen which is removed from \alpha- carbon should be acidic in Nature for enol formation.

CH2 is present between 2- electron-withdrawing group,  It is having acid 'H'

                        

Over the resonance effect decreases because of cross conjugation. So the stability of the enol form decreases.

The two -dicarbonyl compounds have a higher enol content than the two mono carbonyl compounds because hydrogen bonding and conjugation stabilize their enols. The enol content in C (a mono aldehyde) is higher than D because of the reasons outlined above. 

Therefore, option (2) is correct.

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Posted by

Ritika Jonwal

Among the compounds, A and B with molecular formula C_{9}H_{18}O_{3}, A is having higher boiling point than B. The possible structures of A and B are :
Option: 1
Option: 2
Option: 3
Option: 4
 

In (A), extensive inter-molecular H-bonding is possible while in (B) there is no Inter-molecular H-bonding.

Option 4 has this type of arrangement that follows the above conditions.

Therefore, Option(4) is correct.

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Posted by

Kuldeep Maurya

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The increasing order of  \mathrm{pK_{b }} values of the following compounds is :
Option: 1 II<IV<III<I 
Option: 2 II<I<III<IV   
Option: 3I<II<III<IV     
Option: 4 I<II<IV<III
 

 

III contains -M and -I group, hence it is the least basic.

IV contains -I group

I contains +M group and is more basic than II.

Thus order of strength: I > II > IV > III.

Order of \mathrm{pK_{b}} : I < II < IV < III.

Therefore, Option(4) is correct.

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Posted by

Kuldeep Maurya

The correct order of statbility for the following alkoxides is:
Option: 1 C > A > B
Option: 2 C > B > A
Option: 3 B > A > C
Option: 4 B > C > A
 

When a negative charge is delocalised with an electron-withdrawing group like (NO2) then stability increases.

(A) The negative charge is localised

(B) The negative charge is delocalised with the carbon of the alkene

(C) Negative charge is delocalised with NO2 group

So, the order will be

 C > B > A

Therefore, Option(2) is correct.

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Posted by

Ritika Jonwal

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The ammonia (NH_{3}) released on quantitative reaction of 0.6g, urea (NH_{2}CONH_{2}) with sodium hydroxide (NaOH) can be neutralised by:
Option: 1 200 ml of 0.2 N HCl
Option: 2200 ml of 0.4 N HCl
Option: 3100 ml of 0.1N HCl
Option: 4100 ml of 0.2N HCl
 

2 × mole of Urea = mole of NH_{3} ........(1)
mole of NH_{3} = mole of HCl ........(2)
\therefore mole of HCl = 2 × mole of Urea

mole of HCl =2\times \frac{0.6}{60}=0.02mol ...(i)

[We know , mole = M X V = N X n X V]

 \mathrm{100 \ ml\times 0.2N\times 1=0.02\ mol} ...as (i).

Therefore, Option(4) is correct.

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Posted by

Ritika Jonwal

The IUPAC name of the following compound is :
Option: 1 2- nitro-4-hydroxymethyl-5 -aminobenzaldehyde
Option: 2 3-amino-4-hydroxymethyl-2-nitrobenzaldehyde
Option: 3 5-amino-4-hydroxymethyl-2-nitrobenzaldehyde
Option: 4 4-amino-2-formyl-5-hydroxymethyl nitrobenzene

Stepwise -

1. CHO comes first in the seniority table.

2. Numbering 

    Clockwise       -   2+4+5 = 11 Minimum (Correct)

    AnitClockwise -   3+4+6 = 13  

3. Write all substitute Prefix Name with Number

    5-amino

    4-hydroxymethyl

    2-nitro

4. Main group SUffix Name

     benzaldehyde

5. write the Prefix name with Number in alphabetical order.

The IUPAC name:

5-amino-4-hydroxymethyl-2-nitrobenzaldehyde

Therefore, the correct option is (3).

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Posted by

Kuldeep Maurya

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The average molar mass of chlorine is 35.5\: g\: mol^{-1}. The ratio of ^{35}Cl\; to\; ^{37}Cl in naturally occurring chlorine is close to :  
Option: 1 4:1
Option: 2 3:1
Option: 3 2:1
Option: 4 1:1

Given,

The average molar mass of chlorine is 35.5\: g\: mol^{-1}.

 

Let , the ratio of ^{35}Cl\; to\; ^{37}Cl in naturally occurring chlorine is close to x : y

 

Now, we know

\text { Av. molar mass }=\frac{\mathrm{n}_{1} \mathrm{M}_{1}+\mathrm{n}_{2} \mathrm{M}_{2}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)}

So, 

35.5=\frac{x \times 35+y \times 37}{x+y}

1.5 y=-0.5 x

\frac{x}{y}=\frac{3}{1}

Therefore, the correct option is (2).

 

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Posted by

Kuldeep Maurya

The minimum number of moles of O2 required for complete combustion of 1 mole of propane and 2 moles of butane is ______
 

Combustion reaction of 1 mole of propane and 2 moles of butane-

 

So, Total required mol of O2 = 5 + 13 = 18.

Ans = 18

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Posted by

Kuldeep Maurya

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