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31 Year NEET Questions: Some Basic Principles And Techniques- 4 - NEET MCQ


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30 Questions MCQ Test Chemistry 31 Years NEET Chapterwise Solved Papers - 31 Year NEET Questions: Some Basic Principles And Techniques- 4

31 Year NEET Questions: Some Basic Principles And Techniques- 4 for NEET 2024 is part of Chemistry 31 Years NEET Chapterwise Solved Papers preparation. The 31 Year NEET Questions: Some Basic Principles And Techniques- 4 questions and answers have been prepared according to the NEET exam syllabus.The 31 Year NEET Questions: Some Basic Principles And Techniques- 4 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 below.
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31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 1

How many chain isomers could be obtained from the alkane C6 H14 ? [1988]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 1

Five chain isomers are possible which are –

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 2

Which of the following is an optically active compound ? [1988]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 2

The compound containing a chiral carbon atom i.e., (a carbon atom which is attached to four different substituents is known as a chiral carbon atom) is optically active.

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31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 3

The Cl – C – Cl angle in 1,1,2,2- tetrachloroethene and tetrachloromethane respectively will be about[1988]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 3

Tetrachloroethene being an alkene has sp2 -hybridized C– atoms and hence the angle Cl – C – Cl is 120° while in tetrachloromethane, carbon is sp3 hybridized, therefore the angle Cl – C – Cl is 109.5°.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 4

Which of the following possesses a sp-carbon in its structure ? [1989]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 4

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 5

Cyclic hydrocarbon ‘A’ has all the carbon and hydrogen atoms in a single plane. All the carbon carbon bonds have the same length, less than 1.54 Å, but more than 1.34 Å. The C – C – C bond angle will be [1989]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 5

All the properties mentioned in the question suggest that it is a benzene molecule. Since in benzene all carbons are sp2–hybridized, therefore, C – C – C angle is 120°

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 6

Lassaigne’s test is used in qualitative analysis to detect[1989]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 6

Nitrogen, sulphur and halogens are tested in an organic compound by Lassaigne's test.
The organic compound is fused with sodium metal as to convert these elements into ionisable inorganic substances,

The cyanide, sulphide or halide ions can be confirmed in aqueous solution by usual test.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 7

Which one of the following can exhibit cis-trans isomerism ? [1989]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 7

Such isomers, which possess the same molecular and structural formula but differ in the arrangement of atoms around the double bonded carbon atoms are known as geometrical  isomers.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 8

Kjeldahl’s method is used in the estimation of

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 8

Kjeldal's method is suitable for estimating nitrogen in those compounds in which nitrogen is linked to carbon and hydrogen.
The method is not used in case of nitro, azo and azoxy compound. This method is basically used for estimating nitrogen in food fertilizers and agricultural products.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 9

An organic compound X (molecular formula C6H7O2N ) has six carbon atoms in a ring system, two double bonds and a nitro group as substituent, X is [1990]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 9

Hence it is homocyclic (as the ring system is made of one type of atoms, i.e. carbon) but not aromatic.As it does not have (4n +2)π electron required for aromaticity.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 10

In sodium fusion test of organic compounds, the nitrogen of the organic compound is converted into[1991]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 10

Sodium cyanide (Na + C + N → NaCN). (Lassaigne's test)

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 11

The shortest C – C bond distance is found in

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 11

Shortest C – C distance (1.20 Å) is in acetylene.
As acetylene has sp hybridisation, the bond length increases in the order

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 12

An sp3 hybrid orbital contains [1991]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 12

sp3 orbital has 1/4(25%) s-character. & 75% p character.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 13

A straight chain hydrocarbon has the molecular formula C8H10. The hybridization of the carbon atoms from one end of the chain to the other are respectively sp3, sp2, sp2, sp3, sp2, sp2, sp and sp. The structural formula of the hydrocarbon would be : [1991]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 13

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 14

Isomers of a substance must have the same

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 14

Organic compounds having same molecular formula but differ from each  other in physical properties or chemical properties or both are known as isomers.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 15

Which of the following is the most stable carbocation (carboniumion) ? [1991]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 15

Higher the possibility of delocalisation of the positive charge, greater is stability of the species. Thus

Benzyl carbocation is more stable than tertbutyl due to resonance in the former.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 16

A is a lighter phenol and B is an aromatic carboxylic acid. Separation of a mixture of A and B can be carried out easily by using a solution of

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 16

Carboxylic acids dissolve in NaHCO3 but phenols do not.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 17

2-Methyl 2-butene will be represented as  [1992]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 17

2-Methyl-2-butene

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 18

The IUPAC name of [1992] 

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 18

4-Hydroxy-2-methylpent-2-en-1-al

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 19

When the hybridization state of carbon atom changes from sp3 to sp2 and finally to sp, the angle between the hybridized orbitals [1993]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 19

Angle increases progressively sp3 ( 109°28' ), sp2 (120°), sp (180°)

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 20

The most reactive compound for electrophilic nitration is [1992]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 20

Due to + I-effect of the CH3 group, toluene has much higher electron density in the ring than benzene, nitrobenzene and benzoic acid as nitro and carboxylic group show- I-effect and hence toluene is most reactive towards nitration.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 21

Which is the correct symbol relating the two Kekule structures of benzene ? [1993]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 21

Resonance structures are separated by a double headed arrow 

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 22

The restricted rotation about carbon carbon double bond in 2-butene is due to [1993]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 22

Rotation around π bond is not possible. If any attempt is made to rotate one of the carbon atoms, the lobes of π-orbital will no longer remain coplanar i.e no parallel overlap will be possible and thus π-bond will break .
This is known as concept of restricted rotation. In other words the presence of π-bonds makes the position of two carbon atom.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 23

An important chemical method to resolve a racemic mixture makes use of the formation of [1994]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 23

Diastereomers since they have different melting points, boiling points, solubilities etc.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 24

The process of separation of a racemic modification into d and ℓ -enantiomers is called

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 24

Resolution

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 25

Correct increasing order of acidity is as follows:

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 25

Such questions can be solved by considering the relative basic character of their conjugated bases which for H2O, C2H2, H2CO3 and C6H5OH are

More the possibility for the dispersal of the negative charge, weaker will be the base.
Thus the relative basic character of the four bases is

Thus the acidic character of the four corresponding acids wil be

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 26

An example of electrophilic substitution reaction is[1994]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 26

Chlorination of methane proceeds via free radical mechanism. Conversion of methyl chloride to methyl alcohol proceeds via nucleophilic substitution. Formation of ethylene from ethyl alcohol proceeds via dehydration reaction. Nitration of benzene is electrophilic substitution reaction.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 27

 Which of the following IUPAC names is correct for the compound? [1994]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 27

3-Ethyl-2-methylpentane

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 28

Lassaigne’s test for the detection of nitrogen fails in[1994]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 28

Hydrazine (NH2NH2) does not contain carbon and hence on fusion with Na metal, it cannot form NaCN; consequently hydrazine does not show Lassaigne’s test for nitrogen.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 29

The most suitable method for separtion of a 1 : 1 mixture of ortho and para nitrophenols is [1994]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 29

The boiling point of o-nitrophenol is less than para-nitrophenol due to presence of intramolecular hydrogen bonding. Since p-nitrophenol is less volatile in than onitrophenol due to  presence of inter molecular hydrogen bonding hence they can be separated by steam distillation.

31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 30

The first organic compound, synthesized in the laboratory, was [1995]

Detailed Solution for 31 Year NEET Questions: Some Basic Principles And Techniques- 4 - Question 30

The vital force theory suffered first death blow in 1828 when Wohler synthesized the Ist organic compound urea in the laboratory from inorganic compounds reported below :

Later on a further blow to vital force theory was given by Kolbe (1845) who prepared acetic acid, the first organic compound, in laboratory from its elements.

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