Test: Organic Chemistry - Some Basic Principles And Techniques 4 - From Past 28 Years Questions


30 Questions MCQ Test Chemistry 28 Years Past year papers for NEET/AIPMT Class 11 | Test: Organic Chemistry - Some Basic Principles And Techniques 4 - From Past 28 Years Questions


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QUESTION: 1

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

Solution:

Five chain isomers are possible which are –

QUESTION: 2

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

Solution:

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.

QUESTION: 3

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

Solution:

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°.

QUESTION: 4

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

Solution:

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]

Solution:

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°

QUESTION: 6

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

Solution:

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.

QUESTION: 7

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

Solution:

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.

QUESTION: 8

Kjeldahl’s method is used in the estimation of

Solution:

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.

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]

Solution:

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.

QUESTION: 10

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

Solution:

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

QUESTION: 11

The shortest C – C bond distance is found in

Solution:

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

QUESTION: 12

An sp3 hybrid orbital contains [1991]

Solution:

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

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]

Solution:

QUESTION: 14

Isomers of a substance must have the same

Solution:

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

QUESTION: 15

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

Solution:

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.

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

Solution:

Carboxylic acids dissolve in NaHCO3 but phenols do not.

QUESTION: 17

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

Solution:

2-Methyl-2-butene

QUESTION: 18

The IUPAC name of [1992] 

Solution:

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

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]

Solution:

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

QUESTION: 20

The most reactive compound for electrophilic nitration is [1992]

Solution:

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.

QUESTION: 21

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

Solution:

Resonance structures are separated by a double headed arrow 

QUESTION: 22

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

Solution:

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.

QUESTION: 23

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

Solution:

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

QUESTION: 24

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

Solution:

Resolution

QUESTION: 25

Correct increasing order of acidity is as follows:

Solution:

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

QUESTION: 26

An example of electrophilic substitution reaction is[1994]

Solution:

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.

QUESTION: 27

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

Solution:

3-Ethyl-2-methylpentane

QUESTION: 28

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

Solution:

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.

QUESTION: 29

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

Solution:

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.

QUESTION: 30

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

Solution:

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.