NEET Chemistry: Organic Chemistry – Basic Principles and Techniques questions with solutions

Q1. The structure of isobutyl group in an organic compound is:

1. CH3—CH—CH2—CH3
2. CH3—CH2—CH2—CH2—
3. CH3—CH—CH3
4. CH3—CH2—CH3

Answer: CH3—CH—CH2—CH3

The isobutyl group has the structure CH3—CH—CH2—CH3, where a branched methyl group is attached to the second carbon of a butyl chain.

Q2. The general molecular formula, which represents the homologous series of alkanols is

1. C2H2O
2. C2H2n+1O
3. C2H2n+2O
4. C2H2nO2

Answer: C2H2n+2O

The general molecular formula for alkanols (alcohols) is CnH2n+2O, as they are derived from alkanes (CnH2n+2) by replacing one hydrogen atom with a hydroxyl (-OH) group.

Q3. The correct order regarding the electronegativity of hybrid orbitals of carbon is

1. sp > sp2 > sp3
2. sp > sp3 > sp2
3. sp < sp2 < sp3
4. sp > sp2 < sp3

Answer: sp > sp2 > sp3

The electronegativity of carbon increases with the s-character of the hybrid orbital. Since sp has 50% s-character, sp2 has 33.3%, and sp3 has 25%, the correct order is sp > sp2 > sp3.

Q4. The number of possible isomers of the compound with molecular formula C4H8O is

1. 3
2. 5
3. 7
4. 9

Answer: 7

The molecular formula C4H8O corresponds to compounds with one degree of unsaturation. These include aldehydes, ketones, and cyclic alcohols. By considering all structural possibilities, there are 7 isomers.

Q5. Tautomerism will be exhibited by

1. (CH3)2NH
2. (CH3)3CNO
3. R3CNO2
4. RCH2NO2

Answer: RCH2NO2

Tautomerism occurs when a compound has an acidic hydrogen atom adjacent to a functional group capable of delocalization. In RCH2NO2, the α-hydrogen adjacent to the nitro group can participate in keto-enol tautomerism, making it the correct answer.

Q6. Isomers of a substance must have the same

1. Structural formula
2. Physical properties
3. Chemical properties
4. Molecular formula

Answer: Molecular formula

Isomers are compounds that have the same molecular formula but differ in their structural arrangement or spatial orientation. Hence, the molecular formula remains identical.

Q7. Sodium cyanide (Na + C + N → NaCN) is used in Lassaignes test.

1. Sodium cyanide (Na + C + N → NaCN)
2. Sodium sulphide (Na + S → Na2S)
3. Sodium halide (Na + X2 → 2NaX)
4. None of the above

Answer: Sodium cyanide (Na + C + N → NaCN)

In Lassaigne's test, sodium reacts with carbon and nitrogen in the organic compound to form sodium cyanide (NaCN), which is used to detect nitrogen.

Q8. Kjeldahl's method is suitable for estimating nitrogen in those compounds in which nitrogen is linked to carbon and hydrogen.

1. Kjeldahl's method is suitable for estimating nitrogen in those compounds in which nitrogen is linked to carbon and hydrogen.
2. Kjeldahl's method is not used in case of nitro, azo and azoxy compounds.
3. Kjeldahl's method is used for estimating nitrogen in fertilizers and agricultural products.
4. None of the above

Answer: Kjeldahl's method is suitable for estimating nitrogen in those compounds in which nitrogen is linked to carbon and hydrogen.

Kjeldahl's method is specifically suitable for estimating nitrogen in organic compounds where nitrogen is directly bonded to carbon and hydrogen. This excludes nitro, azo, and azoxy compounds.

Q9. Nitrogen, sulphur and halogens are tested in an organic compound by Lassaignes test.

1. Nitrogen, sulphur and halogens are tested in an organic compound by Lassaignes test.
2. Nitrogen and sulphur are tested in an organic compound by Lassaignes test.
3. Halogens are tested in an organic compound by Lassaignes test.
4. None of the above

Answer: Nitrogen, sulphur and halogens are tested in an organic compound by Lassaignes test.

Lassaigne's test is used to detect the presence of nitrogen, sulphur, and halogens in organic compounds by converting them into ionic forms that can be identified through specific reactions.

Q10. Higher the +I effect greater is stability of the species. Thus, (CH3)3C+ > (CH3)2CH+ > C6H5CH2+ > CH3CH2+.

1. (CH3)3C+ > (CH3)2CH+ > C6H5CH2+ > CH3CH2+
2. (CH3)2CH+ > (CH3)3C+ > CH3CH2+ > C6H5CH2+
3. C6H5CH2+ > CH3CH2+ > (CH3)2CH+ > (CH3)3C+
4. CH3CH2+ > C6H5CH2+ > (CH3)2CH+ > (CH3)3C+

Answer: (CH3)3C+ > (CH3)2CH+ > C6H5CH2+ > CH3CH2+

The stability of carbocations increases with the +I (inductive) effect of alkyl groups. Tertiary carbocations like (CH3)3C+ are more stabilized due to the presence of three electron-donating methyl groups, followed by secondary, benzyl, and primary carbocations.

Q11. With respect to the conformers of ethane, which of the following statements is true?

1. Bond angle changes but bond length remains same
2. Both bond angle and bond length change
3. Both bond angles and bond length remains same
4. Bond angle remains same but bond length changes

Answer: Both bond angles and bond length remains same

In the conformers of ethane, the bond angles and bond lengths remain unchanged because the rotation about the C-C single bond does not affect these structural parameters.

Q12. The correct statement regarding the comparison of staggered and eclipsed conformation of ethane is

1. The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain
2. The eclipsed conformation of ethane is more stable than staggered conformation, because eclipsed conformation has no torsional strain
3. The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain
4. The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain

Answer: The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain

The staggered conformation of ethane is more stable than the eclipsed conformation because it minimizes torsional strain by maximizing the distance between hydrogen atoms on adjacent carbons.

Q13. Geometrical isomers differ in:

1. position of functional group
2. position of atoms
3. spatial arrangement of atoms
4. length of carbon chain

Answer: spatial arrangement of atoms

Geometrical isomers differ in the spatial arrangement of atoms or groups around a double bond or a ring structure, leading to cis and trans forms.

Q14. Geometrical isomers differ in spatial arrangement of atoms. Which of the following is an example?

1. CH3—CH=CH—CH3
2. CH3—CH2—CH=CH2
3. CH3—CH—CH2—CH3
4. CH3—CH2—CH2—CH3

Answer: CH3—CH=CH—CH3

Geometrical isomerism arises in compounds with restricted rotation around a double bond and different groups attached to the doubly bonded carbons. CH3—CH=CH—CH3 exhibits this property, forming cis and trans isomers.

Q15. Electrophiles have high affinity for electrons. They attack at the site where electron-density is highest. Electron donating groups increases the electron density. The electron donating tendency decreases in the order:

1. -OH > -CH3 > -H > -Cl
2. -CH3 > -OH > -Cl > -H
3. -Cl > -H > -CH3 > -OH
4. -H > -Cl > -OH > -CH3

Answer: -OH > -CH3 > -H > -Cl

Electron-donating groups increase electron density through inductive or resonance effects. The -OH group donates electrons via resonance, followed by -CH3 through inductive effects, while -H and -Cl are less effective. Hence, the correct order is -OH > -CH3 > -H > -Cl.

Q16. An example of a sigma bonded organometallic compound is:

1. Grignard's reagent
2. Ferrocene
3. Cobaltocene
4. Ruthenocene

Answer: Grignard's reagent

Grignard's reagent (RMgX) contains a sigma bond between the carbon atom and the metal (magnesium), making it a sigma-bonded organometallic compound. The other options involve pi-bonding interactions.

Q17. Which of the following acids does not exhibit optical isomerism?

1. Maleic acid
2. α-amino acids
3. Lactic acid
4. Tartaric acid

Answer: Maleic acid

Maleic acid does not exhibit optical isomerism because it lacks a chiral center, unlike the other options which have asymmetric carbon atoms.

Q18. If there is no rotation of plane polarised light by a compound in a specific solvent, though to be chiral, it may mean that:

1. the compound is certainly meso
2. there is no compound in the solvent
3. the compound may be a racemic mixture
4. the compound is certainly achiral

Answer: the compound may be a racemic mixture

A racemic mixture contains equal amounts of enantiomers, which cancel out each other's optical activity, resulting in no rotation of plane-polarized light.

Q19. How many stereoisomers does this molecule have? CH3CH=CHCH2CHBrCH3

1. 4
2. 6
3. 8
4. 2

Answer: 4

The molecule has one double bond with cis/trans isomerism and one stereocenter, leading to 2 × 2 = 4 stereoisomers.

Q20. Which of the following is an optically active compound?

1. 1-Butanol
2. 1-Propanol
3. 2-Chlorobutane
4. 4-Hydroxyheptane

Answer: 2-Chlorobutane

2-Chlorobutane has a chiral center, making it optically active, as it can rotate plane-polarized light.