NEET Chemistry: Hydrocarbons questions with solutions

Q1. Which type of isomerism is shown by But-2-yne and cyclobutene?

1. Ring chain isomerism
2. Chain isomerism
3. Position isomerism
4. Functional isomerism

Answer: Ring chain isomerism

But-2-yne and cyclobutene have the same molecular formula, C4H6, but differ in whether the carbon atoms form a ring. This open-chain vs cyclic difference is ring-chain isomerism.

Q2. Method used to separate Petrol from crude oil is:

1. Distillation
2. chromatography
3. substallisation
4. fractional distillation

Answer: fractional distillation

Fractional distillation separates crude oil into useful fractions because its components have different boiling points. As the vapors rise through the column, each hydrocarbon condenses at a different level.

Q3. A mixture of oxygen and ethyne is burnt for welding. Why a mixture of ethyne and air is not used?

1. ethene can not burn in air
2. incomplete combustion takes place
3. ethene burn in air with explosion
4. none of these

Answer: incomplete combustion takes place

Air contains only about 21% oxygen, so ethyne does not get enough oxygen for complete combustion. This leads to incomplete combustion, producing a sooty, less hot flame unsuitable for welding.

Q4. What is called black gold?

1. Petroleum
2. Natural gas
3. coal tar
4. coal

Answer: Petroleum

Petroleum is called black gold because it is dark in color and extremely valuable as a major energy source and raw material. Its economic importance is why it gets this nickname.

Q5. In the hydrocarbon CH3—CH=CH—CH2—C≡CH, the state of hybridization of carbons 1, 3 and 5 are in the following sequence:

1. sp2, sp3, sp3
2. sp, sp3, sp2
3. sp2, sp2, sp3
4. sp3, sp2, sp

Answer: sp3, sp2, sp

Carbon 1 (CH3) is sp3 hybridized as it forms four sigma bonds. Carbon 3 (CH=) is sp2 hybridized due to one double bond and two sigma bonds. Carbon 5 (C≡) is sp hybridized because it forms one triple bond and one sigma bond.

Q6. The alkane that gives only one mono-chloro product on chlorination with Cl2 in presence of diffused sunlight is

1. Isopentane
2. 2,2-dimethylbutane
3. neopentane
4. n-pentane

Answer: neopentane

Neopentane (C) has all its hydrogen atoms equivalent due to its symmetrical structure, so chlorination produces only one mono-chloro product.

Q7. Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is

1. CH≡CH
2. CH2=CH2
3. CH4
4. CH3−CH3

Answer: CH4

The hydrocarbon (A) must be methane (CH4) because it undergoes substitution with bromine to form CH3Br. In the Wurtz reaction, CH3Br reacts with sodium to form ethane (C2H6), which is a gaseous hydrocarbon with less than four carbon atoms.

Q8. Which of the compounds with molecular formula C5H10 yields acetone on ozonolysis?

1. 3-methyl-1-butene
2. cyclopentane
3. 2-methyl-1-butene
4. 2-methyl-2-butene

Answer: 2-methyl-2-butene

2-methyl-2-butene undergoes ozonolysis to yield two molecules of acetone due to its symmetrical structure with a double bond at the central carbon.

Q9. CH3 − CH = CH − CH2 + HBr → A. A (predominantly) is:

1. CH3 − CH − CH2 − CH2Br
2. CH3 − C − CH2 − CH3 | CH3
3. CH3 − CH − CH − CH3 | Br CH3
4. CH3 − CH − CH − CH3 | CH3 Br

Answer: CH3 − CH − CH − CH3 | Br CH3

The reaction follows Markovnikov's rule, where the H from HBr adds to the carbon with more hydrogens, and Br adds to the carbon with fewer hydrogens. This results in the formation of CH3 − CH − CH − CH3 with Br attached to the second carbon.

Q10. Which one of the following alkenes will react faster with H2 under catalytic hydrogenation conditions? (R = Alkyl Substituent)

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

Answer: CH3 − CH = CH2

The alkene CH3 − CH = CH2 (option B) is less substituted compared to the others, making it more reactive towards catalytic hydrogenation due to lower steric hindrance around the double bond.

Q11. Reaction of HBr with propene in the presence of peroxide gives:

1. isopropyl bromide
2. 3-bromyl propane
3. allyl bromide
4. n-propyl bromide

Answer: n-propyl bromide

In the presence of peroxide, HBr adds to propene via the anti-Markovnikov rule, leading to the formation of n-propyl bromide as the major product.

Q12. The compound CH3 − C ≡ CH − CH3 on reaction with NaIO4 in the presence of KMnO4 gives:

1. CH3CHO + CO2
2. CH3COCH3
3. CH3COCH3 + CH3COOH
4. CH3COCH3 + CH3CHO

Answer: CH3COCH3 + CH3CHO

The compound undergoes oxidative cleavage of the triple bond in the presence of NaIO4 and KMnO4, producing CH3COCH3 (acetone) and CH3CHO (acetaldehyde).

Q13. Which of the following alkenes is more stable due to hyperconjugation?

1. cis-2-butene
2. trans-2-butene
3. 2-methylpropene
4. 1-butene

Answer: 2-methylpropene

2-methylpropene is more stable due to hyperconjugation as it has more alkyl groups attached to the double-bonded carbon atoms, increasing electron donation and stability.

Q14. C6H6 + CH2 = CH2 → AlCl3 → ?

1. C6H5CH2CH3
2. C6H5CH3
3. C6H5CH2CH2CH3
4. C6H5CH2CH2CH2CH3

Answer: C6H5CH2CH3

The reaction involves the Friedel-Crafts alkylation of benzene (C6H6) with ethene (CH2=CH2) in the presence of AlCl3, which acts as a Lewis acid catalyst. This results in the formation of ethylbenzene (C6H5CH2CH3).

Q15. The reaction of toluene with Cl2 in presence of FeCl3 gives 'X' and reaction in presence of light gives 'Y'. Thus, 'X' and 'Y' are:

1. X = Benzal chloride, Y = o-Chlorotoluene
2. X = m-Chlorotoluene, Y = p-Chlorotoluene
3. X = o- and p-Chlorotoluene, Y = Trichloromethyl-benzene
4. X = Benzyl chloride, Y = m-Chlorotoluene

Answer: X = o- and p-Chlorotoluene, Y = Trichloromethyl-benzene

In the presence of FeCl3, toluene undergoes electrophilic substitution to form o- and p-chlorotoluene (X). In the presence of light, a free radical substitution occurs at the methyl group, forming trichloromethyl-benzene (Y).

Q16. Benzene reacts with CH3Cl in the presence of anhydrous AlCl3 to form:

1. Chlorobenzene
2. Benzylchloride
3. Xylene
4. Toluene

Answer: Toluene

Benzene reacts with CH3Cl in the presence of anhydrous AlCl3 via a Friedel-Crafts alkylation reaction, forming toluene as the product.

Q17. Benzene reacts with n-propyl chloride in the presence of anhydrous AlCl3 to give:

1. 3-Propyl-1-chlorobenzene
2. n-Propylbenzene
3. No reaction
4. Isopropylbenzene

Answer: Isopropylbenzene

In the presence of anhydrous AlCl3, a Friedel-Crafts alkylation occurs. The n-propyl carbocation rearranges to a more stable isopropyl carbocation, leading to the formation of isopropylbenzene.

Q18. Consider the following reaction: PhBr →(alc.KOH)→ X →(H2SO4, room temperature)→ Y →(H2O, heat)→ Z The product Z is:

1. CH3CH2–O–CH2–CH3
2. CH3–CH2–O–SO3H
3. CH3CH2OH
4. CH2=CH2

Answer: CH2=CH2

The reaction involves elimination of HBr from PhBr using alcoholic KOH to form ethene (CH2=CH2). Subsequent steps do not alter the ethene structure significantly, leading to Z as CH2=CH2.

Q19. The cylindrical shape of an alkyne is due to the fact that it has

1. three sigma C–C bonds
2. two sigma C–C and one π C–C bond
3. three π C–C bonds
4. one sigma C–C and two π C–C bonds

Answer: one sigma C–C and two π C–C bonds

The cylindrical shape of an alkyne arises from the linear geometry around the carbon-carbon triple bond, which consists of one sigma bond and two π bonds. This arrangement ensures the bond angles are 180°.

Q20. The distance between the two adjacent carbon atoms is largest in

1. Benzene
2. Ethene
3. Butane
4. Ethyne

Answer: Butane

In butane, the carbon-carbon bond is a single bond (sigma bond), which is longer than the double bond in ethene, the triple bond in ethyne, or the delocalized bonds in benzene.