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Hydrocarbons MCQs for NEET — Chemistry Questions with Answers

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1-butyne on reaction with hot alkaline ${KMnO}_{4}$ gives:

${CH}_{3} {CH}_{2} {CH}_{2} COOH$

${CH}_{3} {CH}_{2} COOH$ + ${CO}_{2}$

${CH}_{3} {CH}_{2} COOH$

${CH}_{3} {CH}_{2} COOH$ + HCOOH

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Benzene reacts with ${CH}_{3} Cl$ in the presence of anhy. ${AlCl}_{3}$ to form

toluene

chlorobenzene

benzylchloride

xylene

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Explanation

The reaction of benzene with methyl chloride (CH3Cl) in the presence of anhydrous aluminum chloride (AlCl3) follows the Friedel-Crafts alkylation mechanism, where a methyl group is introduced onto the benzene ring to form toluene.

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$R - CH = {CH}_{2} \to_{C_{2} H_{5} OH}^{Na / {NH}_{3} (l)}$ ${RCH}_{2} {CH}_{3}$ is called:

Clemmensen reduction

Fisher-Spier reduction

Birch reduction

Arndt-Eistert reduction

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Explanation

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The cylindrical shape of an alkyne is due to

three sigma $C - C$ bonds

three $π$ $C - C$ bonds

two sigma $C - C$ and one $π$ $C - C$ bonds

one sigma $C - C$ and two $π$ $C - C$ bonds

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Explanation

(d) In alkynes $C \equiv C$ is present, in such bond one sigma and two $π$ -bonds are present. Sigma bond is formed by sp-hybrid orbitals whereas $π$ -bonds are formed by unhybridised orbitals.

Hence, it shows cylindrical shape.

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The shortest $C - C$ bond distance is found in

diamond

ethane

benzene

acetylene

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Explanation

(d) Acetylene have shortest $C - C$ bond length because it have $C \equiv C$ triple bond. The bond length follows the following order

$C - C$ > $C \equiv C$ > $C \equiv C$

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The distance between two adjacent cabron atoms is largest in

benzene

ethene

butane

ethyne

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Explanation

(c) The $C - C$ bond length is maximum for single bond, so butane have largest $C - C$ bond length because it contains carbon-carbon single bond.

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Products of the following reaction

${CH}_{3} C \equiv C . {CH}_{2} {CH}_{3} \to_{(ii) Hydrolysis}^{(i) O_{3}}$ ... are

${CH}_{3} CHO$ + ${CH}_{3} {CH}_{2} CHO$

${CH}_{3} COOH$ + ${CH}_{3} {COCH}_{3}$

${CH}_{3} COOH$ + $HOOC . {CH}_{2} {CH}_{3}$

${CH}_{3} COOH$ + ${CO}_{2}$

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Which of the following compounds with molecular formula, $C_{5} H_{10}$ yields acetone on ozonolysis?

2-methyl-2-butene

3-methyl-1-butene

Cyclopentane

2-methyl-1-butene

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Explanation

(a) 2-methyl-2-butene (molecular formula $C_{5} H_{10}$ ) yields acetone on ozonolysis.

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Ozonolysis ( $O_{3}$ , $H_{2} O$ ) of, ${CH}_{3} - \underset{{CH}_{3}}{\underset{|}{CH}} - C \equiv C - {CH}_{3}$ gives :

${CH}_{3} \underset{{CH}_{3}}{\underset{|}{CH}} COOH$ + ${CH}_{3} COOH$

${CH}_{3} \underset{{CH}_{3}}{\underset{|}{CH}} CHO$ + ${CH}_{3} CHO$

${CH}_{3} \underset{{CH}_{3}}{\underset{|}{CH}} CHO$ + ${CH}_{3} COOH$

none of the above

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Explanation

The aldehydes formed are oxidized by $H_{2} O_{2}$ formed during hydrolysis.

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When 3,3-dimethyl-2-butanol is heated with conc. $H_{2} {SO}_{4}$ , the major product obtained is

2,3-dimethyl-2-butene

3,3-dimethyl-2-butene

2,3-dimethyl-1-butene

cis and trans-isomers of 2,3-dimethyl-2-butene

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