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A hydrocarbon

\mathrm{C}_{5} \mathrm{H}_{10}

does not react with chlorine in ...

Jun 5, 2024

Answer

The hydrocarbon is cyclopentane (C₅H₁₀)

Solution

The given hydrocarbon has the formula

C_5H_{10}

. This formula suggests that it could be either an alkene or a cycloalkane

The hydrocarbon does not react with chlorine in the dark, which indicates that it is not an alkene, as alkenes typically react with chlorine even in the absence of light

The hydrocarbon forms a single monochloro compound

C_5H_9Cl

in bright sunlight. This suggests that the hydrocarbon is a cycloalkane, as cycloalkanes can undergo substitution reactions with chlorine in the presence of light

Since the hydrocarbon forms only one monochloro compound, it must be a symmetrical cycloalkane. The only symmetrical cycloalkane with the formula

C_5H_{10}

is cyclopentane

Key Concept

Identification of hydrocarbons based on their chemical reactions

Explanation

The hydrocarbon

C_5H_{10}

is identified as cyclopentane because it does not react with chlorine in the dark (indicating it is not an alkene) and forms a single monochloro compound in bright sunlight (indicating it is a symmetrical cycloalkane).

solve 164 165 166 167 168 169 170

Sure, Rakshit! Let's solve the questions one by one. Question 164 Explain why the treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols, but in the presence of alcoholic KOH, alkenes are the major products.

Solution

Aqueous KOH: In aqueous solution, KOH dissociates completely to give

OH^-

ions, which act as a nucleophile. This nucleophile attacks the carbon atom bonded to the chlorine atom in the alkyl chloride, leading to a substitution reaction and forming an alcohol

Alcoholic KOH: In alcoholic solution, KOH provides

OH^-

ions, but the solvent (alcohol) favors the elimination reaction. The

OH^-

ion abstracts a proton from a β-carbon atom, leading to the formation of a double bond (alkene) and the elimination of HCl

164 Answer

Aqueous KOH leads to alcohols via nucleophilic substitution, while alcoholic KOH leads to alkenes via elimination.

Key Concept

Reaction conditions determine the product: substitution in aqueous KOH and elimination in alcoholic KOH.

Explanation

Aqueous KOH provides nucleophilic

OH^-

ions for substitution, while alcoholic KOH favors elimination due to the solvent effect.

Question 165 An allyl halide of formula $CH_2Cl$ on treatment with potassium tertiary butoxide gives two isomeric alkenes, $(CH_n)$. Name the isomeric alkenes.

Solution

Allyl halide: The given allyl halide is

CH_2=CH-CH_2Cl

Potassium tertiary butoxide: This strong base induces elimination, leading to the formation of two isomeric alkenes

Isomeric alkenes: The two possible alkenes are 1,3-butadiene (

CH_2=CH-CH=CH_2

) and 1-butene (

CH_2=CH-CH_2-CH_3

)

165 Answer

The isomeric alkenes are 1,3-butadiene and 1-butene.

Key Concept

Elimination reactions can produce multiple isomeric alkenes.

Explanation

Potassium tertiary butoxide induces elimination, leading to the formation of 1,3-butadiene and 1-butene from the allyl halide.

Question 166 Primary alkyl halide $C_4H_3Br$ (A) reacted with alcoholic $KOH$ to give compound (B). Compound $A$ reacted with $HBr$ to give $(C)$ which is an isomer of $(A)$. When (A) is reacted with sodium metal, it gives compound $(D), C_3H_4$ which is different from the compound formed when $n$-butyl bromide reacted with sodium. Give the structural formula of (A) and write the equations for all reactions.

Solution

Primary alkyl halide (A): The given formula

C_4H_3Br

suggests 1-bromo-2-butene

Reaction with alcoholic KOH: This leads to elimination, forming 1,3-butadiene (B)

Reaction with HBr: This forms 2-bromo-2-butene (C), an isomer of (A)

Reaction with sodium: This forms 1-butyne (D), different from the product of

n

-butyl bromide with sodium, which forms butane

166 Answer

(A) is 1-bromo-2-butene, (B) is 1,3-butadiene, (C) is 2-bromo-2-butene, and (D) is 1-butyne.

Key Concept

Structural isomers and elimination reactions.

Explanation

The reactions involve elimination and substitution, leading to different structural isomers and products.

Question 167 Compound 'A' with molecular formula $C_3H_2Br$ is treated with aqueous $KOH$ solution. The rate of $S_N$ reaction depends upon the concentration of the compound 'A' only. When another optically active isomer 'B' of this compound was treated with aqueous $KOH$ solution, the rate of reaction was found to be dependent on the concentration of compound and $KOH$ both.(i) Write down the structural formula of both compounds 'A' and 'B'.(ii) Out of these two compounds, which one will be converted to the product with an inversion of configuration.

Solution

Compound 'A': The given formula

C_3H_2Br

suggests 2-bromopropane

Compound 'B': The optically active isomer is 1-bromo-2-propanol

S_N1

reaction: Compound 'A' undergoes

S_N1

reaction, which depends only on the concentration of 'A'

S_N2

reaction: Compound 'B' undergoes

S_N2

reaction, which depends on the concentration of both 'B' and

KOH

Inversion of configuration:

S_N2

reaction leads to inversion of configuration, so compound 'B' will be converted with inversion

167 Answer

(i) 'A' is 2-bromopropane, 'B' is 1-bromo-2-propanol. (ii) Compound 'B' will be converted with inversion of configuration.

Key Concept

S_N1

and

S_N2

reaction mechanisms.

Explanation

S_N1

depends on the concentration of the substrate, while

S_N2

depends on both the substrate and nucleophile, leading to inversion of configuration.

Question 168 An alkyl halide (A), on reaction with magnesium in dry ether followed by treatment with ethanol, gives methyl butane. Write the possible structures of $A$.

Solution

Grignard reagent formation: The reaction with magnesium in dry ether forms a Grignard reagent

Reaction with ethanol: The Grignard reagent reacts with ethanol to form methyl butane

Possible structures: The alkyl halide (A) could be 1-bromo-2-methylpropane or 2-bromo-2-methylpropane

168 Answer

The possible structures of

A

are 1-bromo-2-methylpropane and 2-bromo-2-methylpropane.

Key Concept

Grignard reagent formation and subsequent reaction with ethanol.

Explanation

The Grignard reagent reacts with ethanol to form methyl butane, indicating the possible structures of the alkyl halide.

Question 169 Tert-butyl bromide was treated separately with (i) $C_2H_5ONa / C_2H_5OH$ and (ii) Hot $C_2H_5OH$. In each case, two products (A and B) were obtained. Whereas in one case (A) was the major product but in another case, (B) was the major product. Identify the products (A) and (B) and comment upon the mechanism of the reaction.

Solution

Reaction with

C_2H_5ONa / C_2H_5OH

: This leads to elimination, forming 2-methylpropene (A) as the major product

Reaction with hot

C_2H_5OH

: This leads to substitution, forming tert-butyl ethyl ether (B) as the major product

Mechanism: The reaction with

C_2H_5ONa / C_2H_5OH

follows an E2 mechanism, while the reaction with hot

C_2H_5OH

follows an

S_N1

mechanism

169 Answer

(A) is 2-methylpropene, (B) is tert-butyl ethyl ether.

Key Concept

E2 and

S_N1

reaction mechanisms.

Explanation

The reaction conditions determine the mechanism: E2 for elimination and

S_N1

for substitution.

Question 170 n-Butane on photochemical chlorination at $298 K$ gives a mixture of two isomeric butyl chlorides (A and B). The major product (A) on Wurtz reaction with $Na$ in dry ether gives compound (B). Identify the compounds (A), (B), and (C).

Solution

Photochemical chlorination: This forms 1-chlorobutane (A) and 2-chlorobutane (B)

Wurtz reaction: 1-chlorobutane (A) reacts with sodium to form n-octane (C)

Isomeric butyl chlorides: The two isomeric butyl chlorides are 1-chlorobutane and 2-chlorobutane

170 Answer

(A) is 1-chlorobutane, (B) is 2-chlorobutane, (C) is n-octane.

Key Concept

Photochemical chlorination and Wurtz reaction.

Explanation

The reaction conditions lead to the formation of isomeric butyl chlorides and subsequent formation of n-octane via the Wurtz reaction.