Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 3

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Weekly Quiz Competition

Question 1
4 / -1

Chlorobenzene is prepared commercially by

A
Raschig process

B
Deacon process

C
Etard’s process

D
Dow’s process

Solution

The Raschig process is a chemical process for the production of hydroxylamine, developed by German chemist Friedrich Raschig. The main step in this process, patented by Raschig in 1887, is the reduction of nitrite with bisulfite towards hydroxylamine disulfonate, which is hydrolysed to hydroxylammonium sulfate.
Question 2
4 / -1

Which of the following is major product of following reaction?
CH₃CH = CH₂ + HBr

A
2-Bromo propene

B
2-Bromo propane

C
1-Bromo propene

D
1-Bromo propane

Solution

The correct answer is Option D.
As HBr in the presence of a peroxide follows antimarkonikoff rule where the negative group should attach to the carbon having less no of hydrogen atoms.
Question 3
4 / -1

Which of the following compound will not give addition reaction with bromine?

A
Propene

B
1-Butene

C
Iso butane

D
Iso butene

Solution

The correct answer is Option C.
For Addition reaction π bond is required .Iso butane has no pi Bond .
Question 4
4 / -1

RCl + NaI R-I+NaCl

This reaction is known as:

A
Wurtz reaction

B
Frankland reaction

C
Fittig reaction

D
Finkelstein reaction

Solution

The correct answer is option D

where, X=Cl or Br.
This reaction is known as the Finkelstein reaction. Finkelstein's reaction is used to prepare alkyl iodides starting from alkyl chlorides and alkyl bromides.
Question 5
4 / -1

What is the order of reactivity of following halogen acids on alcohols?
HI, HBr, HCl

A
HI < HBr < HCl

B
HI > HBr > HCl

C
HI > HCl > HBr

D
HBr > HCl> HI

Solution

The correct answer is Option B.
Longer the bond length, lesser will be dissociation energy and hence, more reactivity.
Among halogen acids, bond length increases from HCl to HI.
∴ Order of reactivity of halogen acids towards alcohol is
HI > HBr > HCl
Question 6
4 / -1

Which of the following is formed by Swarts Reaction?

A
Alkyl iodide

B
Alkyl fluoride

C
Alkyl bromide

D
Alkyl chloride

Solution

The correct answer is Option B.
When alkyl bromide or iodide are treated with metallic fluorides like AgF,HgF2, etc.They form alkyl fluoride. This reaction is called Swarts reaction and is an important method for the preparation of fluoro alkanes.
Question 7
4 / -1

On reacting ethene with HCl we get:

A
Chloroethane

B
Chloropropane

C
Chloromethane

D
None of these

Solution

hydrogen chloride adds to ethene to make chloroethane:
CH₂=CH₂ + HCl → CH₃CH₂Cl
Question 8
4 / -1

Which of the following is formed in reaction mechanism of halogenation of alkanes in presence of UV light?

A
Carbocation

B
Carbanion

C
Diazonium ion

D
Free radical

Solution

Halogenation of alkanes in presence of UV light follows the free radical mechanis
Question 9
4 / -1

Which of the following is intermediate in Sandmeyer’s reaction?

A
Diazonium salt

B
Hydronium ion

C
Ammonium salt

D
Nitronium ion

Solution

The correct answer is option A
Diazonium salt
Aryl halides can be prepared by the decomposition of aryl diazonium salts in presence of copper halides solution dissolved in the corresponding halogen acid, the diazo group is replaced by a halogen atom. This reaction is known as the Sandmeyer reaction.
Question 10
4 / -1

Which of the following is the major product of following reaction?
CH₃CH = CH₂ + HBr in the presence of peroxide gives:

A
2 – bromopropane

B
1 – bromopropene

C
1 – bromopropane

D
2 – bromopropene

Solution

When 1-propene reacts with hydrogen bromide in the presence of peroxide, the more electronegative atom which is bromine gets attached to the carbon atom which has more number of hydrogen atoms. This reaction is named as Anti-Markovnikov's reaction.