Haloalkanes & Haloarenes Test

Result

Haloalkanes & Haloarenes Test - 11

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Weekly Quiz Competition

Question 1
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When (S)-2-bromopentane is brominated, several 2,3-dibromopentane are formed, which of the following is not formed?

A

B

C

D
Option B,C are Correct

Solution

During bromination of (S)-2-bromopentane, the configuration at second carbon is not affected . It remains R configuration. The configuration at the third carbon atom can be either S or R .

In options B and C, the configuration at the second carbon atom is R. Hence, they cannot be formed.
Question 2
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The yield of alkyl bromide obtained as a result of heating the dry silver salt of carboxylic acid with bromine in CCI₄ is

A
1 °>3 °>2 °bromides

B
3 °>2 °>1 °bromides

C
1 °>2 °>3 °bromides

D
3 °>1 °>2 °bromides

Solution

The reaction follows free radical mechanism and alkyl free radical is formed in the propagation step as

Hence, stability of alkyl free radical (3 °>2 °>1 °) determine the reactivity.
Question 3
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Which is incorrect about Hunsdiecker 's reaction?

A
Only Cl₂ can give alkyl halide.

B
I₂ will give ester when treated with RCOOAg.

C
The reaction proceeds through free radical.

D
F₂ cannot give alkyl halide.

Solution

Except F₂ , almost all halogens react with RCOOAg giving alkyl halide via Hunsdiecker reaction .
With l₂ if RCOOAg is in excess, R —I formed in the first step reacts further with unreacted salt to give ester as
R —COOAg + R —I →R —COOR + AgI
Question 4
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The major product of the following reaction is

A

B

C

D

Solution

Free radical bromination occur. Preferably at highest degree carbon where most stable free radical is formed.
Question 5
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Racemic mixture is obtained due to the halogenation of

A
n-pentane

B
isopentane

C
neopentane

D
Both (a) and (b)

Solution

If free radical halogenation generate a chiral carbon, racemic mixture of halides are always formed due to equal probability of halogenation from both sides of planar free radical.
Question 6
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The reaction of SOCI₂ on alkanols to form alkyl chlorides gives good yields because

A
alkyl chlorides are immiscible with SOCI₂

B
the other products of the reaction are gaseous and escape out

C
alcohol and SOCI₂ are soluble in water

D
the reaction does not occur via intermediate formation of an alkyi chlorosulphite

Solution

The gaseous byproducts escape out on its own continuously driving the reaction in forward direction.
Question 7
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Addition of bromine on propene in the presence of brine yields a mixture of

A
CH₃ CHCICH₂ Br and CH₃ CHBrCH₂ CI

B
CH₃ CHCICH₂ CI and CH₃ CHBrCH₂ Br

C
CH₃ CHCICH₂ Br and CH₃ CHBrCH₂ Br

D
CH₃ CHCICH₂ Cl and CH₃ CHBrCH₂ Cl

Solution

Nucleophilic attack in step-ll occur at the carbon atom which can better accommodate the positive charge. Hence, attack of Br^- or Cl^- in second step occur at 2 °carbon rather that at 1 °carbon.
Question 8
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One or More than One Options Correct Type

Direction : This section contains multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q. Which of the following reagents can bring about free radical chlorination of propane?

A
SOCI₂

B
SO₂ CI₂

C
none

D
PCI₃

Solution

Both SO₂ CI₂ and Cl₂ undergo homolytic bond fission when heated or irradiated with light.
Question 9
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What is the order of SN₂ reaction of the alkyl halide?

A
RCl >RBr >RF >RI

B
RI >RBr >RCl >RF

C
R-F >R-Cl >RBr >RI

D
RBr >RI >RCl >RF

Solution

Answer: (b) The order of SN₂ reaction of the alkyl halide is RI >RBr >RCl >RF.
Explanation: Iodine is a good nucleophile and a good leaving group. Thus, it eliminates easily from an alkyl halide favouring SN₂ elimination reaction
Question 10
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Consider the following reaction,

Q.

The expected product(s) is/are

A

B

C

D
All are correct

Solution

NBS in CCI₄ brings about allylic brom ination by free radical mechanism: