Haloalkanes and Haloarenes Test 5

Result

Haloalkanes and Haloarenes Test 5

Score
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Rank
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TIME Taken - -

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Weekly Quiz Competition

Question 1
1 / -0

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 2
1 / -0

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 3
1 / -0

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 4
1 / -0

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 5
1 / -0

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 6
1 / -0

Addition of bromine on propene in the presence of brine yields a mixture of

A
CH₃CHCICH₂Br and CH₃CHBrCH₂CI

B
CH₃CHCICH₂CI and CH₃CHBrCH₂Br

C
CH₃CHCICH₂Br and CH₃CHBrCH₂Br

D
CH₃CHCICH₂Cl and CH₃CHBrCH₂Cl

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.