Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 78

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$$CH\equiv CH\overset{H_2O/Hg^{2+}}{\underset{H_2SO_4}{\rightarrow}} X \overset{LiAlH_4}{\rightarrow} Y \overset{P_4/Br_2}{\rightarrow} Z$$ where Z is:

A
Ethylene bromide

B
Ethanol

C
Ethyl bromide

D
Ethylidene bromide

Solution
Question 2
1 / -0

The order of decreasing reactivities of these alcohols towards nucleophilic substitution with $$HBr$$ is:

A
$$III> I> IV> II$$

B
$$III> I> II> IV$$

C
$$I> III> IV> II$$

D
$$I> III> II> IV$$
Question 3
1 / -0

When hydrochloric acid gas is treated with propene in the presence of benzoyl peroxide, it gives:

A
2-chloropropane

B
allyl chloride

C
no reaction

D
n-propyl chloride
Question 4
1 / -0

Which of the following cannot shows $$S_{N}1$$ reaction?

A

B

C

D

Solution
Question 5
1 / -0

Which of the following compounds are optically active?

A

B

C

D
Question 6
1 / -0

2-Bromopentane is heated wtih potassium ethoxide in ethanol. The major product is:

A
trans-2-pentene

B
2-ethoxypentane

C
1-pentane

D
cis-2-pentene
Question 7
1 / -0

Substitution takes place at the position :

A
1

B
2

C
3

D
Both (A) and (C)
Question 8
1 / -0

Which is optically inactive:

A

B

C
Both of these

D
None of these
Question 9
1 / -0

$$S_N1$$ mechanism for the hydrolysos of an alkyl halide involves the formation of intermediate:

A
Free radical

B
Carbocation

C
Carbanion

D
none of these

Solution

$$S_{N}1$$ is a 2 step mechanism. The first step involves the breaking of the $$C-X$$ bond to form a $$carbocation\ intermediate,$$ while the second step involves the attack of nucleophile on carbocation.

Option B is correct.
Question 10
1 / -0

1-Bromopropane and 2-bromopropane on treatment with sodium in presence of ether gives:

A
n-hexane

B
2,3-dimethylbutane

C
2-methylpentane

D
a mixture of all these different alkanes '

Solution

According to the $$\rightarrow$$ $$2RX + 2Na \xrightarrow[]{ether} R-R+ 2NaX$$

$$CH_3-CH_2-CH_2-Br+CH_3-CH(Br)-CH_3 \xrightarrow[Na]{ether} \underset {\displaystyle {2-methylbutane}}{CH_3 -CH_2-CH_2-\underset{\underset{\,\,\,\, \displaystyle {\,\,\,\,CH_3}}|}{CH}-CH_3}$$

$$+\underset{\displaystyle {hexane}}{CH_3-CH_2-CH_2-CH_2-CH_2-CH_3} + \underset{\displaystyle {2,3-dimethylbutane}}{CH_3-\underset{\underset{\,\,\,\, \displaystyle {\,\,\,\,CH_3}}|}{CH}- \underset{\underset{\,\,\,\, \displaystyle {\,\,\,\,CH_3}}|}{CH}-CH_3}$$

Limitation:-
1. Wurtz reaction does not give odd number of carbon chain product

$$Hence \;the \;correct \;option \;is\; D.$$