Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

The IUPAC name of haloalkane is always written as a ______ word.

A
single

B
double

C
triple

D
none of these

Solution

Haloalkanes are always written as single words.

Eg: 1-Chlorobutane, 2-Bromopropane, etc.
Question 2
1 / -0

Heating many alkyl chlorides or bromides in water effects their conversion into alcohol through a $$ S_{N}$$1 reaction. Order each of the following sets of compounds with respect to solvolytic reactivity.

A
$$i>ii>iii$$

B
$$ii>i>iii$$

C
$$iii>ii>i$$

D
$$iii>i>ii$$

Solution
Question 3
1 / -0

The ability of rotation of plane polarized light either towards right or left is called as:

A
optical activity

B
chirality

C
superimposibility

D
non superimposibility

Solution

The ability of a compound to rotate the plane polarized light either towards right or left is called optical activity. If it rotates the plane polarized light right it is called dextro form or (+) form of the compound.If it rotates the plane polarized light left it is called leavo form or (-) form of the compound.
Question 4
1 / -0

Which of the following haloalkanes is most reactive?

A
1-chloropropane

B
1-bromopropane

C
2-chloropropane

D
2-bromopropane

Solution

Among chlorine and bromine chlorine is more reactive as it is better leaving group than bromine.

On the other hand, which has the open side, it is the open side. In the given option 1-chloro has the open side so it is more reactive than 2-chloro.

So, option A is the correct answer.
Question 5
1 / -0

An optically active compound:

A
rotates the plane polarised light

B
changes the direction of polarised light

C
do not allow plane polarised light to pass through

D
none of the above

Solution

The compound that is capable of rotating the plane of vibration of the polarized light is termed as the optically active compound.
Question 6
1 / -0

The conversion (Image) Can be effected using

A
$$Br_2/CCl_4$$

B
$$Br_2/H_2O$$

C
$$Br_2/Fe$$

D
$$Br_2/benzoyl peroxide$$

Solution
Question 7
1 / -0

Which is the correct increasing order of electronegativity of halides?

A
$$F,Cl,Br,I$$

B
$$I,Br,Cl,F$$

C
$$Br,Cl,F,I$$

D
$$Cl,Br,I,F$$

Solution

The bond between carbon and the halogen atom (C-X) is polar. Since halogen atoms are more electronegative than carbon, there is partial negative charge on halogen atom and positive charge on carbon. This makes the molecule dipolar and polar nature of C-X bond is increase with increase in the electronegativity difference between carbon and halogen atom. Hence the order of polarity of $$C-X$$ bond is $$C-F > C-Cl > C-Br > > C-I$$
Question 8
1 / -0

Which of the following is used for metal cleaning and finishing?

A
$$CHCl_{3}$$

B
$$CHI_{3}$$

C
$$CH_{2}Cl_{2}$$

D
$$C_{6}H_{6}$$

Solution

Chloroform is used as solvent and as anaesthetic.
$$CHI_3$$ is used as antiseptic.
$$CH_2Cl_2$$ is used for metal cleaning and finishing.
$$C_6H_6$$ is used for printing and lithography, paint, rubber, dry cleaning, adhesives and coatings, detergents.
Question 9
1 / -0

Arrange the following alkyl halides in decreasing order of the rate of $$\beta $$-elimination reaction with alcoholic KOH.

$${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }Br\\ \quad \quad \quad \quad \quad (i)$$ $${ CH }_{ 3 }-CH{ -CH }_{ 2 }Br\quad \\ \quad \quad \quad \ |\\ \quad \quad \ { CH }_{ 3 }\\ \quad \quad (ii)$$ $${ CH }_{ 3 }-{ CH }_{ 2 }{ -CH }-{ CH }_{ 3 }\quad \\ \quad \quad \quad \quad \quad \quad |\\ \quad \quad \quad \quad \quad \quad Br\\ \quad \quad \quad \quad \quad (iii)$$

A
$$(ii)\quad >\quad (iii)\quad >\quad (i)$$

B
$$(iii)\quad >\quad (i)\quad >\quad (ii)$$

C
$$(i)\quad >\quad (ii)\quad >\quad (iii)$$

D
$$(ii)\quad >\quad (i)\quad >\quad (i)$$

Solution

$$CH_3-CH_2-CH_2-Br$$ $$CH_3-\underset {|}{CH}-CH_2-Br$$ $$CH_3-CH_2-\underset{|}{CH}-CH_3$$
(i) $$CH_3$$ $$Br$$
(ii) (iii)
$$(iii) > (i) > (ii)$$
Decreasing order of rate of $$\beta$$-elimination reaction with alcoholic $$KOH$$ .
Question 10
1 / -0

Which type of isomerism can not be shown by benzaldoxime ?

A
Optical isomerism

B
Functional group isomerism

C
Geometrical isomerism

D
Configuration isomerism

Solution

Geometrical isomerism

In the syn-form,
H and OH are present on the same side.
In the anti-form,
H and OH are present on the opposite side.
The isomerism shown by Benzaldoxime is $$geometrical$$. Benzaldoxime exists in the synform and anti-form.