Hydrocarbons Test

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Hydrocarbons Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

Reaction of alkanes with halogens is explosive in case of ?

A
$$F_2$$

B
$$Cl_2$$

C
$$Br_2$$

D
$$I_2$$

Solution

The order of reactivity of halogens is: $$F_{2}>Cl_{2}>Br_{2}>I_{2}$$ . Therefore reaction of alkanes with fluorine is highly explosive in nature.
Question 2
1 / -0

The name of $$CH_2(Cl)C(Br)C(Br)CH_2Cl$$ according to IUPAC system is:

A
2,3-dibromo-1,4-dichlorobut-2-ene

B
1,4-dichloro-2,3-dibromobut-2-ene

C
dichlorodibromobutene

D
dichlorodibromobutane

Solution

IUPAC name of given compound is 2,3-dibromo-1,4-dichlorobut-2-ene. The unsaturated hydrocarbon has a $$C_{2}= C_{3}$$, two bromine atoms at $$C_{2}$$ & $$C_{3}$$ and chlorine atoms are present as substituents at $$C_{1}$$ & $$C_{4}$$, and bromine gets proirity over chlorine in substuient naming.
Question 3
1 / -0

When methane is mixed with excess chlorine and the mixture is exposed to sunlight, the final product is:

A
$$CH_3Cl$$

B
$$CH_2Cl_2$$

C
$$CHCl_3$$

D
$$CCl_4$$

Solution

When a mixture of methane and chlorine is exposed to ultraviolet light - typically sunlight - a substitution reaction occurs and the organic product is chloromethane.
$$ CH_{4} + Cl_{2} \rightarrow CH_{3}Cl + HCl$$
However, when excess chlorine is present, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms. That means that you could get any of chloromethane, dichloromethane, trichloromethane or tetrachloromethane.
$$CH_{4} + Cl_{2} \rightarrow CH_{3}Cl + HCl$$
$$CH_{3}Cl + Cl_{2} \rightarrow CH_{2}Cl_{2} + HCl$$
$$CH_{2}Cl_{2} + Cl_{2} \rightarrow CHCl_{3} + HCl$$
$$CHCl_{3} + Cl_{2} \rightarrow CCl_{4} + HCl$$
Question 4
1 / -0

Structural formula of benzene is

A

B

C

D

Solution

benzene is a cyclic compound with alternate double bonds having molecular formula $$C_6H_6$$. looking at the options, C is correct
Question 5
1 / -0

The compound with highest boiling point is:

A
n-hexane

B
n-pentane

C
2,2-dimethylpropane

D
2-methylbutane

Solution

A straight-chain alkane has a higher boiling point than the branched one.

In straight chain alkane, the surface area of contact is more and hence the Vander Waal forces are stronger.

Also, more are the no. of carbon atoms in the straight-chain, more is the surface area of the molecule and higher is the boiling point.

Therefore n-hexane has the highest boiling point.

Option A is correct.
Question 6
1 / -0

Structure represents -

A
Isopropylidene Bromide

B
Active amylene Iodide

C
Isobutylene glycol

D
Isobutylene
Question 7
1 / -0

If octane number of gasoline sample is 70, then this sample will be similar to:

A
70% n-heptane + 30% n-octane

B
70% n-octane + 30% n-heptane

C
70% isooctane + 30% n-heptane

D
70% n-heptane + 30% isooctane

Solution

Octane number = 70 means gasoline contain 70% isoctane for which octane number = 100 and 30% of heptane for which octane number = 0.
Question 8
1 / -0

The structure shows :

A
Decane

B
Propane

C
Pentane

D
Butane
Question 9
1 / -0

(E)-3-bromo-3-hexene when treated with $$CH_{3}O^{\circleddash }$$ in $$CH_{3}OH$$ gives

A
3- hexyne

B
2-hexyne

C
2,3-hexadiene

D
2,4-hexadiene

Solution

$$ (E)-3-$$bromo$$-3-$$hexene when treated with $$CH_3O^-$$ in $$CH_3OH$$ gives $$2,3-$$hexadiene. A molecule of HBr is eliminated as shown in the above reaction.
Question 10
1 / -0

No. of isomers $$C_4H_8$$ :

A
4

B
3

C
2

D
5