Hydrocarbons Test

Result

Hydrocarbons Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Ethane is formed during the formation of chloromethane by chlorination of methane because:

A
higher members of the hydrocarbons are generally formed during reactions

B
two methyl free radicals may combine during chlorination to give ethane

C
two chloromethane molecules react to form ethane

D
chlorine free radical reacts with methane to give ethane.

Solution

Chlorination of methane is a free radical reaction. The free radicals formed during the reaction are $$Cl^{\bullet}$$ and $$CH_3^{\bullet}$$. During the process if two methyl radicals collapse they form ethane. So, ethane is also formed as a byproduct.
$$2{Cl}^{\bullet} \rightarrow {Cl}_{2}$$
$${{H}_{3}C}^{\bullet} + {Cl}^{\bullet} \longrightarrow C{H}_{3}Cl$$
$${{H}_{3}C}^{\bullet} + {{H}_{3}C}^{\bullet} \longrightarrow C{H}_{3}-C{H}_{3}$$
Question 2
1 / -0

A
A

B
B

C
C

D
D

Solution
Question 3
1 / -0

The IUPAC name of the given compound :

A
1-Cyclohexyl-3-methylpent-1-ene

B
3-Methyl-5-cyclohexylpent-1-ene

C
1-Cyclohexyl-3-ethylbut-1-ene

D
1-Cyclohexyl-3,4-dimethyl-but-1-ene

Solution

The ring becomes the substitute as the side chain contains a double bond.

Double bond at carbon $$1$$ and a methyl group on carbon $$3$$.

So, $$IUPAC$$ name is 1-Cyclohenxyl-3-methylpent-1-ene
Question 4
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$$CH_3-C\equiv C$$ MgBr can be prepared by the reaction of _________.

A
$$CH_3-C\equiv C-Br$$ with $$MgBr_2$$

B
$$CH_3-C\equiv CH$$ with $$MgB_2$$

C
$$CH_2-C\equiv CH$$ with KBr and Mg metal

D
$$CH_3-C\equiv CH$$ with $$CH_3MgBr$$

Solution

$$CH_3-C\equiv C-H+CH_3MgBr\rightarrow CH_3-C\equiv CMgBr +CH_4$$.

Option D is correct.
Question 5
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What is the carbon-carbon bond length in benzene?

A
$$ 1.20 \mathring { A }$$ and $$1.31 \mathring { A } $$

B
$$ 1.39 \mathring { A } $$

C
$$ 1.39 \mathring { A }$$ and $$ 1.20 \mathring { A } $$

D
$$ 1.20 \mathring { A } $$

Solution

All $$C-C$$ and $$C=C$$ bond lengths are the same in benzene. Hence, $$C-C$$ bond length in benzene is 1.39 $$\mathring { A } $$.
Question 6
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Nitration and chlorination of benzene are:

A
nucleophilic and electrophilic substitution respectively

B
electrophilic and nucleophilic substitution respectively

C
electrophilic substitution in both the reactions

D
nucleophilic substitution in both the reactions.

Solution

Since nitration and halogenation, both are carried out by the help of electrophile $${N{O}_{2}}^{+} \; \& \; {Cl}^{+}$$. Hence, both reactions are electrophilic substitution reactions.
Question 7
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Although benzene is highly unsaturated it does not undergo addition reactions. The explanation of this can be suggested as:

A
$$\pi$$-electrons of benzene ring are delocalised

B
since $$\pi$$-electrons are present inside the ring, addition cannot take place

C
cyclic structures do not show addition reactions

D
benzene is not a reactive compound.

Solution

(A) $$\pi$$-electrons of the benzene ring are delocalised.
There are delocalised $$\pi$$-electrons above and below the plane of the ring. The presence of the delocalised $$\pi$$-electrons makes benzene particularly stable. Benzene resists addition reactions because that would involve breaking the delocalisation and losing that stability.
Question 8
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What is the order of reactivity of hydrogen atoms attached to the carbon atom in an alkane for free radical substitution?

A
$$3^o >1^o >2^o$$

B
$$2^o >1^o >3^o$$

C
$$3^o >2^o >1^o$$

D
$$1^o >2^o >3^o$$

Solution

REF: free radical will have $$ 9 \alpha H.$$

$$ CH_{3}-CH_{2}-CH_{3}$$ free radical will have $$ 6 \alpha H $$

$$ CH_{3}-CH_{3}-H$$ free radical will have $$ 3\alpha H $$

As no.of $$ \alpha -H $$ increase stability of free radical increase
reactivity order of H $$ 3^{\circ} > 2^{\circ} > 1^{\circ}$$

Hence, te correct option is $$\text{C}$$
Question 9
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One mole of 1,2-dibromopropane on treatment with $$X$$ moles of $$NaNH_2$$ followed by treatment with ethyl bromide gave a 2-pentyne. The value of $$X$$ is:

A
one

B
two

C
three

D
four

Solution

$$C{H}_{3}CH(Br)C{H}_{2}(Br) \xrightarrow{2NaN{H}_{2}} C{H}_{3}-C \equiv CH \xrightarrow{NaN{H}_{2}} {C{H}_{3}-C \equiv C}^{-}{Na}^{+} \xrightarrow{{C}_{2}{H}_{5}Br} C{H}_{3}C \equiv CC{H}_{2}C{H}_{3}$$
Hence, $$X=3$$.
Question 10
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What happens when calcium carbide is treated with water?

A
Ethane is formed.

B
Methane and ethane are formed.

C
Ethyne is formed.

D
Ethene and ethyne are formed.

Solution

If we mix calcium carbide with water ethyne is released due to endothermic reaction between them. This is the industrial method of producing ethyne.
$$Ca^{2+}{[{C}\equiv C]}^{2-} + 2{H}_{2}O \longrightarrow \underset{Ethyne}{CH \equiv CH} + Ca{(OH)}_{2}$$