Hydrocarbons Test

Result

Hydrocarbons Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

The end product $$C$$ in the given reaction is:

A

B

C

D

Solution
Question 2
1 / -0

Freon 12 is manufactured from $$C{ Cl }_{ 4 }$$ by:

A
Wurtz reaction

B
Swarts reaction

C
Fittig reaction

D
Wurtz-Fittig reaction

E
Sandmeyer reaction

Solution

Freon-12 is made by $$C{ Cl }_{ 4 }$$ by the Swarts reaction as shown below:

$$\begin{matrix} 3C{ Cl }_{ 4 } \\ \quad \end{matrix}\begin{matrix} + \\ \quad \end{matrix}\begin{matrix} 2Sb{ F }_{ 3 } \\ \quad \end{matrix}\begin{matrix} \longrightarrow \\ \quad \end{matrix}\begin{matrix} 3C{ Cl }_{ 2 }{ F }_{ 2 } \\ \left( Freon-12 \right) \end{matrix}\begin{matrix} + \\ \quad \end{matrix}\begin{matrix} 2Sb{ Cl }_{ 3 } \\ \quad \end{matrix}$$
Question 3
1 / -0

A gas decolourised by $$KMnO_4$$ solution but gives no precipitate with ammoniacal cuprous chloride is:

A
ethane

B
methane

C
ethene

D
acetylene

Solution

A gas decolourised by $$KMnO_4$$ solution but gives no precipitate with ammoniacal cuprous chloride is ethene $$\displaystyle CH_2=CH_2$$. Since it contains carbon-carbon double bond, it reacts with $$KMnO_4$$ to form vicinal diol. Hence, $$KMnO_4$$ is decolourised.
$$ CH_2=CH_2 + H_2O + (O) \xrightarrow{ \text {alkaline} KMnO_4} HO-CH_2-CH_2-OH$$
Note: Acetylene gives a precipitate with ammoniacal cuprous chloride.
Question 4
1 / -0

The alkene formed as a major product in the above elimination reactions is:

A

B
$$CH_2=CH_2$$

C

D

Solution
Question 5
1 / -0

The reagent(s) used for the conversion of benzene diazonium hydrogensulfate to benzene is/ are:

A
$$H_{2}O$$

B
$$H_{3}PO_{2} + H_{2}O$$

C
$$H_{2}SO_{4} + H_{2}O$$

D
$$CuCl/ HCl$$

Solution

Certain mild reducing agents like hypophosphorous acid(phosphonic acid) reduce the diazonium salts to arenes and themselves get oxidised to phosphorous acid and ethanal.
Question 6
1 / -0

Which of the following will give alkane?

A
$$RN{ H }_{ 2 } + { R }^{ ' }MgX \longrightarrow$$

B
$$RC{ H }_{ 2 }OH+{ R }^{ ' }MgX\longrightarrow $$

C
$$R-Mg-X +H_2O\longrightarrow $$

D
All of the above

Solution

(i) Grignard reagent, all compounds having acidic (reactive-$$H$$) hydrogen will give alkane.

(ii) $$RMgX$$ which reacts with $${ H }_{ 2 }O$$ to give alkane.
Question 7
1 / -0

Which of the following compounds can be best prepared by Wurtz - reaction?

A
Iso-butane

B
n-butane

C
n-pentane

D
Iso-pentane

Solution

Due to stability of $$2^o-$$ radical, it is easy to synthesize Iso-butane,
The Wurtz reaction is a coupling reaction in organic chemistry, whereby two alkyl halides are reacted with sodium metal in dry ether solution to form a higher alkane: $$2R–X + 2Na \rightarrow R–R + 2Na^{+} X^{-}$$
Question 8
1 / -0

The hydrocarbon which can react with sodium in liquid ammonia is:

A
Styrene

B
Acetylene

C
Propylene

D
Pentane

Solution

The hydrocarbon which can react with sodium in liquid ammonia is acetylene.
$$ \displaystyle H-C \equiv C-H +Na \xrightarrow [NH_3]{liq} \underset { \displaystyle \text { monosodium acetylide } }{ H-C \equiv CNa }\xrightarrow [120^o]{NaNH_2} \underset { \displaystyle \text { Disodium acetylide} }{ NaC \equiv CNa}$$
This reaction is given by terminal alkynes as $$ \displaystyle R-C \equiv C-H $$ contains acidic hydrogen atom. Thus,$$ \displaystyle CH_3-C \equiv C-H $$ will also give the same reaction.
.
Question 9
1 / -0

What will be the major product, when 2-methyl butane undergoes bromination:

A
1-bromo-2-methyl butane

B
2-bromo-2-methyl butane

C
2-bromo-3-methyl butane

D
1-bromo-3-methyl butane

Solution

We know that halogenation of alkanes follows free radical substitution in the presence of UV light. In the process of bromination of 2-methyl butane, bromine gets attached to that carbon atom which has the least steric hindrance and the extraction of H atom is relatively easy. This follows the reactivity order of:
tertiary hydrogen > secondary hydrogen > primary hydrogen

Therefore, the major product after bromination of 2-methyl butane will be
2-bromo-2-methyl butane.
Question 10
1 / -0

Chloroform reacts with oxygen in the presence of light to give:

A
carbon tetrachloride

B
carbonyl chloride

C
methyl chloride

D
methylene dichloride

E
acetaldehyde

Solution

Chloroform reacts with oxygen in the presence of light to give a poisonous compound carbonyl chloride, which is also known by the name phosgene.
$$CHCl_{3} +\dfrac {1}{2}O_{2}\xrightarrow {hv} \underset {Carbonyl\ chloride (poisonous)}{COCl_{2}} + HCl$$
That's why chloroforms generally stored in dark coloured bottles filled upto brim.