Hydrocarbons Test

Result

Hydrocarbons Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

$$R C-C=R \xrightarrow[Lindlar catalsyst]{H_2}$$?

A

B

C
Both (1) and (2)

D
$$R - CH_2 - CH_2 - R$$

Solution

By using Lindlar catalyst
we get cis-alsene
$$\therefore$$ we will get
Question 2
1 / -0

Aromatic hydrocarbons are the derivatives :

A
Benzene

B
Methane

C
Paraffins

D
Alkanes

Solution

Paraffins are alkanes.

Aromatic compounds are derivatives of benzene.

Aromatic compounds are made from benzene and its derivatives. It is a six-membered ring with three double bonds.

Option A is correct.
Question 3
1 / -0

Identify final product in the following :

A

B

C

D

E

Solution
Question 4
1 / -0

The final product of chlorination of methane in the sun light is ?

A
$$CH_{3}C_{1}$$

B
$$CH_{2}Cl_{2}$$

C
$$CHCI_{3}$$

D
$$CCI_{4}$$

Solution

Chlorination of methane follows free radial mechanism. When chlorine is present in small quantity then only single substitution takes place, but if chlorine is present in excess then all the hydrogen is substituted by chlorine. So tetra substituted product is formed. The following is the reaction of chlorination of methane:
Question 5
1 / -0

Choose which one from the given options is an incorrect statement.

A
Tertiary ($$3^0$$) butyl is 1,1-Dimethylethyl

B
Neopentane is 1,1-Dimethylpropane

C
Isopentane is 2-methylbutane

D
The correct IUPAC numbering of following compound is

Solution

$$\text{So option B is incorrect.}$$
Question 6
1 / -0

$$C_{6}H_{6}$$ is very good industrial solvent for:

A
$$NaCl$$

B
$$MgCl_{2}$$

C
$$CaCO_{3}$$

D
fats

Solution

Benzene is an organic solvent. (non polar solevnt) covalent compounds are soluble in non polar solvents. In the given options all are ionic except fats. Fats are soluble in Benzene.
Question 7
1 / -0

The ratio of sigma and pi bonds in benzene is :

A
$$4:1$$

B
$$2:3$$

C
$$6:1$$

D
$$1:1$$

Solution

Benzene has 12 sigma bonds and 3 pi bonds so sigma to pi ratio is 4:1.
Question 8
1 / -0

Which of the following reacts with water to give ethane?

A
$$CH_{4}$$

B
$$C_{2}H_{5}MgBr$$

C
$$C_{2}H_{5}OH$$

D
$$C_{2}H_{5}OC_{2}H_{5}$$

Solution

Hydrolysis of Grignard's reagent ($$RMgX$$) produces alkane, where R is alkyl or aryl group and X is halide.
Thus $$C_2H_5MgBr$$ on hydrolysis produces ethane, $$-MgBr$$ is replaced by hydrogen atom of water and $$MgBrOH$$ is also formed.
Question 9
1 / -0

The IUPAC name of the following compound is:

$$CH_{2} = CH - CH (CH_{3})_{2}$$

A
1,1-dimethyl-2-propene

B
3-methyl-1-butene

C
2-vinyl propane

D
1-isopropyl ethylene

Solution

$$\overset {1}{C}H_{2} = \overset {2}{C}H - \overset {3}{\underset {\displaystyle {\underset {\displaystyle {CH_3}}{|}}}{C}}H -\overset {4}{C}H_{3}$$

The IUPAC name of the given compound is 3-methyl-1-butene.
Question 10
1 / -0

The reaction $$C_6H_6 + CH_3 \xrightarrow[anhydrous]{AlCl_3} HCl + C_6H_5CH_3$$ is:

A
Friedel-Crafts alkylation

B
addition reaction

C
Friedel-Crafts acylation

D
Friedel -Craft's benzylation

Solution

The reaction of benzene in the presence of $$AlCl_3$$ in anhydrous conditions is Friedel-Crafts alkylation reaction. It is a substitution reaction. Hydrogen is substituted by an alkyl group.

Friedel-Crafts acylation is the addition of $$-COR$$ group.
Friedel-Crafts benzylation is the addition of a benzene group.