Hydrocarbons Test

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Hydrocarbons Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

Correct order of boiling point is:

A

B

C

D

Solution

There are 3 important trends to consider:

1) The relative strength of the four intermolecular forces is: Ionic > Hydrogen bonding > dipole-dipole > Van der Waals dispersion forces. The influence of each of these attractive forces will depend on the functional groups present.
2) Boiling points increase as the number of carbons is increased.
3) Branching decreases the boiling point.

Hence, option $$C$$ is correct.
Question 2
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$$X$$ is an optically active alkane having lowest molecular mass, predict the structure of the major product obtained on monochlorination of $$X$$.

A
$$CH_3-CH_2-CH_2-\underset{Cl}{\underset{|}{\overset{CH_3}{\overset{|}C}}}-CH_2-CH_3$$

B
$$CH_3-CH_2-CH_2-\overset{CH_3}{\overset{|}{C}H}-\underset{\!\!\!Cl}{\underset{|}{C}H}-CH_3$$

C
$$CH_3-CH_2-CH_2-\overset{CH_3}{\overset{|}{C}H}-CH_2-CH_2-Cl$$

D
$$Cl-CH_2-CH_2-CH_2-\overset{CH_3}{\overset{|}{C}H}-CH_2-CH_3$$

Solution

Halogenation usually occurs at the tertiary position of the alkanes.

The halogenation agents employed are $$CCl_4$$, dichlorine monoxide, $$PCl_5$$, phosgene, hypochlorite etc. in all these cases a chain initiating catalyst is required usually peroxides or UV light.

Regioselectivity of electrophillic halogenation usually takes place at the tertiary carbon atom, because tertiary carbon radical is more stable.

Option A is correct.
Question 3
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Which of the following aromatic compound gives sulphonation reaction very easily?

A
Chlorobenzene

B
Nitrobenzene

C
Toluene

D
Benzene

Solution

Toluene gives a sulphonation reaction very quickly because sulphonation is ESR, so the ring must be electron-rich as nitro and chloro groups are electron-withdrawing groups due to which ring gets deactivated while the $$CH_3$$ group is electron-donating makes ring electron-rich.
Question 4
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Two metals $$X$$ and $$Y$$ have $$3$$ and $$2$$ electrons in their valence shells $$'M'$$ and $$'N'$$ respectively. These metals can form binary compounds $$A$$ and $$B$$ with the same non-metal carbon. Both $$A$$ and $$B$$ can undergo hydrolysis forming hydrocarbons $$C$$ and $$D$$. Identify $$C$$ and $$D$$.

A
$${C}_{6}{H}_{6}, {C}_{3}{H}_{4}$$

B
$${C}_{3}{H}_{6}, {C}_{3}{H}_{4}$$

C
$${CH}_{4}, {C}_{2}{H}_{2}$$

D
$${C}_{4}{H}_{10}, {C}_{3}{H}_{6}$$

Solution

The two metals are aluminium $$Al$$ and calcium $$Ca$$. As they have 3 and 2 electrons in their valance shells and can form binary compounds carbides $$Al_4C_3$$ and $$CaC_2$$.
$$Al_4C_3 + H_2O \rightarrow H_4C$$
$$CaC_2 + H_2O \rightarrow C_2H_2$$
Question 5
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Compound B in the above reaction is:

A

B

C

D
Question 6
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An aqueous mixture of sodium ethanoate and sodium propanoate is electrolyzed. The product of the reaction is __________.

A
$$CH_3CH_2CH_3$$

B
$$CH_3CH_2CH_2CH_3$$

C
$$CH_3CH_3$$

D
all of these
Solution
In the electrolysis of an aqueous mixture of sodium ethanoate and sodium propanoate, the following reactions occur at the anode:

2CH₃COO• → 2CH₃• + 2CO₂
2CH₃CH₂COO• → 2CH₃CH₂• + 2CO₂
- 2CH₃• → CH₃CH₃
- 2CH₃CH₂• → CH₃CH₂CH₂CH₃
- CH₃• + CH₃CH₂• → CH₃CH₂CH₃
Hence, option D is correct.
Question 7
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In the given reaction, which of the following steps is wrong?

A
Step 1

B
Step 2

C
Step 3

D
None of the above

Solution

Step $$3$$ is wrong. The last step will first brominate the double bond. The product will be as shown:
Question 8
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A hydrocarbon on ozonolysis gives one mole each of compounds $$A$$ and $$B$$. Both $$A$$ and $$B$$ on reduction with sodium borohydride gives compounds $$C$$ and $$D$$ belonging to same homologous series. $$C$$ has no position isomer whereas $$D$$ has position isomer. $$A$$ and $$B$$ do not belong to the same homologous series. However, both $$A$$ and $$B$$ on oxidation with acidified $$KMn{O}_{4}$$ give same compound. Identify the hydrocarbon.

A
3- Methyl-2-butene

B
2,3-Dimethyl-1-butene

C
2-Methyl-2-butene

D
3,4-Dimethyl-1-butene

Solution

Ozonolysis of alkenes form ketone or aldehydes. Their reduction with $$NaBH_4$$ gives alcohols. $$C$$ has no position isomer whereas $$D$$ has position isomer.Thus $$C$$ is a primary alcohol and $$D$$ is a secondary alcohol. $$A$$ is an aldehyde and $$B$$ is a ketone. And the compound is $$2-methyl-2-butene$$
$$CH_3-(CH_3)C=CH-CH_3 \overset{(O_3)}{\rightarrow} \underset{[A]}{CH_3-(CH_3)C=O} + \underset{[B]}{O=CH-CH_3} \overset {NaBH_4}{\rightarrow} \underset{[C]}{CH_3-(CH_3)CH-OH} +\underset{[D]} {HO-CH_2-CH_3} \overset {KMnO_4}{\rightarrow} {CH_3-(CH_3)CH-OH} + {HOOC-CH_3}$$
Question 9
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The arrangement of following compounds:
$$i$$. Bromomethane
$$ii$$. Bromoform
$$iii$$. Chloromethane
$$iv$$. Dibromomethane
In the increasing order of their boiling point is :

A
$$iii< i< iv< ii$$

B
$$iv< iii< i< ii$$

C
$$ii< iii< i< iv$$

D
$$i< ii< iii< iv$$

Solution

As the molecular mass increases boiling point increases due to increased polar character. Hence,
$$iii<i<iv<ii$$
Question 10
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Which of the following solvents should be added to the organic solvent containing the pentane in order to remove the impurities in a separate layer that will not mix with the organic solvent?

A
$$CCl_{4}$$

B
$$C_{6}H_{6}$$

C
$$HF$$

D
$$SH_{2}$$

Solution

Due to its symmetrical structure, carbon tetrachloride ($$CCl_4$$) is non-polar. Its non-polar nature makes it a suitable solvent for dissolving non-polar compounds. Carbon tetrachloride is convenient as a solvent for NMR because it contains no protons, however, its poor dissolving power limits its usefulness. As a solvent, it is well suited for dissolving other non-polar compounds, iodine, fats, and oils.
$$CCl_4$$ can undergo various reactions like dimerization to Hexachloroethane, reduction to chloroform.
Hence, the correct option is $$\text{A}$$