Hydrocarbons Test

Result

Hydrocarbons Test - 70

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Weekly Quiz Competition

Question 1
1 / -0

Give the structure of X.

A

B

C

D

Solution

Complete Reaction that takes place as mentioned above:

$$\mathrm{^{CH_3}_{CH_3}>CH-OH \ \ \xrightarrow{P+Br_2} \ \ ^{CH_3}_{CH_3}>CH-Br \ \ \xrightarrow{Na} \ \ ^{CH_3}_{CH_3}>CH-CH<^{CH_3}_{CH_3}}$$

So, $$\mathrm{X \ is \ ^{CH_3}_{CH_3}>CH-CH<^{CH_3}_{CH_3}}$$

Hence, Option "C" is the correct answer.
Question 2
1 / -0

When but-2-yne is treated with dil. $$ H_2SO_4/HgSO_4$$, the product formed is:

A
butan-1-ol

B
butan-2-ol

C
acetone

D
butanone

Solution
Question 3
1 / -0

In order to get propane gas, which of the following should be subjected to sodalime decarboxylation?

A
Sodium formate

B
Mixture of sodium acetate and sodium ethanoate

C
Sodium butyrate

D
Sodium propanoate

Solution

Reaction:
$$\mathrm{CH_3-CH_2-CH_2-COONa \xrightarrow[Sodalime]{\Delta} \underset{Propane \ gas }{CH_3-CH_2-CH_3} + CO_2}$$

So, in order to get propane gas, sodium butanoate should be subjected to sodalime decarboxylation.

Hence, Option "C" is the correct answer.
Question 4
1 / -0

2.84 g of methyl iodide was completely converted into methyl magnesium iodide and the product was decomposed by an excess of ethanol.
The volume of the gaseous hydrocarbon produced at NTP will be

A
22.4 liter

B
22400 ml

C
0.488 liter

D
0.244 liter

Solution

Moles of methyl iodide in 2.84 g of methyl iodide = $$\cfrac{2.84}{142}$$ = 0.02 moles

Reaction: $$\mathrm{CH_3I +Mg \xrightarrow{Dry \ Ether} CH_3MgI}$$

So, the moles of methyl magnesium iodide produced are 0.02 moles.

Next Reaction: $$\mathrm{CH_3MgI + ROH \rightarrow CH_4 + (RO)MgI}$$

So, the moles of hydrocarbon formed are 0.02 moles.

At STP, 1 mole of gas occupies 22.4 L of volume.
$$\implies$$ 0.02 mole of methane will occupy 0.488 L of volume.

Hence, Option "C" is the correct answer.
Question 5
1 / -0

Which of the following is the correct sequence of steps in the halogenation of an alkane?

A
Propagation, initiation, termination

B
Initiation, termination, propagation

C
Initiation, propagation, termination

D
Propagation, termination, initiation

Solution

Steps in the halogenation of alkane are in order : initiation, propagation and termination.
Hence, Option "C" is the correct answer.
Question 6
1 / -0

The order the reactivity of the alkenes,

$$\underset{I} {(CH_{3})_{2} C = CH_{2}}, \ \underset {II}{CH_{3}CH = CH_{2}}, \ \underset {III}{H_{2}C = CH_{2}}$$ when subjected to acid catalyzed hydration is

A
$$I > II > III$$

B
$$I > III > II$$

C
$$III > II > I$$

D
$$II > I > III$$

Solution
Question 7
1 / -0

In the given reactions:

A
A is an aldehyde; B is ketone and C is an alkene

B
A, B and C are alkenes

C
A, B and C are alkanes

D
A and B are alkenes while C is an alkane
Question 8
1 / -0

Order of reactivity of given four alkenes for hydrogenation reaction will be:
I. $$CH_{2} = CH_{2}$$
II. $$CH_{3} - CH = CH_{2}$$
IV. $$CH_{3} - CH = CH - CH_{3}$$.

A
$$III > IV > II > I$$

B
$$I > II > IV > III$$

C
$$I > II > III > IV$$

D
$$II > I > III > IV$$

Solution

$$concept: \text{less crowded alkenes adsorb }H_2\text{ with faster rate.}$$

from the question
$$I\rightarrow $$ no substitution
$$II\rightarrow $$ mono substituted alkene
$$III\rightarrow $$ di substituted alkene( same side)
$$IV\rightarrow $$ di substituted alkene( both side)

Thus, the relative rate of catalytic hydrogenation of the given alkenes is

$$I>II>IV>III$$ $$(\because \text{less the substitution, less will be crowding and more will be the rate.})$$

Option B is correct.
Question 9
1 / -0

In which of the following cases, the correct product is not written?

A
$$CH_{3} - CH = CH_{2} \xrightarrow {BrCl} CH_{3} - \overset {\overset {Cl}{|}}{C}H - \overset {\overset {Br}{|}}{C}H_{2}$$

B
$$(CH_{3})_{2} CH = CH_{2} \underset {H_{2}O}{\xrightarrow {Cl_{2}}} (CH_{3})_{2} CH(OH)CH_{2}Cl$$

C
$$(CH_{3})_{3} C - CH = CH_{2} \xrightarrow {Hydroboration} (CH_{3})_{3} C - CH_{2}CH_{2}OH$$

D
$$CCl_{3} CH = CH_{2} \xrightarrow {HCl} CCl_{3} - CH_{2} - CH_{2} - Cl$$

Solution
Question 10
1 / -0

Which one of the following statements is wrong?

A
Aromatic compounds are richer in carbon content

B
Aromatic compounds burn with sooty flame

C
Aromatic compounds are generally unstable

D
Aromatic compounds show substitution reactions

Solution

Option C is correct.

EXPLANATION:

For option A :
Aromatic compounds generally contains minimum 6 - carbon
atoms . Since, it richer in carbon content.

For option B:
Aromatic compound burn with smoky flame because
they have a ring structure of carbon atom. This causes
incomplete combustion of the carbon chain. Thus ,Aromatic
compounds burn with smoky flame.

For option C :
Aromatic compounds are resonance stabilised .
Hence, these are generally stable.
Since, the given statement is wrong.

For option D:
Aromatic compounds generally participate in
many substitution reactions.Since, the given statement
is also correct.

Among all options, option C is the only wrong option.