Hydrocarbons Test

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Hydrocarbons Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

Nodal planes of $$\pi$$ bonds in benzene are located in :

A
molecular plane

B
one in a molecular plane and two in a plane perpendicular to the molecular plane which contains C C one in the molecular plane and two in the plane perpendicular to the molecular plane which contains C Cone in the molecular plane and two in plane perpendicular to molecular plane which contains C C $$\sigma$$ bonds.

C
one in molecular plane and two in plane perpendicular to molecular plane which contain C C $$\sigma$$ bond and C - H $$\sigma$$ bond.

D
perpendicular to molecular plane which bisect benzene ring in two equal half

Solution

Nodal planes of π bonds in benzene are located in the molecular plane.
Question 2
1 / -0

Monochlorination of ethylbenzene $$(PhCH_2CH_3)$$ with $$Cl_2$$ under heat produces ________.

A
$$PhCH_2CH_2Cl$$

B
$$PhCHClCH_3$$

C
both (A) and (B) in equal amounts

D
none of (B) and less of (A)

Solution

More stable due to resonance and hyper conjugate.
Hence none of (B) and less of (A) for free radial will go in resonance.
Question 3
1 / -0

In benzene, carbon uses _____ p-orbitals for hybridization.

A
0

B
1

C
2

D
3

Solution

In benzene, each carbon atom is $${sp}^{2}$$- hybridized; hence, carbon uses only two p-orbitals for hybridization.
Question 4
1 / -0

Vicinal dihalides undergo double dehydrohalogenation to give terminal alkyne. How many moles of $$NaNH_2$$ are used in the overall reaction ?

A
One

B
Two

C
Three

D
Four

Solution

In the first step, one mole $$ NaNH_2$$ is the base in an elimination reaction to give the alkenyl bromide.

In the second reaction, likewise a second equivalent of $$NaNH_2$$ performs a second elimination reaction to form the alkyne.

For a terminal alkyne, any excess $$NaNH_2$$ will remove the acidic hydrogen from terminal $$C-H$$ and give the alkynyl anion. So if a terminal alkyne is formed, three equivalents of $$NaNH_2$$ will be consumed; the alkyne is protonated upon workup, usually by adding water.
Question 5
1 / -0

Which content(s) of middle oil separate on cooling?

A
Naphthalene

B
Phenol

C
Benzene

D
Pyridine

Solution

The content of middle oil separate on cooling is naphthalene.

Hence the correct option is A.
Question 6
1 / -0

Write hybridisation of carbon atoms as indicated :

A
$$a=sp^2,b=sp^3,c=sp^3,d=sp^2$$

B
$$b=sp^2,a=sp^3,c=sp^3,d=sp^2$$

C
$$c=sp^2,b=sp^3,a=sp^3,d=sp^2$$

D
$$d=sp^2,c=sp^3,a=sp^3,b=sp^2$$

E
none of these

Solution

Carbon (a) forms 2 sigma bond and 1 pi bond , So it is sp$$^2$$
Carbon (b) forms 4 sigma bond , so it is sp$$^3$$
Carbon (c) forms 4 sigma bond , so it is sp$$^3$$
Carbon (d) forms 1 pi bond and 2 sigma bond , So it is sp$$^2$$
Question 7
1 / -0

Among the following groups the group that deactivates the benzene ring for further electrophilic substitution is:

A
Methyl

B
Amino

C
Hydroxyl

D
Chloro

Solution

Halogen atoms are highly deactivating in nature as these are highly electronegative so they are electron withdrawing in nature and increases the positive charge on the carbon atom of benzene ring.
Question 8
1 / -0

In the reaction $$CH_3COONa\, +\, NaOH\, \xrightarrow[Heat]{CaO}\, X\, +\, Na_2CO_3$$, product 'X' would be :

A
$$CH_3CHO$$

B
$$Ch_3COOH$$

C
$$CH_4$$

D
none of these

Solution

The laboratory method for preparing methane is:
$$CH_{3}COONa + NaOH \xrightarrow[CaO]{heat} CH_{4} + Na_{2}CO_{3}$$
Thus, X = methane; $$CH_{4} $$
Question 9
1 / -0

Which of the following alkanes can be synthesized by the Wurtz reaction in good yield?

A
$$(CH_3)_2CH - CH_2 - CH (CH_3)_2$$

B
$$(CH_3)_2CH - CH_2 - CH_2 - CH(CH_3)_2$$

C
$$CH_3 - CH_2 - C(CH_3)_2CH_2 - CH_3$$

D
$$(CH_3)_3C - CH_2 - CH_2 - CH_3$$

Solution

$$\bf{Hint:}$$ The Wurtz reaction gives maximum yield when we use symmetric alkanes.

$$\bf{Correct \ answer:}$$ Option $$B$$

$$\bf{Explanation \ for \ correct \ option:}$$

An even number of symmetrical alkyl halides can easily undergo Wurtz reaction as these form symmetrical alkanes which have uniform physical or chemical properties.
A general form of Wurtz reaction is

$$\bf{2RX + 2Na \rightarrow R-R + 2NaX}$$

Here, $$R$$ denotes any alkyl group, and $$X$$ is a halogen atom.

Asymmetric alkanes are not favorable to prepare by Wurtz reaction as the mixture of different products formed which are very difficult to seperate.
In the given options, $$B$$ is a symmetric alkane that can be easily prepared by the Wurtz reaction.

$$\bf{Explanation \ for \ incorrect \ options:}$$

All given alkanes except $$B$$ are not symmetrical with the odd number of carbon atoms in the parent chain, hence, these are not favoured to give a good yield by Wurtz reaction.
Question 10
1 / -0

Which of the following reagents may be used to distinguish between ethene and ethyne ?

A
Ammoniacal $$AgNO_3$$

B
$$KMnO_4$$

C
$$Br_2/CCl_4$$

D
$$AlCl_3$$

Solution

Ammonical silver nitrate solution(Tollen's reagent) is used to distinguish between ethene and ethyne.
Ethyne + tollens reagent --------> silver acetylide (white ppt) + $$NH_{3} $$ + $$H_{2}O$$
Ethene + tollens reagent ---------> no reaction