Hydrocarbons Test

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Hydrocarbons Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

Among the following compounds 2, 3-dimethylexane is :

A
i

B
ii

C
iii

D
iv

Solution

Simple nomenclature of alkane.
Question 2
1 / -0

Methane with the Molecular formula $"CH_4"$ has

A
$4$ Covalent bonds

B
$8$ Covalent bonds

C
$6$ Covalent bonds

D
$2$ Covalent bonds

Solution

Structure of methane $H-\overset{\displaystyle H}{\overset{|}{\underset{\displaystyle H}{\underset{|}{C}}}}-H$
It has $4$ covalent bonds.
Question 3
1 / -0

The boiling point will increase in the order of:

A
n-butane< iso-butane

B
n-butane <= iso-butane

C
n-butane= iso-butane

D
iso-butane< n-butane

Solution

Straight chain isomers have higher boiling point than the branched ones. Thus boiling point of n-butane is higher than iso-butane. Option (D) is correct.
Question 4
1 / -0

The number of total product produced upon monochloriation of (+) 2-chlorobutane is ___________.

A
3

B
4

C
5

D
6

Solution

B. 4
2-Chlorobutane as reactant will produce four ${ C }_{ 4 }{ H }_{ 8 }{ Cl }_{ 2 }$ .The products are-
${ ClCH }_{ 2 }{ CHClCH }_{ 2 }{ CH }_{ 3 }+{ CH }_{ 3 }{ CCl }_{ 2 }{ CH }_{ 2 }{ CH }_{ 3 }+\\ { CH }_{ 3 }{ CHClCHClCH }_{ 3 }+{ CH }_{ 3 }{ CHClCH }_{ 2 }{ CH }_{ 2 }Cl$
Question 5
1 / -0

Addition of $HBr$ on propene produce:

A
$CH_3 CH_2 CH_2 Br$

B
$CH_3 BrCH_2 CH_3$

C
$CH_3-\underset{Br\,\,}{\underset{|}{C}H}-CH_3$

D
none of these

Solution

The addition of $HBr$ to propene takes place according to Markownikoff's rule. According to this rule negative part of reagent adds to carbon atom having lesser number of hydrogen atoms.
$CH_3CH=CH_2+HBr\rightarrow CH_3-\underset{Br\,\,\,}{\underset{|}{C}H}-CH_3$
Question 6
1 / -0

The hydrocarbon obtained by treating sodium ethanoate with soda lime is _________.

A
ethane

B
methane

C
propane

D
butane

Solution

${ CH }_{ 3 }COONa+NaOH\rightarrow { CH }_{ 4 }+{ Na }_{ 2 }{ CO }_{ 3 }$

$CH_3COONa$ =Sodium ethanoate
$NaOH$ =Soda lime
$CH_4$ =Methane
$Na_2CO_3$ =Sodium carbonate
Question 7
1 / -0

The reaction of one mol of bromine with ethyne yields:

A
$BrCH_2-CH_2Br$

B
$BrCH=CHBr$

C
$Br_2CH-CHBr_2$

D
$CH_3-CH_2Br$

Solution

Halogens, especially chlorine and bromine on reaction with alkynes readily produces tetra-halogen derivatives of alkane. The reaction is carried out in inert solvent like carbon tetrachloride.

Due to the presence of two n-bonds, each molecule of the alkyne can react with two molecules of the halogen.
One mole of Bromine with ethyne gives 1,2-dibromoethene.
$HC\equiv CH + Br_2 \xrightarrow{CCl_4} H(Br)C = C(Br)H$
Question 8
1 / -0

Ethylidene dichloride is obtained by the reaction of excess of HCl with.

A
Ethylene

B
Acetylene

C
Propane

D
Methane

Solution

$\underset{acetylene}{CH\equiv CH}+HCl\rightarrow \underset{vinyl chloride}{CH_2=CHCl}\overset{HCl}{\rightarrow} \underset{ethylidene dichloride}{CH_3CHCl_2}$
[Alkenes due to presence of only $1^o$ of unsaturation, form only monochloro derivative.]
Question 9
1 / -0

X compound reacts with Na to give $CH_3CH_2CH_2CH_3$ . The compound X is:

A
$CH_3 CH_2 OH$

B
$CH_3 CH_2 - Cl$

C
$CH_3 - CH_3$

D
$CH_3 - CH_2 - CH_2 - CH_2 - OH$

Solution

$\underset{ethyl chloride}{CH_3 CH_2} Cl+2Na + Cl CH_2 CH_3 \rightarrow CH_3 - CH_2 - CH_2 CH_3 + 2 NaCl$
This is Wurtz reaction.
Question 10
1 / -0

The compound formed when ethyl alcohol is treated with conc. ${H}_{2}{SO}_{4}$ at ${170}^{o}C$ is _______.

A
ethane

B
ethene

C
ethyne

D
ethanal

Solution

$Ethanol$ can be dehydrated to give $ethene$ by heating it with an excess of concentrated sulphuric acid at about 170C.
$CH_3CH_2-OH \overset {Conc. H_2SO_4 170^oC}{\rightarrow} CH_2=CH_2 + H_2O$

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Re-Attempt Weekly Quiz Competition

Hydrocarbons Test - 20

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Hydrocarbons Test - 20

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Question 1

i

ii

iii

iv

Solution

Question 2

444 Covalent bonds

888 Covalent bonds

666 Covalent bonds

222 Covalent bonds

Solution

Question 3

n-butane< iso-butane

n-butane <= iso-butane

n-butane= iso-butane

iso-butane< n-butane

Solution

Question 4

3

4

5

6

Solution

Question 5

CH3CH2CH2BrCH_3 CH_2 CH_2 BrCH3​CH2​CH2​Br

CH3BrCH2CH3CH_3 BrCH_2 CH_3CH3​BrCH2​CH3​

CH3−C∣HBr −CH3CH_3-\underset{Br\,\,}{\underset{|}{C}H}-CH_3CH3​−Br∣C​H​−CH3​

none of these

Solution

Question 6

ethane

methane

propane

butane

Solution

Question 7

BrCH2−CH2BrBrCH_2-CH_2BrBrCH2​−CH2​Br

BrCH=CHBrBrCH=CHBrBrCH=CHBr

Br2CH−CHBr2Br_2CH-CHBr_2Br2​CH−CHBr2​

CH3−CH2BrCH_3-CH_2BrCH3​−CH2​Br

Solution

Question 8

Ethylene

Acetylene

Propane

Methane

Solution

Question 9

CH3CH2OHCH_3 CH_2 OHCH3​CH2​OH

CH3CH2−ClCH_3 CH_2 - ClCH3​CH2​−Cl

CH3−CH3CH_3 - CH_3CH3​−CH3​

CH3−CH2−CH2−CH2−OHCH_3 - CH_2 - CH_2 - CH_2 - OHCH3​−CH2​−CH2​−CH2​−OH

Solution

Question 10

ethane

ethene

ethyne

ethanal

Solution

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$4$ Covalent bonds

$8$ Covalent bonds

$6$ Covalent bonds

$2$ Covalent bonds

$CH_3 CH_2 CH_2 Br$

$CH_3 BrCH_2 CH_3$

$CH_3-\underset{Br\,\,}{\underset{|}{C}H}-CH_3$

$BrCH_2-CH_2Br$

$BrCH=CHBr$

$Br_2CH-CHBr_2$

$CH_3-CH_2Br$

$CH_3 CH_2 OH$

$CH_3 CH_2 - Cl$

$CH_3 - CH_3$

$CH_3 - CH_2 - CH_2 - CH_2 - OH$