MCQ Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 2

Reduction of benzene is presence of

$Ni$ gives:

0%
cyclohexane
0%
heterocyclic
0%
alicyclic
0%
unsaturated Aliphatic

Explanation

Hydrogenation of benzene is also possible using a palladium catalyst or a rhodium

$(Rh)$ catalyst or

$Ni$ catalyst and cyclohexane forms as a product.

Alkane which is liquid at room temperature is:

0%
${ C }_{ 2 }{ H }_{ 6 }$
0%
${ C }_{ 3 }{ H }_{ 8 }$
0%
${ C }_{ 4 }{ H }_{ 10 }$
0%
${ C }_{ 8 }{ H }_{ 18 }$

Explanation

Alkane from

$C_1 - C_4$ are gases.

Alkane from

$C_5 - C_{17}$ are liquids.

Alkane from

$C_{18}$ onwards is solids.

Hence, alkane

$C_8H_{18}$ is liquid at room temperature and the correct option is (D).

Which of the following is the most stable cycloalkane?

0%
Cyclopropane
0%
Cyclobutane
0%
Cyclopentane
0%
Cyclohexane

Explanation

Explanation:

Six membered rings are the most stable. As the ring size decreases the stability decreases. This stability of the six-membered ring is due to the chair conformation of cyclohexane as shown in the figure.

Step 1: Identification of ring size in each option

A) Cyclopropane: three-membered ring.

B) Cyclobutane: four-membered ring.

C) Cyclopentane: five-membered ring.

D) Cyclohexane: Six membered ring.

Step 2: So stability order will be $D>C>B>A$

Final answer: Cyclohexane

Hence, option $D$ is correct.

${ CH }_{ 2 }={ CH }_{ 2 }+{ H }_{ 2 } \xrightarrow { Ni/{ 300 }^{ 0 }C } { \ CH }_{ 3 }-{ CH }_{ 3 }$

The above reaction is known as:

0%
Kolbe's
0%
Wurtz
0%
Sabatier - Senderens
0%
Berthelot

Explanation

Sabatier - Senderens

$R - CH = CH_2 + H_2 \xrightarrow{Ni/ \Delta} R - CH_2 - CH_3$

Berthelot reaction

The reaction of ammonia with phenol hypochlorite to give indophenol.

$C$ is correct

On heating sodium propionate with sodium hydroxide and quicklime, the gas evolved is :

0%
$C_{2}H_{2}$
0%
$C_{2}H_{6}$
0%
$CH_{4}$
0%
$C_{2}H_{4}$

Explanation

$CH_3CH_2COONa +NaOH\xrightarrow{CaO} C_2H_6 + Na_2CO_3$

On heating sodium propanoate with sodium hydroxide and quicklime, the gas evolved is ethane.

Option B is correct.

Which of the following compounds is formed, when n-hexane is heated at

$500^{0}C$ and 10 - 20 atm in presence of

$Cr_{2}O_{3}$ ?

0%
Cyclohexane
0%
Benzene
0%
2-Methyl pentane
0%
Cyclobutane

Explanation

Aromatisation: Conversion of aliphatic compounds into aromatic compound is known as aromatisation. In this process

$Cr_2O_3$ is used as catalyst under high temperature and pressure.

$CH_{3}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-CH_{3} \xrightarrow{Cr_{2}O_{3} at 10-20 atm} C_{6}H_{6}$

Order of rate of reaction of ethane with halogens is:

0%
$I > Br > Cl > F$
0%
$Cl > I > Br > F$
0%
$Cl > Br > F > I$
0%
$F > Cl > Br > I$

Explanation

The reactivity of the halogens decreases in the following order:

$F_2 > Cl_2 > Br_2 > I_2$

Fluorine is so explosively reactive it is difficult to control, and iodine is generally unreactive. Chlorination and bromination are normally exothermic.

Chlorination of ethane in presence of diffused light is an example of:

0%
dehydration
0%
elimination reaction
0%
substitution reaction
0%
addition reaction

Explanation

Alkanes participate in substitution reactions like halogenation, nitration.

Halogenation reaction:

$C_{2}H_{6} \xrightarrow []{Cl_2 / h\nu} C_{2}H_{5}Cl +HCl$

$H$ is substituted by

$Cl$ . It is a substitution reaction.

Option C is correct.

Which of the following alkane has the least boiling point?

0%
n-butane
0%
n-pentane
0%
isopentane
0%
neopentane

Torsional strain in eclipsed conformations of a molecule is due to:

0%
repulsion between the eclipsed carbons
0%
repulsion between aligned electron pairs of the eclipsed bond
0%
repulsions betwen aligned ‘H’
0%
repulsions between C-H bonds of eclipsed conformation

The hybridisation of carbon atoms in ethane doesn’t change during :

0%
pyrolysis
0%
combustion
0%
halogenation
0%
both halogenation and combustion

Explanation

A. Pyrolysis : The pyrolysis product of ethane is ethylene. In ethylene molecule, the carbon atom is

$sp^{2}$ hybridized.

B. Combustion : In combustion reaction, ethane is turned into carbon dioxide, which is also

$sp^{2}$ hybridized.

C. Halogenation : The halogenation reaction is a replacement type of reaction where one hydrogen atom of ethane gets replaced by a halogen atom. So, the hybridisation state does not change during the reaction. So, the correct answer is (C).

Statement- 1: Alkanes participate in substitution reactions.

Statement- 2: Alkanes are saturated hydrocarbons.

0%
Both statement- 1 and statement-2 are correct statement-2 is a correct explanation of statement-1.
0%
Both statement-1 and statement-2 are correct statement-2 is not a correct explanation of statement-1.
0%
Statement-1 is correct , statement-2 is incorrect.
0%
Statement-1 is incorrect , statement-2 is correct.

Explanation

statement-1: Alkanes participate in substitution reactions.

The example includes free radical halogenation reaction in which H atom of an alkane is replaced with a halogen atom.

$\displaystyle CH_3CH_3 \xrightarrow [light] {Cl_2} CH_3CH_2Cl$

statement-2: Alkanes are saturated hydrocarbons.

Saturated hydrocarbons are hydrocarbons that contain only single bonds between carbon atoms. They are called saturated because each carbon atom is bonded to as many hydrogen atoms as possible.

In other words, the carbon atoms are saturated with hydrogen

Thus both statement-1 and statement-2 are correct statement-2 is a correct explanation of statement-1.

The deviation of bond angle in cyclopropane:

0%
$49^{0}.5'$
0%
$24^{0}.44'$
0%
$-5^{0}.16'$
0%
$-9^{0}.33'$

Explanation

Deviation of the bond angle in cyclopropane is 109.5 - 60

$=$ 49.5

Identify A and B in the following reaction sequence:

${ C }_{ 2 }{ H }_{ 6 }\xrightarrow { A } { C }_{ 2 }{ H }_{ 5 }Cl\xrightarrow { Zn/THF } B$

0%
$Cl_{2}/uv$ light and $C_{4} H_{10}$
0%
$PCl_{3}$ and $C_{2} H_{4}$
0%
$HCl$ and $C_{2}H_{6}$
0%
$Cl_{2}$ and $C_{2} H_{2}$

Explanation

The reactions are as follows:
Halogenation reaction:

${ C }_{ 2 }{ H }_{ 6 }\xrightarrow [ ]{ Cl_2/hv } { C }_{ 2 }{ H }_{ 5 }Cl +HCl$

Frankland's reaction:

${ C }_{ 2 }{ H }_{ 5 }Cl+Zn+{ C }_{ 2 }{ H }_{ 5 }Cl\rightarrow C_2H_5-C_2H_5+ZnCl_2$

Option A is correct.

When cyclohexane is poured on water, it floats because cyclohexane is :

0%
In boat form
0%
In chair form
0%
In crown form
0%
Less dense than water

Alkanes can be iodinated in the presence of:

0%
$HI$
0%
$I_{2}$ and P
0%
$HIO_{3}$
0%
$PI_{3}$

Explanation

This is because iodination is reversible and thus needs a reducing agent to drive the reaction to forward direction according to Le Chatelier's principle.

The reaction is as follow:

$CH_{4} + I_{2} \leftrightarrow CH_{3}I + HI$

The HI formed in a product makes the reaction reversible. Thus,

$HIO_{3}$ is added which reduce HI to

$I_{2}$ and stop the reversible reaction.

$HIO_{3} + 5HI \rightarrow 3H_{2}O + 3I_{2}$

Option C is correct.

In which of the following pair both angle strain and torsional strain is present?

0%
Cyclopropane, Cyclobutane
0%
Cyclopropane, Cyclohexane
0%
Cyclobutane, Cyclohexane
0%
Cyclohexane, Cyclopentane

A single substitution of H atom in an alkane of molar mass by 72g / mole on chlorination produces only one product. The alkane is :

0%
n- pentane
0%
2- methyl butane
0%
2,2-dimethyl propane
0%
n- butane

The most strained cycloalkane is :

0%
cyclopropane
0%
cyclobutane
0%
cyclopentane
0%
cyclohexane

Explanation

In cyclopropane,the C-C-C bond angles are 60° whereas tetrahedral 109.5° bond angles are expected.

The intense angle strain leads to nonlinear orbital overlap of its

$sp^3$ orbitals.

Because of the bond's instability, cyclopropane is more reactive than other alkanes. Since any three points make a plane and cyclopropane has only three carbons, cyclopropane is planar.

The H-C-H bond angle is 115° whereas 106° is expected as in the

$CH_2$ groups of propane.

Option A is correct.

Which of the following statement(s) is/are not correct?

0%
Alkanes are non polar in nature
0%
Alkanes are soluble in non-polar solvents
0%
Straight chain alkanes have lesser boiling point than corresponding branched chain isomers
0%
Alkanes exhibit alternations in melting point

The structure of 1,3-dicyclohexyl propane is:

Condition for maximum yield of

$C_{2} H_{5} Cl$ is:

0%
$C_{2} H_{6} (Excess) + Cl_{2} \xrightarrow[]{UV light}$
0%
$C_{2}H_{6}+ Cl_{2}\xrightarrow[R.T]{}$
0%
$C_{2} H_{6} + Cl_{2} (Excess) \xrightarrow[]{UV light}$
0%
$C_{2} H_{6} + Cl_{2}\xrightarrow[Dark]{}$

Explanation

Halogenation reaction:

$C_{2}{H}_{6} \xrightarrow []{Cl_2/hv} C_{2}H_{5}Cl$

Bond angle in planar ring of cyclopropane is

0%
$90^{0}$
0%
$108^{0}$
0%
$60^{0}$
0%
$120^{0}$

Bromination of an alkane as compared to chlorination proceeds _____________.

0%
at a slower rate
0%
at a faster rate
0%
with equal rate
0%
with equal rate (or) different rate depending upon the temperature

Explanation

Bromination of an alkane occurs at a slower rate than chlorination of an alkane. As we know that halogenation involves the formation of a carbon-halogen bond. Now, the ease of halogenation will certainly depend on the carbon-halogen bond strength.

Bond formation energy of

$C-Cl$ is higher than that of

$C-Br$ bond. So, the ease of bromination is lower than the ease of chlorination.

So, the correct answer is option (A).

2.84 gms of methyl iodide was completely converted into

$CH_{3} MgI$ and the product was decomposed by the excess of ethanol. The volume of the gaseous hydrocarbon produced at NTP will be_________.

0%
22.4 lit.
0%
2240 lit.
0%
0.448 lit.
0%
224 lit.

Explanation

In this reaction, one mole of Grignard reagent reacts with one mole of alcohol to produce hydrocarbon.

$CH_3I \rightarrow CH_3MgI +C_2H_5OH\rightarrow CH_4 + C_2H_5OMgI$

Mass of methyl iodide = 2.84 gm (given)
No. of mole of methyl iodide = 2.84/142 =0.02

So hydrocarbon produced = 0.02 mole

So the volume of produced hydrocarbon = 0.448 lit
(one mole produced 22.4 lit).

Option C is correct.

Which one of the following is reduced with

$Zn-Hg$ and

$HCl$ to give alkane ?

0%
Ethyl acetate
0%
Acetic acid
0%
Acetamide
0%
Butan-2-one

The reaction

${ C }_{ 12 }{ H }_{ 26 }\rightarrow { C }_{ 6 }{ H }_{ 12 }+{ C }_{ 6 }{ H }_{ 14 }$ represents :

0%
Substitution
0%
synthesis
0%
Cracking
0%
Addition.

The compound obtained by the reaction of methanol and

$CH_{3} MgX$ is :

0%
propanone
0%
methane
0%
ethane
0%
ethanol

Explanation

$CH_3MgX$ is the grignard reagent.
Grignard reagent first of all reacts with the active hydrogen if it is present on the reactant after that it further goes for the substitution reaction.
Active hydrogen is the hydrogen which is attached with high electronegative elements like oxygen, nitrogen, chlorine.
For the given reaction:

$CH_3MgX$ +

$CH_3OH$

$\to$

$CH_4$ +

$CH_3OMgX$

Which of the following will produce methane on hydrolysis?

0%
$BeC_{2}$
0%
$Al_{3}C_{4}$
0%
$Al_{4}C_{3}$
0%
$Be_{2}C$

Explanation

$BeC_2 and Al_3C_4$ not even exist.
These two carbides are called methanides. They usually prepare methane on hydrolysis.

$H_{2}C=CH_{2}+H_{2}\xrightarrow[]{Ni/300^{0}C}C_{2}H_{6}$ to get back the ethene, ethane must be:

0%
dehydrogenated in the presence of $Cu$ at $300^{0}C$
0%
heated to $450^{0}C$ in the presence of $HNO_3$ vapoar
0%
heated $170^{0}C$ in the presence of conc. $H_{2} SO_{4}$
0%
heated to $750^{0}C$

On halogenation, an alkane (

$C_{5} H_{12}$ ) gives only one mono halogenated product. The alkane is:

0%
n- pentane
0%
2- methyl butane
0%
2,2-dimethyl propane
0%
None of these

Which of the following compounds will form a hydrocarbon on reaction with a Grignard reagent?

0%
Ethanol
0%
Ethanal
0%
Propanone
0%
None of the above

Explanation

A Grignard reagent reacts with compounds containing active H atom, such as water, ammonia, amines, alcohols, alkynes etc., to form a hydrocarbon.

It does not form hydrocarbon with ethanal and propanone.

$RMgX +R'O-H \rightarrow R-H +MgXOR'$

Option A is correct.

Methane can not be prepared by :

0%
Wurtz reaction
0%
Decarboxylation
0%
Kolbes reaction
0%
Sabatier - senderens reaction

Explanation

Hint: The formula of methane is $CH_4$ . It has a single carbon atom.

Step 1: Wurtz Reaction

Wurtz reaction is-

$CH_3Cl+2Na+CH_3Cl\rightarrow C_2H_6+2NaCl$

Here, the product will have more than one carbon. So, it can not form methane.

Step 2: Decarboxylation

In a decarboxylation reaction, a carboxylic acid undergoes decarboxylation to form an alkane.

$CH_3COOH\rightarrow CH_4+CO_2$

So, here methane can be produced.

Step 3: Kolbe reaction

In Kolbe reaction,

$2$ moles of the potassium salt of ester react with water to form alkane, carbon dioxide, hydrogen gas.

$2CH_3COOK+H_2O\rightarrow C_2H_6+2KOH+2CO_2+H_2$

So, here methane can not be produced.

Step 4: Sabatier - senderens reaction

Sabatier senderens reaction is-

$C_nH_{2n}+H_2\rightarrow C_nH_{2n+2}+heat$

Here, an alkene gets converted into an alkane that has the same number of carbon atoms. So it can not produce an alkane with only one carbon as alkenes have two or more carbons.

Final Step: Correct options (A),(C), and (D).

2-Methyl butane on reacting with bromine in presence of sunlight gives mainly :

0%
1-Bromo-2-methyl butane
0%
2-Bromo-2-methyl butane
0%
2-Bromo-3-methyl butane
0%
1-Bromo-3-methyl butane

Explanation

$CH_3CH(CH_3)CH_2CH_3+Br_2 \overset{sunlight} {\longrightarrow}H_3CC(Br)(CH_3)CH_2CH_3$

The reaction proceeds via the free radical mechanism.

Since the radical formed in this case is tertiary, as it is more stable than primary and secondary, 2-Bromo-2-methyl butane is formed.

Option B is correct.

The IUPAC name of given compound is :

0%
2 - methyl propane
0%
iso butane
0%
n-butane
0%
propane

The

$C-C$ bond length of the following molecules is in the order:

0%
$CH_{2} >C_{2}H_{4}> C_{6}H_{6}< C_{2}H_{2}$
0%
$C_{2}H_{2}< C_{2}H_{4}< C_{6}H_{6}< C_{2}H_{6}$
0%
$C_{2}H_{6}< C_{2}H_{2}< C_{6}H_{6}< C_{2}H_{4}$
0%
$C_{2}H_{4}< C_{2}H_{6}< C_{2}H_{2}< C_{6}H_{6}$

Explanation

Bond length is related to bond order, when more electrons participate in bond formation the bond is shorter. By approximation, the bond lengths of two different atoms are the sum of the two individual covalent radii.

The bond length of carbon-carbon triple bond is shorter than carbon-carbon double bond and is shorter than resonance bonding in benzene and is shorter than carbon-carbon single bond.

Therefore,

$C_2H_2< C_2H_4< C_6H_6< C_2H_6$

Difference in the molecular formulae of

$D$ and

$C$ is:

0%
${ C }_{ 2 }{ H }_{ 3 }$
0%
${ C }_{ 2 }{ H }_{ 4 }$
0%
${ CH }_{ 2 }$
0%
${ C }_{ 3 }{ H }_{ 6 }$

Explanation

$RMgX \xrightarrow {Methanol} \underset {A } {CH_3CH_3} \xrightarrow [h \nu] {1 \ mole \ of \ chlorine} \underset {B} {CH_3CH_2-Cl}$

$\underset {B} {CH_3CH_2-Cl}\xrightarrow [Ethanol] {Zn-Cu} \underset {C} {CH_3CH_3}$

$\underset {B} {CH_3CH_2-Cl}\xrightarrow {Na/ dry ether} \underset {D}{CH_3CH_2CH_2CH_3}$
Difference in the molecular formulae of

$D$ and

$C$ is

${ C }_{ 2 }{ H }_{ 4 }$ .

IUPAC name of neo-pentane is

0%
2-ethylpentane
0%
2,2-dimethylpentane
0%
2,2-dimethylpropane
0%
2-methylpropane

IUPAC name is

0%
3, 4-dimethyl -3- propylnonane
0%
4-ethyl -4, 5-dimethyldecane
0%
6, 7-dimethyl -7-ethylnonane
0%
6, 7-dimethyl-1-7-ethyldecane

Which of the following has the maximum

$C - H$ bond length?

0%
$C_{2}H_{4}$
0%
$C_{2}H_{2}$
0%
$C_{2}H_{6}$
0%
$C_{6}H_{6}$

Explanation

Ethane

$C_2H_6$ has maximum

$C-H$ bond length.
Bond length decreases with the increase in s character since an s-orbital is smaller than a

$p-$ orbital.

Compound	Hybridization	$C-H$ bond length ( $A^o$ )
Alkane	$sp^3$	1.093
Alkene	$sp^2$	1.087
Alkyne	$sp$	1.057

The ratio of sigma and pi bonds in benzene is :

0%
$3 :1$
0%
$4 : 1$
0%
$2 : 1$
0%
$1 : 1$

Explanation

The molecular formula of benzene is

$C_{6}H_{6}$ .

Benzene molecule contains 12

$\sigma$ and 3

$\pi$ bonds.

Hence, the ratio is

$4:1$ .

The IUPAC name of the given structure:

0%
2,2 - dimethylbutane
0%
isohexane
0%
2,3 - dimethylbutane
0%
di-isohexane

Explanation

$\Rightarrow$ methyl substituent at C-2 and C-3
So, IUPAC name of the given structure is 2,3-dimethylbutane

The IUPAC name of the given compound is:

0%
3 - ethylpentane
0%
1-ethylpentane
0%
2-ethylpentane
0%
3-methylpentane

Explanation

3-Methylpentane is a branched-chain alkane with the molecular formula

$C_6H_14$

It is a structural isomer of hexane composed of a methyl group bonded to the third carbon atom in a pentane chain.

$\Rightarrow$ 1 methyl substituent c-3.
so the name of given compound will be 3-methyl Pentane

The ratio of pi to sigma bonds in benzene is:

0%
3 : 1
0%
1 : 6
0%
1 : 4
0%
1 : 2

Explanation

no of

$\pi$ Bonds (c-c)=3
no of

$\sigma$ Bonds (c-c)=6
no of

$\sigma$ bonds (c-h)=6
total

$\sigma$ bonds=6+6=12
ratio of

$\pi$ and

$\sigma =\frac{3}{12}=1:4$

Which of the following is the chain propagation step in chlorination of methane?

0%
$CH_{3}+Cl\rightarrow CH_{3}Cl$
0%
$CH_{3}+Cl-Cl\rightarrow CH_{3}-Cl+Cl$
0%
$Cl-Cl\rightarrow Cl+Cl$
0%
$Cl+Cl\rightarrow Cl_{2}$

Explanation

Chain initiation

$Cl_{2}\overset{h2}{\rightarrow}\dot{C}l+C\dot{l}$
chain propagation:

$CH_{3}+Cl-Cl\rightarrow CH_{3}Cl+C\dot{l}$

$CH_{3}Cl+Cl-Cl\rightarrow CH_{2}Cl+C\dot{l}$

$CH_{2}Cl_{2}+Cl-Cl\rightarrow CHCl_{3}+C\dot{l}$

$CHCl_{3}+Cl-Cl\rightarrow\ CCl_{4}+C\dot{l}$

The IUPAC name of the given compound is :

0%
3, 4, 4-trimethylheptane
0%
3, 4, 4- trimethyloctane
0%
2-butyl-2-methyl-3-ethylbutane
0%
2-ethyl-3,3-dimethylheptane

What is the IUPAC name of the given compound?

0%
2-Methylbutane
0%
3-Methylbutane
0%
Pentane
0%
Isopentane

Explanation

Methyl substituent at 2nd carbon.

The chain is of four carbon.

IUPAC name of the given compound is 2-methyl butane.

Hence, option

$A$ is correct.

The

$C -H$ bond distance is the longest in:

0%
$C_{2}H_{6}$
0%
$C_{2}H_{2}Br_{2}$
0%
$C_{2}H_{4}$
0%
$C_{2}H_{2}$

Explanation

$\Rightarrow$ Higher the

$s -$ character means that the Bonded pair electrons are held more closely, ensuring a smaller Bond length and a stronger bond.
So,

$sp$ has the least bond length and it is the strongest

$sp^3$ has longest bond length and weakest.
So,

$A$ is the correct answer.

The IUPAC name of the given structure is :

0%
tertiarybutane
0%
2,2 - dimethylpropane
0%
neopentane
0%
neobutane

Explanation

$\Rightarrow$ 2 methyl groups are substituted at c-2
2-2 dimethylpropane.

The correct IUPAC name of

$CH_{3}-CH_{2}-CH(CH_{3})-CH(C_{2}H_{5})_{2}$ is:

0%
4-ethyl-3-methylhexane
0%
3-ethyl-4-methylhexane
0%
4-methyl-3-ethylhexane
0%
2,4 - diethylpentane

Explanation

Solution:

List of some rules for nomenclature

First we have to find out the long chain and name it.
Then look for the side chains and functional groups.
If the chains of equal length are engaging for selection of the parent chain, the selection goes in the series to:

(i) the greatest number of side chains having by the parent chain.

(ii) the chain whose substituent have the lowest numbers.

(iii) the greatest number of carbon atoms in smaller side chain having by the chain.

(iv) the least branched side chains having by the chain.

The IUPAC name of the compound

$CH_{3}-CH_{2}-CH(CH_{3})-CH(C_{2}H_{5})_{2}$ is 3-ethyl-4-methylhexane.

Hence, the correct option is (B).

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 2 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 2 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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