MCQ Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 5

$CH_3-\overset{CH_3}{\overset{|}{C}H}-CH_3$ (excess)+

$Br_2\xrightarrow{hv} CH_3-\overset{CH_3\!\!\!\!}{\overset{|}{\underset{Br\!\!}{\underset{|}C}}}-CH_3+CH_3-\overset{CH_3}{\overset{|}{C}H}-CH_2-Br$

The relative reactivity of

$1^oH, 2^oH$ and

$3^oH$ in bromination reaction has been found to be 1:82:1600 respectively. In the reaction the percentage yields of the products (A) and (B) are expected to be

0%
99.4%, 0.6%
0%
50%, 50%
0%
0.6%, 99.4%
0%
80%, 20%

2-Bromo-3-phenylpropane can be synthesized by:

0%
$C_6H_5CH_2CH(OH)CH_3+PBr_3 \rightarrow$
0%
$C_6H_5CH=CHCH_3+HBr+ benzoyl peroxide\rightarrow$
0%
$C_6H_5CH_2CH_2CH_3+Br_2+ light\rightarrow$
0%
none of these

Explanation

2-Bromo-3-phenylpropane can be synthesized by reactions A and
B. In reaction A, -OH group is replaced with Br.
(A)

$C_6H_5CH_2CH(OH)CH_3+PBr_3 \rightarrow C_6H_5CH_2CH(Br)CH_3$
In the reaction B, a molecule of HBr is added across C=C bond.
(B)

$C_6H_5CH=CHCH_3+HBr+ benzoyl peroxide\rightarrow C_6H_5CH_2CH(Br)CH_3$
In the reaction C, 1-Bromo-1-phenylpropane is obtained.
(C)

$C_6H_5CH_2CH_2CH_3+Br_2+ light\rightarrow C_6H_5CHBrCH_2CH_3$

Select true statement(s):

0%
Stereochemical outcome of a radical substitution and a radical addition reaction is identical.
0%
I2 does not react with ethane at room temperature even though I2 is more easily cleaved homolytically

than the other halogens.
0%
The rate of bromination of methane is decreased if HBr is added to the reaction mixture.
0%
Allylic chloride adds halogens faster than the corresponding vinylic chloride.

Explanation

The statement (A) is false and the statements (B), (C) and (D) are true.

(A) Stereochemical outcome of a radical substitution and a radical addition reaction are different.
(B)

$I_2$ does not react with ethane at room temperature even though

$I_2$ is more easily cleaved homolytically than the other halogens.

$HBr$ is added to the reaction mixture.

(D) Allylic chloride adds halogens faster than the corresponding vinylic chloride.

An optically active hydrocarbon(X) on catalytic hydrogenation gives an optically inactive compound (Y), C

$_6$ H

$_{14}$ . The hydrocarbon (X) is-

0%
3-methyl-1-pentene
0%
3-methyl-2-pentene
0%
2-ethyl-1-butene
0%
3-methylcyclopentene

Explanation

The optically active hydrocarbon X is 3-methyl-1-pentene

$CH_2=CH-CH(CH_3)CH_2CH_3$ . On catalytic hydrogenation, it forms, 3-methyl pentane

$CH_3CH_2CH(CH_3)CH_2CH_3$ , which is optically inactive.

Name of the compund given below is :-

0%
2,3-Diethyl heptane
0%
5-Ethyl-6-methyl octane
0%
4-Ethyl-3-methyl octane
0%
3-Methyl-4-ethyl octane

The incorrect IUPAC name is:

The correct IUPAC name of the given compound is:

0%
$2-ethyl-3-methylpentane$
0%
$3,4-dimethylhexane$
0%
$2-Sec-butyl$ $butane$
0%
$2,3-dimethyl$ $butane$

Explanation

The correct IUPAC name is

$3,4-dimethylhexane.$
The parent hydrocarbon contains 6 carbon atoms. Hence, it is called hexane.
Two methyl groups are present as substituents at third and fourth carbon atoms.
The numbering from either side gives the same locants to methyl groups.

The IUPAC name of the compound is:

0%
2, 3-Dimethyl hexane
0%
2-Ethyl-4-methyl pentane
0%
3-Ethyl-2-methyl pentane
0%
2, 4-Dimethyl hexane

Which of the following compounds represents 2,2,3-trimethyl hexane ?

0%
$CH_{3}C(CH_{3})_{2}CH_{2}CH_{2}CH(CH_{3})_{2}$
0%
$CH_{3}C(CH_{3})_{2}CH_{2}CH(CH)CH_{2}CH_{3}$
0%
$CH_{3}C(CH_{3})_{2}CH(CH_{3})CH_{2}CH_{2}CH_{3}$
0%
$CH_{3}C(CH_{3})_{2}CH_{2}C(CH_{3})_{2}CH_{3}$

Explanation

The compound of the option

$C$ represents 2,2,3-trimethyl hexane.

It is

$CH_{3}C(CH_{3})_{2}CH(CH_{3})CH_{2}CH_{2}CH_{3}$ . The parent hydrocarbon contains

$6\ C$ atoms. Hence, it is hexane. three methyl groups are present as substituents on second, second and third carbon atoms.

Which of the following halogen reacts with alkanes with explosive violence?

0%
$F_2$
0%
$Cl_2$
0%
$Br_2$
0%
$l_2$

Which alkane will give only one monochlorinated product?

0%
$CH_3CH_2CH_3$
0%
$CH_3CH_2CH_2CH_3$
0%
$(CH_3)_4C$
0%
$CH_3(CH_2)_3CH_3$

On heating sodium propionate with sodalime the hydrocarbon obtained is:

0%
butane
0%
ethane
0%
propane
0%
ethene

Explanation

Sodium propionate reacts with soda lime (NaOH+CaO) to form ethane.

The reaction can be given as follows:

$\underset{sodium\ propionate}{C_2H_{5}COONa} + NaOH \xrightarrow {CaO/\Delta} CH_{3}-CH_{3} + Na_{2}CO_{3}$

Option

$(B)$ is correct.

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

0%
Methyl
0%
Amino
0%
Hydroxyl
0%
Chloro

On mixing a certain alkane with chlorine and irradiating it with ultraviolet light, it forms only one monochloroalkane. This alkane could be :

0%
neopentane
0%
propane
0%
pentane
0%
isopentane

Explanation

Neopentane, also called 2,2-dimethylpropane, is a double-branched-chain alkane with five carbon atoms with formula;

$CH_3-C(CH_3)_2-CH_3$ . Thus it gives only one mono chloroalkane on chlorination.

Which of the following statement is correct in relation to the halogenation of alkane?

0%
The reactivity of chlorine is less than bromine towards alkanes.
0%
For photochemical chlorination of methane, CI is formed in slowest step.
0%
Free radicals are pyramidal intermediate, stabilised by hyperconjugation and resonance.
0%
Bromine has much higher regioselectivity than chlorine in abstracting $3^{\circ}$ hydrogen.

Benzene reacts with n - propyl chloride in the presence of anhydrous

$AlCl_3$ to give predominantly:

0%
n - propylbenzene
0%
isopropylbenzene
0%
3 - propyl - 1 - chlorobenzene
0%
no reaction

Identify the aromatic compound?

Explanation

Hint - Aromatic compounds obeys Huckel Rule

$\bf{Explanation-}$

Conditions for aromatic compounds:

1. It must be planar(flat), cyclic, and conjugated

$\pi-$ electrons system.

2. It must obey Huckel's rule.

Huckel's rule for aromatic compound:

Number of

$\pi-$ electrons

$=(4n+2)$ . Where, n = 0, 1, 2, .....

$\bullet$ Positive chage has no

$\pi-$ electrons while negative charge has

$2\ \pi-$ electrons.

In options A, C, and D, the number of

$\pi-$ electrons are

$4,4$ and

$8$ respectively. Hence, they do not follow the

$(4n+2)$ rule of aromaticity.

In option

$B$ , the number of

$\pi-$ electrons are

$6$ . Hence, it follows the

$(4n+2)$ rule and since, it is planar and conjugated so it follows all the conditions of aromaticity.

$\bf{Conclusion-}$ Hence, option

$B$ is correct.

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

0%
$(CH_3)_2CH - CH_2 - CH (CH_3)_2$
0%
$(CH_3)_2CH - CH_2 - CH_2 - CH(CH_3)_2$
0%
$CH_3 - CH_2 - C(CH_3)_2CH_2 - CH_3$
0%
$(CH_3)_3C - CH_2 - CH_2 - CH_3$

Explanation

$\bf{Hint:}$ 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.

The reaction least likely to occur is:

Explanation

In the presence of ultraviolet light (but without a catalyst present), hot benzene will also undergo an addition reaction with chlorine or bromine. The ring delocalization is permanently broken and a chlorine or bromine atom adds on to each carbon atom.

$C_6H_6 + Cl_2\xrightarrow{heat} C_6H_5Cl$ .

The increasing order of stability of the following free radicals is:

0%
$(CH_2)_2\, \dot{C}H\, <\, (CH_3)_3\, \dot{C}\, <\, (C_6H_5)_2\, \dot{C}H\, <\, (C_6H_5)_3\, \dot{C}$
0%
$(C_6H_5)_2\, \dot{C}\, <\, (C_6H_5)_2\, \dot{C}H\, <\, (CH_3)_3\, \dot{C}\, <\, (CH_3)_2\, \dot{C}H$
0%
$(C_6H_5)_2\, \dot{C}H\, <\, (C_6H_5)_3\, \dot{C}H\, <\, (CH_3)_3\, \dot{C}\, <\, (CH_3)_2\, \dot{C}H$
0%
$(CH_3)_2\, \dot{C}H\, <\, (CH_3)_3\, \dot{C}\, <\, (C_6H_5)_3\, \dot{C}H\, <\, (C_6H_5)_2\, \dot{C}H$

Which statements is/are correct about photochemical chlorination of the above compound?

$Ph-CH_2-\overset{CH_3}{\overset{|}{C}H} - CH_3\xrightarrow[(monochlorination)]{Cl_2/h\upsilon }$

0%
The major product will have chiral carbon atom but is optically inactive.
0%
The intermediate free radical of the major product is resonance stabilised.
0%
The intermediate free radical is tertiary for major product.
0%
The intermediate free radical is planar, and stabilised by only hyper conjugation.

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

0%
3
0%
4
0%
5
0%
6

Explanation

B. 4
2-Chlorobutane as reactant will produce four

${ C }_{ 4 }{ H }_{ 8 }{ Cl }_{ 2 }$ .The products are-

${ ClCH }_{ 2 }{ CHClCH }_{ 2 }{ CH }_{ 3 }+{ CH }_{ 3 }{ CCl }_{ 2 }{ CH }_{ 2 }{ CH }_{ 3 }+\\ { CH }_{ 3 }{ CHClCHClCH }_{ 3 }+{ CH }_{ 3 }{ CHClCH }_{ 2 }{ CH }_{ 2 }Cl$

Which reaction would not give propene?

0%
Adding excess hot concentrated sulfuric acid to $propan-1-ol$
0%
Adding warm aqueous sodium hydroxide to $2-bromopropane$
0%
Adding warm ethanolic sodium hydroxide to $1-bromopropane$
0%
Passing $propan-2-ol$ vapour over heated aluminium oxide

Explanation

Adding warm aqueous sodium hydroxide to

$2$ −bromopropane will give Propane-

$2$ -ol.

${H}_{3}C-\overset{\underset{|}{Br}}{CH}-C{H}_{3} + NaOH \longrightarrow {H}_{3}C-\overset{\underset{|}{OH}}{CH}-C{H}_{3} + NaBr$

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

0%
1-Bromo-3-methylbutane
0%
1-Bromo-2-methylbutane
0%
2-Bromo-3-methylbutane
0%
2-Bromo-2-methylbutane

l-bromo-3-chloro cyclobutane will react with two moles of Na in ether producing:

The general molecular formula of alkynes is______________.

0%
$C_nH_{2n}$
0%
$C_nH_{2n\, -\, 4}$
0%
$C_nH_{2n\, -\, 2}$
0%
$C_nH_{2n\, +\, 2}$

Explanation

The general formula of alkynes is ;

$C_{n}H_{2n-2}$

In an alkane if the value of n = 19, it is:

0%
solid
0%
liquid
0%
gaseous
0%
none of the above

Explanation

The homologous of alkanes larger than hexadecane are solids, solid alkanes are normally soft, with low melting points. These characteristics are due to strong repulsive forces generated between electrons on neighboring atoms, which are in close proximity in crystalline solids. The strong repulsive forces counterbalance the weak van der Waals forces of attraction, alkanes are almost completely insoluble in water. If n=19, then the alkane

$C_{19}H_{40}$ is a solid.

Among the following compounds 2, 3-dimethylexane is :

0%
i
0%
ii
0%
iii
0%
iv

Methane with the Molecular formula

$"CH_4"$ has

0%
$4$ Covalent bonds
0%
$8$ Covalent bonds
0%
$6$ Covalent bonds
0%
$2$ Covalent bonds

Explanation

Structure of methane

$H-\overset{\displaystyle H}{\overset{|}{\underset{\displaystyle H}{\underset{|}{C}}}}-H$

It has

$4$ covalent bonds.

Calcium carbide reacts with water to produce a gas which is used for ripening of fruits like bananas. The gas is a hydrocarbon containing a triple bond in it. The general molecular formula of the homologous series to which the gas belongs is :

0%
$C_nH_{2n+1}OH$
0%
$C_nH_{2n+2}$
0%
$C_nH_{2n}$
0%
$C_nH_{2n-2}$

Explanation

Calcium carbide reacts with water to produce acetylene gas(

$CH\equiv CH$ ) which is an alkyne. Thus, the general formula of alkynes is;

$C_{n}H_{2n-2}$ .

Which of the following is properly matched?

0%
Kerosene - $C_5\, -\, C_{10}$
0%
Diesel oil - $C_5\, -\, C_6$
0%
Petrol - $C_7\, -\, C_9$
0%
All are correct

Explanation

The number of carbon atoms in petrol is

$C_{7}-C_{9}$ .

The boiling point will increase in the order of:

0%
n-butane< iso-butane
0%
n-butane <= iso-butane
0%
n-butane= iso-butane
0%
iso-butane< n-butane

Equal weights

${C}_{2}{H}_{6}$ and hydrogen are mixed in an empty container at

${0}^{o}C$ . The fraction of total pressure exerted by hydrogen is-

0%
$1:2$
0%
$1:1$
0%
$1:16$
0%
$15:16$

Explanation

Consider x gm be the weight of

$C_{2}H_{6}$ and hydrogen.
No. of moles of hydrogen = x/2
No. of moles of

$C_{2}H_{6}$ = x/30
Mole fraction of hydrogen =( x/2)/(x/2+x/30) = 15/16
The pressure exerted by hydrogen = mole fraction of hydrogen * total pressure
= 15/16 * 1 = 15/16

Which of the following having delocalised electrons?

0%
Benzene
0%
Cyclohexane
0%
$CH_4$
0%
$C_2H_6$

Explanation

Presence of delocalised

$\pi$ electrons is characteristic of aromatic compounds. Benzene is an aromatic compound having 6 delocalised electrons.

X compound reacts with Na to give

$CH_3CH_2CH_2CH_3$ . The compound X is:

0%
$CH_3 CH_2 OH$
0%
$CH_3 CH_2 - Cl$
0%
$CH_3 - CH_3$
0%
$CH_3 - CH_2 - CH_2 - CH_2 - OH$

Explanation

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

$H - C \equiv C - H$ is a structure of :

0%
acetaldehyde
0%
ethanal
0%
ethane
0%
acetylene

Explanation

The molecule of acetylene

$C_2 H_2$ is represented by

$H - C \equiv C - H$

For which of the following compounds the least number of a mono-chlorinated product is possible?

0%
$CH_3 CH_2 CH_2 CH_2 CH_3$
0%
$CH_3 - \underset{CH_3}{\underset{|}{C}H} - CH_2 CH_3$
0%
$CH_3 -\!\!\!\!\! \underset{\,\,\,\,CH_3}{\underset{|}{\overset{\quad CH_3}{\overset{|}{C}}} \!\!\!\!-}\, H$
0%
$CH_3 -\!\!\!\!\! \underset{\,\,\,\,CH_3}{\underset{|}{\overset{\quad CH_3}{\overset{|}{C}}} \!\!\!\!-} \,CH_3$

Explanation

$CH_3 -\!\!\!\!\! \underset{\,\,\,\,CH_3}{\underset{|}{\overset{\quad CH_3}{\overset{|}{C}}} \!\!\!\!-} \,CH_3$

will give only one type of mono chlorinated product because all the hydrogen atoms are of same type.

The compound formed when ethyl alcohol is treated with conc.

${H}_{2}{SO}_{4}$ at

${170}^{o}C$ is _______.

0%
ethane
0%
ethene
0%
ethyne
0%
ethanal

Explanation

$Ethanol$ can be dehydrated to give

$ethene$ by heating it with an excess of concentrated sulphuric acid at about 170C.

$CH_3CH_2-OH \overset {Conc. H_2SO_4 170^oC}{\rightarrow} CH_2=CH_2 + H_2O$

Which of the following compounds on hydrolysis produces methane?

0%
$Ca{CN}_{2}$
0%
$Ca{C}_{2}$
0%
${Al}_{4}{C}_{3}$
0%
$SiC$

Explanation

Aluminium carbide on hydrolysis produces methane, and aluminium hydroxide.

$Al_4C_3 + 12H_2O\rightarrow 4Al(OH)_3 + 3CH_4$
Hydrolysis is a reaction that takes place between a substance and water, as a result of which the substance and water break down, and new compounds form.

Ethylene is a/an :

0%
saturated hydrocarbon
0%
alkene
0%
paraffin
0%
aromatic compound

Explanation

Ethylene is an alkene (unsaturated hydrocarbon) with the structure;

$CH_{2}= CH_{2}$

Which one of the following lists the alkane in order of decreasing boiling point?

0%
Octane, Methane, Butane, Ethane
0%
Methane, Ethane, Propane, Butane
0%
Hexane, Octane, Propane, Methane
0%
Hexane, Pentane, Propane, Ethane

Explanation

As the number of carbon atoms in alkanes increases the number of different isomers increases dramatically. To give an indication of the variation in boiling points of different isomers of the same alkane, cycloalkanes have boiling points typically around

$10^0C$ to

$20^0C$ higher than the corresponding linear alkane. Boiling points of alkanes increase with an increase in the number of n. A straight-chain alkane will have a boiling point higher than a branched-chain alkane due to the greater surface area in contact, thus the greater van der Waals forces, between adjacent molecules. Thus Hexane>Pentane>propane>ethane.

Methyl acetylene is also known as :

0%
butyne
0%
propyne
0%
ethyne
0%
pentyne

Explanation

Propyne (methylacetylene) is an alkyne with the chemical formula as follow:

$CH_{3}-C\equiv CH$

_________ on heating with alcoholic potash loses one molecule of hydrogen halide to give alkene.

0%
Halogens
0%
Haloalkanes
0%
Haloakynes
0%
None of the above

What is the reagent used and appropriate temperature required to prepare ethene from ethyl alcohol?

0%
Dilute $H_2SO_4 , 160^\circ C$
0%
Concentrated $H_2SO_4 , 170^\circ C$
0%
Dilute $HCl, 170 ^\circ C$
0%
Concentrated $HCl , 160^\circ C$

Explanation

Ethene is prepared by the action ofconc.

$H_2SO_4$ on ethanol.

In dehydration of alcohol, a molecule of ________ is eliminated from an alcohol molecule by heating the alcohol.

0%
carbon
0%
hydrogen
0%
water
0%
oxygen

Which compounds are aromatic?

0%
(i) & (iii)
0%
(ii) & (iv)
0%
(i) & (ii)
0%
(iii) & (iv)

Explanation

$\bf{Hint:}$ An aromatic compound is a planer and conjugated cyclic system accompanied by delocalized

$pi$ electrons.

$\bf{Correct Answer:}$ Option

$A$

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

Huckel's rule says that a compound is aromatic if it follows the following conditions.

The compound must be cyclic.
It should be planar and it should have a conjugated system.
It should have $(4n+2) \pi$ electrons, where $n = 0, 1, 2, 3...$

Compound $(i)$ has a planar structure and a conjugated system which follow Huckel's rule of $(4n+2) \pi$ electrons
if $n = 1$ , then $(4n+2) \pi$ = $[(4\times 1+2) \pi]$ = $6\pi$ electrons
Hence it is an aromatic compound.

Compound $(iii)$ has a planar structure and a conjugated system which follow Huckel's rule of $(4n+2) \pi$ electrons
if $n = 1$ , then $(4n+2) \pi$ = $[(4\times 1+2) \pi]$ = $6\pi$ electrons
Hence it is an aromatic compound.

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

Compound $(ii)$ has no conjugation due to absence of $\pi$ electron cloud like in compound $(i)$ and is not planar. Also, it does not follow Huckel's rule of $(4n+2) \pi$ electrons, hence it is not aromatic.

Compound $(iv)$ has a cyclic structure that has no conjugation due to absence of $\pi$ electron cloud like in compound $(i)$ and it could not follow Huckel's rule of $(4n+2) \pi$ electrons, hence it is not aromatic.

For production of alkene, alcohol is heated in presence of:

0%
conc. $H_2SO_4$
0%
conc. $H_3PO_4$
0%
anhydrous $ZnCl_2$
0%
all of the above

Explanation

Alkene is prepared by heating an alcohol with concentrated mineral acid, such as concentrated

$H_2SO_4$ or concentrated

$H_3PO_4$ . Anhydrous zinc chloride can also be used as a dehydrating agent.

Which of the following represents alkynes?

0%
-C-C
0%
-C=C-
0%
-C $\equiv$ C -
0%
none of these

Alkenes can be easily obtained by _______ of alkynes.

0%
Hydrogenation
0%
Oxidation
0%
Reduction
0%
Halogenation

Explanation

Alkenes can be easily obtained by hydrogenation of alkynes. An alkyne on controlled hydrogenation with hydrogen in the presence of Ni or Pd at

$200^o$ C gives a corresponding alkane.

Alkene can be produced by:

0%
dehydration of alcohols
0%
dehydrohalogenation of alkyl halides
0%
dehalogenation of haloalkanes
0%
all of above

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 5 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 5 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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