MCQ Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 11

Which of the following statements regarding 1,2-dimethyl cyclopentane and 1,3-dimethyl cyclopentane is/are correct?

0%
In both , cis form is meso,while trans form is resolvable
0%
In both , trans form is meso,while cis form is resolvable
0%
In both,cis and tans forms are meso.
0%
In both,cis and tans forms are resolvable.

Compound

$(E)$ is:

0%
Propyne
0%
Butyne
0%
cis-But $-2-$ yne
0%
trnas-But $-2-$ yne

Explanation

Compound

$(E)$ , answer is

$(b)$

1693160_1768241_ans_8aab26a4f33a44b789ec41b0036940c8.png

Acidic hydrogen is present in:

0%
Ethyne
0%
Ethene
0%
Benzene
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Ethane

Explanation

$H$ atom attached to

$sp-C$ atom is more acidic than

$sp^2-C$ atom, which in turn is more acidic than one attached to

$sp^3-C$ atom.

Which is the correct IUPAC name for

$CH_3CH(C_2H_5)CH_2-Br$

0%
1-Bromo-2-ethylpropane
0%
1-Bromo-2-ethyl2-methylethane
0%
1-Bromo-2-methylbutane
0%
2-Methyl-1-bromobutane

Bromine reacts with saturated hydrocarbon at room temperature in the

0%
absence of sunlight
0%
presence of water
0%
presence of sunlight
0%
presence of hydrochloric acid

The number of single and double bonds present in benzene are

0%
$9$ and $6$
0%
$9$ and $3$
0%
$12$ and $3$
0%
$12$ and $6$

IUPAC name of iso-butane is :

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2-methyl butane
0%
2-methyl propane
0%
2-ethyl butane
0%
2-butyne

Which of the following groups do not show addition reactions?

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Alkanes
0%
Alkaloids
0%
Cycloalkenes
0%
Ketones

The IUPAC of the hydrocarbon represented by the Swastik sign is

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Neononane
0%
Tetraethylcarbon
0%
$2$ -Ethylpentane
0%
$3,3$ -Diethylpentane

The IUPAC for the compound is

0%
$2$ -Cyclopentylpropane
0%
$1,1$ -Dimethyl- $1$ -cyclopentylmethane
0%
( $1$ -Methylethyl)cyclopentane
0%
$1$ -Cyclopentylpropane

The IUPAC name of

$CH_{3} - \underset{C_{2}H_{5}\!\!\!\!}{\underset{|} {\overset{H} {\overset{|} {C}}}} - \underset{CH_{3}\!\!\!\!\!\!}{\underset{|} {\overset{C_4H_9} {\overset{|} {C}}}} - CH_{3}$ is

0%
$2$ -Butyl- $2$ -methyl- $2$ -ethylbutane
0%
$2$ -Ethyl- $3,3$ -dimethylheptane
0%
$3 , 4 , 4$ -Trimethylheptane
0%
$3,4 ,4$ -Trimethyloctane.

What is the better way to prepare

$3$ -ethyl-

$3$ -heptene?

0%
Make a Wittig reagent from $1$ -bromobutane and combine that with $3$ -pentanone
0%
Make a Wittig reagent from $3$ -bromopentane and combine that with butanal
0%
It does not matter because they will work equally well
0%
It does not matter because neither method will be successful

What is the ratio of

$\pi-$ and

$\sigma$ - bonds in benzene?

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

Explanation

The ratio of σ and π bonds in benzene is 4:1. Benzene has molecular formula

$C_{6}H_{6}$ o there is the presence of one double bond in between alternate carbon atoms and each carbon is attached to two carbon atoms by means of a single bond and a double bond and a hydrogen atom by a single bond. This we can observe by seeing its structure So there are a total of 12 σ bonds and 3 π bonds in benzene. It is 4:1.

Hence option

A

is correct.

Which of the following will show aromatic behaviour?

Explanation

Only option B has a conjugated system of

$6\pi$ - electrons and is planar. While cyclopentadienyl cation has only

$4\pi$ - electrons even though planar so it is anti-aromatic.

An organic compound on treatment with

$Br_2$ in

$CCl_4$ gives bromoderivative of an alkene. The compound will be

0%
$CH_3-CH=CH_2$
0%
$CH_3CH=CHCH_3$
0%
$HC\equiv CH$
0%
$H_2C=CH_2$

Explanation

Solution:

On adding $Br_2$ to $C_2H_2$ i.e. Acetylene in $CCl_4$ gives Bromo derivative of an alkene.

$CH\equiv CH\xrightarrow[Br_2]{CCl_4}\ BrCH=CHBr$

option C is correct

Bracetylene gives 1,2-dibromoethaen-a bromoderivative of an alkene

Among the following the aromatic compounds is

Explanation

Only

$(A)$ has

$2 \pi-$ electrons and planar so is aromatic. All others have

$4 \pi$ electrons and hence are anti-aromatic.

Elimination reaction of 2-bromo-pentane to form pent-2-ene is:
a)

$\beta$ - Elimination reaction
b) Follows Zaitsev rule
c) Dehydrohalogenation reaction
d) Dehydration reaction

0%
a), c), d)
0%
b), c), d)
0%
a), b), d)
0%
a), b), c)

Explanation

Refer above image,
This reaction is an example of

$\beta$ - elimination.

Hydrogen is removed from

$\beta$ - carbon and halogen from

$\alpha$ - carbon, dehydrohalogenation reaction

Generally in E2 reaction Zaitsev alkene is formed as major product (more stable alkene)

2069537_1974591_ans_6889328621fe406397643efc506b6ba1.png

Assertion: In the Kolbe's electrolysis of an aqueous solution of sodium acetate, hydrogen gas is evolved at the cathode.
Reason:

$H^{+}$ is easily reduced than

$Na^{+}$ .

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Both Assertion and Reason are correct and Reason is the correct explanation of Assertion
0%
Both Assertion and Reason are correct but the Reason is not the correct explanation of Assertion
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The Assertion is correct but the Reason is incorrect
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The Assertion is incorrect but the Reason is correct

Explanation

Kolbes electrolysis:
2

$CH_3COO-Na$ + + 2

$H_2O$

$\to$

$CH_3CH_3$ + 2

$CO_2$ +

$H_2$ +

$2NaOH$

Reaction at cathode

$2Na^++ 2e^-\rightarrow 2Na\xrightarrow {H_2O} 2NaOH+H_2$

In this process, Na reduces

$H^+$ to form hydrogen gas as reduction potential of sodium ion is less than that of hydrogen ion.

Aromatization in alkane can be brought by using:

0%
$Al_{2}O_{3}$
0%
$Fe_{2}O_{3}$
0%
$Cr_{2} O_{3} / Al_{2} O_{3}$ at $600^{o} C$
0%
Conc. $H_{2} SO_{4}$

Explanation

Aromatization in alkane can be brought by using

$Cr_{2} O_{3} / Al_{2} O_{3}$ at

$600^{0} C$ .

This process is known as reforming.

In this reaction, one hydrogen atom is removed from each terminal carbon atom of n-hexane to form cyclohexane. This is cyclization reaction.

This is followed by the removal of three hydrogen molecules to form benzene. This is aromatization reaction.

Option C is correct.

Match the following

List -I (Products)	List -II (Reactants-Reaction)
A) Methane	(1) Sabatier - Senderens reaction of propylene
B) Ethane	(2) Wurtz reaction of methyl iodide
C) Propane	(3) Decarboxylation of sodium acetate
D) Butane	(4) Sodium butanoate by Kolbes electrolysis

(5) Sodium propanoate by Kolbes electrolysis

The correct match is :

0%
A-2 B-3 C-1 D-5
0%
A-4 B-2 C-1 D-5
0%
A-3 B-2 C-1 D-5
0%
A-2 B-3 C-1 D-4

Explanation

Wurtz reaction:

$RX+2Na+XR\rightarrow R-R+2NaX$
Sabatier senderens reaction:

$C_nH_{2n}+ H_2 \rightarrow C_nH_{2n+2} + heat$
Kolbe's electrolysis:

$2RCOOK+2H_2O\rightarrow R-R+2CO_2+2KOH+H_2$
In decarboxylatioin of sodium acetate

$R-H$ form as a product, In all other reaction

$R-R$ forms as a product.
In Sodium propanoate by Kolbes electrolysis reaction occur at anode:

$2C_2H_5COO^{\bullet}\rightarrow 2C_2H_5^{\bullet}+2CO_2$

$2C_2H_5^{\bullet} \rightarrow C_4H_{10}$

DBN is a bicyclic compound which is used as base. What is the major product in the following reaction?

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0%
0%
0%
Both A and B

The total number of stereoisomers of

$(CH_3\! -\! \! CH\! \! =\! \! CH\! -\! \! CH_2 )_3 CH$ is:

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6
0%
4
0%
3
0%
5

Explanation

The compound

$(CH_3\! -\! \! CH\! \! =\! \! CH\! -\! \! CH_2-\! \! \! \! ) _3 CH$ has four geometrical isomers. Which are either of E type or Z type geometrical isomerism.

$\left( { CH }_{ 3 } \right) _{ 3 }C-MgCl$ on reaction with

${ D }_{ 2 }O$ produces :

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$\left( { CH }_{ 3 } \right) _{ 3 } CD$
0%
$\left( { CH }_{ 3 } \right) _{ 3 } OD$
0%
$\left( { CD }_{ 3 } \right)C MgCl$
0%
$\left( { CD }_{ 3 } \right) _{ 3 } OD$

Explanation

$\left( { CH }_{ 3 } \right) _{ 3 }C-MgCl$ is a strong base.

$\left( { CH }_{ 3 } \right) _{ 3 }C-MgCl$ +

$D_2O$

$\longrightarrow$

$\left( { CH }_{ 3 } \right) _{ 3 }C-D$ +

$MgBrOD$ .

The total number of isomeric bromo-chloro-fluoro-iodopropadienes is:

0%
5
0%
6
0%
7
0%
8

Explanation

There are

$6$ isomers possible for the given compound. All of these isomers are optical isomers i.e. d and l isomers.

The nature of the aqueous solution after Kolbe electrolysis is:

0%
acidic
0%
alkaline
0%
neutral
0%
may be acidic, alkaline or neutral

Explanation

The reaction of Kolbe electrolysis:

$2RCOOK+2HOH\xrightarrow{electrolysis} R-R+2CO_2+2KOH+H_2$
After completion of reaction solution will be alkaline due to formation of

$KOH$ (strong base).

In the above reaction, 'A' is a carbide of

$3^{rd}$ period element which forms amphoteric oxide. D can be converted into B by using:

0%
$Ni$
0%
$Na/ether$
0%
$Zn- Hg / HCl$
0%
${ LiAlH }_{ 4 }$

Explanation

$Al_4C_3+H_2O \rightarrow CH_4 \xrightarrow {O_2/Mo_2O_3/500^oC} HCHO$
A B D

Clemmensen reduction:

Aldehyde and ketone are reduced to alkanes in presence of

$Zn-Hg$ and conc.

$HCl$

$HCHO \xrightarrow {Zn-Hg/HCl} CH_4$

The hydrolysis of metal carbide A gives metal hydroxide and methane B.

Methane B on oxidation with

$Mo_2O_3$ at

$500^0C$ gives Formaldehyde D.

The carbonyl group in formaldehyde D can be reduced to the methylene group to obtain methane B by using zinc amalgam in conc HCl. This reaction is known as Clemmensen's reduction.

So, the correct option is 'C'.

The compound with the highest boiling point is

0%
n-hexane
0%
n-pentane
0%
2, 2-dimethylpropane
0%
2-methylbutane

The correct IUPAC name of the following hydrocarbon is :

0%
4,5-diisopropyl-3-methylheptane
0%
3-methyl-4,5-diisopropylheptane
0%
5-ethyl-4-isopropyl-3,6-dimethylheptane
0%
3-Ethyl-4-isopropyl-2,5-dimethylheptane

Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure.
The correct order of their boiling point is:

0%
$I > II > III$
0%
$III > II > I$
0%
$II > III > I$
0%
$III > I > II$

Explanation

The correct order of their boiling point is

$III > II > I$
More the branching in an alkane, lesser will be the surface area, lesser will be the boiling point as weaker will be the intermolecular attractive forces.

The correct IUPAC name of the following hydrocarbon is:

0%
4-ethyl-6-methyl-5-propyloctane
0%
3-methyl-4-propyl-5-ethyloctane
0%
5-Ethyl-3-methyl-4-propyloctane
0%
5-ethyl-4-propyl-3-methyloctane

The major product A is :

Number of secondary carbon atoms present in the above compounds are respectively :

0%
6, 4, 5
0%
4, 5, 6
0%
5, 4, 6
0%
6, 2, 1

The major product obtained on acid catalysed hydration of 3-phenylbut-1-ene is:

0%
3-phenylbutan-2-ol
0%
2-phenylbutan-2-ol
0%
2-phenylbutan-3-ol
0%
2-phenylbutan-1-ol

Major product is

Identify the product formed.

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0%
0%
0%
no reaction

Based upon your understanding of the basic principles of molecular orbitals it should be possible to answer the following question.Consider the six molecular orbitals of benzene shown. How many of these orbitals will contain electrons?

0%
Only one molecular orbitals
0%
Only two molecular orbitals
0%
Only three molecular orbitals
0%
All six molecular orbitals

Heating benzene in a large excess of 80%

$D_{ 4 }{ SO }_{ 2 }$ in

$D_{ 2 }O$ results in what product?

0%
${ C }_{ 6 }{ H }_{ 5 }{ SO }_{ 3 }D$
0%
${ C }_{ 6 }{ H }_{ 5 }OD$
0%
${ C }_{ 6 }{ H }_{ 5 }D$
0%
${ C }_{ 6 }{ D }_{ 6 }$

Explanation

It replaces all H of benzene so product formed is

$C_6D_6$ .

Consider two alternative mechanisms for the chlorination of methane, Which step is crucial in deciding which mechanism is not likely to be observed?
Mechanism I:

$A.\ \:Cl_{2} \rightarrow 2 Cl ^{*}$

$B. \ \:Cl^{*}+CH_{4} \rightarrow HCl+^{*}CH_{3}$

$C. \ \:\dot{C}H_{3}+Cl_{2} \rightarrow CH_{3} Cl+Cl^{*}$
Mechanism II:

$D.\ \:Cl_{2} \rightarrow 2 Cl ^{*}$

$E. \ \:Cl^{*}+CH_{4} \rightarrow CH_{3}Cl+H ^{*}$

$F. \ \:H^{*}+Cl_{2} \rightarrow HCl+Cl^{*}$

0%
$I$
0%
$II$
0%
$I$ and $II$ both
0%
None

Explanation

In chlorination of methane in presence of sunlight is the free radical mechanism where chain elongation step is rate determining step. For that formation of

$Cl^{*}$ free radical is crucial.

Chain elongation process-

$Cl_2+CH_{3}^{*} \rightarrow CH_3Cl+Cl^{*}$

Which of the following will not form alkene as a product?

Explanation

Option A has no

$\beta-H$

In option B, alcohol is not dehydrated with alkali

So options A & C will not form alkene as a product

Hence options A & C are correct.

Consider two alternative mechanism for the chlorination of methane. Which step is crucial in deciding which mechanism is not likely to be observed?
Mechanism I Mechanism II

A ${ Cl }_{ 2 }\longrightarrow { 2Cl }^{ \bullet }$	A ${ Cl }_{ 2 }\longrightarrow { 2Cl }^{ \bullet }$
B ${ Cl }^{ \bullet }+{ CH }_{ 4 }\longrightarrow HCl+_{ }^{ \bullet }{ { CH }_{ 3 } }$	D ${ Cl }^{ \bullet }+{ CH }_{ 4 }\longrightarrow { CH }_{ 3 }Cl+{ H }^{ \bullet }$
C $\overset { \bullet }{ C } { H }_{ 3 }+{ Cl }_{ 2 }\longrightarrow { CH }_{ 3 }Cl+{ Cl }^{ \bullet }$	E ${ H }^{ \bullet }+{ Cl }_{ 2 }\longrightarrow HCl+{ Cl }^{ \bullet }$

0%
A
0%
B
0%
C
0%
D

Explanation

According to the reaction, step B is crucial and it will tell whether mechanism is possible or not.

1. Initial reaction

$Cl_2\xrightarrow{h\nu} 2Cl\bullet$

2. Chain propagation

$Cl\bullet+CH_4\rightarrow HCl+CH_3\bullet$

$CH_3\bullet+Cl_2\rightarrow CH_3Cl+Cl\bullet$

3. Chain termination reactions

$Cl\bullet+Cl\bullet\rightarrow Cl_2$

$CH_3\bullet+CH_3\bullet\rightarrow CH_3-CH_3$

$Cl\bullet+CH_3\rightarrow CH_3Cl$

Which

${ C }_{ 9 }{ H }_{ 12 }$ hydrocarbon would give a single

${ C }_{ 9 }{ H }_{ 11 }{ SO }_{ 3 }H$ product on sulphonation?

How many products will be formed excluding stereo when cis-1,3,5-terimethyl cyclohexene reacts with NBS?

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

Explanation

When cis-1,3,4-trimethylcyclohexene reacts with NBS, bromine is added to the carbon next to the double bond. 3 such carbon positions are possible. So, 3 products are formed. Hence, answer is

$A$ .

Based upon your understanding of the basic principles of molecular orbitals it should be possible to answer the following question. Consider the molecular orbitals of benzene shown. Which of these molecular orbitals has the lowest energy?

0%
Orbital $II$
0%
Orbital $III$
0%
Orbital $IV$
0%
Orbital $V$

Alkene ( P ) & ( Q ) respectively are

How many dichlorocyclohexane would be obtained on chlorination of chlorocyclohexane (including stereoisomers)?

0%
$4$
0%
$6$
0%
$8$
0%
$9$

Explanation

Chlorination of chlorocyclohexane gives total 9 dichlorocyclohexanes.

Chlorination replaces one

$-H$ atom by one

$-Cl$ atom. There are four types of hydrogens with respect to previous

$-Cl$ .

Product 1 - Does not have any chiral carbon so only one product

Product 2 - Two chiral centres so RR, RS, SS isomers possible.

Product 3 - Two chiral centres so RR, RS, SS isomers possible.

Product 4 - Two chiral centres but plane of symmetry exists so only two isomers possible.

The major monochlorination product that forms upon treatment of

$1,1,2-$ trimethyl cyclopentane with

$Cl_{2}/hv$ is:

0%
2-Chloro-1,1,2-trimethyl cyclopentane
0%
1-Chloro-2,2,3-trimethyl cyclopentane
0%
1-Chloro-2,3,3-trimethyl cyclopentane
0%
1-Chloro-1,2,2-trimethyl cyclopentane

Explanation

Reaction of

$Cl_2$ in presence of sunlight with alkane follows radical mechanism. Free radical is most stable at tertiary carbon so treatment of 1,1,2-trimethyl cyclopentane will give 1-chloro-1,2,2-trimethyl cyclopentane as major product.

What is the IUPAC name of above compound ?

0%
(Z)-6-bromo-2-methylhept-5-en-3-yne
0%
(Z)-2-bromo-5,5-dimethylhept-2-en-4-yne
0%
(Z)-2-bromo-6-methylhept-2-en-4-yne
0%
(Z)-2-bromo-5-methylhept-2-en-4-yne

Consider the following carbides:

$\underset {I}{CaC_2} {\;} \underset {II}{Be_2C} {\;} \underset {III}{MgC_2} {\;} \underset {IV}{SrC_2}$
Select the carbide which gives different product on hydrolysis than other carbides.

0%
I
0%
II
0%
III
0%
IV

Explanation

$CaC_2 + 2 H_2O \rightarrow C_2H_2 + Ca(OH)_2$

$Be_2C + 4 H_2O \rightarrow 2 Be(OH)_2 + CH_4$

$MgC_2 + 2 H_2O \rightarrow Mg(OH)_2 + C_2H_2$

$SrC_2+H2O\rightarrow Sr(OH)_2+C_2H_2$

So only option B will give a different product.

Hence option B is correct.

What product(s) is (are) expected during electrolysis of aqueous solution of potassium propanoate ?

0%
Butane
0%
Ethyl ethanoate
0%
Ethyl propanoate
0%
(CH3CO) O - O (OCCH3)

Explanation

$CH_3CH_2COO^{-}\rightarrow CH3CH2COO^{*}$ (ethyl propanoate).Electrolysis

$I^{-}+CO_2\rightarrow CH_3CH_2^{*}\rightarrow CH_3CH_2CH_2CH_3$ (Butane).Coupling.

II (a)

Hence options A & C are correct.

The sequence of reagent which can efficiently bring about the above conversion is (are) ?

0%
$Br_{2}$ -hv/heat
0%
$Br_{2}$ -hv/heat, $C_{2}H_{5}ONa/C_{2}H_{5}OH;NBS-hv;B_{2}H_{6}- THF$ then $CH_{3}COOH$
0%
$Br_{2}$ -hv/heat ; $(CH_{3})_{3}CONa-(CH_{3})_{3}COH HBr-ROOR$
0%
NBS-hv/ $CCl_{4}$

Explanation

Hence options

$B$ and

$C$ are correct.

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

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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