Discharge colour of bromine water

2 Butyl 2 methyl 3 ethyl butane

$$Mg_2C_3, Be_2C, Al_4S_3$$ and $$CaC_2$$

$$Mg_2C_3, Be_2C$$ and $$Al_4C_3$$

$$Be_2C, Al_4C_3$$ and $$CaC_2$$

MCQ Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 3

The IUPAC name of the this compound is:

0%
2-isopropyl pentane
0%
2, 3-dimethyl hexane
0%
isononane
0%
2, 4-dimethyl hexane

Which one of the following is a free radical substitution reaction ?

0%
0%
0%
$CH_3CHO+HCN\rightarrow CH_3-\underset{OH}{\underset{|}{C}H}-CN$
0%

Explanation

Chlorination of toluene in presence of sunlight follows free radical mechanism. Benzyl radical (

$C_6H_5CH_2^+)$ is most stable so benzyl chloride is major product formed.

All other reactions follows ion formation mechanism.

Option D is correct.

IUPAC name of

$CH_{3}-CH(CH_{3})-C\equiv C-CH(CH_{3})-CH_{3}$ is:

0%
2,5-dimethyl-3-hexyne
0%
2,5-dimethyl-2-hexyne
0%
3,5-dimethyl-3-hexyne
0%
2-heptyne

Explanation

1 2 3 4 5 6

$CH_{3}-CH(CH_{3})-C\equiv C-CH(CH_{3})-CH_{3}$

The given compound is alkyne as it contains a triple bond. Its parent chain contains 6 carbon atom.

As methyl is present as substituent group at carbon no. 2 and 5 and a triple bond is present at carbon number 3.

So, IUPAC name of the compound is 2,5-dimethyl-3-hexyne.

Which of the following method is not used in the preparation of acetylene :

0%
Dehydrohalogenation
0%
Dehalogenation
0%
Hydrolysis
0%
Hydrogenation

Explanation

Alkyne can be produced by
Kolbe electrolysis:
acetylene can be produced by electrolysis of aq. solution of sodium/potassium maleate or fumarate.
Dehydrohalogenation:
Preparation of alkyne by heating ethylene bromide with alc. KOH followed by sodamide in ammonia solution.
Synthesis:

$2C+H_2\rightarrow C_2H_2$ at high temp.
and by hydrolysis.

(A) : The C, C bond length in benzene is intermediate between C, C single bond length and C, C double bond length.
(R) : Benzene molecule exhibits resonance.

0%
Both (A) and (R) are true and (R) is the correct explanation of (A)
0%
Both (A) and (R) are true and (R) is not the correct explanation of (A)
0%
(A) is true but (R) is false
0%
(A) is false but (R) is true

The IUPAC name of the given compound is:

0%
3 - ethyl - 4, 4 - dimethylheptane
0%
1,1 - diethyl - 2, 2 - dimethylpentane
0%
4, 4- dimethyl - 5, 5 - diethylpentane
0%
5, 5- diethyl - 4,4 - dimethylpentane

Benzene can :

0%
Discharge colour of $KMnO_{4}$
0%
Discharge colour of bromine water
0%
Polymerise
0%
Undergo substitution reactions

Explanation

"Benzene decolourise bromine water". Discharge colour of

$KMnO_4$ . Both are incorrect and also it doesn't undergo polymerization. Because it does not undergo addition reaction as it is highly saturated due to presence of electron cloud above and below it. (aromatic system is more stable)

But it undergo substitution reactions.

Option D is correct.

Ethyne is not prepared by:

0%
heating iodoform with $Ag$ powder
0%
hydrolysis of calcium carbide
0%
heating $C_{2}H_{2}Cl_{4}$ with zinc dust
0%
heating iodoform with $Na$

Explanation

In all the given reactions Acetylene or ethylene is formed except heating Idoform with Na. As iodoform doesn't even reacts with

$Na$ so this will not produce alkyne.
Option D is correct.

$C_{2}H_{2}\overset{Red hot}{\longrightarrow}A\overset{fuming.H_{2}SO_{4}}{\longrightarrow}B$

'B' is :

0%
Benzene
0%
Toluene
0%
Chlorobenzene
0%
Benzene sulphonic acid

A direct iodination of benzene is not possible because :

0%
Iodine is an oxidising agent
0%
Resulting $C_{6}H_{5}I$ is reduced to $C_{6}H_{6}$ by $HI$
0%
$HI$ is unstable
0%
The ring gets deactivated

Explanation

Correct Answer: Option B

Explanation:

The direct iodination reaction of benzene is,

${{C}_{6}}{{H}_{6}}+{{I}_{2}}\rightleftharpoons {{C}_{6}}{{H}_{5}}I+HI$

This reaction is a reversible reaction. The product $HI$ reacts with iodobenzene to get back benzene.

irect iodination of benzene is not possible because the product ${{C}_{6}}{{H}_{5}}I$ is reduced by $HI$ .

The number of structural isomers possible with molecular formula

$C_{5}H_{8}$ is:

0%
more than 7
0%
6
0%
5
0%
4

Benzene reacts with

$CH_3Cl$ in the presence of anhydrous

$AlCl_3$ to form:

0%
xylene
0%
toluene
0%
chlorobenzene
0%
benzyl chloride

Explanation

Benzene reacts with

$CH_{3}Cl$ in presence of anhydrous

$AlCl_{3}$ to give Toulene. This is fridel crafts Alkylation and an electrophillic substitution.

All alkanes are gases.

0%
True
0%
False

Which of the following reacts with water to give ethane?

0%
$CH_{4}$
0%
$C_{2}H_{5}MgBr$
0%
$C_{2}H_{5}OH$
0%
$C_{2}H_{5}OC_{2}H_{5}$

Explanation

Hydrolysis of Grignard's reagent (

$RMgX$ ) produces alkane, where R is alkyl or aryl group and X is halide.
Thus

$C_2H_5MgBr$ on hydrolysis produces ethane,

$-MgBr$ is replaced by hydrogen atom of water and

$MgBrOH$ is also formed.

The IUPAC name of the following compound is:

$CH_{2} = CH - CH (CH_{3})_{2}$

0%
1,1-dimethyl-2-propene
0%
3-methyl-1-butene
0%
2-vinyl propane
0%
1-isopropyl ethylene

Explanation

$\overset {1}{C}H_{2} = \overset {2}{C}H - \overset {3}{\underset {\displaystyle {\underset {\displaystyle {CH_3}}{|}}}{C}}H -\overset {4}{C}H_{3}$

The IUPAC name of the given compound is 3-methyl-1-butene.

$R C-C=R \xrightarrow[Lindlar catalsyst]{H_2}$ ?

0%
0%
0%
Both (1) and (2)
0%
$R - CH_2 - CH_2 - R$

Explanation

By using Lindlar catalyst
we get cis-alsene

$\therefore$ we will get

Structural formula of ethyne is :

0%
$HC\equiv CH$
0%
$H_{3} C C H$
0%
0%

Explanation

Correct formula of Ethyne is:

$HC\equiv CH$ .

Ethylbenzene reacts with

$Cl_{2}$ in sunlight to give:

0%
1-Chloro-1-phenylethane
0%
1-Chloro-2-phenylethane
0%
4-Chloroethylbenzene
0%
3-Chloroethylbenzene

Explanation

Reaction:

$ph-CH_{2}CH_{3} + Cl_{2} \xrightarrow{sunlight} ph-CH(Cl)-CH_{3}$

This is a substitution reaction where chlorine replaces one hydrogen atom to form 1-chloro-1-phenylethane.

The final product of chlorination of methane in the sun light is ?

0%
$CH_{3}C_{1}$
0%
$CH_{2}Cl_{2}$
0%
$CHCI_{3}$
0%
$CCI_{4}$

Choose which one from the given options is an incorrect statement.

0%
Tertiary ( $3^0$ ) butyl is 1,1-Dimethylethyl
0%
Neopentane is 1,1-Dimethylpropane
0%
Isopentane is 2-methylbutane
0%
The correct IUPAC numbering of following compound is

Explanation

$\text{So option B is incorrect.}$

The unlikely representation(s) of resonating structures of fluorobenzene is

The major monobromination product in the photochemical bromination of

$2,2 -dimethyl butane$ is:

Explanation

The major monobromination product in the photochemical bromination of

$2,2 -dimethyl butane$ is

$3-bromo-2,2-dimethyl butane.$
Bromination occurs selectively at the carbon where more stable free radical is formed.

The correct IUPAC name of the following compounds is :

0%
1, 2-dimethylpropyleyelopropane
0%
2-cyclopropyl-3-methylbutane
0%
2-methyl-3-cyclopropylbutane
0%
isopentylcyclopropane

Select the correct name of the above compound :

0%
2,5-diethyl-6-methyloctane
0%
4,7-diethyl-3-methyloctane
0%
4-ethyl-3, 7-dimethylnonane
0%
6-ethyl-3, 7-dimethylnonane

When cyclohexane is poured on water, it floats because :

0%
cyclohexane is in a 'boat' form
0%
cyclohexane is in a 'chair' form
0%
cyclohexane is in a 'crown' form
0%
cyclohexane is less dense than water

Explanation

Hint: Compounds that are less dense than water float on water.

Explanation:

The substance is less dense than water then the substance will float.
Cyclohexane is a cyclic hydrocarbon containing $6C$ .
Cyclohexane is a non-polar compound and does not dissolve in water and the density is less than water.
As its density is less than water that is why it floats on water.

Correct Option :

$D$ .

Structure represents -

0%
Isopropylidene Bromide
0%
Active amylene Iodide
0%
Isobutylene glycol
0%
Isobutylene

The reaction

$C_6H_6 + CH_3 \xrightarrow[anhydrous]{AlCl_3} HCl + C_6H_5CH_3$ is:

0%
Friedel-Crafts alkylation
0%
addition reaction
0%
Friedel-Crafts acylation
0%
Friedel -Craft's benzylation

Explanation

The reaction of benzene in the presence of

$AlCl_3$ in anhydrous conditions is Friedel-Crafts alkylation reaction. It is a substitution reaction. Hydrogen is substituted by an alkyl group.

Friedel-Crafts acylation is the addition of

$-COR$ group.
Friedel-Crafts benzylation is the addition of a benzene group.

No. of isomers

$C_4H_8$ :

0%
4
0%
3
0%
2
0%
5

When methane is mixed with excess chlorine and the mixture is exposed to sunlight, the final product is:

0%
$CH_3Cl$
0%
$CH_2Cl_2$
0%
$CHCl_3$
0%
$CCl_4$

Explanation

When a mixture of methane and chlorine is exposed to ultraviolet light - typically sunlight - a substitution reaction occurs and the organic product is chloromethane.

$CH_{4} + Cl_{2} \rightarrow CH_{3}Cl + HCl$
However, when excess chlorine is present, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms. That means that you could get any of chloromethane, dichloromethane, trichloromethane or tetrachloromethane.

$CH_{4} + Cl_{2} \rightarrow CH_{3}Cl + HCl$

$CH_{3}Cl + Cl_{2} \rightarrow CH_{2}Cl_{2} + HCl$

$CH_{2}Cl_{2} + Cl_{2} \rightarrow CHCl_{3} + HCl$

$CHCl_{3} + Cl_{2} \rightarrow CCl_{4} + HCl$

The structure shows :

0%
Decane
0%
Propane
0%
Pentane
0%
Butane

$C_6H_{12}$ and

$C_6H_6$ are the examples of aliphatic and aromatic hydrocarbons respectively.

0%
True
0%
False

Explanation

$C_{6}H_{12}$ is an aliphatic compound , whereas

$C_{6}H_{6}$ is an aromatic compound.

Preceding and succeeding homologues of

$C_{10}H_{22}$ respectively are:

0%
$C_7H_{14}, C_{11}H_{22}$
0%
$C_9H_{20}, C_{11}H_{24}$
0%
$C_{11}H_{24}, C_9H_{20}$
0%
$C_9H_{18}, C_9H_{20}$

Explanation

$C_{10}H_{12}$

$\downarrow$

$C_nH_{2n + 2}$

$n = 10$

$n – 1 = 9 \rightarrow C_9H_{20}$

$n + 1 = 11 \rightarrow C_{11}H_{24}$

The replacement of a hydrogen atom in benzene by alkyl group can be brought about with the following reagents :

0%
alkyl chloride and $AlCl_{3}$
0%
alkene and $AlCl_{3}$
0%
alkanol and alkali
0%
alkanol and acid

Give IUPAC name:

0%
3-Ethyl-4,6-dimethyloctane
0%
4,6-dimethyl- 3-Ethyl octane
0%
3,5-dimethyl- 6-Ethyl octane
0%
None of these

Which of the following statement(s) is/are incorrect?

0%
Although benzene contains three double bonds, normally it does not undergo addition reaction.
0%
m-chlorobromobenzene is an isomer of m-bromochlorobenzene.
0%
In benzene, carbon uses all the three p orbitals for hybridization.
0%
An electron donating substitutent in benzene orients the incoming electrophilic group to the meta position.

Give IUPAC name :

0%
3,4,4 Trimethyl heptane
0%
3,4,4 Trimethyl octane
0%
2 Butyl 2 methyl 3 ethyl butane
0%
2 Ethyl 3,3 dimenthyl heptane

Give IUPAC name:

0%
Neononane
0%
Tetra ethylmethane
0%
3-Ethyl pentane
0%
3,3- Diethyl pentane

The ratio of

$\sigma$ and

$\pi$ bonds in benzene is:

0%
$2$
0%
$6$
0%
$4$
0%
$8$

Explanation

Benzene has 12

$\sigma$ bonds and 3

$\pi$ bonds , So their ratio = 4:1

$\dfrac{\sigma}{\pi} = \dfrac{12}{3} = 4:1$

If octane number of gasoline sample is 70, then this sample will be similar to:

0%
70% n-heptane + 30% n-octane
0%
70% n-octane + 30% n-heptane
0%
70% isooctane + 30% n-heptane
0%
70% n-heptane + 30% isooctane

(E)-3-bromo-3-hexene when treated with

$CH_{3}O^{\circleddash }$ in

$CH_{3}OH$ gives

0%
3- hexyne
0%
2-hexyne
0%
2,3-hexadiene
0%
2,4-hexadiene

Explanation

$(E)-3-$ bromo

$-3-$ hexene when treated with

$CH_3O^-$ in

$CH_3OH$ gives

$2,3-$ hexadiene. A molecule of HBr is eliminated as shown in the above reaction.

Which of the following is not the property of benzene?

0%
Characteristic smell
0%
Inflammable
0%
Addition reactions
0%
Colourless

Which of the following have only 2 types of H-atom?

0%
only a
0%
a, b and d
0%
a, c and d
0%
All of them

Explanation

Only compound (a) naphthalene has two different types of hydrogen atoms. One is alpha hydrogen and the other is beta hydrogen.
In compound (b) three types of hydrogen atoms are present

$\alpha, \beta$ and

$\gamma$ .
In compound (c) four types of hydrogen atoms are present

$\alpha, \beta, \gamma$ and methyl group.
In compound (d) four types of hydrogen atoms are present

$\alpha, \beta, \gamma$ and methylene group.

Which among the following product is formed when ethyne undergoes hydration?

0%
Formaldehyde
0%
Formic acid
0%
Acetaldehyde
0%
Acetic acid

Heating many alkyl chlorides or bromides in water effects their conversion into alcohol through a

$\displaystyle S_{N}1$ reaction. Order of the above sets compounds with respect to solvolytic reactivity is :

0%
II > I>III
0%
III < II < I
0%
III > II > I
0%
None

Explanation

$S_N1$ , mechanism RDS step is the formation of the carbocation.

So, the rate of reaction is directly proportional to the stability of carbocation forms.

$Rate\ of\ reaction\propto stability\ of\ carbocation(C^+)$

order of stability of

$C^+$ is (

$3^0>2^0>1^0$ )

Thus, option C is correct.

The correct order of rate of Wurtz reaction is:

0%
$\displaystyle I > II > III> IV$
0%
$\displaystyle II > I > III> IV$
0%
$\displaystyle IV > III > II> I$
0%
In all, the rate of Wurtz reaction is same.

Explanation

Reaction mechanism of Wurtz reaction:

$R-X + 2Na \rightarrow R^- Na^+ +NaX$

$R^- Na^+ + R-X \rightarrow R-R + NaX$

Leaving group stability order is

$I^- > Br^- > Cl^- > F^-$ .

Reactivity order is dependent on the leaving group stability. If the leaving group is stable, reactivity will be more.

So the reactivity order is

$IV>III>II>I$ .

Option C is correct.

Methanides are:

0%
$Mg_2C_3, Be_2C, Al_4S_3$ and $CaC_2$
0%
$Mg_2C_3, Be_2C$ and $Al_4C_3$
0%
$Be_2C, Al_4C_3$ and $CaC_2$
0%
$Be_2C$ and $Al_4C_3$

Explanation

Ionic carbides have discrete carbon anions of the forms

$C^{4-}$ , called methanides since they can be viewed as being derived from methane (

$CH_4$ ). So

$Be_2C$ and

$Al_4C_3$ are methanides as they have

$C^{4-}$ ions.

Which of the following is methanide?

0%
$Be_2C$
0%
$CaC_2$
0%
$Mn_3C$
0%
$Mg_3C_2$

Explanation

Methanides are carbides which on hydrolysis give methane

$(CH_4)$ .

$Be_2C$ and

$Mn_3C$ is a methanide as it produces

$CH_4$ on reacting with

$H_2O$

The compound which is insoluble in cold, conc.

$H_{2}SO_{4}$ is:

0%
n-hexane
0%
$1$ -hexene
0%
$2$ -hexene
0%
ethyl acetate

Explanation

Alkane is insoluble in

$H_2SO_4$ while alkenes and alkynes form bisulphate and get dissolved in dilute

$H_2SO_4$ .

The discovery that chlorofluorocarbons initiate a radical -chain reaction that catalytically destroys ozone won the 1995Nobel prize in chemistry Given the initiation step below which of the following reactions is the most likely to serve as one of the propagation steps?

$CCl_{2}F_{2}+hv\rightarrow ^{*} CClF_{2}+^{*}Cl$

0%
$^{* }Cl+^{* }Cl\rightarrow Cl_{2}$
0%
$^{*}Cl+O_{3}\rightarrow ^{*}OCl+O_{2}$
0%
$^{*}CClF_{2}+^{*}Cl \rightarrow CCl_{2}F_{2}$
0%
$2^{*}CClF_{2}+2O_{3}\rightarrow 2COF_{2}+Cl_{2}+2O_{2}$

Explanation

The ultraviolet light, as it hits the atmosphere, makes CFCs break up to form free radicals like this:

$CCl_2F_2 \rightarrow ^*CClF_2 + Cl^*$

Chlorine free radicals will then go on to react with the ozone in the stratosphere:

$O_3 + Cl^* \rightarrow ClO^* + O_2$ (Propagation step)

The chlorine oxide molecule

$(ClO)$ is very reactive, and it then proceeds to react with ozone to make two oxygen molecules and a

$Cl^*$ free radical:

$ClO^* + O_3 \rightarrow 2O_2 + Cl^*$

That

$Cl^*$ free radical will then go on and react with another ozone molecule, resulting in a chain reaction between the last two equations that theoretically never stops, due to their unreactive nature and thus their inability to be easily removed from the atmosphere. The chlorine atoms aren’t used up, which means they are free to carry on breaking down the ozone.

Which of the following is soluble in methanol:
I : Hexane

II : Cyclohexane

III :

$2-methyl pentane$

IV : Chloroethane

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

Explanation

Like dissolves like.

Chloroethane is being polarly miscible in methanol.

cyclohexane, hexane,

$2-methylpentane$ are non-polar and immiscible with methanol.

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 3 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 3 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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