More reactive than ethyl bromide

Nearly as reactive as methyl chloride

More reactive than isopropyl chloride

MCQ Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 10

The most important method to prepare lower hydrocarbon is

0%
cracking
0%
Sabatier-Senderen's reaction
0%
heating salts of fatty acids with sodalime
0%
direct synthesis

Explanation

The most important method to prepare lower hydrocarbon is cracking.

For eg. :

$\mathrm{C_6H_{14} \xrightarrow{\Delta} \underset{Ethene}{C_2H_4} + \underset{Butane}{C_4H_{10}}}$

Hence, Option "A" is the correct answer.

Ethane can be prepared by

0%
heating sodalime with sodium acetate
0%
electroysis of sodium succinate
0%
electroysis of sodium acetate
0%
all of these

Explanation

Reactions:

$\mathrm{CH_3-COONa \xrightarrow[Sodalime]{\Delta} \underset{Methane}{CH_4} + CO_2}$

B) Sodium succinate on electrolysis produce ethyne.

C) Sodium acetate on electrolysis produce ethane.

Hence, Option "C" is the correct answer.

Action of heat on a mixture of sodium propanoate and sodalime produces

0%
methane
0%
ethane
0%
propene
0%
ethylene

Explanation

Action of heat on a mixture of sodium propanoate and sodalime:

$\mathrm{CH_3CH_2COONa \xrightarrow[Sodalime]{\Delta} CH_3-CH_3 +CO_2}$

So, the product obtained is ethane.

Hence, Option "B" is the correct answer.

The IUPAC name of neo-pentane is:

0%
$2-$ methyl butane
0%
$2,2-$ dimethyl butane
0%
$2-$ methyl propane
0%
$2,2-$ dimethyl propane

Explanation

Neo-pentane is

$\mathrm{CH_3-C(CH_3)(CH_3)-CH_3}$ .

So, IUPAC Nme of Neo-pentane is 2,2-dimethylpropane.

Hence, Option "D" is the correct answer.

Aluminium amalgam used as a reducing agent, is obtained by:

0%
dipping aluminium foil in mercuric chloride solution
0%
mixing aluminium powder with mercury
0%
mixing aluminium nitrate with mercuric oxide
0%
adding mercury to aluminium chloride solution

X is heated with sodalime and gives ethane. X is:

0%
ethanoic acid
0%
methanoic acid
0%
propanoic acid
0%
either (a) or (c)

Explanation

The given reaction is decarboxylation reaction:

$\mathrm{RCOOH \xrightarrow[Sodalime]{\Delta} RH + CO_2}$

This means one carbon is reduced in decarboxylation reaction.

So, X was Propanoic acid.

Hence, Option "C" is the correct answer.

Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene ?

0%
both A and B
0%
only A
0%
only B
0%
none of these

It is necessary to use ______ in the iodonation of alkane.

0%
alcohol
0%
oxidising agent
0%
benzene
0%
reducing agent

Alkenes can be prepared by:

0%
Dehydrohalogenaion of alkyl halide
0%
Sabatier-Sanderen's reduction of alkynes
0%
Dehydration of alcohols with $conc.H_{2}SO_{4}$ at $443\ K$
0%
Treatment of $3^{\circ}$ alkyl halide at $573K$

The compound having only primary hydrogen atoms is:

0%
butane
0%
iso-butene
0%
cyclohexane
0%
$2,3-$ dimethylbutene

Explanation

$\bullet$ Butane has 4 secondary and 6 primary hydrogen atoms.

$\bullet$ Iso-butene has 8 primary hydrogen atoms and 0 secondary hydrogen atoms.

$\bullet$ Cyclohexane has 0 primary hydrogen atoms and 12 secondary hydrogen atoms.

$\bullet$ 2,3-dimethylbutene has 11 hydrogen atoms, 0 secondary hydrogen atoms and 1 tertiary hydrogen atoms.

So, only Iso-butene has only primary hydrogen atoms.

Hence, Option "B" is the correct answer.

On mixing certain alkane with chlorine and irradiating it with ultraviolet light, it forms only one monochloro alkane. The alkane could be:

0%
propane
0%
pentane
0%
iso-pentane
0%
neo-pentane

State, whether the following statements are True or False:
Benzene is an unsaturated hydrocarbon and it shows addition reactions like alkenes.

0%
True
0%
False

State, whether the following statements are True or False:
Benzene has a planar structure.

0%
True
0%
False

Carbon-carbon bond distance in benzene is

0%
Longer than a $C = C$ double bond
0%
Shorter than a $C = C$ double bond
0%
Shorter than a $C = C$ triple bond
0%
Longer than a $C - C$ single bond

The benzene molecule is

0%
Trigonal
0%
Tetrahedral
0%
Planar
0%
Pyramidal

Explanation

The shape of Benzene is a

$planar$ regular hexagon, with bond angles of 120°.

This is easily explained.

It is a regular hexagon because all the bonds are identical. The delocalisation of the electrons means that there aren't alternating double and single bonds.

The centric formula of benzene was proposed by

0%
Dewar
0%
Baeyer and Armstrong
0%
Ladenbarg
0%
Kekule

Explanation

Centric structure of benzene was proposed by

$Armstrong \ and\ Baeyer$ .

The number of sigma and pi bonds in benzene molecule is

0%
$6\sigma$ and $6\pi$
0%
$12\sigma$ and $3\pi$
0%
$6\sigma$ and $9\pi$
0%
$9\sigma$ and $3\pi$

Explanation

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

$C_6H_6$ so there exists a double bond between alternating carbon atoms and each carbon is connected to two carbon atoms through a single bond and a double bond and hydrogen atom by a single bond.

12σ and 3π bonds is present in benzene molecule.

Which gives only monosubstituted product?

0%
o-Dinitrobenzene
0%
m-Dinitrobenzene
0%
p-Dinitrobenzene
0%
Nitrobenzene

Explanation

m-Dinitrobenzene will give only monosubstituted product.

$NO_2$ group is meta directing group and there is only one

$meta-$ position available in case of m-dinitrobenzene.

The carbide which gives propyne on hydrolysis:

0%
$Al_4C_3$
0%
$Mg_2C_3$
0%
$B_4C$
0%
$La_4C_3$

Explanation

The hydrolysis of Magnesium Carbide ( $Mg_2C_3$ ) gives out Propyne. The reaction involved is:

$Mg_2C_3 + 4H_2O\to 2Mg(OH)₂ + C_3H_4$ (Propyne)

Hence, option B is correct.

The carbon-carbon bond length in benzene is:

0%
Same as in $C_{2}H_{4}$
0%
In between $C_{2}H_{6}$ and $C_{2}H_{2}$
0%
In between $C_{2}H_{6}$ and $C_{2}H_{4}$
0%
In between $C_{2}H_{4}$ and $C_{2}H_{2}$

Explanation

We know, that there are only sigma bonds in ethane (C₂H₆), one sigma and one PI bond in ethene (C₂H₄), and one sigma and two PI bonds in acetylene (C₂H₂). Benzene (C₆H₆) is an aromatic compound, and due to resonance, there is partial double bonds everywhere.

Now, hybridization of carbons of alkane is

$sp^3$ , that of alkene is

$sp^2$ , and that of alkyne is

$sp$ . Thus, in

$sp$ , there's an increased s-character, that reduces the bond length.

Hence the correct options is C.

STATEMENT-1 : Alkanes float on the surface of water.
STATEMENT-2 : Density of alkanes is in the range of

$0.6 - 0.9\ g/ml$ , which is lower than water.

0%
Statement -1 is True, Statement-2 True; Statement-2 is a correct explanation for Statement-1.
0%
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
0%
Statement-1 is True, Statement-2 is False
0%
Statement-1 is False, Statement-2 is True

Explanation

Density of alkanes is in the ranges of

$0.6 - 0.9\ g/ml$ , which is lower than water. Hence floats on water.

Option A is correct.

IUPAC name of the compound

0%
4-isopropyl-1-6-methyl octane
0%
3-methyl-5-(1-methylethyl) octane
0%
3-methyl-5-isopropyl octane
0%
3-methyl-4-(1-methylethyl ) octane

A hydrocarbon

$(R)$ has six membered ring in which there is no unsaturation. Two alkyl groups are attached to the ring adjacent to each other. One group has

$3$ carbon atoms with branching at

$1 ^{ st }$ carbon and another has

$4$ carbon atoms. The larger alkyl group has main chain of three carbon atoms of which second carbon is substituted. Correct IUPAC name of compound

$(R)$ is

0%
$1-$ ( $1-$ Methylethyl) $-2-$ ( $1-$ methylpropyl)cyclohexane
0%
$1-$ ( $2-$ Methylethyl) $-2-$ ( $1-$ methylpropyl)cyclohexane
0%
$1-$ ( $1-$ Methylethyl) $-2-$ ( $2-$ methylpropyl)cyclohexane
0%
$1-$ ( $1-$ Methylethyl) $-2-$ butylcyclohexane

Which does not show substitution in benzene ring?

0%
$ClSO_{3}H$
0%
$Conc. H_{2}SO_{4}$ on heating
0%
$Conc. HCl$
0%
$CH_{3}Cl/ AlCl_{3}$

Explanation

Benzene would never react with

$HCl.$ As

$HCl$ as an inorganic acid doesn't have electrophile to replace the

$H$ atom of benzene. Here

$Cl$ act as Lewis base so it doesn't act as an electrophile . Also, addition does not take place if it does it lose its aromaticity.

Total number of isomers of a disubstituted benzene compound is

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

Explanation

There are

$3$ isomers are possible.
1683719_1760982_ans_676d5878dd2f48b3b9234155ebdc70c8.png

Name of the compound given is

0%
5-ethyl-6-methyloctane
0%
4-ethyl-3-methyloctane
0%
3-methyl-4-ethyloctane
0%
2, 3-diethylheptane

Chlorobenzene is

0%
Less reactive than benzyl chloride
0%
More reactive than ethyl bromide
0%
Nearly as reactive as methyl chloride
0%
More reactive than isopropyl chloride

Explanation

Chlorobenzene is less reactive than benzyl chloride.

In chlorobenzene the lone pairs present on

$Cl$ atom get involved in resonance with

$\pi$ electrons of benzene due to which

$C - Cl$ bond acquires double bond character Hence , reactivity decreases.

1643968_1762226_ans_cf4b4131f3ec4f8280d85e3489501c89.PNG

The IUPAC name of the compound is

0%
3, 3-dimethyl-1-cyclohexanol
0%
1,1-dimethyl-3-hydroxy cyclohexane
0%
3,3-dimethyl-1-hydroxy cyclohexane
0%
1, 1-dimethyl-3-cyclohexane

$CH_3 - CH_2 - CH_2Br + KOH (alc.) \rightarrow$ Product Product in above reaction is

0%
$CH_3 - CH = CH_2$
0%
$CH_3 - CH_2 - CH_3$
0%
(a) and (b) both
0%
None of these

Explanation

this is called Elimination reaction E1 andd E2

si the

$CH_3- CH_2-CH_2Br+ KOH(Alc) \rightarrow CH_3-CH= CH_2$

Which of the following compound has maximum boiling point?

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

Explanation

$n-$ hexane because it is linear so it has larger surface area therefore strong Van der Waal force.

Which of the following has maximum boiling point?

0%
iso-octane
0%
$n-$ octane
0%
$2, 2, 3, 3-$ tetramethyl butane
0%
$n-$ butane

Explanation

$n-$ octane has maximum boiling point.

Boiling point depends on molecular mass. Greater the molecular mass higher will be the boiling point.

Boiling point also depends on the structure. If two compounds have some molecular mass then straight chain or linear compounds has higher boiling point.

The IUPAC name of compound is

0%
$(2Z, 4Z)-2,4-$ hexa di-ene
0%
$(2Z, 4E)-2,4-$ hexa di-ene
0%
$(4Z, 4Z)-2,4-$ hexa di-ene
0%
$(2E, 4Z)-2,4-$ hexa di-ene
0%
$(2E, 4E)-2,4-$ hexa di-ene

Explanation

If atom of group of higher priority are on the opposite direction at the double bond of each carbon atom then the configuration is known as

$E$ and if they are in same direction then the configuration is known as

$Z$ configuration. Double bond is present in C number 2 and 4. and priority group in opposite side in both the cases.

Benzene reacts with chlorine to form benzene hexachloride in presence of

0%
Nickel
0%
$AlCl_3$
0%
Bright sunlight
0%
Zinc

The propene reacts with

$HBr$ to form

0%
Ethane
0%
Hexane
0%
1-bromo-propane
0%
2-bromopropane

Which is not found in alkenes?

0%
Chain isomerism
0%
Geometrical isomerism
0%
Metamerism
0%
Position isomerism

Which of the following statements is incorrect

0%
$C_2H_5Br$ reacts with alco. $KOH$ to form $C_2H_5OH$
0%
$C_2H_5Br$ when treated with metallic sodium gives ethane
0%
$C_2H_5Br$ when treated with sodium ethoxide forms diethyl ether
0%
$C_2H_5Br$ with $AgCN$ forms ethyl isocyanide

The correct structure of benzene was proposed by

0%
Faraday
0%
Davy
0%
Kekule
0%
Wohler

The

$C-H$ bond length is minimum in the bond formed by

0%
$sp - s$ overlapping (as in alkynes)
0%
$sp^{2} - s$ overlapping (as in alkynes)
0%
$sp^{3}$ overlapping (as in alkanes)
0%
None of these

Explanation

Here the answer is option A $$sp-s \text{overlapping(as in alkynes)
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)
Alkane	$sp^3$	1.093
Alkene	$sp^2$	1.087
Alkyne	$s p$	1.057

As the number of carbon atoms in a chain increases the boiling point of alkanes

0%
increases
0%
decreases
0%
remains same
0%
may increase or decrease

$CH\equiv CH\xrightarrow[Pressure]{Ni(CN)_{2}}X$ . Here X in the reaction

0%
Benzene
0%
Ethane
0%
Cyclooctatetraene
0%
Cyclohexane

Which alkene is formed from the following reaction

$CH_3CH_2CH_2CH = PPh_3 + 2$ - Butanone

0%
$3$ -Methyl - $3$ -heptene
0%
$4$ -Methyl - $3$ -heptene
0%
$5$ -Methyl - $3$ -heptene
0%
$1$ -Methyl - $5$ -methane

Explanation

$CH_3$

$\mid$

$CH_3CH_2 - C = O + CH_3CH_2CH_2CH = PPh_3$

$2$ -Butanone

$CH_3$

$\mid$

$\rightarrow \, CH_3CH_2C =CHCH_2CH_2CH_3 + Ph_3PO$

$3$ -methyl-

$3$ -heptene

Option A is correct.

Which of the following has acidic hydrogen?

0%
Ethyne
0%
Ethene
0%
Ethane
0%
Benzene

Explanation

$2CH\equiv CH+2Na\rightarrow 2CH\equiv C^{-}Na^{+}+H_{2}$

Reaction of ethyne with sodium metal shows that hydrogen is acidic.

By heating a mixture of

$CHCl_3$ with silver powder , the compound formed is

0%
Acetylene
0%
Silver acetate
0%
Methanol
0%
None of these

Methyl chloride reacts with silver acetate to yield

0%
Acetaldehyde
0%
Acetyl chloride
0%
Methyl acetate
0%
Acetic acid

The number of possible isomers of butene are

0%
$3$
0%
$2$
0%
$4$
0%
$5$
0%
$6$

Explanation

There are

$cis-$ and

$trans-$ isomers of 2-butene is also possible.

So total 4 possible isomers.

1688393_1763158_ans_466d04ba410146698d49da2938074072.jpg

Which of the following is not aromatic?

0%
Benzene
0%
Naphthalene
0%
Pyridine
0%
$1,3,5-$ heptatriene

Explanation

A compound is said to be aromatic if it is planar and there is complete delocalization of

$\pi$ electrons, which is only possible if it is a conjugated cyclic system and the number of electrons used in delocalization is

$(4n + 2). 1, 3, 5-$ heptatriene is not an aromatic compound because complete delocalization of pi electrons is not possible in it and it is not planar.
1689307_1765502_ans_44fb224c5bff42eb9b3adef66d58251a.png

1689307_1765502_ans_44fb224c5bff42eb9b3adef66d58251a.png

Benzene shows

0%
Substitution
0%
Addition
0%
Oxidation
0%
All of these

Silver acetylide when heated with

$HCl$ gives ?

0%
$C_2H_2$
0%
$H_2$
0%
$C_2H_4$
0%
None of these

Explanation

$Ag - C \equiv C - Ag + 2HCl \rightarrow CH \equiv CH + 2AgCl$

The reagent which is used to distinguish between propene
and propyne is

0%
Bromine
0%
Alkaline $KMnO_{4}$
0%
Ammoniacal $AgNO_{3}$
0%
Ozone

Explanation

Propene and propyne can be distinguish by using

$\text{ Ammoniacal silver nitrate test.}$ Propyne has an acidic hydrogen due to which it readily react with

$\text{Ammoniacal silver nitrate }$ gives white precipitate while propene does not give any precipitate.

Which kind of isomerism will butene

$-2$ show?

0%
Geometrical
0%
Optical
0%
Position
0%
None of these

Explanation

$\text{2-Butene}$ shows geometrical isomerism.
1687215_1766296_ans_380ff5c3a2044a26a1411e211b0f84ce.png

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 10 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 10 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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