MCQ Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 6

Ethyne, on reaction with chlorine consequently, forms :

0%
dichloro methane
0%
trichloro ethene
0%
dichloro ethane
0%
tetrachloro ethane

Ethylidene dichloride is obtained by the reaction of excess of HCl with.

0%
Ethylene
0%
Acetylene
0%
Propane
0%
Methane

Explanation

C H \equiv C H a c e t y l e n e + H C l \to C H_{2} = C H C l v i n y l c h l o r i d e H C l \to C H_{3} C H C l_{2} e t h y l i d e n e d i c h l o r i d e

$\underset{acetylene}{CH\equiv CH}+HCl\rightarrow \underset{vinyl chloride}{CH_2=CHCl}\overset{HCl}{\rightarrow} \underset{ethylidene dichloride}{CH_3CHCl_2}$
[Alkenes due to presence of only

1^{o}

$1^o$ of unsaturation, form only monochloro derivative.]

IUPAC name of given compound is :

0%
$3-$ Ethyl $-4, 4-$ dimethyl hexane.
0%
$4-$ Ethyl $-3, 3-$ dimethyl hexane.
0%
$3, 4-$ Diethyl $-4-$ methyl pentane.
0%
$2-$ Methyl $-2, 3-$ diethyl pentane.

Dehydrohalogenation of an alkyl halide is a / an:

0%
Nucleophilic substitution reaction
0%
Elimination reaction
0%
Both nucleophilic substitution and elimination reaction
0%
Rearrangement

Explanation

E x p l a n a t i o n :

$Explanation:$

Dehydrohalogenation is an elimination reaction. In this reaction, hydrogen and halogen atoms get eliminated from alkyl halide to give an alkene or alkynes.

For example: When chloropropane is heated with alcoholic

K O H

$KOH$ gives Propene.

R e a c t i o n : C H_{3} C H_{3} C H_{2} C l + K O H_{a l c o h o l i c} h e a t - - \to C H_{3} C H = C H_{2} + K C l + H_{2} O

$Reaction:\ CH_3CH_3CH_2Cl\ +KOH_{alcoholic} \xrightarrow{heat} CH_3CH=CH_2+KCl+H_2O$

F i n a l A n s w e r :

$Final\ Answer:$ Option

B

$B$ .

Ethyne is normally prepared by the pyrolysis of:

0%
methyne
0%
alcohol
0%
methane
0%
carboxylic acid

Explanation

Ethyne is normally prepared by the pyrolysis of methane. In this procedure, a stream of methane gas is briefly heated to

1500^{o}

$1500^o$ C in an airless chamber. Air must be excluded from the reaction or oxidation (combustion) will occur.

Which of the following molecules has the strongest carbon-to-carbon bond?

0%
$C_2H_2$
0%
$C_2H_4$
0%
$C_2H_6$
0%
$C_3H_8$
0%
$C_4H_{10}$

Explanation

Acetylene

C_{2} H_{2}

$\displaystyle C_2H_2$ or

H - C \equiv C - H

$\displaystyle H - C \equiv C - H$ has strongest carbon to carbon bond as it has triple bond between 2 C atoms.

C_{7} H_{8} 3 C l_{2}, H e a t - ----- \to A F e / B r_{2} - --- \to B Z n / H C l - ---- \to C

$C_{7}H_{8} \xrightarrow {3Cl_{2}, Heat} A\xrightarrow {Fe/ Br_{2}} B\xrightarrow {Zn/ HCl} C$
Here, the compound

C

$C$ is:

0%
3-bromo-2,4,6-trichlorotoluene
0%
o-bromotoluene
0%
p-bromotoluene
0%
m-bromotoluene

Secondary butyl chloride will undergo alkaline hydrolysis in the polar solvent by the mechanism

0%
${S}_{N}2$
0%
${S}_{N}1$
0%
${S}_{N}1$ and ${S}_{N}2$
0%
None of the above

Explanation

S_{N} 1

${S}_{N}1$ mechanism is favored by polar solvents like water, in which the rate of the reaction only depends on the concentration of the alkyl halide.

Which is an important raw material in the chemical industry to produce polyethyelene?

0%
acetylene
0%
ethyne
0%
both a and b
0%
none of the above

The loss of _____ and _______ atoms from the double-bonded carbon atoms leads to alkyne formation.

0%
oxygen, halogen
0%
hydrogen, halogen
0%
oxygen, hydrogen
0%
carbon, halogen

C H_{3} - C B r_{2} - C B r_{2} C H_{3} + 2 Z n \to

$CH_3-CBr_2-CBr_2CH_3+2Zn\rightarrow$

0%
$CH_3-C\equiv C-CH_3$
0%
$CH_3-C\equiv C-CH_2Br$
0%
$CH_3-CBr\equiv C-CH_3$
0%
$CH_2-C\equiv C-CH_3$

Explanation

C H_{3} - C B r_{2} - C B r_{2} - C H_{3} + 2 Z n \to C H_{3} - C \equiv C - C H_{3} + 2 Z n B r_{2}

$CH_3-CBr_2-CBr_2-CH_3+2Zn\rightarrow CH_3-C\equiv C-CH_3+2ZnBr_2$
The above reaction is dehalogenation of vic-Tetrahalogen Compounds. Vicinal tetrahaloalkanes can be dehalogenated with zinc metal in an organometallic reaction to form alkynes. This reaction has the drawback that the halogen compound is itself prepared by halogen addition to alkynes.

The structure of the third member of the alkyne series is:

0%
$H-C\equiv C - H$
0%
$H-C\equiv C - CH_{3}$
0%
$H-C\equiv C - CH_{2}CH_{3}$
0%
$H-C\equiv C - C\equiv C - H$
0%
$H-C - C - CH = C - H_{2}$

Explanation

Alkyne has the general molecular formula

$C_{n}H_{2n}$

and alkyne group start from

$C_{2}H_{2}$ .

Thus third member will have 4 carbon atom and it will be given as follow:

$H-C\equiv C-CH_{2}CH_{3}$

Which of the following is true about the given reaction?

0%
The above reaction is of iodination
0%
$HNO_{3}$ acts as reducing agent in the reaction
0%
Iodine cation is electrophile of the reaction
0%
The reaction involves electrophilic aromatic substitution

Which of the following describes the nature of aromatic hydrocarbons?
I.

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

II.

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

0%
I only
0%
III only
0%
I and II only
0%
II and III only
0%
I, II and III

Statement 1: The double and single bonds in benzene are subject to resonance.
Statement 2: Benzene has de-localized pi electrons that stabilize its structure

0%

Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.
0%

Both Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.
0%

Statement 1 is correct but Statement 2 is not correct.
0%

Statement 1 is not correct but Statement 2 is correct.
0%

Both the Statement 1 and Statement 2 are not correct.

In a series of liquid hydrocarbons, a sample of the diagramed compound would be best identified by which of the following?

	Viscosity	Capillary Action
A	High	High
B	High	Low
C	Low	High
D	Low	Low

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

Methane can be converted into ethane by :

0%
chlorination followed by the reaction with alcoholic $KOH$
0%
chlorination followed by the reaction with aqueous $KOH$
0%
chlorination followed by Wurtz reaction
0%
chlorination followed by decarboxylation

Explanation

Explanation:

To covert methane to ethane, i.e. $CH_4\ to\ C_2H_6,$ methane is treated with chlorine gas in the presence of ultraviolet light to form chloromethane. Chloro methane is treated with sodium metal and dry ether to form the product ethane. This process is also called Wurtz synthesis.

Chemical reactions are as follows:

$CH_4+Cl_2\rightarrow CH_3Cl+HCl$ (in presence of sunlight) this is chlorination

$2CH_3Cl+Na\rightarrow C_2H_6$ (in presence of dry ether) this is wurtz reaction.

So, option C is correct

Which of the following hydrocarbons have the highest boiling point?

0%
$CH_4$
0%
$C_3H_8$
0%
$C_4H_{10}$
0%
$C_5H_{12}$
0%
$C_6H_{14}$

Explanation

Hexane

$\displaystyle C_6H_{14}$ has the highest boiling point as it has maximum number of C atoms and maximum molecular weight.
With increase in the molecular weight, the boiling point of straight chain hydrocarbons increases.

The correct structural formula of benzene is :

Explanation

The correct structural formula of benzene is represented by the option (B). Here one H atom is attached to each C atom. In the 6 member ring, alternate single and double bonds are present. Each C atom is sp

$\displaystyle _2$ hybridised as in ethylene molecule. 3 C

$\displaystyle =$ C double bonds are completely delocalised over entire ring.

IUPAC name of the first member of alkyne is ___________.

0%
ethene
0%
ethyne
0%
ethane
0%
methane

Explanation

Alkynes are carbon compounds with general formula

$C_nH_{2n-2}$

The first member of alkyne series is

$C_2H_2$ (ethyne)

$H-C\equiv C-H$

Alkyne series:

$C_2H_2, C_3H_4, C_4H_6, C_5H_8$ .

In the nomenclature of alkenes, if a substituent, double bond and triple bond all are present in the structure and both multiple bonds are at the same distance from substituent then:

0%
the double bond is preferred over triple bond
0%
the triple bond is preferred over double bond
0%
the substituent is preferred over multiple bonds
0%
none of these

The IUPAC name of given compound is:

$CH_3-C\equiv C-C(CH_3)_2-CH_3$

0%
4,4-dimethyl-2-heptene-5-yne
0%
4,4-dimethyl-2-heptene-6-yne
0%
4,4-dimethyl-5-heptene-2-yne
0%
None of these

Explanation

$CH_3-C\equiv C-C(CH_3)_2-CH_3$

IUPAC name: 4,4-dimethyl-pent-2-yne

Which of the compound does not dissolve in concentrated

$H_2SO_4$ ?

0%
Hexane
0%
Benzene
0%
Ethylene
0%
Aniline

Explanation

$\displaystyle Hexane$ does not dissolve in concentrated

$\displaystyle H_2SO_4$ .

$\displaystyle Hexane$ is nonpolar hydrocarbon. Polar compound such as

$\displaystyle benzene$ ,

$\displaystyle ethylene$ and

$\displaystyle aniline$ dissolve in concentrated

$\displaystyle H_2SO_4$ as they donate their

$\displaystyle \pi$ electrons to

$\displaystyle H^+$ ions of concentrated

$\displaystyle H_2SO_4$ .

Option A is correct.

Which one of the following is the most important halogen in terms of preparation of halogen derivatives of alkane?

0%
Bromine
0%
Chlorine
0%
Iodine
0%
Fluorine

Give IUPAC name of the following compound :

$CH\equiv C-C(CH_3)_3$

0%
3,3- dimethylpropyne
0%
3,3-dimethylbutyne
0%
2,2-dimethylbut-3-yne
0%
None of these

Explanation

$\underset {1}{C}H\equiv \underset{2}{C}- \underset{3}{C}(CH_3)_2-\underset{4}{C}H_3$

The numbering of the compound starts from the triple bond carbon, as alkynes are given higher priority in the order of priority of functional groups when compared to alkanes or single bonds, and methyl substituents.

IUPAC name: 3,3 - dimethylbutyne.

Hence, the correct option is B.

The IUPAC name of the compound

$CH\equiv C-CH_2-CH_2-CH=CH_2$ is:

0%
Hex-1-en-5-yne
0%
Hexyn-5-ene
0%
Hex-5-en-2-yne
0%
Hex-5-en-1-yne

Explanation

$CH\equiv C-CH_2-CH_2-CH=CH_2$

IUPAC name: Hex-1-en-5-yne.

Numbering starts from the double bond bearing carbon, as alkenes are higher priority functional groups than alkynes (triple bond).

s-Butyl bromide is allowed to react with sodium in dry ether, The major product formed is :

s-Butyl bromide is allowed to react with sodium in dry ether.The major product formed is :

Which of the following can be used for the preparation of propane?

0%
$CH_3 CH \equiv CH_2 \xrightarrow [(ii)\ AgNO_3 /NaOH ] { (i)\ B_2 H_6 }$
0%
$CH_3 CH_2 CH_2 Cl \xrightarrow [(ii)H_2O_2]{(i)\ Mg / ether}$
0%
$CH_3 CH_2 CH_2 I \xrightarrow { HI / 150^{\circ}C}$
0%
$CH_3 CH_2 CH_2 COONa \xrightarrow [\triangle]{ NaOH/CaO}$

Explanation

The chemical formula for propane is

$CH_3CH_2CH_3$ .

Decarboxylation is a chemical reaction that removes a carboxyl group and releases carbon dioxide

$CO_2$ from a compound.

Usually, decarboxylation refers to the reaction of carboxylic acids or their derivatives, removing a carbon atom from a carbon chain in the form of

$CO_2$ .

An alkane with one carbon atom less than the parent carbon chain will be formed in this process.

$CH_3CH_2CH_2COONa\ \xrightarrow[\Delta]{NaOH/CaO}\ CH_3CH_2CH_3+Na_2CO_3$

The given reaction involves the removal of a carbon dioxide molecule, therefore, it is a decarboxylation reaction. Therefore, propane is formed by the decarboxylation of sodium propanoate

Hence, option (D) is correct.

Which is not stable at room temperature?

Explanation

Device referred to above image is anti-aromatic , so it is highly unstable at room temperature.

Stability

$=$ Aromatic

$>$ Non-Aromatic

$>$ Anti-aromatic

$.$

1034934_698677_ans_943730f1c3ab4db883c66fb2054cfc20.PNG

$CH\equiv CH\xrightarrow [ { Cu }_{ 2 }{ Cl }_{ 2 } ]{ { NH }_{ 4 }Cl }$ product
Product is:

0%
$Cu-C\equiv C-Cu$
0%
${CH}_{2}=CH-C\equiv CH$
0%
$CH\equiv C-Cu$
0%
$Cu-C\equiv C-{NH}_{4}$

Explanation

It is prepration of organometallic reagent

$CH\equiv CH \xrightarrow [ CH_2Cl_2 ]{ NH_ 4Cl } \underset{Dicopper\ acetylide}{CuC\equiv CCu}$

The complete IUPAC name of the compound:

0%
[R]-1-Bromo-1-phenyl ethane
0%
[S]-1-Bromo-1-phenyl ethane
0%
[E]-1-Bromo-1-phenyl ethane
0%
[Z]-1-Bromo-1-phenyl ethane.

Explanation

From

$CIP$ rules, numbering the substituent.

The substituents are named in alphabetic order. The largest claim of carbon is ethane therefore,

$[R]-1-Bromo-1-phenyl-ethane$

1082799_693597_ans_b734c26bcf224aee976a2c25e1b3f078.png

The product of the above reaction is

The bond order of individual carbon-carbon bonds in benzene is:

0%
one
0%
two
0%
between one and two
0%
one and two, alternately

Explanation

Benzene is a resonance hybrid of two resonating structures. Bon dorder can be calculated as:

$\text {Bond Order} = \dfrac {\text{Number of Resonating bonds}}{\text {Number of resonating structures}} = \dfrac{3}{2} = 1.5$

The correct answer is between one and two.

How many monochloro structural isomers are expected in free radical monochlorination of

$2-$ methylbutane?

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

Explanation

Number of mono chloro isomers depends upon the types of hydrogen atoms present in the substate alkane. The structure of

$2-$ methylbutane is

$\overset {d}{C}H_{3} - \overset {c}{C}H_{2} - \overset {b}{H}\underset {\underset{a}{C}H_3}{\underset{|}{C} -} \overset {a}{C}H_{3}$
Because of the presence of four types of hydrogen, it gives

$four + two (d$ and

$l$ forms)

$=$ six monochloro products on monochlorination as

$CH_3-CH_2\underset{CH_3}{\underset{|}{C}H}-CH_3\xrightarrow[hv]{Cl_2}\underset{ \begin{matrix}\text{d and l forms}\\\text{1-chloro-2-methyl butane}\end{matrix}}{CH_3CH_2\dot{C}H-CH_2Cl}+ \underset{2-chloro-2-methyl\,butane}{CH_3-CH_2-\overset{Cl}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}-}CH_3}$

$+\underset{2-chloro-3-methyl\,butane}{CH_3-\underset{Cl}{\underset{|}{\dot{C}}}H-CH(CH_3)_2}+\underset{1-chloro-3-methyl\,butane}{CH_2ClCH_2-\underset{CH_3}{\underset{|}{C}H}-CH_3}$

Among the following reaction(s) which gives tret - butyl benzene as the major product is/are:

Explanation

The maximum yield of a reaction between alkyl halide and benzene is obtained when it takes place in the presence of $AlCl_3$ .

In option $B$ , there will be a hydride shift in the above image, due to which $tert-butyl$ cation will attack the benzene ring to give $tert-butyl$ $benzene$ as the major product.

$CH_3 - \underset{CH_3}{\underset{|}{C}H} MgCl + CH_3 - \!\!\!\!\overset{CH_3}{\overset{|}{\underset{\,\,\,\,CH_3}{\underset{|}{C} }}\!\!\!-} \,OH \rightarrow$ '

$Q$ '

What is 'Q'?

0%
Isobutane
0%
Isopropane
0%
Tert. butyl chloride
0%
Propane

Explanation

In the given reaction, the reaction between Grignard's reagent and tert-butyl alcohol takes place.

Here, breaking of

$R$ and

$MgX$ bond takes place to give caranion ion. This ion has a high affinity of a proton. The

$MgX$ part substitues

$H$ from

$-OH$ to remove

$H^+$ ion.

This

$H^+$ ions gets attached to the carbanion ion to form Isopropane as a product.

The major product of the reaction of 2-butene with alkaline

$KMnO_4$ solution is

Explanation

Alkenes give diol with alkaline

$KMnO_4$ .

Option D is correct.

The reactivity towards the hydrogenation of the following compounds is :

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

Explanation

Reactivity toward hydrogenation =

$I>II>III>IV$

More stable alkenes require less heat of hydrogenation.

1171747_760049_ans_72a5ea509d97430995b44ec9f24947c7.png

The major product of acid-catalysed dehydration of Pentan

$-3-ol$ is:

0%
$Pent-1-ene$
0%
$2-Methyl-1-butene$
0%
$Pent-2-ene$
0%
$3-Methyl-1-butene$

Arrange the given alkenes in decreasing order of their reactivity towards bromination:

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

Explanation

reactivity towards Bromination depends on e

$^-$ rich alkene since Br

$^+$ is electrophile.

The alkene which have more methyl group attached to the doubly bonded carbon atom will be more electron rich.

Thus the order of the reactivity towards bromination reaction can be given as follow:

$i> ii> iii> iv$

Option A is correct.

Which of the following compounds is insoluble even in hot concentrated

$H_{2}SO_{4}$ ?

0%
Benzene
0%
Hexane
0%
Aniline
0%
Ethylene

Explanation

Hexane is an alkane which is also called as paraffin or less reactive. Thus they are not soluble in concentrated

$H_2SO_4$ .

Also, hexane is non-polar and concentrated

$H_2SO_4$ is polar, thus hexane is not soluble in

$H_2SO_4$ .

$CH\ \equiv \ CH\ \xrightarrow [ Ni{ \left( CN \right) }_{ 2 } ]{ Pressure } X$ . Here

$X$ in the reaction.

0%
Benzene
0%
Ethane
0%
Cycloctatetraene
0%
Cyclohexane

Explanation

3 moles of acetylene when catalysed with

$Ni(CN)_2$ gives benzene.

What is

$A$ ?

In commercial gasolines the type of hydrocarbons which are more desirable, is:

0%
Branch hydrocarbon
0%
Straight chain hydrocarbon
0%
Automatic hydrocarbon such as toluene
0%
Linear unsaturated hydrocarbon

Ethyl benzene react with

$BH_{2}/ FeBr_{3}$ to give:

$Ph - C \equiv C - CH_3 \xrightarrow[]{Hg^{2+}/H^+}A$
The product A is:

Explanation

$Ph - C \equiv C - CH_3 \xrightarrow[]{Hg^{2+}/H^+} \underset {A}{Ph-CO-CH_2-CH_3}$

The product A is propiophenone

$\displaystyle Ph-CO-CH_2-CH_3$ as the intermediate carbocation

$\displaystyle (Ph-C^+=CH-CH_3)$ is stabilised by resonance with benzene ring. In this reaction a molecule of water is added to

$C \equiv C$ triple bond to from an enol which tautomerises to ketone.

$CH_3-C\equiv C$ MgBr can be prepared by the reaction of _________.

0%
$CH_3-C\equiv C-Br$ with $MgBr_2$
0%
$CH_3-C\equiv CH$ with $MgB_2$
0%
$CH_2-C\equiv CH$ with KBr and Mg metal
0%
$CH_3-C\equiv CH$ with $CH_3MgBr$

Explanation

$CH_3-C\equiv C-H+CH_3MgBr\rightarrow CH_3-C\equiv CMgBr +CH_4$ .

Option D is correct.

On monochlorination of

$2-methylbutane$ , the total number of chiral compounds formed is:

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

$CH_{3} - CH_{2} - CH_{2} - CH_{2} - \underset {\underset {CH_{2}}{\parallel}}{C} - CH_{2} - CH_{2} - CH_{3}$ .

0%
$2-propylhex-1-ene$
0%
$2-butylpent-1-ene$
0%
$2-propyl-2-butylethene$
0%
$propyl-1-butylethene$

Explanation

Longest C chain containing C=C has 6 C as:

$\overset { 6 }{ CH_{ 3 } } -\overset { 5 }{ CH_{ 2 } } -\overset { 4 }{ CH_{ 2 } } -\overset { 3 }{ CH_{ 2 } } -\overset { 2 }{ \underset { \overset { \parallel }{ { C{ H }_{ 2 } }^{ 1 } } }{ C } } -\overset { 2' }{ CH_{ 2 } } -\overset { 3' }{ CH_{ 2 } } -\overset { 4' }{ CH_{ 3 } }$

Parent name: hex

functional group: ene

lowest locant number to double bond:1

Substituent position:2

Substituent name: propyl

Therefore the name is: 2-propylhex-1-ene. Option A is correct.

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 6 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 6 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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