Match the column I with Column II and mark the appropriate choice. Column I Column II (A) $$n-Butane \rightarrow 2-Methylpropane$$ (i) Free radical substitution (B) $$CH_4+Cl_2\xrightarrow [ ]{ hv } CH_3Cl$$ (ii) Wurtz reaction (C) $$RCOONa + Soda\,lime \rightarrow RH$$ (iii) Isomerisation (D) $$RX + Na \xrightarrow [ ]{ ether } R-R$$ (iv) Decarboxylation

$$(A)\rightarrow (iii) ,(B)\rightarrow(i) , (C)\rightarrow (iv),(D)\rightarrow (ii)$$

$$(A)\rightarrow (ii) ,(B)\rightarrow(iv) , (C)\rightarrow (i),(D)\rightarrow (iii)$$

$$(A)\rightarrow (i) ,(B)\rightarrow(ii) , (C)\rightarrow (iv),(D)\rightarrow (iii)$$

$$(A)\rightarrow (iv) ,(B)\rightarrow(i) , (C)\rightarrow (iii),(D)\rightarrow (ii)$$

MCQ Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 7

Match the column I with Column II and mark the appropriate choice.

	Column I		Column II
(A)	$n-Butane \rightarrow 2-Methylpropane$	(i)	Free radical substitution
(B)	$CH_4+Cl_2\xrightarrow [ ]{ hv } CH_3Cl$	(ii)	Wurtz reaction
(C)	$RCOONa + Soda\,lime \rightarrow RH$	(iii)	Isomerisation
(D)	$RX + Na \xrightarrow [ ]{ ether } R-R$	(iv)	Decarboxylation

0%
$(A)\rightarrow (iii) ,(B)\rightarrow(i) , (C)\rightarrow (iv),(D)\rightarrow (ii)$
0%
$(A)\rightarrow (ii) ,(B)\rightarrow(iv) , (C)\rightarrow (i),(D)\rightarrow (iii)$
0%
$(A)\rightarrow (i) ,(B)\rightarrow(ii) , (C)\rightarrow (iv),(D)\rightarrow (iii)$
0%
$(A)\rightarrow (iv) ,(B)\rightarrow(i) , (C)\rightarrow (iii),(D)\rightarrow (ii)$

Explanation

(A)

$n-Butane \longrightarrow 2-Methylpropane$

Isomerisation is a reaction where one organic compound converts to another compound with same molecular mass but different structures.

n-Butane and 2-Methylpropane, both are isomers and the reaction is isomerisation

$\rightarrow$ (iii) Isomerisation.

(B)

$C{H}_{4} + {Cl}_{2} \xrightarrow{h \nu} C{H}_{3}Cl$

The above reaction is Free radical substitution which takes place in presence of sunlight

$\rightarrow$ (i) Free radical substitution.

(C)

${RCOO}^{-}{Na}^{+} + \underset{Soda\,lime}{NaOH + CaO} \longrightarrow RH + {Na}_{2}C{O}_{3}$

The above reaction is known as Decarboxylation used to prepare hydrocarbons. Sodium salt of carboxylic acid releases

$C{O}_{2}$ to form alkanes and

${Na}_{2}C{O}_{3} \rightarrow$ (iv) Decarboxylation.

(D)

$RX + Na \xrightarrow{ether} R-R$

The above reaction is termed as Wurtz reaction which unites two alkyl halides to form one alkane in presence of sodium and ether

$\rightarrow$ (ii) Wurtz reaction.

Chlorination of alkanes is a photochemical process. It is initiated by the process of:

0%
heterolysis
0%
homolysis
0%
pyrolysis
0%
hydrolysis

Explanation

Photochemical chlorination of alkane takes place by free radical mechanism which are possible by homolysis of

$Cl - Cl$ bond by sunlight.

${Cl}_{2} \xrightarrow{h \nu} {Cl}^{\bullet} + {Cl}^{\bullet}$

$C{H}_{3}-C{H}_{3} + {Cl}^{\bullet} \longrightarrow \,^{\bullet}{C{H}_{2}CH_3} + HCl$

$Cl-Cl + \,^{\bullet}{C{H}_{2}CH_3} \longrightarrow Cl-{C{H}_{2}CH_3} + {Cl}^{\bullet}$

The cycloalkane formed when

$1,4-dichloropentane$ is heated with sodium in presence of dry ether is ________.

0%
methylcyclopentane
0%
cyclopentane
0%
methylcyclobutane
0%
cyclobutane

Which one is incorrect name?

0%
Propyne
0%
But-2-yne
0%
Pent-3-yne
0%
But-1 -yne

Explanation

The correct answer is option C, because As the structural formula of Pent-3-yne is

$CH_3 - CH_2 - C \equiv C - CH_3$ , According to the IUPAC nomenclature the numbering of the carbon atom is done from that side of the carbon chain from which the triple bond is nearest as in this structure the numbering can also be done from right side. Therefore, the correct IUPAC name of the compound is

$Pent-2-yne$ .

Which one of the following IUPAC name is not correctly matched?

Ethane is formed during the formation of chloromethane by chlorination of methane because:

0%
higher members of the hydrocarbons are generally formed during reactions
0%
two methyl free radicals may combine during chlorination to give ethane
0%
two chloromethane molecules react to form ethane
0%
chlorine free radical reacts with methane to give ethane.

Explanation

Chlorination of methane is a free radical reaction. The free radicals formed during the reaction are

$Cl^{\bullet}$ and

$CH_3^{\bullet}$ . During the process if two methyl radicals collapse they form ethane. So, ethane is also formed as a byproduct.

$2{Cl}^{\bullet} \rightarrow {Cl}_{2}$

${{H}_{3}C}^{\bullet} + {Cl}^{\bullet} \longrightarrow C{H}_{3}Cl$

${{H}_{3}C}^{\bullet} + {{H}_{3}C}^{\bullet} \longrightarrow C{H}_{3}-C{H}_{3}$

Finely divided platinum and palladium commonly known as platinum and palladium black, may be prepared by reducing their,soluble salts with:

0%
$H_{2}O$
0%
$C_{2}H_{5}OH$
0%
$HCHO$
0%
$C_{6}H_{6}$

Cis-dicholoroethylene can be converted to trans-dicholorethylene by irradiations. During the interconversion.

0%
Both the $\pi$ and $\sigma$ bonds are broken and reformed
0%
Only the $\pi$ bond is broken and reformed
0%
The $C-Cl$ and $C-H$ bonds are broken, $H$ and $Cl$ atoms interchanges their positions and new bonds are formed
0%
Both the $\pi$ and $\sigma$ bonds rotate along with the groups

Explanation

Cis-dicholoroethylene can be converted to trans-dicholorethylene by irradiation. During the inter-conversion, only the

$\pi$ bond is broken and reformed. Rotation about

$C-C$ single bond converts cis isomer into trans isomer and vice versa.

Which of the following has the lowest boiling point?

0%
2-Methylbutane
0%
2-Methylpropane
0%
2,2-Dimethylpropane
0%
n-pentane

Explanation

Boiling point increases with increase in molecular mass. For the compounds with the same molecular mass, boiling point decreases with an increase in branching.

The molecular mass of 2-Methylbutane:

$72\,g\,mol^{-1}$

The molecular mass of 2-Methylpropane:

$58\,g\,mol^{-1}$

The molecular mass of 2,2-Dimethylpropane:

$72\,g\,mol^{-1}$

The molecular mass of 2-Methylbutane:

$72\,g\,mol^{-1}$

2-Methylpropane has the lowest molecular mass among all of the given compounds.

Thus, 2-Methylpropane has the lowest boiling point among the given options.

During halogenation of alkanes, the halogens and alkane show a specific trend. Which of the following statements is not correct?

0%
The reactivity of halogens is in the order $F_2>Cl_2>Br_2>I_2$ .
0%
For a given halogen the reactivity of hydrocarbon is in the order of $3^o>2^o>1^o$ .
0%
Bromine is less reactive than chlorine towards a particular alkane.
0%
On chlorination monosubstituted product is formed while on bromination disubstituted products are formed.

The heat of combustion of carbon to

$C{O}_{2}$ is -393.5 kJ/mol. The heat released upon formation of 35.2 g of

$C{O}_{2}$ from catbon and oxygen gas will be :

0%
+315 kJ/mole
0%
-31.5 kJ/mole
0%
-315 kJ/mole
0%
+31.5 kJ/mole

Explanation

Heat of combustion for

$1$ mole of

$CO_2$ is

$-393.5KJ$

Number of moles in

$35.2g$ of

$CO_2= \cfrac {35.2}{44}=0.8$ mole

Thus heat of combustion for

$0.8$ mole of

$CO_2$ is

$-393.5 \times 0.8=-315KJ$

What is the carbon-carbon bond length in benzene?

0%
$1.20 \mathring { A }$ and $1.31 \mathring { A }$
0%
$1.39 \mathring { A }$
0%
$1.39 \mathring { A }$ and $1.20 \mathring { A }$
0%
$1.20 \mathring { A }$

Explanation

All

$C-C$ and

$C=C$ bond lengths are the same in benzene. Hence,

$C-C$ bond length in benzene is 1.39

$\mathring { A }$ .

Nitration and chlorination of benzene are:

0%
nucleophilic and electrophilic substitution respectively
0%
electrophilic and nucleophilic substitution respectively
0%
electrophilic substitution in both the reactions
0%
nucleophilic substitution in both the reactions.

Explanation

Since nitration and halogenation, both are carried out by the help of electrophile

${N{O}_{2}}^{+} \; \& \; {Cl}^{+}$ . Hence, both reactions are electrophilic substitution reactions.

Although benzene is highly unsaturated it does not undergo addition reactions. The explanation of this can be suggested as:

0%
$\pi$ -electrons of benzene ring are delocalised
0%
since $\pi$ -electrons are present inside the ring, addition cannot take place
0%
cyclic structures do not show addition reactions
0%
benzene is not a reactive compound.

Explanation

(A)

$\pi$ -electrons of the benzene ring are delocalised.

There are delocalised

$\pi$ -electrons above and below the plane of the ring. The presence of the delocalised

$\pi$ -electrons makes benzene particularly stable. Benzene resists addition reactions because that would involve breaking the delocalisation and losing that stability.

What happens when calcium carbide is treated with water?

0%
Ethane is formed.
0%
Methane and ethane are formed.
0%
Ethyne is formed.
0%
Ethene and ethyne are formed.

Explanation

If we mix calcium carbide with water ethyne is released due to endothermic reaction between them. This is the industrial method of producing ethyne.

$Ca^{2+}{[{C}\equiv C]}^{2-} + 2{H}_{2}O \longrightarrow \underset{Ethyne}{CH \equiv CH} + Ca{(OH)}_{2}$

Fill in the blanks with appropriate words. Benzene has a planar structure. All carbon atoms in benzene are I hybridised. The ring structure of benzene was proposed by II . It shows III substitution reactions. It reacts with IV in presence of aluminium chloride to form acetophenone.

0%
$I = sp_2 \,\,\, II = Kekule \,\,\, III= electrophilic \,\,\, IV = acetyl\, chloride$
0%
$I = sp \,\,\, II = Dewar \,\,\, III= nucleophilic \,\,\, IV =chloromethane$
0%
$I = sp_3 \,\,\, II = Ladenberg \,\,\, III= electrophilic \,\,\, IV = chloroethane$
0%
$I = sp_2 \,\,\, II = Baeyer \,\,\, III= nucleophilic \,\,\, IV = methyl\, bromide$

Explanation

Benzene has a planar structure. All carbon atoms in benzene are

$s{p}^{2}$ hybridised being bonded to two other carbon atoms and one hydrogen atom. The ring structure of benzene was proposed by Kekule. It shows electrophillic substitution reactions due to electron rich character. It reacts with acetyl chloride in presence of aluminium chloride to form acetophenone and the reaction is known as acylation of benzene.

Hence, A is correct answer

When n-butane is treated with chlorine, how many monochloro products are formed?

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

Explanation

${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CH }_{ 3 }+{ Cl }_{ 2 }\rightarrow { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CH }_{ 2 }{ Cl }+{ CH }_{ 3 }{ CH }_{ 2 }{ - }\underset { \overset { | }{ Cl } }{ CH } -{ CH }_{ 3 }$

What is the order of reactivity of hydrogen atoms attached to the carbon atom in an alkane for free radical substitution?

0%
$3^o >1^o >2^o$
0%
$2^o >1^o >3^o$
0%
$3^o >2^o >1^o$
0%
$1^o >2^o >3^o$

Explanation

REF: free radical will have

$9 \alpha H.$

$CH_{3}-CH_{2}-CH_{3}$ free radical will have

$6 \alpha H$

$CH_{3}-CH_{3}-H$ free radical will have

$3\alpha H$

As no.of

$\alpha -H$ increase stability of free radical increase

reactivity order of H

$3^{\circ} > 2^{\circ} > 1^{\circ}$

Hence, te correct option is

$\text{C}$

1138047_928875_ans_6dfc9f8774b14ba480880807d526a61f.png

One mole of 1,2-dibromopropane on treatment with

$X$ moles of

$NaNH_2$ followed by treatment with ethyl bromide gave a 2-pentyne. The value of

$X$ is:

0%
one
0%
two
0%
three
0%
four

Explanation

$C{H}_{3}CH(Br)C{H}_{2}(Br) \xrightarrow{2NaN{H}_{2}} C{H}_{3}-C \equiv CH \xrightarrow{NaN{H}_{2}} {C{H}_{3}-C \equiv C}^{-}{Na}^{+} \xrightarrow{{C}_{2}{H}_{5}Br} C{H}_{3}C \equiv CC{H}_{2}C{H}_{3}$

Hence,

$X=3$ .

Mark the incorrect statement from the following.

0%
Benzene has a planar structure
0%
Benzene is an unsaturated hydrocarbon and shows addition reactions like alkenes
0%
In benzene carbon uses two p-orbitals for hybridisation
0%
Aromatic hydrocarbons contain high percentage of carbon hence burn with sooty flame

The compound formed when an alcoholic solution of ethylene dibromide is heated with granulated zinc is:

0%
ethene
0%
ethyne
0%
ethane
0%
bromoethane

Explanation

The given reaction is dehalogenation reaction. Zinc removes bromine from ethylene dibromide as zinc bromide and forms ethene.

$\underset{\text{Ethylene dibromide}}{Br{H}_{2}C-C{H}_{2}Br} + Zn \xrightarrow[alcohol]{heat} {H}_{2}C=C{H}_{2} +Zn{Br}_{2}$

For the given Question (1,2,3) consider the following reaction.
How many monohalo derivatives are possible (excluding stereoisomers) ?

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

Halogenation is a substitution reaction, where halogen replaces one or more hydrogens of hydrocarbon.
Chlorine free radical make

$1^{\circ} ,2^{\circ} ,3^{\circ}$ radicals with almost equal ease, whereas bromine free radicals have a clear preference for the formation of tertiary free radicals. So , bromine is less reactive, and more selective whereas chlorine is less selective and more effective.
The relative rate of abstraction of hydrogen by Br is

$\underset{(1600)}{3^\circ} \, > \, \underset{(82)}{2^\circ} \, > \, \underset{(1)}{1^\circ}$
The relative rate of abstraction of hydrogen by Cl is

$\underset{(5)}{3^\circ} \, > \, \underset{(3.8)}{2^\circ} \, > \, \underset{(1)}{1^\circ}$

$CH_3 - \overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}C}}} -CH_2 - \overset{CH_3}{\overset{|}CH} - CH_3 \xrightarrow{Br_2/hv} CH_3 - \overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}C}}} - CH_2 - CH_2 - \overset{CH_3}{\overset{|}{\underset{Br}{\underset{|}C}}} - CH_3$
2,Bromo-2,5,5 trimethyl hexane(x%)
What is the value of x(% yield of product) ?

0%
18 %
0%
82 %
0%
90 %
0%
60 %

Explanation

$CH_3 - \!\!\!\overset{\,\,\,\,CH_3}{\overset{|}{\underset{\,\,\,\,CH_3}{\underset{|}C}}}\!\!\! -CH_2 - \overset{CH_3}{\overset{|}CH} - CH_3 \xrightarrow{Br_2/hv} CH_3 -\!\!\! \overset{\,\,\,\,CH_3}{\overset{|}{\underset{\,\,\,\,CH_3}{\underset{|}C}}}\!\!\! - CH_2 -\!\!\! \overset{\,\,\,\,CH_3}{\overset{|}{\underset{Br}{\underset{|}C}}} - CH_3$

Here, tertiary bromination occurs-

To find the percentage

$(3^o\%)=\cfrac {3^o}{All}\times 100$

As,

$3^o\rightarrow 1\times 1600=1600$

$2^o\rightarrow 2\times 82=164$

$1^o\rightarrow 15\times 1=15$

So, All is

$1600+164+15=1779$

$\therefore \%=\cfrac {1600}{1779}\times 100=90\%$

Which of the following is aromatic in nature?

0%
0%
0%
0%
0%
None of these

Explanation

R.E.F image

Non Aromatic

R.E.F image

Antiaromatic

$4\pi e^{-}s$

R.E.F image

Aromatic

$2\pi e^{-}s$

R.E.F image

Non Aromatic

1138088_933078_ans_ba515be57a654a3994a6d18fb7c85c48.png

Select the chain propagation steps in the free-radical chlorination of methane.
(1)

$Cl_2\, \rightarrow 2Cl^{\cdot}$
(2)

$Cl^\cdot \, + \, CH_4 \, \xrightarrow \, CH_3C1 \,+ \, H^\cdot$
(3)

$C1^\cdot \, + \, CH_4 \, \xrightarrow \, CH_3 ^\cdot \, + \, HCI$
(4)

$H^\cdot \, + \, Cl_2 \, \xrightarrow \, HCl+Cl^\cdot$
(5)

$CH_3 ^\cdot \, + \, Cl_2 \, \xrightarrow \, CH_3CI+Cl^\cdot$

0%
2,3,5
0%
1,3,6
0%
3,5
0%
2,3,4

Explanation

(1)

$Cl_2 \longrightarrow 2Cl^\cdot$ [Initiation]

(2)

$Cl^\cdot+CH_4 \longrightarrow CH_3Cl+H^\cdot$

(3)

$Cl^\cdot + CH_4 \longrightarrow CH_3^\cdot+ HCl$ [Propagation]

(4)

$H^\cdot+Cl_2 \longrightarrow HCl+Cl^\cdot$

(5)

$CH_3^\cdot+Cl_2 \longrightarrow CH_3Cl+Cl^\cdot$ [Propagation]

In the first propagation step (3), a

$Cl^\cdot$ combines with a hydrogen on

$CH_4$ giving

$HCl$ and

$^\cdot CH_3$ . In the

$2^{nd}$ propagation step (5)

$Cl_2$ combines with

$^\cdot CH_3$ to form

$CH_3Cl$ and

$Cl^\cdot$ .

Rank the transition states that occur during the following reaction steps in order of increasing stability

$(least \xrightarrow \,most stable)$
1.

$H_3C \, - \, \overset{+}{OH_2} \, \xrightarrow \, CH_3^{+} \, + \, H_2O$
2.

$(CH_3)_3C \, - \, \overset{+}{OH_2} \xrightarrow \, - \,(CH_3)_3C \, + \, H_2O$
3.

$(CH_3)_2 \, CH \, - \, \overset{+}{OH_2} \xrightarrow \,(CH_3)_2CH^+ \, + \, H_2O$

0%
1 < 2 < 3
0%
2 < 3 < 1
0%
1 < 3 < 2
0%
2 < 1 < 3

Explanation

(1) (Refer to Image 1)

(2) (Refer to Image 2)

(3) (Refer to Image 3)

Now stability of carbocations: tertiary

$>$ secondary

$>$ primary. Therefore transition state corresponding to tertiary carbocation will be most stable. So stability order will be

$1 < 3 <2$ .

1116707_989065_ans_e1e8a98cac6e49f19703835d6194e4a9.png

n-Butane

$\xrightarrow{Cl_2/hv}$
Give the total number of monochioro products (including stereoisomers), which are possible in the above reaction.

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

Explanation

$n-Butane$ gives three products i.e.

$S-2-chlorobutane, R-2-chlorobutane$ and

$1-chlorobutane$ when

$n-Butane$ is treated with

$Cl_2$ in presence of light.

1116682_988977_ans_703e8f1b08f34871be18fb0badf16af5.PNG

An alkane

$(mol.\ wt.\ = 86)$ on bromination gives only two monobromo derivatives (excluding stereoisomers). The alkane is:

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

Explanation

$CH_3-\underset {|}{CH}-CH_2-CH_2-CH_3 \xrightarrow [h\nu]{Br_2} 5$ monobromoderivatives

$CH_3$

$A$

$CH_3$

$CH_3-\underset {|}{\overset {|}{C}}-CH_2-CH_3 \xrightarrow [h\nu]{Br_2} 3$ monobromoderivatives

$CH_3$

$B$

$CH_3-\underset {|}{CH}-\underset {|}{CH}-CH_3 \xrightarrow [h\nu]{Br_2} 2$ monobromoderivatives

$CH_3$

Among them

$A,B$ and

$C$ have only molecular weight

$86$ . Now

$A$ has

$5$ different equivalent

$H$ atoms and

$B$ has

$3$ different equivalent

$H$ atoms, so

$A$ and

$B$ will give

$5$ and

$3$ monobromo derivatives respectively. But

$C$ has

$2$ different equivalent

$H$ atoms, so

$C$ will give

$2$ monobromo derivatives.

$\xrightarrow[hv]{cl_2}$ (x). (x) = total number of di-chloro product S-2 chloro hexane.

0%
$7$
0%
$6$
0%
$16$
0%
none

Explanation

$Cl$

$H_3C-\overset {|}{\underset {|}C}-CH_2-CH_2-CH_2-CH_3 \xrightarrow [h\nu]{Cl_2} (x)$

$H$

The possible product may be

$Cl$

(1)

$H_2\underset {|}{C}-\overset {|}{\underset {|}C^*}-CH_2-CH_2-CH_2-CH_3 \longrightarrow 2^1=2$ stereoisomers

$Cl$

$H$

$Cl$

(2)

$H_3C-\overset {|}{\underset {|}C}-CH_2-CH_2-CH_2-CH_3 \longrightarrow 2^o= 1$ stereoisomer

$Cl$

(3)

$H_3C-_*\overset {|}{\underset {|}C}-_*\overset {|}{\underset {|}C}-CH_2-CH_2-CH_3= 2^2=4$ stereoisomers

$H$

$Cl$

(4)

$H_3C-^*\overset {|}{\underset {|}C}-CH_2-\overset {|}{\underset {|}C^*}-CH_2-CH_3=2^2=4$ stereoisomers

$H$

$Cl$

(5)

$H_3C-_*\overset {|}{\underset {|}C}-CH_2-CH_2-_*\overset {|}{\underset {|}C}-CH_3=$ as it is symmetrical molecule=

$2^{2-1}+2^{2/2-1}=3$ stereoisomers

$H$

$Cl$

(6)

$H_3C-\overset {|}{\underset {|}C}-CH_2-CH_2-CH_2-CH_2-Cl= 2^1=2$ stereoisomers

$H$

So, total number of di chloro product obtained=

$2+1+4+4+3+2=16$ product

Arrange the following compounds in decreasing order of their heats of combustion:

0%
(iii) > (ii) > (i)
0%
(ii) > (i) > (iii)
0%
(iii) > (i) > (ii)
0%
(i) > (ii) > (iii)

Explanation

Heat of combustion is inversely proportional to stability.

Since three membered ring is most unstable due to angular strain so it is having highest heat of combustion.

So the order for heat of combustion is:

$(i) > (ii) > (iii)$

Among the following free radical bromination reactions, select those in which 2 halide is the major product

0%
P, Q, R, S
0%
P, R, U
0%
P, R, S, T
0%
P, Q, R, S, T

Explanation

Solution:- (B) P, R, U

Bromine is more selective.

$Q, S, T$ form

$3°$ halide as major and

$P, R, U$ form

$2°$ halide as major.

Which is not an aromatic compound?

0%
Pyridine
0%
Xylene
0%
Naphthalene
0%
Cyclohexane

Rank the hydrogen atoms

$(H_a, H_b, H_c)$ in the following molecules according to their acidic strengths:

0%
$a > b > c$
0%
$b > a > c$
0%
$b > c > a$
0%
$a > c > b$

Explanation

Removal of

$H_c$ creates a negative charge on

$sp^2$ hybridized carbon which is highly unstable, due to which

$H_c$ will be least acidic.
Removal of

$H_b$ creates a negative charge which is stabilized by resonance due to conjugation with double bond, There is no such factor in case of

$H_a$
Hence the order of acid strength is

$b>a>c$ .

In this reaction

$SbF_5$ acts as:

0%
an acid
0%
a base
0%
a nucleophile
0%
an electrophile

Explanation

$F$ is a good nucleophile, so it will attack on

$SbF_5$ , which is electrophilic in nature and act as electrophile.

1091410_990372_ans_ec2bf24f06d1436e84fcd5033ecb118e.JPG

What is the final product (B) of the sequence

In order to prepare 1- chloro-propane which of the following reactants can be employed?

0%
Propene and HCl in the presence of peroxide
0%
Propene and $Cl_2$ followed by treatment with aq.KOH
0%
Propanol-1 and $SOCl_2$ /pyridine
0%
Any of the above can be used

Explanation

Propanol-1 and

$SOCl_{2}$ with pyridine as a solvent can be used to prepare 1 - chloro - propane. The by products

$SO_{2}$ and

$HCl$ are both gaseous so they go off leaving the actual product.

$(A) \, + \, Cl_2 \, \xrightarrow {hv} \, monochloro \, product$
To maximise the yield of monochloro product in the above reaction?

0%
$Cl_2$ must be added in excess
0%
Reactant (A) must be added in excess
0%
Reaction must be carried out in dark
0%
Reaction must be carried out with equimolar mixture of $Cl_2$ and A

Choose the appropriate product for this reaction.

To carry our above conversion,

$(A)$ and

$(B)$ respectively, are :

0%
$NaNH_{2}, Cl - CH_{2} - CH_{2} - CH_{2} - Br$
0%
$NaNH_{2}, F - CH_{2} - CH_{2} - CH_{2} - Br$
0%
$NaNH_{2}, I - CH_{2} - CH_{2} - CH_{2} - Br$
0%
$NaNH_{2}, I - CH_{2} - CH_{2} - CH_{2} - I$

$CH_3MgI$ will give methane with:

0%
$CH_3CH_3$
0%
$CH_3-CH_2-NH_2$
0%
$CH_3-CO-CH_3$
0%
all of these

In the reaction:

$H-C\equiv CH\xrightarrow [ (i)NaNH_{ 2 }/Liq.NH_{ 3 } ]{ { (ii)CH_{ 3 }CH_{ 2 }Br } } X\xrightarrow [ (i)NaNH_{ 2 }/Liq.NH_{ 3 } ]{ { (ii)CH_{ 3 }CH_{ 2 }Br } } Y$

$X$ and

$Y$ are:

0%
$X=2$ -butyne, $Y=2$ -hexyne
0%
$X=1$ -butyne, $Y=2$ -hexyne
0%
$X=1$ -butyne, $Y=3$ -hexyne
0%
$X=2$ -butyne, $Y=3$ -hexyne

Reaction

$H-C\equiv C-H+HOCl\rightarrow$ product, here product will be:

0%
$ClCH_2-CHO$
0%
$CHO-CHO$
0%
$CH-Cl=CHCl$
0%
$CHCl_2-CHCl_2$

Using anhydrous

$AlCl_3$ as catalyst, which one of the following reaction produces ethylbenzene

$(PhEt)$ ?

0%
$H_2C=CH_2+C_6H_6$
0%
$H_3C-CH_3+C_6H_6$
0%
$H_3C-CH_2OH+C_6H_6$
0%
$CH_3-CH=CH_2+C_6H_6$

Explanation

R.E.F image

$CH_{2} = CH_{2} \overset{AlCl_{3}/HCl}{\rightarrow} CH_{3}-C^{\oplus}H_{2}+AlCl_{4}^{\circleddash }$

1126497_1037029_ans_82423f03247f41b4bfe080cd17674774.png

In the following sequence of reactions:
Toluene

$\overset{KMnO_4}{\rightarrow} A\overset{SOCl_2}{\rightarrow} B\overset{H_2/Pd}{\underset{BaSO_4}{\rightarrow}}C$ .

What is C?

0%
Benzaldehyde
0%
Benzoic acid
0%
Benzoyl chloride
0%
Acetophenone

IUPAC name of the compound is:

0%
butane
0%
$\gamma$ -Butane
0%
cyclo-Butane
0%
n-Butane

How will you prepare ethane from methyl chloride?

0%
Williamson's synthesis
0%
Wurtz synthesis
0%
Hydrogenation of methyl chloride
0%
Dehydrohalogenation of methyl chloride

Explanation

$C_{2}H_{6}$ will be lest peppered by ethane synthesis from

$CH_{3}Cl$

$CH_{3}Cl +2Na\rightarrow CH_{3}^{-}Na^{+}+NaCl$

$CH_{3}^{-}Na^{+}+CH_{3}-Cl\rightarrow C_{2}H_{6}+NaCl$

1125178_1034056_ans_c93644d90116416b9c04aec98f622e94.png

Which of the following compound, on reaction with excess of RMgX, gives 3 moles of alkane (R-H)?

0%
$HC \equiv C -{\underset{OH}{\underset{|}{C}}H} - COOH$
0%
$CH_3 - C \equiv C -{\underset{OH}{\underset{|}{C}}H} - COOH$
0%
$HC \equiv C -{\underset{OCH_3}{\underset{|}{C}}H} - COOH$
0%
$CH_2 = CH -{\underset{OH}{\underset{|}{C}}H} - COOH$

Explanation

$HC \equiv C - \underset{OH}{\underset{|}{C}H} - COOH \xrightarrow{3 R-MgX} \,^{\circleddash}C \equiv C - \underset{\,\,O^{\circleddash}}{\underset{|}{C}}\!\!H - COO^{\circleddash} + 3R-H$

The barrier for rotation about indicated bonds will be maximum in which of these compounds?

0%
X
0%
Y
0%
Z
0%
Same in all

The compound

$B$ formed in the following sequences of reactions is:

$CH_3CH_2CH_2OH\overset{PCl_3}{\rightarrow}A\overset{Alc. KOH}{\rightarrow}B$ .

0%
propyne
0%
propene
0%
propanol
0%
propane

Explanation

$CH_{3}CH_{2}CH_{2}OH\overset{PCl_{3}}{\rightarrow}\underset{propyl\ chloride}{CH_3CH_2CH_2Cl}\overset{alc.KOH}{\rightarrow}\underset{propene}{CH_3CH=CH_2}$

In the first step the propyl alcohol undergo substitution reaction to form propyl chloride as a proudct which in presence of alc. KOH undergoes elimination reaction to form propene as a product.

Write chemical equations for the preparation of following alcohols in the options starting from appropriate alkenes.

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 7 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 7 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.