MCQ Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 8

Which of the following compound is not suitable to obtain from wurtz reaction?

0%
Ethane
0%
Butane
0%
Isobutane
0%
Hexane

Which of the following reactants will yield ethane ?

0%
Methyl bromide and sodium
0%
Ethyl bromide and magnesium
0%
Ethanol and $H_2SO_4$
0%
Ethyl bromide and KCN

H_{3} C - C \equiv C - C H_{3}

$H_3C-C\equiv C-CH_3$ maximum number of linear atoms in given compound is:

0%
4
0%
6
0%
10
0%
8

Explanation

Since one carbon

.

$.$ Carbon triple bond is present

The No

.

$.$ of linear atom

= 6

$=6$

Hence

,

$,$ option

(B)

$(B)$ is correct

.

$.$

Which of the following reactions would give isopropyl benzene as the major product?

0%
a, d
0%
b, c
0%
b, c, d
0%
All of these

Explanation

All of these :

(Intermediate is

> \oplus

$> \oplus$ in all )

1425684_1058572_ans_544244562db84fa698a7d004c7c8e804.png

Arrange the following compounds in increasing order of their reactivity with sodium hydroxide solution:

[o-nitrochlorobenzene, chlorobenzene, benzyl chloride]

0%
$benzylchloride$ < $chlorobenzene$ < $o-nitrochlorobenezene$
0%
$chlorobenzene$ < $o-nitrochlorobenzene$ < $benzyl chloride$
0%
$chlorobenzene$ < $benzyl chloride$ < $o-nitrochlorobenzene$
0%
$benzyl chloride$ < $o-nitrochlorobenzene$ < $chlorobenzene$

Explanation

Reactivity with sodium hydroxide solution increases with deactivating group at benzene more reactivity of the compounds.

=benzylchloride

<

$<$ 'chlorobenzene

<

$<$ o-nitrochlorobenzene

correct Answer

A - -

$\underline{A}$

2007473_1095932_ans_800876f6ad824780a322bb7240c85457.PNG

The least reactive alkene towards dil.

H C l

$HCl$ is :

0%
0%
$CH_3-CH = CH_2$
0%
0%
$CH_2 = CH_2$

Explanation

$\begin{array}{l}\text { The addition reaction of alkenes proceed via } \\\text { a carbocation intermediate }\end{array}$

$\begin{array}{l}\text { Since, D) is only primary carbocation. } So \text { , it has } \\\text { least stable carbocation intermediate. Therefore } \\\text { D) }\mathrm{CH}_{2}=\mathrm{CH}_{2} \text { will be least reactive with dil HCl. }\end{array}$

Nitration of which of the following compound is difficult?

0%
Benzene
0%
Nitrobenzene
0%
Toluene
0%
Phenol

Alkanes having odd carbons cannot be prepared in?

A) Wurtz reaction
B) Frankland reaction
C) Kolbe's electrolysis
D) Sabatier-sendersen reaction

0%
B,C&D
0%
A, B & D
0%
A, C & D
0%
A, B & C

Arrange the following compounds in the descending order of their boiling points

a) n-pentane

b) isopentane

c) neopentane

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

In the following sequence of reaction

$Toluene\xrightarrow [ ]{ KMn{ O }_{ 4 } } A\xrightarrow [ ]{ { SOCl }_{ 2 } } B\xrightarrow [ BaS{ O }_{ 4 } ]{ { H }_{ 2 }/Pd } C$
the product

$C$ is:

0%
${ C }_{ 6 }{ H }_{ 5 }COOH$
0%
$C_{6}H_{5}CH_{3}$
0%
${ C }_{ 6 }H_{5}{ CH }_{ 2 }OH$
0%
$C_{6}H_{5}CHO$

Explanation

$KMnO_4$ oxidizes toulene to benzoic acid.

(Refer to Image)

In the presence of

$Pd/BaSO_4$ (Lindlar's catalyst) Benzoyl chloride gets converted to benzaldehyde. The reaction is called Rosenmund reduction.

1214521_1127954_ans_0af40ba821a04f789a792a9f540d4026.PNG

2-Bromopentane is heated with

$EtO^-Na^+$ in ethanol. The major product obtained is:

0%
2-ethoxypentane.
0%
pent-1-ene.
0%
isobutane.
0%
pent-2-ene.

The number of

${ sp }^{ 2 }$ - s sigma bonds in benzene are :

0%
3
0%
6
0%
12
0%
None of these

Which of the following will allow free rotation about double bond?

[arrows indicate the bond to be considered]

Which of the following molecules has a triple bond?

0%
${CH}_{4}$
0%
${C}_{2}{H}_{4}$
0%
${C}_{2}{H}_{2}$
0%
${O}_{3}$

Explanation

$C_2H_2$ is the correct answer.

$CH_4\Longrightarrow$ alkane

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

$C_2H_4\Longrightarrow$ alkene

$H_2C=CH_2$

$C_2H_2\Longrightarrow$ alkyn

$CH\equiv CH$

$O_3\Longrightarrow O=O^+-O^-\longleftrightarrow O-O^+=O$

The reaction of

$P_4$ with aqueous NaOH gives

0%
$P(OH)_3$
0%
$P_2O_5$
0%
$P(OH)_5$
0%
$PH_3$

Explanation

The reaction of

$P_4$ with aqueous NaOH gives phosphine.

$4.3\ gm$ of an alkane is burnt in sufficient oxygen. The

${ CO }_{ 2 }$ formed reacts completely with

$300\ ml, 2N NaOH$ solution producing

${ Na }_{ 2 }{ CO }_{ 3 }$ . The alkane should be

0%
${ C }_{ 3 }{ H }_{ 8 }$
0%
${ C }_{ 12 }{ H }_{ 26 }$
0%
${ C }_{ 6 }{ H }_{ 14 }$
0%
${ C }_{ 2 }{ H }_{ 6 }$

Correct IUPAC name of the following compound is :

0%
$4-(1,1-dimethylethyl)-5-ethyloctane$
0%
$5-ethyl -4-(1,1-dimethylethyl) octane$
0%
$5-(1,1-dimethylethyl)-4-ethyloctane$
0%
$4-ethyl-5-(1,1-dimethylethyl) octane$

Ethylene is formed by dehydration of:

0%
$CH_3CHO$
0%
$C_2H_5OH$
0%
Propyl alcohol
0%
Ethyl acetate

Explanation

$C{H_3} - C{H_2} - OH\xrightarrow{{conc.\,{H_2}S{O_4}}}C{H_2} = C{H_2} + {H_2}O$

correct option is

$B$

Identify product.

0%
$C_2H_2+2CO_2$
0%
$2KOH+H_2$
0%
$2KOH+2CO_2$
0%
$C_2H_2+H_2$

Which of the following is aromatic in nature

0%
0%
0%
0%
$All \ of \ these$

The number of

$sp^2-s$ sigma bonds is benzene are:

0%
3
0%
6
0%
12
0%
none of these

Explanation

Benzene has molecular formula

$C_6H_6$ that is there is the presence of a 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. So there are a total of

$12$ sigma bonds and

$3$ pi bonds in benzene.

The carbon atoms are bonded through

$sp^2-Sp^2$ overlap while six

$C-H$ bond is formed by

$sp^2-s$ sigma bond.

Thus there is

$6$

$sp^2-s$ sigma bond.

For synthesis of 1 -butene,

$CH_3MgI$ should be treated with :

0%
propene
0%
2 - chloropropene
0%
allyl chloride
0%
ethyl chloride

$CH_{3}-C\equiv C-CH_{3}\xrightarrow[Pd+BaSO_{4}]{H_{2}}P\overset{KMnO_{4}/H^{+}}{\rightarrow}Q.$ . here the product Q is

0%
Alkane
0%
Alkene
0%
Aldehyde
0%
Carboxylic acid

Explanation

$CH_3-C\equiv C-CH_3\xrightarrow{H_2} CH_3CH=CH-CH_3\xrightarrow{H_2} CH_3CH_2CH_2CH_3$

Alkane will be the final product on hydrogenation of alkyne.

Option A is the correct answer.

Double bond equivalent (degree of unsaturation) of

$(A)$ is:

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

Explanation

$H_2/Pt$ causes the catalytic hydrogenation of alkenes. Since 1 mole of the reagent is present, therefore, it will bring about the addition of

$H_2$ at one double bond only. So, in the compound A, two cyclopentane rings are should be present along with a double bond.

Due to the presence of two rings and one double bond in A its double bond equivalent is 3. We can also calculate it using the formula.

Double bond equivalent

$= C+1-\dfrac{H}{2}-\dfrac{X}{2}+\dfrac{N}{2}$

$C=10, \ H=2 \ less \ than \ the \ product=18-2=16$

$DBE=10+1-\dfrac{16}{2}=10+1-8=3$

${ CH }_{ 2 }-CH={ CH }_{ 2 }\xrightarrow { { zn,CH }_{ 3 }OH } product\\ |\quad \quad \quad \quad \quad \quad |\\ Br\quad \quad \quad \quad \quad Br$
The remaining product is ?

0%
0%
${ CH }_{ 2 }=C=C={ CH }_{ 2 }$
0%
${ CH }_{ 2 }=CH=CH={ CH }_{ 2 }$
0%

Which are the correct IUPAC name of the following compound ?

0%
5-butyl-4-is isopropyldecane
0%
5-ethyl-4-propyldecane
0%
5-sec-butyl-4-isopropyldecane
0%
4-(1-methylethyl)-5-(1-methylpropyl)decane

The number of possible alkynes with molecular formula

$C_{5}H_{8}$ is

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

1, 1 -dibromoethane when heated with zine dust produces_____.

0%
ethyl bromide
0%
ethene
0%
vinyl bromide
0%
2 - butene

If two moles of isopropyl chloride reacts with

$Na$ in presence of dry ether. Which alkane is formed?

0%
Hexane
0%
2, 3-Dimethyl butane
0%
Isopentane
0%
Neopentane

Explanation

The above reaction is

$Wurtz\ reaction.$ In the Wurtz reaction two alkyl halides react with each other to form alkane. In the given reaction the name of the product formed is

$2,3-dimethyl \ butane.$

Option B is correct.

1939129_1193486_ans_0a76645a706d4ab299aa84f8662d6605.JPG

IUPAC name of the given compound is:

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

The correct IUPAC name of the following alkane is

$H_3C-CH_2-CH_2-CH_2-CH_2-CH_2-CH_2-CH_3$

0%
1,6-dimethylhexane
0%
Octane
0%
Heptane
0%
Decane

What is the Possible number of alkadienes with molecular formula

$C_5H_8$ ? (Excluding stereoisomers)

0%
4
0%
5
0%
6
0%
2

Explanation

No. of alkadienes with molecular formula

$C_{5}H_{8}$ (excluding stereoisomers) are:

${ H }_{ 3 }C-HC=C=CH-C{ H }_{ 3 }$

${ H }_{ 2 }C=C=CH-C{ H }_{ 2 }-C{ H }_{ 3 }$

${ H }_{ 2 }C=CH-C{ H }_{ 2 }-CH=C{ H }_{ 2 }$

${ H }_{ 2 }C=CH-CH=CH-C{ H }_{ 3 }$

$\quad \quad \quad \quad \quad C{ H }_{ 3 }\\ \hspace{2mm} \quad \quad \quad \quad \quad |\\ { H }_{ 2 }C=C=C-C{ H }_{ 3 }$

$\quad \quad \quad \quad \quad \quad C{ H }_{ 3 }\\ \hspace{2mm} \quad \quad \quad \quad \quad \quad |\\ { H }_{ 2 }C=CH-C=C{ H }_{ 2 }$

So, the total number of alkadienes with molecular formula

$C_{5}H_{8}$ (excluding stereoisomers) is 6.

Ethanol reacts with

$CH_3Mgl$ gives

0%
Ethane
0%
Isoproply alcohol
0%
Acetaldehyde
0%
Methane

Reaction of

$trans-2-phenyl -1-bromocyclopentane$ on reaction with alcoholic

$KOH$ produces (major):

0%
$4-phenylcyclopentene$
0%
$2-phenylcyclopentene$
0%
$1-phenylcyclopentene$
0%
$3-phenylcyclopentene$

An alkane containing atleast_____ carbon atoms is converted into an aromatic hydrocarbon, when heated in presence of

$Cr_{2}O_{3}$ on

$Al_{2}O_{3}$

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

The major product (A) is

The given pair is an example of:

0%
Enantiomers
0%
Diasteremers
0%
Identical
0%
Constitutional isomers

Electrolysis of aq. solution of sodium salt of saturated dicarboxylic acids, give

0%
alkyne
0%
alkene
0%
alkane
0%
none of these

Explanation

$\text{Sodium or potassium salt of a dicarboxylic acid on electrolysis gives an }alkene.$ An alkene is generated when an aqueous solution of sodium or potassium salt of a dibasic acid ( with adjacent carboxylic groups ) is electrolyzed. The electrolysis of sodium succinate gives ethene.

Hence, option B is correct.

1940363_1289100_ans_d203a4d26bb24312b764c41a3028f99e.jpg

In the reaction

$H-C\equiv CH\dfrac { { NaNH }_{ 2 }/liq.{ NH }_{ 3 } }{ { CH }_{ 3 }{ CH }_{ 2 }Br } X\dfrac { { NaNH }_{ 2 }/liq.{ NH }_{ 3 } }{ { CH }_{ 3 }{ CH }_{ 2 }Br } Y$

$X$ and

$Y$ are

0%
$X=1-Butyne ; Y=3-Hexyne$
0%
$X=2-Butyne ; Y=3-Hexyne$
0%
$X=2-Butyne ; Y=2-Hexyne$
0%
$X=1-Butyne ; Y=2-Hexyne$

The IUPAC name of

$(CH_3)_3C-CH=CH_2$ is?

0%
$2, 2$ -dimethylbut- $2$ -ene
0%
$2, 2$ -dimenthylpent- $3$ -ene
0%
$3, 3$ -dimethylbut- $1$ -ene
0%
Hex- $1$ -ene

How many optical isomers are possible on monochlorination of

$2-$ methyl butane

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

How many cyclic and alkene isomers of

$C_4H_8$ are possible? ( Excluding stereoisomers)

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

From which one of the following both ethylene and acetylene could be prepared in a single step?

0%
$CH_{3}CH_{2}OH$
0%
$Br - CH_{2} - CH_{2} - Br$
0%
$CH_{3}CH_{2}Br$
0%
$Br - CH = CH_{2}$
0%
$CH_{3}COOH$

Explanation

Ethylene is obtained through dehalogenation while acetylene is obtained thorough a double dehydrohalogenation.

$(Br)CH_2(Br) \xrightarrow {NaNH_2} CH \equiv CH$

1750104_1311252_ans_415034c6935546c8b4ef165b7015f9fb.gif

Identify major product

${{\text{P}}_{\text{3}}}$ :

By which of the following reagent butanoic acid is converted into butane

0%
Red P / HI
0%
NaOH / CaO
0%
${\text{C}}{{\text{H}}_3}{\text{MgBr}}$
0%
Na/ ether

An alkene of molecular formula

$C_{9}H_{18}$ on ozonolysis gives 2,2-dimethylpropanal and butan-2-one, then the alkene is

0%
2,2,4-trimethylhex-3-ene
0%
2,2,6-trimethylhex-3-ene
0%
2,3,4-trimethylhex-2-ene
0%
2,2,4-trimethylhex-2-ene

The correct order for acid strength of compounds

$CH\equiv CH, CH_{3}-C \equiv CH$ and

$CH_{2} = CH_{2}$ is as follows:

0%
$CH \equiv CH > CH_{2} = CH_{2} > CH_{3} - C \equiv CH$
0%
$HC\equiv CH > CH_{3} - C \equiv CH > CH_{2} = CH_{2}$
0%
$CH_{3} - C \equiv CH > CH_{2} = CH_{2} > HC \equiv CH$
0%
$CH_{3} - C \equiv CH > CH\equiv CH > CH_{2} = CH_{2}$

Explanation

$CH\equiv CH > CH_{3} - C \equiv CH > CH_{2} = CH_{2}$
(Acidic strength order).

More is the s-character, more is the acidic strength.

The s-character is maximum in

$sp$ hybrid carbon atom followed by

$sp^2$ and

$sp^3$ .

$n$ propyl alcohol on reacting with anhydrous alumina at

$623\,K$ forms

0%
Ethane
0%
Propene
0%
propyne
0%
Propane

${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 3 }\xrightarrow [ ]{ 400-{ 600 }^{ o }C } X+Y$

$X$ and

$Y$ are

0%
ethylene, ethane
0%
Hydrogen, ethylene
0%
methane, ethylene
0%
Hydrogen, methane

Which of the following does not give alkane with $R-Mg-X.$

0%
$Ph-OH$
0%
$C_6H_6$
0%
$CH_3COOH$
0%
$HCl$

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 8 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 8 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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