MCQ Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 15

The compound that is most reactive towards electrophilic substitution is:

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Toluene
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Benzene
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Nitrobenzene
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Benzoic acid

Methane and ethane both can be obtained in single step from

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$CH_3I$
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$C_2H_5I$
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$CH_3OH$
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$C_2H_5OH$

Explanation

Methyl iodide on wurtz reaction gives ethane and on reduction gives methane.

$CH_3I + 2H \xrightarrow[]{Zn/HCl} CH_4 + HI$

$CH_3I + 2Na +ICH_3 \xrightarrow[\text{Ether}]{\text{Dry}} CH_3 - CH_3 + 2NaI$

Ortho-dihydroxy benzene is

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Carvacrol
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Resorcinol
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Catechol
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Orcinol

Reaction

$C_2H_5I + C_5H_{11}I + 2Na \rightarrow C_2H_5 - C_5H_{11} + 2NaI$ is called

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Hoffmann's reaction
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Dow's reaction
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Wurtz's reaction
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Riemer - Tiemann's reaction

Acetylene can be prepared from

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Potassium fumarate
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Calcium carbide
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Ethylene bromide
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All of these

Explanation

$\begin{matrix} CH & - & COOK \\ \!\!\!\! || & & \\ CH & - & COOK \end{matrix} + 2H_2O \xrightarrow{Electrolyisis} \underbrace{\begin{matrix} CH \\ \!\!\! ||| \\ CH \end{matrix}}_{Anode}+2CO_{ 2 } +2KOH+\underbrace{H_{2}}_{Carthode}$

$CaC_{2}+2H_{2}O\rightarrow Ca(OH)_{2}+CH\equiv CH$

$\underset{Br}{\underset{|}{CH_{2}}}-\underset{Br}{\underset{|}{CH_{2}}}+2KOH\rightarrow CH\equiv CH+2KBr+2H_{2}O$

The

$IUPAC$ name of

$\begin{matrix} \ \ \ \quad \quad CH_{ 3 }CH_{ 2 }CHCH_{ 2 }CH_{ 2 }CH_{ 3 } \\ | \\ { CH }_{ 3 } \end{matrix}$ is

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$4-$ methylhexane
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$3-$ methylhexane
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$2-$ propylbutane
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$2-$ ethylpentane

1,2,3-trihydroxybenzene is also known as:

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Pyrogallol
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Phloroglucinol
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Resorcinol
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Quinol

Explanation

1,2,3-Trihydroxybenzene, also known as pyrogallic acid or Pyrogallol. it's IUPAC name is benzene-1,2,3-triol.

Hence, option

$A$ is correct.

1638210_1762398_ans_2d14af13109d4f17af90ac243a6b37a0.JPG

$R-CH_{2}-CCl_{2}-R\xrightarrow[]{Reagent}R-C\equiv C-R$

The reagent is

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$Na$
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$HCl$ and $H_{2}O$
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$KOH$ in $C_{2}H_{5}OH$
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$Zn$

Explanation

$R-CH_{2}-CCl_{2}-R\xrightarrow[]{KOH+C_{2}H_{5}OH}R-C\equiv C-R+2HCl$

This reaction is an example of dehydrohalogenation. Hence, alcoholic

$KOH$ is used as a reagent.

Which gives

$CH_4$ when treated with water?

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Silicon carbide
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Calcium carbide
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Aluminium carbide
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Iron carbide

Explanation

Aluminium carbide gives

$CH_4$ when treated with water

$\underset{\text{Aluminium carbide}}{Al_4C_3} + 6H_2O \rightarrow \underset{\text{Methane}}{3CH_4} + 2Al_2O_3$

Which of the following does not give alkane?

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Reaction of $CH_3I$ with Na in ether
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Reaction of sodium acetate with sodalime
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Electrolysis of concentrated sodium acetate solution
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Reaction of ethyl chloride with alco. $KOH$

Explanation

The reaction of ethyl chloride with alcoholic KOH does not give alkane.

$CH_3 - CH_2 - Cl + \underset{\text{(alc.)}}{KOH} \rightarrow \underset{\text{Ethene}}{CH_2 = CH_2} + KCl + H_2O$

Compound (E) is:

Simplest alkyne is represented by:

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$CH$
0%
$CH_{2}$
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$C_{2}H_{2}$
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$C_{2}H_{4}$

Explanation

The simplest acyclic alkynes with only one triple bond and no other functional groups form a homologous series with the general chemical formula

$C_nH_{2n-2}$ Alkynes are traditionally known as acetylenes, although the name acetylene also refers specifically to

$C_2H_2$ known formally as ethyne using IUPAC nomenclature.

Propane cannot be prepared from which reaction?

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$CH_3 - CH = CH_2 \xrightarrow[OH^-]{B_2H_6}$
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$CH_3CH_2CH_2I \xrightarrow[P]{HI}$
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$CH_3CH_2CH_2I \xrightarrow[]{Zn,\ dil . HCl+H_2}$
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None of these

Explanation

Hydroboration of alkanes followed by hydrolysis in basic medium yield alcohols and not the alkanes.

$R - CH = CH_2 \xrightarrow[]{B_2H_6} (R - CH_2 - CH_2)_3 - B \xrightarrow[]{OH^-} R - CH_2 - CH_2OH$

B.P. of branched chain alkanes as compared to straight chain alkanes are

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lower
0%
equal
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higher
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independent of the chain

Necessary conditions for halogenation are

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Cold and dark
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Presence of halogen carrier
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Both (a) and (b)
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None

Explanation

Option C is the correct answer .

Necessary condition for halogenation is presence of halogen carrier.

Halogenation is done in dark without any catalyst then according to free radical mechanism new halide radical will be attached to benzylic carbon. This reaction is carried out in dark so that required energy is supplied in general if it would be carried out in light chlorine free radical would be formed but in dark we use heat and supply energy. For chlorination 1kcal , bromination 16 kcal and iodination 33 kcal.

in presence of catalyst like lewis acid like $AlCl_3$ and $FeCl_3$ electrophilic substitution takes place.

$A\xleftarrow [ \Delta ]{ CU } CH_{ 3 }CH_{ 2 }OH\xrightarrow [ \Delta ]{ AL_{ 2 }O_{ 3 } } B.$ A and B respectively are

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Alkene, alkanal
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Alkyne, alkanal
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Alkanal, alkene
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Alkene, alkyne

Explanation

$CH_3CHO \xleftarrow [ \Delta]{Cu} CH_3CH_2OH \xrightarrow [ \Delta]{Al_2O_3 } CH_2 =CH_2$

Option C is correct.

Point out the wrong statement in relation to the structure of benzene

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It forms only one monosubstitution product
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The $C-C$ bond distance in benzene is uniformly $1.397\mathring{A}$
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It is a resonance hybrid of a number of canonical forms
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It has three delocalised $\pi -$ molecular orbitals

Explanation

(A) Benzene typically undergoes substitution reactions, a reaction not typical of alkenes. Benzene reacts with bromine, in the presence of iron(III) bromide as a catalyst, to give a single monosubstituted product,

$C_6H_5Br$ .This result indicates that all six hydrogen atoms of benzene are chemically equivalent.

$\textbf{(B) }$ The six-membered ring in benzene is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å). The cyclohexatriene contributors would be expected to show alternating bond lengths, the double bonds being shorter (1.34 Å) than the single bonds (1.54 Å).

$\textbf{(C) }$ Benzene is commonly seen in Organic Chemistry and it has a resonance form. Benzene has two resonance structures, showing the placements of the bonds. Another example of resonance is ozone. Ozone is represented by two different Lewis structures.

$\textbf{(D) }$ In benzene, the π electrons are delocalized over all six atoms of the ring. In β-carotene, the π electrons are delocalized over 22 carbon atoms.

So, The answer for ghis question is option B.

Which of the following reactions would give a good yield of hydrocarbon product?

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$RCOOK\xrightarrow [ oxidation ]{ electrolytic }$
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$RCOOAg\xrightarrow [ ]{ {I}_{2} }$
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${CH}_{3}{CH}_{3}\xrightarrow [ ]{ {Cl}_{2}/hv }$
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${Me}_{3}C-Cl\xrightarrow [ ]{ {C}_{2}{H}_{5}OH }$

Explanation

(a) Kolbe's electrolytic reaction gives $(R-R)$
(b) Birnbaum-Simonini reaction gives $(RCOOR+{CO}_{2}+2AgI)$
(c) Free radical halogenation gives ${CH}_{3}{CH}_{2}Cl$
(d) No reaction

Option A is correct.

i. Compound (B)

$(C_6H_{14}) \xrightarrow{Cl_2 + h\nu}$ (Two isomeric
hexyl chloride)

$(C_6H_{13}Cl)$

$\downarrow{Zn + aq. \, CH_3COOH}$
(B)

ii. Alkene (C) and alkene (D)

$(C_6H_{12}) \xrightarrow{H_2 + Pt}$ Compound (B)

The alkenes (C) and (D) are :

$CH_4$ can be prepared by the reaction of

$H_2O$ with :

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$Mg_2C_3$
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$CaC_2$
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$Be_2C$
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$Al_4C_3$

Fifteen millilitres of gaseous hydrocarbon (A) was required for complete combustion of

$357 \, ml$ of air (

$21$ % oxygen by volume) and gaseous products occupied

$327\, ml$ (all volumes being measured at STP).
The value of N is :

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2
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3
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4
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5

Explanation

Let the formula of hydrocarbon A is

$C_xH_y$ \times.

$\displaystyle C_{x}H_{y}(g)\, +\, \left (x\, +\, \frac{y}{4}\right)\, O_{2}\, (g)\, \rightarrow\, x\, CO_{2}(g)\, +\, \frac{y}{2}\, H_{2}O(l)$

$1\, ml\, \left (x\, +\, \frac{y}{4}\right)\, ml\,\,\, x\, ml$

$15\, ml\, \left (x\, +\, \frac{y}{4}\right)\, ml\,\,\, 15x\, ml$

Volume of

$O_{2}\, =\, \displaystyle \frac{357\,\times \, 21}{100}\, =\, 75$ ml; volume of

$N_{2}$ = 357- 75 = 282 ml

Volume of

$CO_{2}$ = 327- 282 = 45 ml

$\therefore$ 15x = 45, x = 3

$\therefore \begin{pmatrix}15\, \left (x\, +\,\displaystyle \frac{y}{4}\right )=\, 75\\ x\, +\, \displaystyle \frac{y}{4}\, =\, 5\end{pmatrix}$

3 +

$\displaystyle \frac{y}{4}\, =\, 5$ , on solving, we get y = 8.

Formula of (A) =

$C_{3}H_{8}$

1652847_1767950_ans_ce843d4ba7fd4eeeb13f3c4297881075.png

Olefins can be hydrogenated by

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Zinc and $HCl$
0%
Nascent hydrogen
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Raney $Ni$ and $H_{2}$
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Lithium hydride in ether

Explanation

Olefins can be hydrogenated by the presence of Raney

$Ni$ catalyst and

$H_{2}$ gas.

$\underset{Ethene}{CH_{2}=CH_{2}} + H_{2}\xrightarrow[300^{o}C]{Ni}\underset{Ethane}{CH_{3}-CH_{3}}$

Compound (A)

$(C_6H_{14}) \xrightarrow{Cl_2 + h\nu}$ (Five isomeric
hexyl chloride )

$(C_6H_{13} Cl)$

$\downarrow{Zn + aq. \, CH_3COOH}$
(A)

Compound (A) is :

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$2,2$ -Dimethyl butane
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$2$ -Methyl pentane
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$2,2$ -Dimethyl propane
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$3$ -Methyl pentane

$C_5H_{10}$ is the formula of:

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Pentane
0%
Pentene
0%
Pentyne
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Pentadiene

Of the following compounds, whose ozonolysis proves the Kekule structure of benzene?

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Benzene
0%
Toluene
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o-Xylene
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p-Xylene

Explanation

o-Xylene on ozonolysis gives glyoxal: methylglyoxal: dimethyl glyoxal in the ratio

$3:2:1$ , which proves the existence of resonance in benzene and hence proves the Kekule structure of benzene.

Complete the following reaction:
(i)

$C_2H_5 I+H_2\xrightarrow [C_2H_5OH]{Zn -Cu}X$
(ii)

$CH_3COONa +H_2O \xrightarrow []{Electrolysis} Y$
(iii)

$CH_2CH_2CH_2CH_2CH_2CH_3 \xrightarrow [773\ K]{Cr_2O_3 /Al_2 O_3}Z$

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$X=C_2 H_6, Y=C_2 H_6, Z=C_6H_6$
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$X=CH_4, Y=CH_3 COOH, Z=CH_3CH_3$
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$X=C_2 H_5, Y=CH_4 , Z=C_4H_{10}$
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$X=C_2 H_6, Y=CH_4 , Z=C_5H_{10}$

Which of the following compound is chiral?

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$3-$ methyl $-3-$ hexene
0%
$4-$ chloro-1-methylcyclohexane
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$2-$ phenylpentane
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$1,3-$ diisopropylbenzene

Which of the following compounds is not aromatic?

Explanation

$(A)$ is having a cyclic aromatic compound with

$2\ \pi$ electrons.

$(B)$ and

$(D)$ have a cyclic compound of six

$\pi-$ electrons and hence aromatic while

$(C)$ has

$4 \pi-$ electrons and hence antiaromatic.

Which of the following does not give a white precipitate with

$AgNO_3$ solution?

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Propyne
0%
$1-$ Butyne
0%
$2-$ Butyne
0%
$1-$ Pentyne.

Explanation

$2-$ Butyne does not give a white precipitate with

$AgNO_3$ solution since it is a non-terminal alkyne. Non terminal alkyne does not give a white precipitate with

$AgNO_3$ solution.

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 15 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 15 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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