MCQ Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 12

The balance chemical equations for the following :

Reaction between 1, 2 - dibromoethane & alcoholic potassium hydroxide.

0%
$CH_2Br-CH_2Br \xrightarrow {alc. KOH} CH_3-CH_2Br \xrightarrow {NaNH_2} CH \equiv CH$
0%
$CH_2Br-CH_2Br \xrightarrow {alc. KOH} CH_2Br-CHBr \xrightarrow {NaNH_2} CH \equiv CH$
0%
$CH_2Br-CH_2Br \xrightarrow {alc. KOH} CH_2=CHBr \xrightarrow {NaNH_2} CH \equiv CH$
0%
$CH_3-CH_2Br \xrightarrow {NaNH_2} CH_2 = CH_2$

Explanation

Reaction:

$CH_2Br-CH_2Br \xrightarrow {alc. KOH} CH_2=CHBr \xrightarrow {NaNH_2} CH \equiv CH$ ethyne

The given reaction is used to form alkyne from alkyl dihalide. It uses reducing agents like KOH and sodamide. It first undergoes

$\beta$ elimination reaction to form vinyl halide which further undergoes dehydrohalogenation to form alkyne.

$Et{NH}_{2}+MeMgI\xrightarrow [ in\quad the\quad presence\quad of\quad pyridine ]{ Heated\quad in\quad high\quad temp. } Gas (A)$
The volume of gas

$(A)$ obtained at S.T.P when

$0.45gm$ of

$Et{NH}_{2}$ reacts with

$MeMgI$ is:

0%
$224ml$
0%
$22.4ml$
0%
$448ml$
0%
$44.8ml$

Explanation

One mole of

$Et{NH}_{2}$ reacts with

$2$ mol of

$MeMgI$ to give

$2$ mol of

${CH}_{4}(g)$ . Molecular mass of

$Et{NH}_{2}$ is

$45gm$ .
Therefore,

$45gm$ of

$Et{NH}_{2}$ gives

$2\times 22.4$ litres of

${CH}_{4}$ at S.T.P

$0.45gm$ of

$Et{NH}_{2}$ gives

$\cfrac{2\times 22.4\times 0.45}{45}=0.448$ litres

$=448ml$

What are the numerical values of N and M?

0%
6,6
0%
5,4
0%
4,4
0%
3,3

Explanation

The four possibilities for two chlorines in

$C_3H_6Cl_2$ is for the Cl's to be 1,1, 2,2, 1,2, or 1,3.
i.e. 1,1-dichloropropane, 2,2-dichloropropane, 1,2-dichloropropane, and 1,3-dichloropropane.

Which of the following statements is/are correct?

0%
Molecular formula or molecular mass of a gaseous hydrocarbon can be determined even without knowing their percentage composition by audiometry
0%
In Lassaigne's test, N and S both present in the organic compounds are converted into $\overset{\circleddash}{C}NS$ ion
0%
$K_2SO_4$ and $CuSO_4$ are added in Kjeldahl's method. $K_2SO_4$ acts as a catalyst while $CuSO_4$ raises the boiling point of $H_2SO_4$
0%
Layer test is used to distinguish $Cl^{\circleddash}$ and $Br^{\circleddash}$ ions

Explanation

$K_2SO_4$ raises the boiling point of

$H_2SO_4$ while

$CuSO_4$ acts as a catalyst.

D. Layer test is used to distinguish

$Br^{\circleddash}$ and

$I^{\circleddash}$ ion.

Hence, option (C) and (D) are wrong and (A) and (B) are correct.

Hence the correct options are A and B.

trans-3, 6-Dimethyl oct-4-ene (A) exists in two diastereomers (I) and (II). Which statement is true about (I) and (II).
If the stereochemistry about the double bond in (A) is trans, two diastereomers (I) and (II) exist as shown in. Which statement is correct ?

0%
(I) is optically inactive due to the presence of plane of symmetry.
0%
(I) is optically inactive due to the presence of centre of symmetry.
0%
(II) is optically inactive due to the presence of plane of symmetry.
0%
(II) is optically inactive due to the presence of centre of symmetry.

$CH_3-(CH_3)CH-CH_2-CH_3 \xrightarrow {Cl_2,\ hv} N (\text{isomeric products})$

$\ C_5H_{11}Cl \xrightarrow {\text{fractional distillation}} M(\text{isomeric products})$

What are N and M (N and M are number of products formed) ?

0%
6, 6
0%
6, 4
0%
4, 4
0%
3, 3

Assertion : Chlorination of allylic hydrogen is difficult than vinylic hydrogen.
Reason : Allyl radical is stabilised by resonance.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct
0%
Both Assertion and Reason are incorrect

Give the structure of reactants :

0%
A: Benzaldehyde B: Acetylene
0%
A: Benzyl alcohol B: Ethanol
0%
A: Benzoic acid B: Acetylene
0%
A: Phenol B: Acetylene

Which of the following statement(s) is correct?

0%
The compound (B) is
0%
The compound (B) is
0%
The compound (A) on reaction with HI + Red P gives
0%
The compound (A) on oxidation with $MnO_2$ gives

Explanation

$POCl_3$ is a dehydrating agent and it removes water molecule from alcohols. The more substituted alkene is formed as the major product. Hence the major product is 1−methylcyclohexene.

Hence, the correct option is

$\text{A}$

1491172_249226_ans_ddeaa88f2cbd4aeeb3e8bfa179ca3187.png

What is the major product formed when methylenecyclohexane reacts with NBS?

0%
1 bromo 1 bromomethyl cylohexane
0%
1 bromo 2 methylene cyclohexane
0%
1-(bromomethyl)cyclohexene
0%
none of these

Correct order of boiling point is:

Explanation

There are 3 important trends to consider:

1) The relative strength of the four intermolecular forces is: Ionic > Hydrogen bonding > dipole-dipole > Van der Waals dispersion forces. The influence of each of these attractive forces will depend on the functional groups present.

2) Boiling points increase as the number of carbons is increased.

3) Branching decreases the boiling point.

Hence, option

$C$ is correct.

Consider the reaction

$\left[CH_3CH_2CH_2-\!\!\!\!\!\underset{\,\,\,\,CH_3}{\underset{|}{\overset{\quad CH_3}{\overset{\!\!\!+|}{N}}}\!\!\!\!-}CH_2CH_3\right]OH^- \xrightarrow{Heat}$

Which of the following is formed in major amount?

0%
CH $_2$ =CH $_2$
0%
CH $_3$ CH=CH $_2$
0%
Both (A) and (B) in equal amount
0%
None, as no reaction takes place

$\left[CH_3CH_2CH_2-\!\!\!\!\!\!\underset{\,\,\,\,\,CH_3}{\underset{|}{\overset{\,\,\,\,\,\,\,CH_3}{\overset{\!\!\!\!\!+|}{N}}}\!\!\!\!\!-}CH_2CH_3\right]OH^- \xrightarrow{Heat}$

Consider the reaction Which of the following is formed in major amount?

0%
$CH_2=CH_2$
0%
$CH_3CH_2=CH_2$
0%
Both (1) and (2) in equal amount
0%
None, as no reaction takes place

Explanation

$\left[CH_3CH_2CH_2-\!\!\!\!\underset{\,\,\,CH_3}{\underset{|}{\overset{\,\,\,\,CH_3}{\overset{\!\!\!\!\!+|}{N}}}\!\!\!-}CH_2CH_3\right]OH^- \xrightarrow{Heat} CH_3CH_2CH_2-\!\!\!\!\underset{\,\,\,CH_3}{\underset{|}{\overset{\,\,\,\,CH_3}{\overset{|}{N}}}\!\!\!H}+ CH_2=CH_2$

The given reaction is Hofmann elimination process where a quaternary amine is heated to create a tertiary amine and an alkene.

With the given quaternary ammonium hydroxyl anion, the major alkene product is the least substituted. Thus the major alkene formed is ethene.

Arrange the following in increasing/decreasing order of boiling point :

$\underset{I}{CH_3CH_2CH_2CH_2CH_3}, \ \, \underset{II}{(CH_3)_2 CHCH_2CH_3},\ \ \underset{III}{(CH_3)_4C}$

0%
I > II > III
0%
II > I > III
0%
III > I > II
0%
III < II < I

Explanation

The decrease order of the boiling point is

$\underset{I}{CH_3CH_2CH_2CH_2CH_3}>\,\, \underset{II}{(CH_3)_2 CHCH_2CH_3} >\,\, \underset{III}{(CH_3)_4C}$ .
As the extent of branching increases, the boiling point decreases.

With the help of the following equations and data choose the wrong statement.

0%
Thermal cracking of (CH) bond of methane occurs at 1500 K and that of (CH) bond of ethane breaks at 800900 K.
0%
During homolysis of ethane at high temperature, the (CC) bond breaks more radily than (CH) bonds.
0%
During the cracking of n-butane, reaction (IV) occurs more readily than reaction (V).
0%
Formation of $CH_3CH_2$ radical by reaction (II) will take place at lower temperature than the formation of $C_2H_5$ radical by reaction (IV).

Explanation

The option (D) represents wrong statement.
Formation of

$CH_3CH_2$ radical by reaction (II) will take place at lower temperature than the formation of

$C_2H_5$ radical by reaction (IV).

Which of the following reactions will not give four-membered cyclic compound?

0%
0%
0%
$2CH_2\, =\, CH_2\, \xrightarrow{hv}\, C$
0%

Chloroform reacts with oxygen in the presence of light to give:

0%
carbon tetrachloride
0%
carbonyl chloride
0%
methyl chloride
0%
methylene dichloride
0%
acetaldehyde

Explanation

Chloroform reacts with oxygen in the presence of light to give a poisonous compound carbonyl chloride, which is also known by the name phosgene.

$CHCl_{3} +\dfrac {1}{2}O_{2}\xrightarrow {hv} \underset {Carbonyl\ chloride (poisonous)}{COCl_{2}} + HCl$
That's why chloroforms generally stored in dark coloured bottles filled upto brim.

Which of the following elimination reactions will occur to give but-1-ene as the major product?

Explanation

The reactions of the options (B) and (C) will occur to give but-1-ene as the major product.
The reaction (B) is an example of Hoffmann elimination in which quaternary ammonium salt is heated with sodium ethoxide/ ethanol, less substituted alkene is the major product.
The reaction (C) is an example of dehydrochlorination. An alkyl halide is heated with bulky base such as potassium tert butoxide

$\displaystyle Me_3COK$ , less substituted alkene is the major product.
The options (A) and (D) will give more substituted alkene (2-butene) as the major product.

$C_2H_6+Cl_2 \xrightarrow{sunlight}Q$

$\downarrow aq. KOH$

$C_2H_5OH \xrightarrow[K_2Cr_2O_7]{[O]} R \xrightarrow{C_2H_5OH} S$

Identify Q and S.

0%
Q = Chloroethene, S = Ester
0%
Q = Chloroethene, S = Ethanoic acid
0%
Q = Chloroethane, S = Ester
0%
Q = Chloroethane, S = Ethanoic acid

Explanation

Reaction:

$C_2H_6 + Cl_2 \rightarrow C_2H_5Cl \rightarrow C_2H_5OH \xrightarrow {oxidation} CH_3COOH \xrightarrow {alcohol} esters$

$X$ is an optically active alkane having lowest molecular mass, predict the structure of the major product obtained on monochlorination of

$X$ .

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

Explanation

Halogenation usually occurs at the tertiary position of the alkanes.

The halogenation agents employed are

$CCl_4$ , dichlorine monoxide,

$PCl_5$ , phosgene, hypochlorite etc. in all these cases a chain initiating catalyst is required usually peroxides or UV light.

Regioselectivity of electrophillic halogenation usually takes place at the tertiary carbon atom, because tertiary carbon radical is more stable.

Option A is correct.

Arrange the reactions given below in proper sequence for the conversion of ethane to methane.
(a) Treatment with sodium hydroxide
(b) Treatment with aqueous

$KOH$
(c) Treatement with

$NaOH+CaO$
(d) Halogenation
(e) Reaction with acidified

${K}_{2}{Cr}_{2}{O}_{7}$

0%
$e\ a\ d\ b\ c$
0%
$d\ b\ e\ a\ c$
0%
$e\ c\ b\ a\ d$
0%
$d\ e\ a\ b\ c$

Explanation

for the conversion of ethane to methane.
(d) Halogenation

$\Rightarrow CH_3-CH_3 \rightarrow CH_3-CH_2-Cl$
(e) Reaction with acidified

${K}_{2}{Cr}_{2}{O}_{7}$

$\Rightarrow CH_3-CH_2Cl \rightarrow CH_3-COOH$
(a) Treatment with sodium hydroxide

$\Rightarrow CH_3-COOH+ NaOH \rightarrow CH_3-COONa$
(b) Treatment with aqueous

$KOH$

$\Rightarrow CH_3-COONa \rightarrow CH_4(aq)$
(c) Treatement with

$NaOH+CaO$

$\Rightarrow CH_4(aq)\rightarrow CH_4$

The common name of 2-Butyne is :

0%
acetylene
0%
methyl acetylene
0%
ethyl acetylene
0%
dimethyl acetylene

Explanation

2-Butyne is an alkyne with chemical formula

$CH_3-C \equiv C-CH_3$ it is a colorless, liquid at standard temperature and pressure.

Its common name is Dimethylacetylene.

Option D is correct.

Which of the following aromatic compound gives sulphonation reaction very easily?

0%
Chlorobenzene
0%
Nitrobenzene
0%
Toluene
0%
Benzene

Explanation

Toluene gives a sulphonation reaction very quickly because sulphonation is ESR, so the ring must be electron-rich as nitro and chloro groups are electron-withdrawing groups due to which ring gets deactivated while the

$CH_3$ group is electron-donating makes ring electron-rich.

Two metals

$X$ and

$Y$ have

$3$ and

$2$ electrons in their valence shells

$'M'$ and

$'N'$ respectively. These metals can form binary compounds

$A$ and

$B$ with the same non-metal carbon. Both

$A$ and

$B$ can undergo hydrolysis forming hydrocarbons

$C$ and

$D$ . Identify

$C$ and

$D$ .

0%
${C}_{6}{H}_{6}, {C}_{3}{H}_{4}$
0%
${C}_{3}{H}_{6}, {C}_{3}{H}_{4}$
0%
${CH}_{4}, {C}_{2}{H}_{2}$
0%
${C}_{4}{H}_{10}, {C}_{3}{H}_{6}$

Explanation

The two metals are aluminium

$Al$ and calcium

$Ca$ . As they have 3 and 2 electrons in their valance shells and can form binary compounds carbides

$Al_4C_3$ and

$CaC_2$ .

$Al_4C_3 + H_2O \rightarrow H_4C$

$CaC_2 + H_2O \rightarrow C_2H_2$

Which of the following solvents should be added to the organic solvent containing the pentane in order to remove the impurities in a separate layer that will not mix with the organic solvent?

0%
$CCl_{4}$
0%
$C_{6}H_{6}$
0%
$HF$
0%
$SH_{2}$

Explanation

Due to its symmetrical structure, carbon tetrachloride ( $CCl_4$ ) is non-polar. Its non-polar nature makes it a suitable solvent for dissolving non-polar compounds. Carbon tetrachloride is convenient as a solvent for NMR because it contains no protons, however, its poor dissolving power limits its usefulness. As a solvent, it is well suited for dissolving other non-polar compounds, iodine, fats, and oils.

$CCl_4$ can undergo various reactions like dimerization to Hexachloroethane, reduction to chloroform.

Hence, the correct option is $\text{A}$

A hydrocarbon on ozonolysis gives one mole each of compounds

$A$ and

$B$ . Both

$A$ and

$B$ on reduction with sodium borohydride gives compounds

$C$ and

$D$ belonging to same homologous series.

$C$ has no position isomer whereas

$D$ has position isomer.

$A$ and

$B$ do not belong to the same homologous series. However, both

$A$ and

$B$ on oxidation with acidified

$KMn{O}_{4}$ give same compound. Identify the hydrocarbon.

0%
3- Methyl-2-butene
0%
2,3-Dimethyl-1-butene
0%
2-Methyl-2-butene
0%
3,4-Dimethyl-1-butene

Explanation

Ozonolysis of alkenes form ketone or aldehydes. Their reduction with

$NaBH_4$ gives alcohols.

$C$ has no position isomer whereas

$D$ has position isomer.Thus

$C$ is a primary alcohol and

$D$ is a secondary alcohol.

$A$ is an aldehyde and

$B$ is a ketone. And the compound is

$2-methyl-2-butene$

$CH_3-(CH_3)C=CH-CH_3 \overset{(O_3)}{\rightarrow} \underset{[A]}{CH_3-(CH_3)C=O} + \underset{[B]}{O=CH-CH_3} \overset {NaBH_4}{\rightarrow} \underset{[C]}{CH_3-(CH_3)CH-OH} +\underset{[D]} {HO-CH_2-CH_3} \overset {KMnO_4}{\rightarrow} {CH_3-(CH_3)CH-OH} + {HOOC-CH_3}$

Which of the following is NOT correct?

0%
$LiOH$ is a weaker base than $NaOH$ .
0%
Salts of $Be$ undergo hydrolysis.
0%
$Ca{ \left( HC{ O }_{ 3 } \right) }_{ 2 }$ is soluble in water.
0%
Hydrolysis of beryllium carbide gives acetylene.

Explanation

Hydrolysis of Beryllium carbide does not give acetylene, it gives methane.

$Be_2C + 4 H_2O \rightarrow 2Be(OH)_2 + CH_4$

The product formed when toulene is heated in light with

${Cl}_{2}$ and in absence of halogen carrier is:

0%
Chlorobenzene
0%
Gammexene
0%
Benzotrichloride
0%
DDT

Which of the following statements is false regarding the reaction between

$Cl_2$ and

$C_2H_6$ ?

0%
It is a substitution reaction
0%
The reaction will give a single product of $C_2H_5Cl$
0%
The reaction mechanism involves free radicals
0%
The reaction can be initiated with either sunlight or heat

Explanation

$C_2H_6 + Cl_2\rightarrow C_2H_5Cl + HCl$

This reaction is a substitution reaction, involving chlorine free radicals, and this reaction can be initiated with either sunlight or heat.

This reaction produces ethyl chloride and Hydrochloric acid.

Which of the following compounds is the slowest to react the nitrosonium ion

$(NO^+)$ ?

Explanation

The presence of electron donating groups on benzene ring will increase the compounds reactivity with

$NO^+$ .

Therefore in the given options A has no electron donating group attached .So,it is the slowest to react with

$NO^+$

Acetylene can be obtained by the reaction:

0%
$HCOOK \xrightarrow{Electrolysis}$
0%
$CHI_3 + Ag \xrightarrow{\Delta}$
0%
$CH_3 CH_2 OH \xrightarrow[443 K]{Conc. H_2SO_4}$
0%
$Be_2C+ H_2O \rightarrow$

Explanation

Hint: Acetylene is the common name of the simplest alkyne.

Explanation:

■ In the first reaction, Potassium formate undergoes electrolysis to give hydrogen, Carbon dioxide and Potassium hydroxide.

■ In second reaction, Iodoform reacts with silver to give acetylene and silver iodide.

■ In the third reaction, Ethanol reacts with concentrated sulphuric acid to give ethene and water.

■ In fourth reaction, Beryllium carbide reacts with water to give methane and beryllium oxide.

Final answer:

The correct answer is option $(B).$

The arrangement of following compounds:

$i$ . Bromomethane

$ii$ . Bromoform

$iii$ . Chloromethane

$iv$ . Dibromomethane
In the increasing order of their boiling point is :

0%
$iii< i< iv< ii$
0%
$iv< iii< i< ii$
0%
$ii< iii< i< iv$
0%
$i< ii< iii< iv$

Explanation

As the molecular mass increases boiling point increases due to increased polar character. Hence,

$iii<i<iv<ii$

Which one of the following is not formed when a mixture of methyl bromide and bromobenzene is heated with sodium metal in the presence of dry ether?

0%
Ethane
0%
Diphenyl
0%
Propane
0%
Toluene

Explanation

Propane is not formed when a mixture of methyl bromide and bromobenzene is heated with sodium metal in the presence of dry Ether.
Two moles of methyl bromide in presence of

$Na/ether$ give ethane.
Two moles of bromobenzene in presence of

$Na/ether$ give diphenyl.
One mole of methyl bromide and one mole of bromobenzene in presence of

$Na/ether$ give toluene.

$\displaystyle R-Br + R'-Br +2Na \rightarrow R-R' + 2 NaBr$

Which of the following does not give isopropylbenzene as a product upon reaction with benzene?

0%
$(CH_3)_2CHCl/AlCl_3$
0%
$CH_3CH_2CH_2Cl/AlCl_3$
0%
$CH_3CH=CH_2/H_3PO_4$
0%
$(CH_3)_2C=CH_2/H_3PO_4$

Large molecule alkane on heating at a high temperature forms low molecular alkane and alkene by a process called as___________.

0%
Cracking
0%
Dehydrogenation
0%
Hydration
0%
Both a and b

The end product

$C$ in the given reaction is:

Freon 12 is manufactured from

$C{ Cl }_{ 4 }$ by:

0%
Wurtz reaction
0%
Swarts reaction
0%
Fittig reaction
0%
Wurtz-Fittig reaction
0%
Sandmeyer reaction

Explanation

Freon-12 is made by

$C{ Cl }_{ 4 }$ by the Swarts reaction as shown below:

$\begin{matrix} 3C{ Cl }_{ 4 } \\ \quad \end{matrix}\begin{matrix} + \\ \quad \end{matrix}\begin{matrix} 2Sb{ F }_{ 3 } \\ \quad \end{matrix}\begin{matrix} \longrightarrow \\ \quad \end{matrix}\begin{matrix} 3C{ Cl }_{ 2 }{ F }_{ 2 } \\ \left( Freon-12 \right) \end{matrix}\begin{matrix} + \\ \quad \end{matrix}\begin{matrix} 2Sb{ Cl }_{ 3 } \\ \quad \end{matrix}$

The alkene formed as a major product in the above elimination reactions is:

0%
0%
$CH_2=CH_2$
0%
0%

The major product formed in the oxidation of acetylene by alkaline

$KMnO_{4}$ is ________.

0%
Ethanol
0%
Acetic acid
0%
Formic acid
0%
Oxalic acid

Explanation

Alkaline

$KMnO_4$ acts as oxidising agent in alkaline medium.

On reaction with acetylene (

$C_2H_2$ ) in hot (

$25^o C$ ) gives oxalic acid and in cold gives formic acid as:

In hot:

$\underset { acetylene }{ { C }_{ 2 }{ H }_{ 2 } } +\underset { hot(25\ C) }{ KMn{ O }_{ 4 } } \rightarrow \underset { oxalic\quad acid }{ HOOC-COOH }$

In cold:

$\underset { acetylene }{ { C }_{ 2 }{ H }_{ 2 } } +\underset { cold }{ KMn{ O }_{ 4 } } \rightarrow \underset { formic\quad acid }{ HCOOH }$

Pink colour of

$KMnO_4$ is declorised

Therefore option D is the correct answer.

Which of the following reactions does not occur in practice?

0%
$Ph_{3}CH + CH_{3}(CH_{2})_{3}Na\rightarrow Ph_{3}CNa + CH_{3} (CH_{2})_{2} CH_{3}$
0%
0%
$CH_3\overset{O}{\overset{||}{C}}CH_2\overset{O}{\overset{||}{C}}CH_3+EtONa \longrightarrow CH_3\overset{O}{\overset{||}{C}}CH =\overset{ONa}{\overset{|}{C}C}H_3+EtOH$
0%

Only two isomeric monochloro derivatives are possible for:

0%
n-Butane
0%
2,4-Dimethylpentane
0%
Benzene
0%
2-Methyl propane

Which of the following compounds can be best prepared by Wurtz - reaction?

0%
Iso-butane
0%
n-butane
0%
n-pentane
0%
Iso-pentane

Explanation

Due to stability of

$2^o-$ radical, it is easy to synthesize Iso-butane,

The Wurtz reaction is a coupling reaction in organic chemistry, whereby two alkyl halides are reacted with sodium metal in dry ether solution to form a higher alkane:

$2R–X + 2Na \rightarrow R–R + 2Na^{+} X^{-}$

Which of the following compounds can be best prepared by Wurtz-reaction?

0%
Iso-butane
0%
n-butane
0%
n-pentane
0%
Iso-pentane

When

$n-$ hexane is heated with anhydrous

$AlCl_{3}$ and

$HCl$ gas, the major products obtained is:

0%
$1$ -chlorohexane
0%
$2$ -chlorohexane
0%
$3$ -chlorohexane
0%
hex- $3$ -ene
0%
mixture of $2$ -methylpentane and $3$ -methylpentane

Explanation

When

$n-hexane$ is heated with anhydrous

$AlCl_{3}$ and

$HCl$ gas, at

$573\ K$ under a pressure of about

$30-35$ atmosphere, it isomerises to give a mixture of branched chain alkanes.

$\underset{n-hexane}{CH_3(CH_2)_4CH_3}\xrightarrow[Heat]{AlCl_3.HCl}\underset{2-methyl\, pentane}{CH_3-\overset{CH_3}{\overset{|}{C}H}CH_2CH_2CH_3}+\underset{3-methyl \, pentane}{H_3CCH_2-\underset{CH_3}{\underset{|}{C}H}CH_2CH_3}$

What will be the major product, when 2-methyl butane undergoes bromination:

0%
1-bromo-2-methyl butane
0%
2-bromo-2-methyl butane
0%
2-bromo-3-methyl butane
0%
1-bromo-3-methyl butane

Which one of the following undergoes nucleophilic aromatic substitution at the fastest rate?

Explanation

As electron withdrawing nature increases,the reactivity towards nucleophilic aromatic substitution also increases.

Therefore,

$p-$ chloro nitrobenzene has the highest reactivity.

The energy required to vapourise one mole of benzene at it's boiling point is

$31.2\ kJ$ . For how long a

$100\ w$ electric heater has to be operated in order to vaporize a

$100\ g$ sample of benzene at it's boiling temperature?

0%
$\approx 743\ sec$
0%
$400\ sec$
0%
$3.12\times 10^{4} sec$
0%
$\approx 371\ sec$

Explanation

Power =100

$\cfrac { joule }{ sec }$

For one mole = (31.2X1000)

$\cfrac { joule }{ sec }$

for 100gm

$\rightarrow$ = (31.2X1000) X

$\cfrac { 100 }{ 78.11 }$ joule

Equal to 399.4 seconds.

Of the isomeric hexanes, the isomers that give the minimum and maximum number of monochloro derivatives are respectively :

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2,3-dimethylbutane and n-hexane
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3-methylpentane and 2,3-dimethylbutane
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2,3-dimethylbutane and 2-methylpentane
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2-methylpentane and 2,2-dimethylbutane

Towards catalytic hydrogenation, the least reactive compound is:

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${ CH }_{ 2 }{ \ =\ CH }_{ 2 }$
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${ CH }_{ 3 }\ -\ { CH }{ \ =\ CH }_{ 2 }$
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${ H }_{ 3 }C\ -\ { CH }{ \ =\ CH\ -\ CH }_{ 3 }$
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$({ CH }_{ 3 }{ ) }_{ 2 }C\ =\ C({ CH }_{ 3 }{ ) }_{ 2 }$

Explanation

As a number of

$\alpha$ -Hydrogen increases reactivity towards catalytic hydrogenation increases.

$CH_{2}$ =

$CH_{2}$ have zero

$\alpha$ -Hydrogen. So, it is least reactive towards catalytic hydrogenation.

Major product is:

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 12 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 12 - MCQExams.com

0:0:2

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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