Hydrocarbons — Page 3 | JEE Chemistry

Q21

5 L of an alkane requires 25 L of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is :

A Ethane

B Propane

C Butane

D Isobutane

Correct Answer

Option B

Solution

We know, conbustion of Hydrocarbon Cx Hy + (x +

{y \over 4}

)O2 $\to$ x CO2 +

{y \over 2}

H2O So, to consume 1 mole of Cx Hy we need (x +

{y \over 4}

) mole of O2 gas.

As both Cx Hy and O2 are gas, So avogadro's law is applicable on them.

At constant temperature and pressure according to avogadro's law, volume $\propto$ mole So, for 5L of Cx Hy the volume of O2 = 5(x +

{y \over 4}

) According to the question, 5(x +

{y \over 4}

) = 25

\therefore\,\,\,

(x +

{y \over 4}

) = 5 . . . . . (1) For Ethane (C2 H6), x = 2 and y = 6

\therefore\,\,\,

x +

{y \over 4}

= 2 +

{6 \over 4}

$\ne$ 5

\therefore\,\,\,

C2 H6 can't be the required alkane. For Propane (C3 H8), x = 3. and y = 8

\therefore\,\,\,

x +

{y \over 4}

= 3 +

{8 \over 4}

= 5

\therefore\,\,\,

Propane (C3 H8) is the right alkane. Similarly for Butane (C4 H10) and Isobutane (C4 H10) you can check x +

{y \over 4}

$\ne$ 5.

Q22

The correct order for acid strength of compounds : CH

\equiv

CH, CH3–C

\equiv

CH and CH2 = CH2 is as follows

A CH

\equiv

CH > CH2 = CH2 > CH3 – C

\equiv

CH

B CH3 – C

\equiv

CH > CH2 = CH2 > HC

\equiv

CH

C HC

\equiv

CH > CH3 – C

\equiv

CH > CH2 = CH2

D CH3 – CH

\equiv

CH > CH

\equiv

CH > CH2 = CH2

Correct Answer

Option C

Solution

Among CH $\equiv$ CH and CH2 = CH2, in CH $\equiv$ CH compound H atom is attached with sp hybridised C atom and in CH2 = CH2 compound H atom is attached with sp2 hybridised C atom.

As we know, electronegitivity of sp carbon is more compare to sp2 carbon atom and that H is more acidic which is attached with more electronegative carbon atom.

So CH $\equiv$ CH is more acidic than CH2 = CH2.

Among CH $\equiv$ CH and CH3–C $\equiv$ CH, in CH $\equiv$ CH both the H atom is connected to sp carbon atom but in CH3–C $\equiv$ CH, with one sp carbon atom H atom attached and with other sp carbon atom -CH3 group attached which have +I effect and we know +I effect decreases acidic strength.

So, CH $\equiv$ CH is more acidic than CH3–C $\equiv$ CH.

So correct order is HC $\equiv$ CH > CH3 – C $\equiv$ CH > CH2 = CH2

Q23

Given below are two statements : Statement I : 2-methylbutane on oxidation with KMnO4 gives 2-methylbutan-2-ol. Statement II : n-alkanes can be easily oxidised to corresponding alcohols with KMnO4. Choose the correct option :

A Both statement I and statement II are incorrect

B Both statement I and statement II are correct

C Statement I is correct but statement II is incorrect

D Statement I is incorrect but statement II is correct

Correct Answer

Option C

Solution

Alkanes having tertiary H can be oxidized to corresponding alcohols by KMnO4. whereas ordinary alkanes resist oxidation.

Q24

Given below are two statements : Statement I : One mole of propyne reacts with excess of sodium to liberate half a mole of

\mathrm{H}_2

gas. Statement II : Four g of propyne reacts with

\mathrm{NaNH}_2

to liberate

\mathrm{NH}_3

gas which occupies 224 mL at STP. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both Statement I and Statement II are incorrect

B Statement I is incorrect but Statement II is correct

C Both Statement I and Statement II are correct

D Statement I is correct but Statement II is incorrect

Correct Answer

Option D

Solution

Statement I “One mole of propyne reacts with excess of sodium to liberate half a mole of $\mathrm{H}_2$ gas.”

Reaction and Stoichiometry Propyne (terminal alkyne): $\mathrm{CH_3C \equiv CH}$ .

Reaction with sodium: $2\,\mathrm{CH_3C \equiv CH} \;+\; 2\,\mathrm{Na} \;\longrightarrow\; 2\,(\mathrm{CH_3C \equiv C^-Na^+}) \;+\; \mathrm{H_2}.$ From this balanced equation, 2 moles of propyne produce 1 mole of $\mathrm{H_2}$ .

Hence, 1 mole of propyne will produce $\tfrac{1}{2}$ mole of $\mathrm{H_2}$ .

$\boxed{\text{Statement I is correct.}}$ Statement II “Four grams of propyne reacts with $\mathrm{NaNH_2}$ to liberate $\mathrm{NH_3}$ gas which occupies 224 mL at STP.”

Analysis Moles of propyne Molecular mass of propyne ( $\mathrm{C_3H_4}$ ): $3 \times 12 + 4 \times 1 = 36 + 4 = 40\,\mathrm{g/mol}.$ Four grams of propyne is: $\dfrac{4\,\mathrm{g}}{40\,\mathrm{g/mol}} = 0.1\,\mathrm{mol}.$ Reaction with $\mathrm{NaNH_2}$ For a terminal alkyne: $\mathrm{CH_3C \equiv CH} \;+\; \mathrm{NaNH_2} \;\longrightarrow\; \mathrm{CH_3C \equiv C^-Na^+} \;+\; \mathrm{NH_3}.$ 1 mole of propyne produces 1 mole of $\mathrm{NH_3}$ .

Moles of $\mathrm{NH_3}$ produced With $0.1\,\mathrm{mol}$ of propyne, we get $0.1\,\mathrm{mol}$ of $\mathrm{NH_3}$ .

Volume of $\mathrm{NH_3}$ at STP 1 mole of any ideal gas at STP $\approx 22.4\,\mathrm{L} = 22400\,\mathrm{mL}.$ $0.1\,\mathrm{mol}$ of $\mathrm{NH_3}$ occupies $0.1 \times 22.4\,\mathrm{L} = 2.24\,\mathrm{L} = 2240\,\mathrm{mL}.$ However, Statement II says the liberated $\mathrm{NH_3}$ occupies only 224 mL at STP, which corresponds to $0.01\,\mathrm{mol}$ of $\mathrm{NH_3}$ , not $0.1\,\mathrm{mol}$ .

Therefore, the statement’s volume is off by a factor of 10 and is thus incorrect if the reaction goes to completion in a typical way.

$\boxed{\text{Statement II is incorrect.}}$ Conclusion Statement I is correct.

Statement II is incorrect.

Hence, the best choice is: $\boxed{\text{Option D: Statement I is correct but Statement II is incorrect.}}$

Q25

Presence of which reagent will affect the reversibility of the following reaction, and change it to a irreversible reaction : CH4 + I2

\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{{\mathop{\rm Re}\nolimits} versible}^{hv}}

CH3

-

I + HI

A HOCl

B dilute HNO2

C Liquid NH3

D Concentrated HIO3

Correct Answer

Option D

Solution

Iodination of alkane is reversible reaction.

It can be irreversible in the presence of strong oxidising agent like conc.

HNO3 or conc.

HIO3.

Q26

Match the with .tg .tg Name reaction Product obtainable

List - I		List - II
(A)	Swarts reaction	(I)	Ethyl benzene
(B)	Sandmeyer's reaction	(II)	Ethyl iodide
(C)	Wurtz Fittig reaction	(III)	Cyanobenzene
(D)	Finkelstein reaction	(IV)	Ethyl fluoride

A A-II, B-III, C-I, D-IV

B A-II, B-I, C-III, D-IV

C A-IV, B-I, C-III, D-II

D A-IV, B-III, C-I, D-II

Correct Answer

Option D

Solution

.tg .tg LIST-i Name reaction LIST-II Product obtainable (A) Swarts reaction (I) $\mathrm{Et}-\mathrm{I} \xrightarrow[\mathrm{DMF}]{\mathrm{KF}} \mathrm{Et}-\mathrm{F}$ (B) Sandmeyer's reaction (II) $\mathrm{PhN}_2^{\oplus} \mathrm{Cl}^{-} \xrightarrow{\mathrm{CuCN} / \mathrm{KCN}} \mathrm{PhCN}+\mathrm{N}_2$ (C) Wurtz Fittig reaction (III)

\mathrm{Ph}-\mathrm{Cl}+\mathrm{EtCl} \xrightarrow[\text{ ether }]{\mathrm{Na}}

\mathrm{Ph}-\mathrm{Et}+\mathrm{Ph}-\mathrm{Ph}+\mathrm{Et}-\mathrm{Et}

(D) Finkelstein reaction (IV) $\mathrm{Et}-\mathrm{Cl} \xrightarrow[\text{ acetone }]{\mathrm{NaI}} \mathrm{Et}-\mathrm{I}+\mathrm{NaCl}$

Q27

Acetylene does not react with :

A

\mathrm{Na}

B ammoniacal

\mathrm{AgNO}_3

C

\mathrm{HCl}

D

\mathrm{NaOH}

Correct Answer

Option D

Solution

1. Acetylene reacts with sodium :

\mathrm{2CH} \equiv \mathrm{CH} + 2\mathrm{Na} \longrightarrow 2\mathrm{CH} \equiv \mathrm{C}^{-}\mathrm{Na}^{+} + \mathrm{H}_2

Here, acetylene reacts with sodium to produce sodium acetylide and hydrogen gas.

2.

Acetylene reacts with ammoniacal silver nitrate :

\mathrm{CH} \equiv \mathrm{CH} + 2\mathrm{AgNO}_3 \longrightarrow \mathrm{Ag}_2\mathrm{C}_2 + 2\mathrm{HNO}_3

In this reaction, acetylene reacts with ammoniacal silver nitrate to form a white precipitate of silver acetylide and nitric acid.

3.

Acetylene reaction with hydrochloric acid in the presence of a $\mathrm{Hg}^{2+}$ catalyst (oxymercuration-demercuration) :

\mathrm{CH} \equiv \mathrm{CH} + \mathrm{Hg(OAc)_2} + \mathrm{H}_2\mathrm{O} \longrightarrow \mathrm{CH}_2=\mathrm{CH}-\mathrm{OH} + \mathrm{HgOAc}

\mathrm{CH}_2=\mathrm{CH}-\mathrm{OH} + \mathrm{NaBH}_4 \longrightarrow \mathrm{CH}_2=\mathrm{CH}-\mathrm{H} + \mathrm{B(OH)_3} + \mathrm{NaOAc}

Acetylene does not readily react with $\mathrm{HCl}$ .

However, in the presence of a catalyst like $\mathrm{Hg}^{2+}$ , a specific type of electrophilic addition known as oxymercuration can occur.

The result is vinyl chloride.

Note that $\mathrm{NaBH}_4$ is used to reduce the mercurinium ion intermediate and to replace the $\mathrm{Hg}$ group with a hydrogen atom.

4.

Acetylene does not react with sodium hydroxide :

\mathrm{CH} \equiv \mathrm{CH} + \mathrm{NaOH} \longrightarrow \text{ no reaction }

According to the principles of the Hard and Soft Acids and Bases (HSAB) theory, soft acids prefer to bind with soft bases and hard acids prefer to bind with hard bases.

In this case, $\mathrm{OH}^-$ is a hard base, while the $\mathrm{C}$ in acetylene is a soft acid, so they are not particularly reactive with each other.

5.

Acetylene reacts with sodium amide :

\mathrm{CH} \equiv \mathrm{CH} + \mathrm{NaNH}_2 \longrightarrow \mathrm{CH} \equiv \mathrm{C}^{-}\mathrm{Na}^{+} + \mathrm{NH}_3

Here, sodium amide acts as a strong base and nucleophile, deprotonating the acetylene to form sodium acetylide and ammonia.

Q28

Benzene on nitration gives nitrobenzene in presence of HNO3 and H2SO4 mixture, where :

A both H2SO4 and HNO3 act as a bases

B HNO3 acts as an acid and H2SO4 acts as a base

C both H2SO4 and HNO3 act as an acids

D HNO3 acts as a base and H2SO4 acts as an acid

Correct Answer

Option D

Solution

Reagent for nitration of Benzene

\mathop {{H_2}S{O_4}}\limits_{(Acid)} + \mathop {HN{O_3}}\limits_{(Base)} \mathbin{\lower.3ex\hbox{\overset{\textstyle\rightarrow}{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}}} HSO_4^\Theta + {H_2}\mathop N\limits^ \oplus {O_3}

{H_2}\mathop N\limits^ \oplus {O_3} \mathbin{\lower.3ex\hbox{\overset{\textstyle\rightarrow}{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}}} {H_2}O + \mathop N\limits^ \oplus {O_2}

Q29

Match with : (Chemicals)

List - I		List - II
(a)	Alcoholic potassium hydroxide	(i)	Electrodes in batteries
(b)	Pd/BaSO4	(ii)	Obtained by addition reaction
(c)	BHC (Benzene hexachloride)	(iii)	Used for $\beta$ - elimination reaction
(d)	Polyacetylene List - II (Use / Preparation / Constituent)	(iv)	Lindlar's catalyst Choose the most appropriate match :

A (a) - (ii), (b) - (i), (c) - (iv), (d) - (iii)

B (a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)

C (a) - (iii), (b) - (i), (c) - (iv), (d) - (ii)

D (a) - (ii), (b) - (iv), (c) - (i), (d) - (iii)

Correct Answer

Option B

Solution

Alc.

KOH causes elimination Pd / BaSO4 – Lindlar’s catalyst BHC is obtained by the addition reaction of Cl2 with benzene in presence of U.V.

Thin film of polyacetylene can be used as electrode in batteries.

Q30

On mixing a certain alkane with chlorine and irradiating it with ultravioletlight, it forms only one monochloroalkane. This alkane could be

A pentane

B isopentane

C neopentane

D propane

Correct Answer

Option C

Solution

In neopentane all the

H

atoms are same

\left( {{1^ \circ }} \right).