Hydrocarbons

JEE Chemistry · 35 questions · Page 2 of 4 · Click an option or "Show Solution" to reveal answer

Q11

Butene-1 may be converted to butane by reaction with

A Sn - HCI

B Zn - Hg

C Pd/H2

D Zn - HCI

Correct Answer

Option C

Solution

Alkenes combine with hydrogen under pressure and in presence of a catalyst

(Ni, Pt

Pd)

and form alkanes.

Q12

At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 345 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is:

A C3H8

B C4H8

C C4H10

D C3H6

Correct Answer

Option A

Solution

{C_x}{H_{y\left( g \right)}} + \left( {{{4x + y} \over 4}} \right){O_{2\left( g \right)}}

\to xC{O_{2\left( g \right)}} + {y \over 2}{H_2}O\left( l \right)

Volume of

{O_2}

used

= 375 \times {{20} \over {100}} = 75ml

$\therefore$ From the reaction of combination

1\,\,ml\,{C_x}{H_y}\,\,

requires

= {{4x + y} \over 4}ml\,{O_2}

15\left( {{{4x + y} \over 4}} \right) = 75

So,

4x+y=20

x=3

y=8

$\therefore$

{C_3}{H_8}

Q13

When 2-butyne is treate with H2/Lindlar's catalyst, compound X is produced as the major product and when treated with Na/liq. NH3 it produces Y as the major product Which of the following statements is correct ?

A X will have higher dipole moment and higher boiling point than Y.

B Y will have higher dipole moment and higher boiling point than X.

C X will have lower dipole moment and lower boiling point than Y.

D Y will have higher dipole momet and lower boiling point than X.

Correct Answer

Option A

Solution

Here dipole moment of Y = 0 cis-isomer(X) will have higher dipole moment and higher boiling point than trans (Y).

Q14

Given below are two statements : Statement I : The presence of weaker

\pi

-bonds make alkenes less stable than alkanes. Statement II : The strength of the double bond is greater than that of carbon-carbon single bond. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are correct.

B Both Statement I and Statement II are incorrect.

C Statement I is correct but Statement II is incorrect.

D Statement I is incorrect but Statement II is correct.

Correct Answer

Option A

Solution

The $\pi$ -bond present is alkenes is weaker than $\sigma$ bond present in alkanes.

That makes alkenes less stable than alkanes.

Therefore, statement-I is correct.

Carbon-carbon double bond is stronger than Carbon-carbon single bond because more energy is required to break 1 sigma and 1 pi bond than to break 1 sigma bond only.

Therefore, statement-II is also correct.

Q15

The incorrect statement regarding ethyne is

A The carbon - carbon bonds in ethyne is weaker than that in ethene

B Both carbons are sp hybridised

C The

\mathrm{C}-\mathrm{C}

bonds in ethyne is shorter than that in ethene

D Ethyne is linear

Correct Answer

Option A

Solution

To evaluate which statement regarding ethyne is incorrect, it's necessary to review the characteristics of the chemical bonds in ethyne (acetylene,

\mathrm{C_2H_2}

) compared to ethene (ethylene,

\mathrm{C_2H_4}

Option A: The statement that the carbon-carbon bonds in ethyne are weaker than that in ethene is incorrect.

In ethyne, the carbon-carbon triple bond is composed of one sigma ( $\sigma$ ) bond and two pi ( $\pi$ ) bonds.

This configuration makes the triple bond in ethyne much stronger and shorter compared to the double bond in ethene, which consists of one $\sigma$ bond and one $\pi$ bond.

Thus, the carbon-carbon bond in ethyne is actually stronger than that in ethene.

Option B: The statement that both carbons are sp hybridised in ethyne is correct.

In ethyne, each carbon atom is bonded to the other carbon atom and a single hydrogen atom, necessitating the sp hybridisation to form a linear molecule with $180^\circ$ angles between the bonds.

Option C: The statement that the

\mathrm{C-C}

bonds in ethyne are shorter than that in ethene is correct.

The presence of a triple bond between the carbon atoms in ethyne results in a shorter bond length compared to the double-bonded carbons in ethene.

This is because the additional pi bonds in a triple bond pull the carbon atoms closer together.

Option D: The statement that ethyne is linear is correct.

Due to the sp hybridisation of carbon atoms in ethyne, the molecule adopts a linear geometry with bond angles of $180^\circ$ .

Therefore, the incorrect statement regarding ethyne is Option A: The carbon-carbon bonds in ethyne are weaker than that in ethene.

Q16

Given below are two statements : Statement (I) :

\mathrm{S}_{\mathrm{N}} 2

reactions are 'stereospecific', indicating that they result in the formation of only one stereo-isomer as the product. Statement (II) :

\mathrm{S}_{\mathrm{N}} 1

reactions generally result in formation of product as racemic mixtures. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are false

B Statement I is false but Statement II is true

C Statement I is true but Statement II is false

D Both Statement I and Statement II are true

Correct Answer

Option D

Solution

Statement (I) explains a characteristic of

\mathrm{S}_{\mathrm{N}}2

(Substitution Nucleophilic Bimolecular) reactions. In

\mathrm{S}_{\mathrm{N}}2

reactions, the attack of the nucleophile and the departure of the leaving group occur simultaneously in a single step.

This reaction has a back-side attack mechanism, where the nucleophile attacks the electrophilic carbon from the opposite side of the leaving group.

This back-side attack leads to the inversion of configuration at the carbon atom undergoing substitution, a phenomenon often referred to as 'Walden inversion'.

Thus,

\mathrm{S}_{\mathrm{N}}2

reactions are stereospecific, leading to the formation of a product with a specific configuration (either inversion or retention of configuration, but typically inversion in the case of

\mathrm{S}_{\mathrm{N}}2

). Therefore, Statement I is true. Statement (II) concerns

\mathrm{S}_{\mathrm{N}}1

(Substitution Nucleophilic Unimolecular) reactions, which proceed in two steps: first, the leaving group departs, forming a carbocation intermediate, and then the nucleophile attacks.

The attack of the nucleophile can occur from either side of the planar carbocation, resulting in the formation of both enantiomers of the product.

When the reactant is chiral, the product is usually a racemic mixture, consisting of equal amounts of both enantiomers, assuming that there is no steric or electronic preference for the attack on the carbocation.

This makes

\mathrm{S}_{\mathrm{N}}1

reactions non-stereospecific.

Therefore, Statement II is true as well.

Based on the explanations provided, the correct answer is Option D: Both Statement I and Statement II are true.

Q17

Given below are two statements: Statement I : Ozonolysis followed by treatment with

\mathrm{Zn}, \mathrm{H}_2 \mathrm{O}

of cis-2-butene gives ethanal. Statement II : The product obtained by ozonolysis followed by treatment with

\mathrm{Zn}, \mathrm{H}_2 \mathrm{O}

of 3, 6-dimethyloct-4-ene has no chiral carbon atom. In the light of the above statements, choose the correct answer from the options given below

A Statement I is false but Statement II are true

B Both Statement I and Statement II are False

C Statement I is true but Statement II is false

D Both Statement I and Statement II are true

Correct Answer

Option C

Solution

St-I : Correct statement St-II : In correct statement because product has chiral centre.

Q18

Given below are two statements : Statement (I) : Neopentane forms only one monosubstituted derivative. Statement (II) : Melting point of neopentane is higher than n-pentane. In the light of the above statements, choose the most appropriate answer from the options given below :

A Statement I is incorrect but Statement II is correct

B Both Statement I and Statement II are correct

C Statement I is correct but Statement II is incorrect

D Both Statement I and Statement II are incorrect

Correct Answer

Option B

Solution

Both Statement I and Statement II are correct.

Q19

2–Hexyne gives trans–2–Hexene on treatment with :

A Pt/H2

B Li/NH3

C Pd/BaSO4

D LiAlH4

Correct Answer

Option B

Solution

Anti addition of hydrogen atoms to the the triple bond occurs when alkynes are reduced with sodium (or lithium) metal in ammonia, ethylamine, or alcohol at low temperatures.

This reaction called, a dissolving metal reduction, produces an

(E)

- or

trans

-alkene. Sodium in liq.

N{H_3}

is used as a source of electrons in the reduction of an alkyne to a

trans

alkene.

Q20

Polysubstitution is a major drawback in:

A Reimer Tiemann reaction

B Friedel Craft's alkylation

C Acetylation of aniline

D Friedel Craft's acylation

Correct Answer

Option B

Solution

Polysubstitution is a major drawback of Friedal–Craft alkylation. –CH3 group in highly activating group due to +H effect, so after first time addition of -CH3 group in benzene ring the reactivity of the benzene ring increases.

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