Hydrocarbons — Page 4 | NEET Chemistry

Q31

Which of the following is Lindlar catalyst?

A Partially deactivated palladised charcoal

B Zinc chloride and HCl

C Sodium and Liquid NH3

D Cold dilute solution of KMnO4

Correct Answer

Option A

Solution

Lindlar's catalyst $\Rightarrow$ Pd/CaCO3 + (CH3COO)2 Pb + quinolene

Q32

Which one of the following has the minimum boiling point?

A n-butane

B isobutane

C 1- butene

D 1- butyne

Correct Answer

Option B

Solution

NOTE : Among isomeric alkanes, the straight chain isomer has higher boiling point than the branched chain isomer.

The greater the branching of the chain, the lower is the boiling point.

Further due to the presence of $\pi$ electrons, these molecules are slightly polar and hence have higher boiling points than the corresponding alkanes.

Thus

B.pt.

follows the order alkynes

>

alkene

>

alkanes (straight chain) > branched chain alkanes.

Q33

Which types of isomerism is shown by 2,3-dichlorobutane?

A Diastereo

B Optical

C Geometric

D Structural

Correct Answer

Option B

Solution

2,3

-dichloro butane will exhibit optical isomerism due to the presence of two asymmetric carbon atom.

Q34

In the following sequence of reactions, the alkene affords the compound ‘B’ CH3 - CH = CH - CH3

\overset{{{O_3}}}\longrightarrow

A

\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Zn}^{H{}_2O}}

B The compound B is

A CH3CH2CHO

B CH3COCH3

C CH3CH2COCH3

D CH3CHO

Correct Answer

Option D

Solution

Completing the sequence of given reactions, Thus

'B'\,\,

is

\,\,C{H_3}CHO

Hence

(d)

is correct answer.

Q35

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

or

Pd)

and form alkanes.

Q36

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.

Q37

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.

Q38

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.

Q39

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.

Q40

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