An alkene on vigorous oxidation with ${\rm{KMn}}{{\rm{O}}_4}$ gives only acetic acid. The alkene is

${\rm{C}}{{\rm{H}}_3}{\rm{CH}} = {\rm{CHC}}{{\rm{H}}_3}$

${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{CH}} = {\rm{C}}{{\rm{H}}_2}$

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{C}} = {\rm{C}}{{\rm{H}}_2}$

${\rm{C}}{{\rm{H}}_3}{\rm{CH}} = {\rm{C}}{{\rm{H}}_2}$

Which is the most suitable reagent among the following to distinguish compound (3) from rest of the compounds? ${\rm{i)}}\,\,\,\,C{H_3} - C \equiv C - C{H_3}$ ${\rm{ii)}}\,\,\,\,C{H_3} - C{H_2} - C{H_2} - C{H_3}$ ${\rm{iii)}}\,\,\,\,C{H_3} - C{H_2}C \equiv CH$ ${\rm{iv)}}\,\,\,\,C{H_3} - CH = C{H_2}.$

Bromine in carbon tetrachloride

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NEET • CHEMISTRY • HYDROCARBONS

Hydrocarbons practice · Mixed difficulty

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Showing 1-30 of 229 questions

1 Easy

1 of 229

Remembering +4 | -1

The reaction of alkanes with halogen is explosive in the case of:

View answer and explanation

Correct option: A

{\mathrm{F}_2}

Option A (

{\mathrm{F}_2}

) is correct because the reaction of alkanes with fluorine is highly exothermic and explosive.

2 Easy

2 of 229

Understanding +4 | -1

A mixture of sodium benzoate and soda lime on heating produces

View answer and explanation

Correct option: B

{C_6}{H_6}

Option B (

{C_6}{H_6}

) is correct because sodium benzoate and soda lime on heating undergo decarboxylation to form benzene.

3 Hard

3 of 229

Application +4 | -1

The major product of the following reaction is

C{H_3}C \equiv CH\xrightarrow[{(ii)\;DI}]{{(i)\;DCl\;(1\;equiv.)}}

View answer and explanation

Correct option: D

C{H_3}C(I)(Cl)CH{D_2}

The correct answer is

C{H_3}C(I)(Cl)CH{D_2}

because DCl adds first according to Markovnikov's rule, followed by DI.

4 Hard

4 of 229

Analyzing +4 | -1

S-I : Heavy metal acetylides can be used to purify alkynes
S-II: Terminal alkynes form acetylides which are soluble in acids.

View answer and explanation

Correct option: A

S-I

\&

S-II are true

Option A is correct because heavy metal acetylides are used to purify alkynes by selectively precipitating them, and terminal alkynes do form acetylides that are soluble in acids, allowing for their regeneration.

5 Medium

5 of 229

Analyzing +4 | -1

Select the incorrect statement regarding A

C{H_3} - C \equiv CH + {H_2}OA

View answer and explanation

Correct option: D

Reduces Tollen’s reagent

Option D is correct because the product A, which is 2-propanone (acetone), does not reduce Tollen's reagent. Ketones generally do not reduce Tollen's reagent, while aldehydes do.

6 Medium

6 of 229

Remembering +4 | -1

Kharasch effect is based on

View answer and explanation

Correct option: A

Free radical addition

Option A (Free radical addition) is correct because the Kharasch effect, also known as the peroxide effect, involves a free radical addition mechanism.

7 Medium

7 of 229

Analyzing +4 | -1

A compound

X\left( {{\mathrm{C}_5}{\mathrm{H}_8}} \right)

reacts with ammoniacal

\mathrm{AgN}{\mathrm{O}_3}

to give a white precipitate, and on oxidation with hot alkaline

\mathrm{KMn}{\mathrm{O}_4}

gives the acid,

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CHCOOH}

. Therefore,

X

View answer and explanation

Correct option: C

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CH} - \mathrm{C} \equiv \mathrm{CH}

Option C is correct because the compound X reacts with ammoniacal silver nitrate to give a white precipitate, indicating it is a terminal alkyne. Upon oxidation, it yields isobutyric acid, which confirms the structure of X as 3-methyl-1-butyne.

8 Easy

8 of 229

Understanding +4 | -1

The reactivity of halogens towards any given alkane is in order of

View answer and explanation

Correct option: B

{F_2} > C{l_2} > B{r_2} > {I_2}

Option B (

{F_2} > C{l_2} > B{r_2} > {I_2}

) is correct because the reactivity of halogens towards alkanes decreases down the group due to decreasing bond dissociation energy.

9 Hard

9 of 229

Understanding +4 | -1

The gas

'A'

is having very low reactivity reaches to stratosphere. It is non-toxic and non-flammable but dissociated by UV—radiations in stratosphere. The intermediates formed initially from the gas

'A'

are

View answer and explanation

Correct option: D

\mathop \mathrm{C}\limits^ \bullet \mathrm{l + C}\mathop {{\mathrm{F}_2}}\limits^ \bullet \mathrm{Cl}

Option D (

\mathop \mathrm{C}\limits^ \bullet \mathrm{l + C}\mathop {{{{F}}_2}}\limits^ \bullet \mathrm{Cl}

) is correct because the gas 'A' is a CFC (Chlorofluorocarbon), which upon UV radiation dissociates into chlorine free radical and a fluorocarbon free radical.

10 Easy

10 of 229

Application +4 | -1

The treatment of benzene with benzoyl chloride in the presence of

\mathrm{AlC}{\mathrm{l}_3}

gives

View answer and explanation

Correct option: B

Benzophenone

Benzophenone is the correct answer because the reaction of benzene with benzoyl chloride in the presence of

\mathrm{AlC}{\mathrm{l}_3}

is a Friedel-Crafts acylation, which adds a benzoyl group to benzene.

11 Easy

11 of 229

Application +4 | -1

Electrolysis of cold concentrated aqueous solution of potassium succinate yields

View answer and explanation

Correct option: C

ethene

Option C (ethene) is correct because the electrolysis of potassium succinate leads to the formation of ethene via the Kolbe electrolysis mechanism.

12 Easy

12 of 229

Application +4 | -1

Isopropyl bromide on Wurtz reaction gives

View answer and explanation

Correct option: C

2,3-dimethyl butane

Option C (2,3-dimethyl butane) is correct because the Wurtz reaction involves coupling two alkyl halides, and in this case, two isopropyl groups combine to form 2,3-dimethylbutane.

13 Easy

13 of 229

Understanding +4 | -1

Anti-addition of hydrogen atoms to the triple bond takes place by

View answer and explanation

Correct option: B

Li/N{H_3}

Option B (

Li/N{H_3}

) is correct because the reduction of alkynes with lithium in liquid ammonia results in trans-addition of hydrogen atoms.

14 Hard

14 of 229

Analyzing +4 | -1

The isomer of

\mathrm{C}_6 \mathrm{H}_{14},

which will give maximum number of monochloroderivative is

View answer and explanation

Correct option: B

2-methylpentane

Option B (2-methylpentane) is correct because it has the most non-equivalent hydrogen atoms, leading to the maximum number of monochloro derivatives.

15 Medium

15 of 229

Understanding +4 | -1

Alkene-1 on hydroboration followed with action of

{\mathrm{H}_2}{\mathrm{O}_2}

gives:

View answer and explanation

Correct option: B

Alkanol-1

Option B (Alkanol-1) is correct because hydroboration-oxidation of an alkene results in anti-Markovnikov addition of water, leading to the formation of a primary alcohol (alkanol-1).

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16 Medium

16 of 229

Understanding +4 | -1

Which of the following reagent is not involved in the halogenation of benzene

View answer and explanation

Correct option: C

B{r_2}/CC{l_4}

Option C (

B{r_2}/CC{l_4}

) is correct because

B{r_2}/CC{l_4}

is used for the addition reaction with alkenes, not for the halogenation of benzene.

17 Easy

17 of 229

Analyzing +4 | -1

Which of the following compounds contain

{1^0},{2^0},{3^0}

as well as

{4^0}

carbon atoms

View answer and explanation

Correct option: D

2,2,3-trimethyl pentane

Option D (2,2,3-trimethyl pentane) is correct because it contains primary, secondary, tertiary, and quaternary carbon atoms.

18 Medium

18 of 229

Application +4 | -1

What volume of

\mathrm{C}{\mathrm{H}_4}

at NTP is formed when 20.5 g of

\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}

a is treated with sodalime?

View answer and explanation

Correct option: D

5.6 litre

Option D (5.6 litre) is correct because 20.5 g of CH3COONa produces 0.25 moles of methane, which occupies 5.6 liters at NTP.

19 Easy

19 of 229

Analyzing +4 | -1

An alkene on vigorous oxidation with

\mathrm{KMn}{\mathrm{O}_4}

gives only acetic acid. The alkene is

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{H}_3}\mathrm{CH} = \mathrm{CHC}{\mathrm{H}_3}

Option B (

\mathrm{C}{\mathrm{H}_3}\mathrm{CH} = \mathrm{CHC}{\mathrm{H}_3}

) is correct because vigorous oxidation with

\mathrm{KMn}{\mathrm{O}_4}

yields only acetic acid.

20 Easy

20 of 229

Understanding +4 | -1

Which one of the following is not an oxidation product of alkane

View answer and explanation

Correct option: D

ether

Option D (ether) is correct because ethers are not direct oxidation products of alkanes. Oxidation of alkanes typically yields alcohols, aldehydes, and carboxylic acids.

21 Hard

21 of 229

Understanding +4 | -1

The Baeyer angle strain is minimum in

View answer and explanation

Correct option: C

Cyclopentane

Option D (Cyclohexane) is correct because cyclohexane exists in a chair conformation, which minimizes angle strain, making it the most stable cycloalkane among the given options.

22 Medium

22 of 229

Remembering +4 | -1

On passing benzene vapour through a tube at

700 - 800^\circ C

or through molten lead we get:

View answer and explanation

Correct option: A

Diphenyl

Option A (Diphenyl) is correct because passing benzene vapor through a hot tube or molten lead causes dimerization to form diphenyl.

23 Medium

23 of 229

Analyzing +4 | -1

Which among the following is very strong

o - ,\,p -

directing group?

View answer and explanation

Correct option: D

- \mathrm{NHR}

Option D (

-\mathrm{NHR}

) is correct because alkylated amines are stronger activating groups than alcohols, halides, or unsubstituted amines due to increased electron donation.

24 Medium

24 of 229

Analyzing +4 | -1

When

\mathrm{Ca}{\mathrm{C}_2}

was hydrolysed a gas was obtained. It had a garlic odour due to phosgene present as impurity. The gas was passed through ammoniacal solution of

\mathrm{C}{\mathrm{u}_2}\mathrm{C}{\mathrm{l}_2}

, ared ppt. was obtained. The gas was:

View answer and explanation

Correct option: C

Acetylene

Option C is correct because acetylene (ethyne) reacts with cuprous chloride in an ammoniacal solution to form a red precipitate of copper acetylide. The garlic odor is due to phosphine impurity in the calcium carbide.

25 Easy

25 of 229

Application +4 | -1

Major product of the following reaction is

C{H_3}C{H_2}\mathop C\limits_{\mathop {Cl}\limits^| } HC{H_3}\xrightarrow[\Delta ]{{AlC.KOH}}

View answer and explanation

Correct option: B

but-2-ene

Option B (but-2-ene) is correct because the major product of the dehydrohalogenation reaction is the more substituted alkene (Zaitsev's rule).

26 Hard

26 of 229

Application +4 | -1

At constant temperature and pressure, the percentage of butane by volume in a mixture of butane, propane and propyne is

40 \%.

What will be the volume of

\mathrm{CO}_{2(\mathrm{g})}

(in ml) under same pressure and temperature produced, when 150 ml of gaseous mixture is burnt in excess of

\mathrm{O}_2?

View answer and explanation

Correct option: C

510 ml

Option C (510 ml) is correct because it is calculated based on the combustion equations of butane, propane, and propyne, considering the given percentage of butane.

27 Hard

27 of 229

Analyzing +4 | -1

Which is the most suitable reagent among the following to distinguish compound (3) from rest of the compounds?

\mathrm{i)}\,\,\,\,C{H_3} - C \equiv C - C{H_3}

\mathrm{ii)}\,\,\,\,C{H_3} - C{H_2} - C{H_2} - C{H_3}

\mathrm{iii)}\,\,\,\,C{H_3} - C{H_2}C \equiv CH

\mathrm{iv)}\,\,\,\,C{H_3} - CH = C{H_2}.

View answer and explanation

Correct option: D

Ammonical silver nitrate

Option D is correct because ammoniacal silver nitrate reacts specifically with terminal alkynes (like compound iii) to form a precipitate, while it does not react with internal alkynes, alkanes, or alkenes.

28 Medium

28 of 229

Analyzing +4 | -1

Which of the following will not produce ethane?

View answer and explanation

Correct option: B

Reduction of

\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}

with HI and red P

Option B (Reduction of

\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}

with HI and red P) is correct because the reduction of acetone (propanone) with HI and red P yields propane, not ethane.

29 Medium

29 of 229

Application +4 | -1

In the reaction sequence, the product

Y

{C_6}{H_6}\xrightarrow[{AlC{l_3}}]{{C{H_3}Cl}}X\xrightarrow{{KMn{O_4}}}Y

View answer and explanation

Correct option: C

benzoic acid

Option A is correct because the reaction requires deprotonation of both acidic hydrogens of ethyne using

NaNH_2

, followed by alkylation with excess ethyl iodide to form Hex-3-yne.

30 Hard

30 of 229

Analyzing +4 | -1

(A): 1-butene on reaction with

\mathrm{HBr}

in the presence of a peroxide, produces 1-bromobutane
(R): It involves the formation of a primary radical.

View answer and explanation

Correct option: C

A is true, but R is false

Option C (A is true, but R is false) is correct because 1-butene reacts with HBr in the presence of a peroxide to produce 1-bromobutane (anti-Markovnikov addition), but this occurs via a secondary radical intermediate, not a primary radical.

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