Match the reagent from Column I which on reaction with CH 3 - CH = CH 2 gives some product given in Column II as per the codes given below: Column I Column II (i) O 3 / Zn + H 2 O (a) Acetic acid and CO 2 (ii) KMnO 4 / H + (b) Propan - 1 - ol (iii) KMnO 4 / OH - (c) Propan - 2 - ol (iv) H 2 O / H + (d) Acetaldehyde and formaldehyde (v) B 2 H 6 / NaOH and H 2 O 2 (e) Propan - 1 , 2 - diol

(i) O 3 / Zn + H 2 O (d) Ozonolysis of propene gives acetaldehyde and formaldehyde. During ozonolysis alkenes and alkynes form organic compounds in which the multiple carbon–carbon bond has been replaced by a carbonyl group. (ii) KMnO 4 / H + (a) On reacting with KMnO 4 in acidic medium, propene gives CH 3 COOH and CO 2 . CH 3 - CH = CH 2 → KMnO 4 / H + CH 3 COOH + CO 2 When propene is bubbled through hot aqueous solution of potassium permanganate, ethanoic acid and formic acid are formed. The formic acid decomposes to water and carbon dioxide. (iii) KMnO 4 / OH - (e) Propene changes the colour of alkaline KMnO 4 to form Propane - 1 , 2 - diol . CH 3 - CH = CH 2 + H 2 O + O → KMnO 4 CH 3 CH OH CH 2 OH Formation of vicinal diols takes place when alkenes react with alkaline potassium permanganate. (iv) H 2 O / H + (c) Addition of H 2 O to propene gives Propan - 2 - ol . CH 3 - CH = CH 2 + H + / - OH → CH 3 - CH OH - CH 3 Propene is an unsymmetrical alkene. The addition of water to propene takes place according to Markovnikov's Rule which results in the formation of secondary alcohol rather than primary alcohol. (v) B 2 H 6 / NaOH and H 2 O 2 (b) Propan - 1 - ol is formed. The reaction proceeds in an Anti-Markovnikov manner, where the hydrogen (from BH 3 ) attaches to the more substituted carbon and the boron attaches to the least substituted carbon in the alkene double bond. Column I Column II (i) O 3 / Zn + H 2 O (d) Ozonolysis of propene gives acetaldehyde and formaldehyde. (ii) KMnO 4 / H + (a) On reacting with KMnO 4 in acidic medium, propene gives CH 3 COOH and CO 2 . CH 3 - CH = CH 2 → KMnO 4 / H + CH 3 COOH + CO 2 (iii) KMnO 4 / OH - (e) Propene changes the colour of alkaline KMnO 4 to form Propane - 1 , 2 - diol . CH 3 - CH = CH 2 + H 2 O + O → KMnO 4 CH 3 CH ( OH ) CH 2 OH (iv) H 2 O / H + (c) Addition of H 2 O to propene gives Propan - 2 - ol . CH 3 - CH = CH 2 + H + / - OH → CH 3 - CH ( OH ) - CH 3 (v) B 2 H 6 / NaOH and H 2 O 2 (b) Propan - 1 - ol is formed.

Match the hydrocarbons in Column I with the boiling points given in column II. Column I Column II (i) n-Pentane (a) 282 . 5 K (ii) iso-Pentane (b) 309 K (iii) neo-Pentane (c) 301 K

(i) n -Pentane (b) 309 K In n -Pentane there is no branching and surface area. Hence, there is more Vander Waal’s forces. Therefore, boiling point is high. (ii) iso-Pentane (c) 301 K Iso-pentane, also called 2 -methyl butane, is a branched-chain saturated hydrocarbon and an extremely volatile, flammable liquid at room temperature and pressure. (iii) neo-Pentane (a) 282 . 5 K Neo-pentane is a gas at room temperature and has spherical structure. In this compound two side chains are there having the same molar mass. Therefore, its boiling point is lowest amongst all three compounds. Column I Column II (i) n-Pentane (b) 309 K (ii) iso-Pentane (c) 301 K (iii) neo-Pentane (a) 282 . 5 K

Match the following reactants in Column I with the corresponding reaction products in Column II. Column I Column II (i) Benzene + Cl 2 → AlCl 3 (a) Benzoic acid (ii) Benzene + CH 3 Cl → AlCl 3 (b) Methyl phenyl ketone (iii) Benzene + CH 3 COCl → AlCl 3 (c) Toluene (iv) Toluene → KMnO 4 / NaOH (d) Chlorobenzene (e) Benzene hexachloride

(i) Benzene + Cl 2 → AlCl 3 (d) Chlorobenzene AICl 3 is a catalyst which activates the chlorine molecule to show heterolytic cleavage. AICl 3 is an electron deficient molecule and forms [ AlCl 4 ] - , and Cl + when reacts with Cl 2 . This Cl + electrophile attacks on electron rich benzene ring. (ii) Benzene + CH 3 Cl → AlCl 3 (c) Toluene It involves the alkylation of an aromatic ring with an alkyl halide using a strong Lewis acid, such as aluminium chloride. (iii) Benzene + CH 3 COCl → AlCl 3 (b) Methyl phenyl ketone It is the reaction of an arene with acyl chlorides or anhydrides using a strong Lewis acid catalyst. This reaction proceeds via electrophilic aromatic substitution to form monoacylated products. (iv) Toluene → KMnO 4 / NaOH (a) Benzoic acid KMnO 4 is a strong oxidising agent. It oxidises toluene to form benzoic acid. Column I Column II (i) Benzene + Cl 2 → AlCl 3 (d) Chlorobenzene (ii) Benzene + CH 3 Cl → AlCl 3 (c) Toluene (iii) Benzene + CH 3 COCl → AlCl 3 (b) Methyl phenyl ketone (iv) Toluene → KMnO 4 / NaOH (a) Benzoic acid

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An alkane $C_{8} H_{18}$ is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of tertiary bromide. Write the structure of alkane and the tertiary bromide.

Important Questions on Hydrocarbons

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 13 - Hydrocarbons>Short Answer Type>Q 37

The ring systems having following characteristics are aromatic.
(i) Planar ring containing conjugated $π$ bonds.
(ii) Complete delocalisation of the $π$ -electrons in ring system i.e. each atom in the ring has unhybridised p-orbital, and
(iii) Presence of $(4 n + 2)$ $π$ -electrons in the ring where n is an integer $(n = 0, 1, 2 . .)$ [Huckel rule].

Using this information classify the following compounds as aromatic/ nonaromatic.

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HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 13 - Hydrocarbons>Short Answer Type>Q 38

Which of the following compounds are aromatic according to Huckel’s rule?

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EASY

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 13 - Hydrocarbons>Matching Type>Q 40

Match the reagent from Column I which on reaction with ${CH}_{3} - CH = {CH}_{2}$ gives some product given in Column II as per the codes given below:

	Column I		Column II
(i)	$O_{3} / Zn + H_{2} O$	(a)	Acetic acid and ${CO}_{2}$
(ii)	${KMnO}_{4} / H^{+}$	(b)	$Propan - 1 - ol$
(iii)	${KMnO}_{4} / {OH}^{-}$	(c)	$Propan - 2 - ol$
(iv)	$H_{2} O / H^{+}$	(d)	Acetaldehyde and formaldehyde
(v)	$B_{2} H_{6} / NaOH and H_{2} O_{2}$	(e)	$Propan - 1, 2 - diol$

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 13 - Hydrocarbons>Matching Type>Q 41

Match the hydrocarbons in Column I with the boiling points given in column II.

	Column I		Column II
(i)	n-Pentane	(a)	$282.5 K$
(ii)	iso-Pentane	(b)	$309 K$
(iii)	neo-Pentane	(c)	$301 K$

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 13 - Hydrocarbons>Matching Type>Q 42

Match the following reactants in Column I with the corresponding reaction products in Column II.

	Column I		Column II
(i)	Benzene $+ {Cl}_{2} \overset{{AlCl}_{3}}{\to}$	(a)	Benzoic acid
(ii)	Benzene $+ {CH}_{3} Cl \overset{{AlCl}_{3}}{\to}$	(b)	Methyl phenyl ketone
(iii)	Benzene $+ {CH}_{3} COCl \overset{{AlCl}_{3}}{\to}$	(c)	Toluene
(iv)	Toluene $\overset{{KMnO}_{4} / NaOH}{\to}$	(d)	Chlorobenzene
		(e)	Benzene hexachloride

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11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 13 - Hydrocarbons>Long Answer Type>Q 48

An alkyl halide

C_{5} H_{11} Br

(A)

reacts with ethanolic

KOH

to give an alkene

' B'

. which reacts with

{Br}_{2}

to give a compound

' C'

which on dehydrobromination gives an alkyne

' D'

. On treatment with sodium metal In liquid ammonia one mole of

' D'

gives one mole of the sodium salt of

' D'

and half a mole of hydrogen gas. Complete hydrogenation of

' D'

yields a straight chain alkane. Identify

A, B, C

and

D

. Give the reactions involved.

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11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 13 - Hydrocarbons>Long Answer Type>Q 49

896 mL

vapour of a hydrocarbon

‘ A ’

having carbon

87.80 %

and hydrogen

12.19 %

weighs

3.28 g

at STP. Hydrogenation of

' A'

gives

2

-methylpentane. Also, A on hydration In the presence of

H_{2} {SO}_{4}

and

{HgSO}_{4}

gives a ketone

‘ B ’

having molecular formula

C_{6} H_{12} O

. The ketone

' B'

gives a positive iodoform test. Find the structure of

A

and give the reactions Involved.

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 13 - Hydrocarbons>Long Answer Type>Q 50

An unsaturated hydrocarbon

‘ A ’

adds two molecules of

H_{2}

and on reductive ozonolysis gives butane-

1, 4

-dial, ethanal and propanone. Give the structure of

‘ A ’

, write its IUPAC name and explain the reactions involved.

An alkane C8H18 is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of tertiary bromide. Write the structure of alkane and the tertiary bromide.

Important Questions on Hydrocarbons

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An alkane $C_{8} H_{18}$ is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of tertiary bromide. Write the structure of alkane and the tertiary bromide.