Name: Mechanism of Addition Reactions of Alkenes
Uploaded: 2019-07-19
Duration: 11 min 16 s
Description: The most common reaction for an alkene is the addition reaction to a C=C double bond. Watch the video to learn more about the addition reactions of alkenes.

Question

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)	$Propan - 1, 2 - diol$

Accepted Answer

(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During ozonolysis alkenes and alkynes form organic compounds in which the multiple carbon–carbon bond has been replaced by a carbonyl group.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 ${CO}_{2}$ . ${CH}_{3} - CH = {CH}_{2} \overset{{KMnO}_{4} / H^{+}}{\to} {CH}_{3} COOH + {CO}_{2}$

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 $Propane - 1, 2 - diol$ . ${CH}_{3} - CH = {CH}_{2} + H_{2} O + O \overset{{KMnO}_{4}}{\to} {CH}_{3} CH (OH) {CH}_{2} OH$

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 $Propan - 2 - ol$ . ${CH}_{3} - CH = {CH}_{2} + H^{+} /^{-} OH \to {CH}_{3} - CH (OH) - {CH}_{3}$

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 ${CO}_{2}$ .

${CH}_{3} - CH = {CH}_{2} \overset{{KMnO}_{4} / H^{+}}{\to} {CH}_{3} COOH + {CO}_{2}$

(iii) ${KMnO}_{4} / {OH}^{-}$ (e)
Propene changes the colour of alkaline ${KMnO}_{4}$ to form $Propane - 1, 2 - diol$ .

${CH}_{3} - CH = {CH}_{2} + H_{2} O + O \overset{{KMnO}_{4}}{\to} {CH}_{3} CH (OH) {CH}_{2} OH$

(iv) $H_{2} O / H^{+}$ (c)
Addition of $H_{2} O$ to propene gives $Propan - 2 - ol$ .

${CH}_{3} - CH = {CH}_{2} + H^{+} /^{-} OH \to {CH}_{3} - CH (OH) - {CH}_{3}$

(v) $B_{2} H_{6} / NaOH and H_{2} O_{2}$ (b)
$Propan - 1 - ol$ is formed.

NCERT Solutions for Chapter: Hydrocarbons, Exercise 1: Matching Type

NCERT Chemistry Solutions for Exercise - NCERT Solutions for Chapter: Hydrocarbons, Exercise 1: Matching Type

Questions from NCERT Solutions for Chapter: Hydrocarbons, Exercise 1: Matching Type with Hints & Solutions

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	Column I		Column II
(i)	n-Pentane	(a)	$282.5 K$
(ii)	iso-Pentane	(b)	${KMnO}_{4} / H^{+}$
(iii)	neo-Pentane	(c)	${KMnO}_{4}$

	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