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

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.

Accepted Answer

The reactions involved in identification of $A, B, C$ and $D$ are as follows:

Alkyl halide $C_{5} H_{11} Br$ $(A)$ on heating with alcoholic potash (potassium hydroxide dissolved in alcohol, say, ethanol) eliminate one molecule of halogen acid to form pentene $(B)$ . This reaction is known as dehydrohalogenation i.e., removal of halogen acid.

$\underset{\underset{(A)}{1 - Bromopentane}}{C_{5} H_{11} Br} + \underset{}{alc. KOH} \to \underset{\underset{(B)}{Pentene}}{C_{5} H_{10}} + HBr$

Halogens like bromine or chlorine add up to pentene $(B)$ to form vicinal dihalides i.e., $1, 2$ -dibromopentane $(C)$ . The reddish orange colour of bromine solution in carbon tetrachloride is discharged when bromine adds up to an unsaturation site. This reaction is used as a test for unsaturation.

$\underset{\underset{(B)}{Pentene}}{C_{5} H_{10}} + {Br}_{2} \overset{{CCl}_{4}}{\to} \underset{\underset{(C)}{1, 2 - dibromopentane}}{C_{5} H_{10} {Br}_{2}}$

$1, 2$ -dibromopentane $(C)$ on treatment with alcoholic potassium hydroxide undergo dehydrohalogenation. One molecule of hydrogen halide is eliminated to form alkenyl halide which on treatment with sodamide gives pentyne $' B'$ .

$\underset{\underset{(C)}{1, 2 - dibromopentane}}{C_{5} H_{10} {Br}_{2}} + alc. KOH \to \underset{\underset{(D)}{Pentyne}}{C_{5} H_{8}} + 2 HBr$

Hydrogenation of alkyne $(D)$ gives straight chain alkane hence, all the compounds $A, B, C$ and $D$ must be straight chain compounds. Pentyne reacts with sodium to form sodium pentynide and hydrogen gas. Alkyne gives sodium alkynide which proves $D$ is terminal alkyne.

$2 \underset{\underset{(D)}{Pentyne}}{C_{5} H_{8}} + 2 Na \to \underset{sodium Pentynide}{2 C_{5} H_{7} Na} + H_{2} ↑$

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