A B C DHere- the compound C will be -

The correct option is B WestronThe compound (C) is Westron. Ethylidene bromide A CH_3 CHBr_2 #160; acetylene B HC ≡ CH #160; Westron C CHCl_2 CHCl_2 ...

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Standard XII

Chemistry

Introduction to Electrophilic Substitution

A B C DHere, ...

Question

$A A l c - K O H - --------- \to B 2 {C l}_{2} - ------ \to C {C l}_{4} C C a {O H}_{2} - ------- \to D$
Here, the compound $C$ will be :

Lewisite

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Westron

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Acetylene tetrachloride

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Both b and c

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Solution

The correct option is B Westron
The compound (C) is Westron.
$C H_{3} - C H B r_{2} Ethylidene bromide A A l c - K O H - --------- \to N a N H_{2} H C \equiv C H acetylene B 2 {C l}_{2} - ------ \to C {C l}_{4} C H C l_{2} - C H C l_{2} Westron C C a {O H}_{2} - ------- \to C H C l - C l_{2} Westrosol D$
Dehydrobromination of ethylidene bromide (compound A) gives acetylene (compound B). Chlorination of acetylene gives westron (compound C). Treatment with calcium hydroxide forms westrosol (compound D).

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product, here product will be:

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AAlc−KOH−−−−−−−−−−→B2Cl2−−−−−−−→CCl4CCaOH2−−−−−−−−→DHere, the compound C will be :

$A A l c - K O H - --------- \to B 2 {C l}_{2} - ------ \to C {C l}_{4} C C a {O H}_{2} - ------- \to D$
Here, the compound $C$ will be :