Complete each synthesis by giving missing starting material, reagent, or products:

Question

Consider following reaction :
$Fe(OH)_2 + 2e^- \rightarrow Fe(s) + 2OH^-$ , $E^{\circ} = -0.88 \text{ V}$
$AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq)$ , $E^{\circ} = 0.07 \text{ V}$
Select the correct statement.
(1) $E^{\circ}_{cell} = -0.95 \text{V}$
(2) $E^{\circ}_{cell}$ is an extensive property
(3) $Fe$ is getting reduced
(4) Net cell reaction: $Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightarrow Fe(OH)_2 + 2Ag(s) + 2Br^-(aq)$

Answer 1

(a) Hydroxylamine (NH₂OH) is formed from an aldehyde (O) and hydroxylamine (HO-NH₂) catalyzed by acid (H⁺).

$ O + HO-NH_2 \xrightarrow{H^+} NH_2OH $

(b) Ethene (CH₂=CH₂) undergoes ozonolysis followed by reductive workup to yield formaldehyde (HCHO). Heating ($ \Delta $) does not cause further reaction.

$ CH_2=CH_2 \xrightarrow{(i) O_3 (ii) Zn-H_2O} 2HCHO \xrightarrow{\Delta} \text{No further change} $

(c) An alcohol (OH) is converted to an alkyl chloride (Cl) using thionyl chloride (SOCl₂) and heating ($ \Delta $).

$ OH \xrightarrow{SOCl_2} Cl $

(d) An aldehyde (CHO) is converted to a carboxylic acid (COOH) using sodium cyanide (NaCN) and hydrochloric acid (HCl).

$ CHO \xrightarrow{NaCN/HCl} COOH $

(e) Chlorobenzene (Cl₂C₆H₅) is methylated to produce a methylated chlorobenzene derivative (Cl₂C₆H₄CH₃).

$ Cl_2C_6H_5 \xrightarrow{CH_3} Cl_2C_6H_4CH_3 $

Complete each synthesis by giving missing starting material, reagent, or products:

Show Hint

Solution and Explanation

Top Questions on Electrochemistry

Questions Asked in CBSE Class XII exam