Question

$N{a}_2{C}_2\stackrel{HCl}{\to }\underset{\left(\text{excess}\right)C{H}_{3}I}{\overset{\left(\text{excess}\right)NaN{H}_2}{\to }}\underset{Pd-BaS{O}_4}{\overset{H_2}{\to }}\underset{ii.Zn-H_{2}O}{\overset{i.O_3}{\to }}$

Final product can be obtained by:

• A. $Me-CH=CH-Me\underset{ii.H^{+}/H_{2}O}{\overset{i.mCPBA}{\to }}\stackrel{HI{O}_4}{\to }$
• B. $H_{2}C=C{H}_2\underset{ii.H_2{O}_2,O{H}^{-}}{\overset{i.B_2{H}_6,THF}{\to }}\underset{\Delta }{\overset{Cu}{\to }}$
• C. $Me-COOH\stackrel{SOC{l}_2}{\to }\stackrel{LiAlH(OC{H}_3{)}_3}{\to }$
• D. $Me-Br\underset{DMSO}{\overset{KCN}{\to }}\underset{ii.H^{+}/H_{2}O}{\overset{i.SnC{l}_2/HCl}{\to }}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $Me-CH=CH-Me\underset{ii.H^{+}/H_{2}O}{\overset{i.mCPBA}{\to }}\stackrel{HI{O}_4}{\to }$
B $H_{2}C=C{H}_2\underset{ii.H_2{O}_2,O{H}^{-}}{\overset{i.B_2{H}_6,THF}{\to }}\underset{\Delta }{\overset{Cu}{\to }}$
C $Me-COOH\stackrel{SOC{l}_2}{\to }\stackrel{LiAlH(OC{H}_3{)}_3}{\to }$
D $Me-Br\underset{DMSO}{\overset{KCN}{\to }}\underset{ii.H^{+}/H_{2}O}{\overset{i.SnC{l}_2/HCl}{\to }}$

$N{a}_2{C}_2\stackrel{HCl}{\to }HC\equiv CH\underset{\left(\text{excess}\right)C{H}_{3}I}{\overset{\left(\text{excess}\right)NaN{H}_2}{\to }}Me-C\equiv C-Me\underset{Pd-BaS{O}_4}{\overset{H_2}{\to }}H\stackrel{Me}{\stackrel{|}{C}}=\stackrel{Me}{\stackrel{|}{C}}H\underset{ii.Zn-H_{2}O}{\overset{i.O_3}{\to }}\stackrel{Me}{\stackrel{|}{C}}=O$


$a)Me-CH=CH-Me\underset{ii.H^{+}/H_{2}O}{\overset{i.mCPBA}{\to }}Me-\stackrel{OH}{\stackrel{|}{C}}H-\stackrel{OH}{\stackrel{|}{C}}H-Me\stackrel{HI{O}_4}{\to }Me-\stackrel{O}{\stackrel{||}{C}}H$

$b)H_{2}C=C{H}_2\underset{ii.H_2{O}_2,O{H}^{-}}{\overset{i.B_2{H}_6,THF}{\to }}{H}_{3}C-\stackrel{OH}{\stackrel{|}{C}}{H}_2\underset{\Delta }{\overset{Cu}{\to }}{H}_{3}C-\stackrel{O}{\stackrel{||}{C}}H$

$c)Me-COOH\stackrel{SOC{l}_2}{\to }{H}_{3}C-\stackrel{O}{\stackrel{||}{C}}-Cl\stackrel{LiAlH(OC{H}_3{)}_3}{\to }{H}_{3}C-\stackrel{O}{\stackrel{||}{C}}H$

$d)Me-Br\underset{DMSO}{\overset{KCN}{\to }}{H}_{3}C-CN\underset{ii.H^{+}/H_{2}O}{\overset{i.SnC{l}_2/HCl}{\to }}{H}_{3}C-\stackrel{O}{\stackrel{||}{C}}H$