the red coloured compound formed in victor-meyer's test for ethyl alcohol is
a) CH₃-C(NO₂)=NO^-Na⁺ 
b) CH₃-CH(NO)-NO₂ ^-Na⁺
​c) CH₃-CH(NO₂)-NO
d) (CH₃)₂-C(NO₂)-NO
 

Dear Student,

The reagents used in victor meyer's test are P+I₂, AgNO_​2, HNO_2, NaOH and the colour result from this are,
Primary alcohol = red colour
Secondary alcohol = blue colour 
Teritiary alcohol = no reaction

So the given ethyl alcohol belongs to primary alcohol category, 

$C{H}_3-C{H}_2-OH \stackrel{P+I_2}{\to } C{H}_3-C{H}_2-I\stackrel{AgN{O}_2}{\to }C{H}_3-C{H}_2-N{O}_2\stackrel{NaN{O}_2+dil.HCl(0-5^{o}C)}{\to }C{H}_3-\underset{N{O}_2}{C}=N-OH \stackrel{NaOH}{\to }C{H}_3-C\left(N{O}_2\right)=N{O}^{-}N{a}^{+}$

Hence the answer is Option (A)

Regards.