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The compound that will not form a yellow precipitate on heating with an alkaline solution of iodine is: 
1. CH₃CH(OH)CH₃                                 
2. CH₃CH₂CH(OH)CH₃
3. CH₃OH                                               
4. CH₃CH₂OH

Glycerol on oxidation with conc HNO₃ gives:
1. tartronic acid                               
2. mesoxalic acid
3. oxalic acid                                   
4. glyceric acid

The number of  moles of HI consumed by the given compound is:

$\underset{{\mathrm{CH}}_2{\mathrm{Cl}}_2}{\overset{{\mathrm{Br}}_2}{\to }}$ W. Product W is:
1. 
2. 
3. 
4. 

Find missing reagents.
1. x = LiAlH₄, y = NaBH₄
2. x = LiAIH₄ / AlCl₃, y = LiAlH₄
3. x = LiAIH₄, Y = LiAIH₄ / AlCl₃
4. x = H₂ / Ni, y = H₂ / Pt

Which are not cleaved by HIO₄?
I : glycerol
II : glycol
III: 1, 3-propanediol
IV: methoxy-2-propanol
1. I, II, III, IV
2. I, II
3. II, III
4. III, IV

1. Dichloromethyl cation \((^{+}CHCl_{2})\)
2. Formyl cation \((^{+}CHO)\)
3. Dichloromethyl anion \((^{-}CHCl_{2})\)
4. Dichlorocarbene (:CCl₂)

The compound among the following that will not be soluble in
sodium hydrogen carbonate is:

1-propanol and 2-propanol can be best distinguished
by:
1. Oxidation with alkaline KMnO₄ followed by
reaction with Fehling solution.
2. Oxidation with acidic dichromate followed by
reaction with Fehling solution
3. Oxidation by heating with copper followed by
reaction with Fehling solution.
4. Oxidation with conc. H₂SO₄
followed by reaction with Fehling solution

$\stackrel{{\mathrm{H}}^{+}/∆}{\to }A\left(major\right)$
The product ‘A’ is

$\underset{\left(Predominant\right)}{\left(\mathrm{B}\right)}\stackrel{PB{r}_{\mathit{3}}}{\leftarrow }$$\stackrel{\mathrm{HBr}}{\to }\underset{\left(Predominant\right)}{\left(A\right)}$
(A) and (B) are respectively :