$\stackrel{\mathrm{PCC}}{\to }$(A)$\underset{{\mathrm{H}}^{+}}{\overset{\mathrm{OH}-\mathrm{CH}2-\mathrm{CH}2-\mathrm{OH}}{\to }}$(B)$\stackrel{\left(1\right)<mspace\; width="1em"></mspace>\mathrm{MeMgBr}<mspace\; width="1em"></mspace>\left(2\right)<mspace\; width="1em"></mspace>{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }$
$$(C)$\stackrel{{\mathrm{NaBH}}_4\mathrm{EtOH}}{\to }$(D) Product

(D) in above reaction Is:

(a) 

(b) 

(c) 

(d) 

(A) _{→ B + C; (B) and (C) both gives +ve iodoform test.Compound (A) is:-}
C₅H₁₀O       H₃O+


(1) CH₃-CH=CH-O-CH₂-CH₃

            H
            l

(2)CH₃-C-O-CH₂-CH₃
            l
           CH₃

(3)CH₃-C-O-CH₂-CH₃
            ll
           CH₂

(4)Both (b) and (c)

A solution of Ph₃CCO₂ in conc. H₂SO₄ gives (X) when poured into methanol X is :-
                O
               ll
(1) Ph₃C-C-O-CH₃

                   O
                 ll
(2) Ph₂CH-C-O-CH<sub>3

(3) Ph3C-OCH3</sub>

(4) Ph₃C-CH₃

$\underset{\mathrm{Pyridine}}{\overset{\mathrm{TsCl}}{\to }}$(A)$\stackrel{{\mathrm{LiAlH}}_4}{\to }$(B)    

Product (B) of the above reaction is:

The reducing agent among the following is -

1. CrO₃/H⁺

2. KMnO₄

3. LiAlH₄

4. O₃

 $CHO$
$<mspace\; width="1em"></mspace>|$
$CH-OH<mspace\; width="1em"></mspace>\stackrel{2Hl{O}_4}{\to }$
$<mspace\; width="1em"></mspace>|$
$C{H}_2-OH$Products obtained in the above reaction are :

(1) HCHO, HCO₂H

(2) HCHO, 2HCO₂H

(3) CO₂, 2HCO₂H

(4) CO₂, HCHO, HCO₂H

0.092 g of a compound with the molecular formula C₃H₈O₃ reaction with an excess of CH₃MgI gives 67.00 mL of methane at STP. The number of active hydrogen atoms present in a molecule of the compound is:
(1) one

(2) two

(3) three

(4) four

Optically active 2-octanol rapidly loses its optical activity when exposed to :

Product Z of above reaction is :

Which of the following compounds is hemiacetal?
 

1
2.
3.
4.	All of these