$\stackrel{{\mathrm{H}}_2{\mathrm{CrO}}_4}{\to }$(A) Product (A) is:   

(a) No reaction

(b) 

(c) 

(d) 

Consider the following alcohols,

(I)  (II) 

(III) (IV) 

The order of decreasing reactivities of these alcohols towards nucleophilic substitution with HBr is:-

(a) III>I>IV>II

(b) III>I>II>IV

(c) I>III>IV>II

(d) I>III>II>IV

$\underset{\left(2\right) {\mathrm{H}}^{+}, {\mathrm{H}}_2\mathrm{O} }{\overset{\left(1\right) \mathrm{EtLi} (3.5 \mathrm{eq})}{\to }} \left(\mathrm{A}\right) 65\% \mathrm{Yield}$; Chemist added extra 0.5 mole of Et-Li in above reaction to obtain product (A), which is ?

Which is the best reagent to convert isopropyl alcohol to isopropyl bromide ?

 $C{H}_3 C{H}_3$
$| |$
$C{H}_3-CH-OH\stackrel{?}{\to } C{H}_3-CH-Br$

(1) HBr

(2) SOBr₂

(3) Br₂

(4) CH₃MgBr

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

Which of the following compounds are not oxidized by HIO₄?

(1)   (2)  (3) 

(4)  (5)  (6)  

(7) 

1. 5,6,7     
2. 4,5,6,7     
3. 6,7     
4. 3,4,5,6,7

Succinic acid $\stackrel{∆}{\to }$(A)$\underset{∆}{\overset{{\mathrm{NH}}_3}{\to }}$(B)$\underset{\mathrm{KOH}}{\overset{{\mathrm{Br}}_2}{\to }}$(C); Product (C) will be:

(1) $\begin{array}{ccc}\underset{|}{{\mathrm{CH}}_2}& -& {\mathrm{CO}}_2\mathrm{H}\\ {\mathrm{CH}}_2& -& \begin{array}{cc}{\mathrm{CH}}_2& \begin{array}{cc}-& {\mathrm{NH}}_2\end{array}\end{array}\end{array}$

(2) $\begin{array}{ccc}\underset{|}{{\mathrm{CH}}_2}& -& {\mathrm{CO}}_2\mathrm{H}\\ {\mathrm{CH}}_2& -& {\mathrm{NH}}_2\end{array}$

(3) $\begin{array}{ccc}\underset{|}{{\mathrm{CH}}_2}& -& {{\mathrm{CO}}^{-}}_2{\mathrm{K}}^{+}\\ {\mathrm{CH}}_2& -& {\mathrm{NH}}_2\end{array}$

(4) $\begin{array}{ccc}\underset{|}{{\mathrm{CH}}_2}& -& {\mathrm{CO}}_2\mathrm{H}\\ {\mathrm{CH}}_2& -& \begin{array}{cc}{\mathrm{CH}}_2& \begin{array}{cc}-& \mathrm{Br}\end{array}\end{array}\end{array}$

$\stackrel{\mathrm{PCC}}{\to }$(A)$\underset{{\mathrm{H}}^{+}}{\overset{\mathrm{OH}-\mathrm{CH}2-\mathrm{CH}2-\mathrm{OH}}{\to }}$(B)$\stackrel{\left(1\right) \mathrm{MeMgBr} \left(2\right) {\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }$
$$(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)