MCQ
What will happen if $LiAl{H_4}$ is added to an ester
- ✓Two units of alcohol are obtained
- BOne unit of alcohol and one unit of acid is obtained
- CTwo units of acids are obtained
- DNone of these
✓
Answer
Correct option: A.
Two units of alcohol are obtained
a
(a) $R - \mathop {\mathop C\limits^{||} }\limits^O - R'\xrightarrow{{LiAl{H_4}}}\mathop {R - C{H_2}OH + R'OH}\limits_{{\text{Two}}\,{\text{units}}\,{\text{of}}\,{\text{alcohols}}} $
(a) $R - \mathop {\mathop C\limits^{||} }\limits^O - R'\xrightarrow{{LiAl{H_4}}}\mathop {R - C{H_2}OH + R'OH}\limits_{{\text{Two}}\,{\text{units}}\,{\text{of}}\,{\text{alcohols}}} $
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Which of the following is not formed when $H_2S$ reacts with acidic $K_2Cr_2O_7$ solution ?
→The standard reduction potential data at $25^{\circ} C$ is given below.
→$E ^0\left( Fe ^{3+} . Fe ^{2+}\right)=+0.77 V $
$E ^0\left( Fe ^{2+} . Fe \right)=-0.44 V $
$E ^{\circ}\left( Cu ^{2+} . Cu \right)=+0.34 V $
$E ^0\left( Cu ^{+} . Cu \right)=+0.52 V $
$E ^{\circ}\left( O _2( g )+4 H ^{+}+4 e ^{-} \rightarrow 2 H _2 O \right)=+1.23 V $
$E ^{\circ}\left( O _2( g )+2 H _2 O +4 e ^{-} \rightarrow 4 OH \right)=+0.40 V $
$E ^0\left( Cr ^{3+} . Cr \right)=-0.74 V $
$E ^{\circ}\left( Cr ^{2+} . Cr \right)=-0.91 V$
Match $E ^{\circ}$ of the rebox pair in List $I$ with the values given in List $II$ and select the correct answer using the code given below the lists:
| List $I$ | List $II$ |
| $P.$ $\quad E ^{\circ}\left( Fe ^{3+}, Fe \right)$ | $1.$ $\quad-0.18 V$ |
| $Q.$ $\quad E ^{\circ}\left(4 H _2 O \rightleftharpoons 4 H ^{+}+4 OH ^{-}\right)$ | $2.$ $\quad-0.4 V$ |
| $R.$ $\quad E ^{\circ}\left( Cu ^{2+}+ Cu \rightarrow 2 Cu ^{+}\right)$ | $3.$ $\quad-0.04 V$ |
| $S.$ $\quad E ^{\circ}\left( Cr ^{3+}, Cr ^{+2}\right)$ | $4.$ $\quad-0.83 V$ |
Codes: $ \quad P \quad Q \quad R \quad S $
Identify the correct formula of oleum from the following
Product obtained in above reactions $(1), (2)$ & $(3)$ is
→Which reaction cannot be used for the production of halogen acid
$(X)$ and $(Y)$ are
→In which of reactions final product is $NOT$ a ketone
→For the reaction:
→$RCH _2 Br + I ^{-} \stackrel{\text { Acetone }}{\longrightarrow} \underset{\text { major }}{ RCH _2 I + Br ^{-}}$
The correct statement is :
Which of the following noble gas does not have an octet of electrons in its outermost shell
$\text{CH}_3\text{−CH}\text{−CH}−\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ | \\ \ \ \ \ \ \ \ \ \ \text{OH}\ \ \ \ \ \ \text{OH}$
→IUPAC name of the given compound is: