MCQ
A substance forms zwitter ion. It can have functional groups
- A$ - N{H_2}, - COOH$
- B$ - N{H_2}, - S{O_3}H$
- ✓Both
- DNone of these
✓
Answer
Correct option: C.
Both
c
(c) A substance forms Zwitter ion. It can have functional groups $ - N{H_2}COOH$ and $ - N{H_2}$, $ - S{O_3}H.$
(c) A substance forms Zwitter ion. It can have functional groups $ - N{H_2}COOH$ and $ - N{H_2}$, $ - S{O_3}H.$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Isobutyraldehyde on reaction with formaldehyde and $K _{2} CO _{3}$ gives compound $'A'.$ Compound $'A'$ reacts with $KCN$ and yields compound $'B',$ which on hydrolysis gives a stable compound $'C'.$ The compound $'C'$ is .... .
→At higher temperature, iodoform reaction is given by
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 $
When a reaction is progressing
Vinegar obtained from sugarcane has
Acetone reacts with ${I_2}$ in presence of $NaOH$ to form
→Given $D$-Glucose The compound($s$), which on reaction with $HNO _3$ will give the product having degree of rotation, $[\alpha]_{ D }=-52.7^{\circ}$ is (are)
→Which of the following compounds can form alcohol with $NaNO_2/HCl$ ?
→In given compounds which one does not give glyoxal as one of the product on ozonolysis ?
Which of the following reagent must be used in given interconversion ?
