MCQ
Which compound gives yellow ppt. with iodine and alkali
- A$2-$ hydroxy ethane
- ✓Acetophenone
- CMethyl acetone
- DAcetamide
✓
Answer
Correct option: B.
Acetophenone
b
(b) $\mathop {C{H_3} - \mathop {\mathop C\limits_{||} }\limits_O {C_6}{H_5}}\limits_{{\text{Acetophenone}}} \mathop {\xrightarrow{{NaOH}}}\limits_{{I_2}} \mathop {CH{I_3}}\limits_{{\text{Yellow ppt}}{\text{.}}} + $
(b) $\mathop {C{H_3} - \mathop {\mathop C\limits_{||} }\limits_O {C_6}{H_5}}\limits_{{\text{Acetophenone}}} \mathop {\xrightarrow{{NaOH}}}\limits_{{I_2}} \mathop {CH{I_3}}\limits_{{\text{Yellow ppt}}{\text{.}}} + $
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
Which electronic arrangement shows ground state of an element
$CH_2=CH-CH_2-CH_3 \xrightarrow{{{H^ \oplus }/\Delta }} Major \,product$ Identify in above reaction
→Which of the following compound can exist in two different stereoisomeric form
Azo-dyes are prepared from :
If the bidentate ligand ethane $1, 2-$ diamine $(en)$ in progressively added in green aqueous solution of $[Ni(H_2O)_6]Cl_2$ in the molar ratios $en : Ni$ as $1 : 1, 2 : 1, 3 : 1$ then their associated spin only magnetic moment (in $B.M.\, unit$) are respectively
→Select the true statement from the following for metal carbonyls ?
Given below are two statements: one is labelled as Assertion $(A)$ : and the other is labelled as Reason $(R)$. Assertion $(A)$ : Both rhombic and monoclinic sulphur exist as $S_5$ while oxygen exists as $O_2$.
→Reason $(R)$ : Oxygen forms $\mathrm{p} \pi-\mathrm{p} \pi$ multiple bonds with itself and other elements having small size and high electronegativity like $\mathrm{C}, \mathrm{N}$, which is not possible for sulphur.
In the light of the above statements, choose the most appropriate answer from the options given below :
Given $pH $ of a solution $ A$ is $3$ and it is mixed with another solution $B$ having $pH\, 2$. If both mixed then resultant $pH$ of the solution will be
→The total number of lone pairs of electrons in $N _2 O _3$ is
→Isotonic solutions are those which have the: