What combination of acid chloride or anhydride and arene would you choose to prepare the given compound?
- A
- B
- C
- D
Explore More
Similar Questions
Match List-$I$ with List-$II$.
List-$I$ (Name of reaction) List-$II$ (Reagent used) $A$. Hell-Volhard-Zelinsky reaction $I$. $NaOH + I_2$ $B$. Iodoform reaction $II$. $(i) CrO_2Cl_2, CS_2; (ii) H_2O$ $C$. Etard reaction $III$. $(i) Br_2 / \text{red phosphorus}; (ii) H_2O$ $D$. Gatterman-Koch reaction $IV$. $CO, HCl, \text{anhyd. } AlCl_3$
Choose the correct answer from the options given below:
| List-$I$ (Name of reaction) | List-$II$ (Reagent used) |
| $A$. Hell-Volhard-Zelinsky reaction | $I$. $NaOH + I_2$ |
| $B$. Iodoform reaction | $II$. $(i) CrO_2Cl_2, CS_2; (ii) H_2O$ |
| $C$. Etard reaction | $III$. $(i) Br_2 / \text{red phosphorus}; (ii) H_2O$ |
| $D$. Gatterman-Koch reaction | $IV$. $CO, HCl, \text{anhyd. } AlCl_3$ |
Choose the correct answer from the options given below:
The product of the following reaction is :
Medium
View SolutionThe acid strength of the active methylene group in
$(a)$ $CH_3COCH_2COOC_2H_5$
$(b)$ $CH_3COCH_2COCH_3$
$(c)$ $C_2H_5OOCCH_2COOC_2H_5$
decreases as:
$(a)$ $CH_3COCH_2COOC_2H_5$
$(b)$ $CH_3COCH_2COCH_3$
$(c)$ $C_2H_5OOCCH_2COOC_2H_5$
decreases as:
What will be $X$ and $Y$ in the following reactions?
Difficult
View SolutionMatch Column-$I$ with Column-$II$ based on the $pK_a$ values of the given carboxylic acids:
Column-$I$ Column-$II$ $(A)$ $CF_3COOH$ $(i)$ $3.41$ $(B)$ Benzoic acid $(ii)$ $4.76$ $(C)$ $CH_3COOH$ $(iii)$ $0.23$ $(D)$ $4$-nitrobenzoic acid $(iv)$ $4.20$
| Column-$I$ | Column-$II$ |
| $(A)$ $CF_3COOH$ | $(i)$ $3.41$ |
| $(B)$ Benzoic acid | $(ii)$ $4.76$ |
| $(C)$ $CH_3COOH$ | $(iii)$ $0.23$ |
| $(D)$ $4$-nitrobenzoic acid | $(iv)$ $4.20$ |
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo