MCQ
The major product obtained when $B{r_2}/Fe$ is treated with is
- ✓
- B
- C
- D
✓
Answer
Correct option: A.
a
(a) The phenyl ring having $H -N<$ group is activated while another one is deactivated due to $ - \mathop {\mathop C\limits_{||} }\limits_O - $, so electrophilic aromatic bromination will occur at para position with respect to $H -N<$ group inactivated ring.
(a) The phenyl ring having $H -N<$ group is activated while another one is deactivated due to $ - \mathop {\mathop C\limits_{||} }\limits_O - $, so electrophilic aromatic bromination will occur at para position with respect to $H -N<$ group inactivated ring.
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
If reaction between $A$ and $B$ to give $C$shows first order kinetics in $A$ and second order in $B$, the rate equation can be written as
→| List - I (Carbohydrate) | List - II (Linkage Source) |
| (A) Amylose | (I) $\quad \beta- C _1- C _4$, plant |
| (B) Cellulose | (II) $\alpha- C _1- C _4$, animal |
| (C) Glycogen | (III) $\alpha- C _1- C _4$, $\alpha- C _1- C _6$, plant |
| (D) Amylopectin | (IV) $\alpha- C _1- C _4$, plant |
Two Faraday of electricity is passed through a solution of $CuSO_4.$ The mass of copper deposited at the cathode is ......... $g$. (at. mass of $Cu = 63.5\, amu$)
→$[Image]$
→Consider the above reaction. The percentage yield of amide product is ...... . (Round off to the Nearest Integer). (Given : Atomic mass : $C : 12.0\, u , H : 1.0 \,u,N : 14.0\, u , O : 16.0\, u , Cl : 35.5\, u )$
One among the following compounds will not give effervescence with sodium carbonate
The spectrum of $He$ is expected to be similar to
→Consider the kinetic data given in the following table for the reaction $A + B + C \rightarrow$ Product.
→| Experiment No. | $\begin{array}{c}{[ A ]} \\ \left( mol dm ^{-3}\right)\end{array}$ | $\begin{array}{c}{[ B ]} \\ \left( mol dm ^{-3}\right)\end{array}$ | $\begin{array}{c}{[ C]} \\ \left( mol dm ^{-3}\right)\end{array}$ | Rate of reaction $\left( mol dm ^{-3} s ^{-1}\right)$ |
| $1$ | $0.2$ | $0.1$ | $0.1$ | $6.0 \times 10^{-5}$ |
| $2$ | $0.2$ | $0.2$ | $0.1$ | $6.0 \times 10^{-5}$ |
| $3$ | $0.2$ | $0.1$ | $0.2$ | $1.2 \times 10^{-4}$ |
| $4$ | $0.3$ | $0.1$ | $0.1$ | $9.0 \times 10^{-5}$ |
The rate of the reaction for $[ A ]=0.15 mol dm ^{-3},[ B ]=0.25 mol dm ^{-3}$ and $[ C ]=0.15 mol dm ^{-3}$ is found to be $Y \times 10^{-5} mol dm d ^{-3} s ^{-1}$. The value of $Y$ i. . . . . . .
'A' salt which gives green flame with bunsen burner, on reaction with dil. $H_2SO_4$ produces gas $(2)$ and white turbidity $(Z)$. The saturated solution of gas $(2)$ on reaction with $H_2S$ gas produces '$Z$' then select $CORRECT$ statement
→Match List $I$ with List $II$
→$1-$ Bromopropane is reacted with reagents in List $I$ to give product in List $II$
| List$I-$ Reagent | List$II-$Product |
| $A$ $KOH (\text { alc) }$ | $I$ Nitrile |
| $B$ $KCN \text { (alc) }$ | $II$ Ester |
| $C$ $AgNO _2$ | $III$ Alkene |
| $D$ $H _3 CCOOAg$ | $IV$ Nitroalkane |
Following reaction occurring in an automobile $2{C_8}{H_{18}}(g) + 25{O_2}(g) \to 16C{O_2}(g) + 18{H_2}O(g).$ The sign of $\Delta H,\,\Delta S$ and $\Delta G$ would be
→