MCQ
Product $(A)$ is
- A
- ✓
- C
- D
✓
Answer
Correct option: B.
b
$(b)$ In presence of pair leaving group such as (fluorine), reaction proceed through stable carbanion in transition state.
$(b)$ In presence of pair leaving group such as (fluorine), reaction proceed through stable carbanion in transition state.
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 of following is in capable to show iodoform test ?
Which of the following has greatest reducing power
Which of the following will show optical isomers?
→$(I)$ cis $- [Co(NH_3)_2(en)_2]^{3+}$ $(II)$ trans $-[IrCl_2(C_2O_4)_2]^{3-}$
$(III)\, [Rh(en)_3]^{3+}$ $(IV)$ cis $-[Ir(H_2O)_3Cl_3$
Among the given oxides of nitrogen; $N _{2} O , N _{2} O _{3}$, $N _{2} O _{4}$ and $N _{2} O _{5}$, the number of compound/(s) having $N - N$ bond is ...
→In the reaction, the acid obtained will be
→$CH_3CHO + HCN \to CH_3CH(OH)CN \xrightarrow{H-OH} CH_3CH(OH)COOH$
Complex compound $(s)$ having even number of space (stereo) isomers is/are :
(where $AA-$ symmetrical bidentate ligand and $a,b,c,d,e-$ monodentate ligands)
→(where $AA-$ symmetrical bidentate ligand and $a,b,c,d,e-$ monodentate ligands)
Relation between $K_1$ and $K_2$ is
→Given below are two statements:
→Statement$-I$ $: \alpha$ and $\beta$ forms of sulphur can change reversibly between themselves with slow heating or slow cooling.
Statement$-II :$ At room temperature the stable crystalline form of sulphur is monoclinic sulphur.
In the light of the above statements, choose the correct answer from the options given below:
$3.65\,gms$ of $HCl$ is dissolved in $16.2\,gms$ of water. The mole fraction of $HCl$ in the resulting solution is
→Which one among the following non-metals liquid at ${25^o}C$
→