MCQ
Among the following the weakest base is
- A$C_6H_5CH_2NH_2$
- B$C_6H_5CH_2NHCH_3$
- C$NO_2CH_2NH_2$
- ✓$CH_3NHCHO$
✓
Answer
Correct option: D.
$CH_3NHCHO$
d
It is weakest base because the lone pair of electrons of the $N$ atom is delocalized on account of resonance with the carbonyl group since $N O_{2}$ has a poweful $-I$ effect, the correct of basic strength is
It is weakest base because the lone pair of electrons of the $N$ atom is delocalized on account of resonance with the carbonyl group since $N O_{2}$ has a poweful $-I$ effect, the correct of basic strength is
$C_{6} H_{5} N H C H_{3}>C_{6} H_{5} C H_{2} N H_{2}>O_{2} N C H_{2} N H_{2}>C H_{3} N H C H O$
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 ice and water are at equilibrium then which is correct. To obtain high amount of water
→Ice $ \rightleftharpoons $ Water
Silver nitrate is mainly used
The degree of hydrolysis in hydrolytic equilibrum ${A^ - } + {H_2}O$ $\rightleftharpoons$ $HA + O{H^ - }$ at salt concentration of $0.001 \,\,M$ is $\left( {{K_a} = 1 \times {{10}^{ - 5}}} \right)$
→${C_6}{H_5}C{H_2}Cl + KCN(aq.) \to X + Y$
→Compounds $X$ and $Y $ are
Consider the reaction
→$(i)\,\,(CH_3)_2CH-CH_2Br \xrightarrow{{{C_2}{H_5}OH}} (CH_3)_2CH-CH_2OC_2H_5 + HBr$
$(ii)\,\,(CH_3)_2CH-CH_2-Br \xrightarrow{{{C_2}{H_5}{O^- }}} (CH_3)_2CH-CH_2OC_2H_5 + Br^-$
The mechanisms of reaction $(i)$ and $(ii)$ are respectively
The molecular electronic configuration of $H_2^ - $ ion is
→A $0.004\, M$ solution of $N{a_2}S{O_4}$ is isotonic with a $0.010 \,M$ solution of glucose at same temperature. The apparent degree of disociation of $N{a_2}S{O_4}$ is.....$\%$
→The tautorner of $II$ is
→Given below are two statements :
→Statement $(I)$ : The oxidation state of an element in a particular compound is the charge acquired by its atom on the basis of electron gain enthalpy consideration from other atoms in the molecule.
Statement $(II)$ : $\mathrm{p} \pi-\mathrm{p} \pi$ bond formation is more prevalent in second period elements over other periods.
In the light of the above statements, choose the most appropriate answer from the options given below :
Which pair is different from the others