Relation between (A) and (B) is * 20 c O || C H_3 - C - N H_2 KOBr (A) * 20 c O || C H_3 - C - N H_2 LiAlH_4 (B)

Relation between (A) and (B) is * 20 c O || C H_3 - C - N H_2 KOB…

Download App

MCQ

Relation between $(A)$ and $(B)$ is

$\begin{array}{*{20}{c}}
O \\
{||} \\
{C{H_3} - C - N{H_2}}
\end{array}$ $\xrightarrow{{KOBr}}$ $(A)$

$\begin{array}{*{20}{c}}
O \\
{||} \\
{C{H_3} - C - N{H_2}}
\end{array}$ $\xrightarrow{{LiAlH_4}}$ $(B)$

A
Identical
B
Functional isomer
✓
Homologous
D
Positional isomers

✓

Answer

Correct option: C.

Homologous

c
$\begin{array}{*{20}{c}}
O \\
{||} \\
{C{H_3} - C - N{H_2}}
\end{array}$ $\xrightarrow[{(Hoffmann\,\,broamide)}]{{KOBr}}$ $\mathop {C{H_3} - N{H_2}}\limits_{(A)} $

$\begin{array}{*{20}{c}}
O \\
{||} \\
{C{H_3} - C - N{H_2}}
\end{array}$ $\xrightarrow[{(\operatorname{Re} duction)}]{{LiAl{H_4}}}$ $\mathop {C{H_3} - C{H_2} - N{H_2}}\limits_{(B)} $

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Which of the following Carbocation will not undergo rearrangement

→

$25\, g$ of a solute of molar mass $250\, g\, mol^{-1}$ is dissolved in $100\, ml$ of water to obtain a solution whose density is $1.25\, g\, ml^{-1}$. The molarity and molality of the solution are respectively

→

$PC{l_5} + {P_4}{O_{10}}\xrightarrow{\Delta }$ oxygen containing compound $(X)$ .

How many moles of an oxy acid are formed on hydrolysis of $1$ mole $(X)$ .

→

For the following reaction

$2 X + Y \xrightarrow{i} P$

the rate of reaction is $\frac{ d [ P ]}{ dt }=k[ X ]$. Two moles of $X$ are mixed with one mole of $Y$ to make $1.0 L$ of solution. At $50 s , 0.5$ mole of $Y$ is left in the reaction mixture. The correct statement(s) about the reaction is(are)

(Use: $\ln 2=0.693$ )

$(A)$ The rate constant, $k$, of the reaction is $13.86 \times 10^{-4} s ^{-1}$.

$(B)$ Half-life of $X$ is $50 s$.

$(C)$ At $50 s ,-\frac{ d [ X ]}{ dt }=13.86 \times 10^{-3} mol L ^{-1} s ^{-1}$.

$(D)$ At $100 s ,-\frac{ d [ Y ]}{ dt }=3.46 \times 10^{-3} mol L ^{-1} s ^{-1}$.

→

$\begin{array}{*{20}{c}}
{C{H_3} - CH - CH = C{H_2} + HBr \to X(product)} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_{3\,\,\,\,\,\,\,\,\,\,\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$

What is the major product i.e. $X$ in the reaction?

→

The conversion, $PhCN \to PhCOCH_3,$ can be achieved most conveniently by reaction with

→

$(A)$ and $(B)$ in the following reactions are