In the Gabriel's phthalimide synthesis, phthalimide is treated fi… - Vidyadip

MCQ

In the Gabriel's phthalimide synthesis, phthalimide is treated first with

✓
${C_2}{H_5}I/KOH$
B
Ethanolic $Na$
C
Ethanol and ${H_2}S{O_4}$
D
Ether and $LiAl{H_4}$

✓

Answer

Correct option: A.

${C_2}{H_5}I/KOH$

a
Gabriel phthalimide is firstly treated with alcoholic Kor fo form potassium phthalimide which on further reaction with ethyl iodide gives Ne ethyl phthalimide, The product $N-$ethyl phthalimite is hydrolysed with dilute $H C$ lo form primary amine, ie. ethylamine. Th is reaction is known as Gabriel phthalimide synthesis.

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

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

Similar questions

The wave function $(\Psi)$ of $2 s$ is given by

$\Psi_{2 s}=\frac{1}{2 \sqrt{2 \pi}}\left(\frac{1}{a_0}\right)^{1 / 2}\left(2-\frac{ r }{ a _0}\right) e ^{- r / 2 a _0}$

At $r=r_0$, radial node is formed. Thus, $r_0$ in terms of $a_0$

The relatively high boiling point of $HF$ is due to

If the rate of the reaction is equal to the rate constant, the order of the reaction is

Copper sulphate solution reacts with $KCN$ to give

The $'d'$ orbital with the orientation along $X$ and $Y$ axis is called.

Which is the best description of the behaviour of bromine in the reaction given below ${H_2}O + B{r_2} \to HOBr + HBr$

Which of the following has the lowest barrier to rotation about the indicated bond ?

Arrhenius equation is represented as

$k = A{e^{ - {E_a}/RT}}$

The activation energy $E_a$ of the reaction can be calculated by plotting

$pH$ of a solution produced when an aqueous solution of $pH$ $6$ is mixed with an equal volume of an aqueous solution of $pH$ $3 $ is about

Oxygen is more electronegative, than sulphur, yet $H_2S$ is acidic in nature while $H_2O$ is neutral. because