MCQ
A compound is treated with $NaN{H_2}$ to give sodium salt. Identify the compound
- ✓${C_2}{H_2}$
- B${C_6}{H_6}$
- C${C_2}{H_6}$
- D${C_2}{H_4}$
✓
Answer
Correct option: A.
${C_2}{H_2}$
a
(a)$CH \equiv CH + NaN{H_2} \to CH \equiv {C^ - }N{a^ + } + N{H_3}$
(a)$CH \equiv CH + NaN{H_2} \to CH \equiv {C^ - }N{a^ + } + N{H_3}$
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
Strong electrolytes are those which
On combustion, carbon forms two oxides $ CO $ and $C{O_2}$, Heat of formation of $C{O_2}$ gas is $94.3 \,kcal $ and that of $ CO $ is $26.0\, kcal.$ Heat of combustion of carbon is.......$kcal$
→For the reaction $AB(g)\, \rightleftharpoons \,A(g) + B(g),\,AB$ is $33.3\%$ dissociated at a total equilibrium pressure of $P.$ Therefore, $P$ is correctly related to $K_p$ by which of the following option
→Product $(A)$ obtained is
→For the reaction ${H_2}(g) + B{r_2}(g) \to 2HBr(g)$, the experimental data suggest, rate $ = K[{H_2}]{[B{r_2}]^{1/2}}$. The molecularity and order of the reaction are respectively
→Which of the following reactions will not give propane
For the reaction $A + B \to $products, doubling the concentration of $A$ the rate of the reaction is doubled, but on doubling the concentration of $B$ rate remains unaltered. The over all order of the reaction is
→Which of the reagent is used to convert $2$ -Butanone into propanoic acid :-
→The major product of the reaction is
The radius of the $2^{\text {nd }}$ orbit of $Li ^{2+}$ is $x$. The expected radius of the $3^{\text {rd }}$ orbit of $Be ^{3+}$ is
→