MCQ
$HO - C{H_2} - C{H_2} - C{H_2} - C{H_2} - CH = C{H_2}\xrightarrow{{B{r_2}}}{\text{Product ?}}$
- ✓
- B
- C
- D
✓
Answer
Correct option: A.
a
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
Consider the following values of $IE(ev)$ for elements $W$ and $X$
→| Element | $IE_1$ | $IE_2$ | $IE_3$ | $IE_4$ |
| $W$ | $10.5$ | $15.5$ | $24.9$ | $79.8$ |
| $X$ | $8$ | $14.8$ | $78.9$ | $105.8$ |
Other two elements $Y$ and $Z$ have outer electronic configuration $ns^2$ $np^4$ and $ns^2$ $np^5$ respectively. According to given information which of the following compound $(s)$ is/are not possible
$(a)\, W_2Y_3\, (b)\, X_2Y_3\, (c)\, WZ_2\, (d)\, XZ_2$
When light is directed at the metal surface, the emitted electrons
Natural gas contains mainly
The e.m.f. of the cell $Zn|Z{n^{2 + }}(0.01\,M)||F{e^{2 + }}(0.001\,M)|Fe$ at $298\,K$ is $0.2905 $ then the value of equilibrium for the cell reaction is
→The number of $2p$ electrons having spin quantum number $s = - 1/2$ are
→Di-imide $(N_2H_4)$ is used to reduce doubble bond of
→How many geometrical isomers are possible for the above compound ?
In the following structure, the double bonds are marked as $I, II, III$ and $IV$ Geometrical isomerism is not possible at site $(s)$
→Galvanization is applying a coating of :
The major aromatic product $C$ in the following reaction sequence will be :
→