Consider the given reaction: $CH_3-CH=CH-CH_2-OH \xrightarrow[S_N1]{HBr} [P]$. In the given reaction,the product $[P]$ is:
- A$CH_3-CH=CH-CH_2-Br$
- B$CH_3-CH(Br)-CH=CH_2$
- C$CH_2=CH-CH=CH_2$
- D$CH_3-CH(Br)-CH_2-CH_2-OH$
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Arrange the following in order of ease of dehydrohalogenation:
$(i)$ Bromocyclohexane
(ii) $3-$Bromocyclohexene
(iii) $3-$Bromocyclohexa$-1,4-$diene
(iv) Bromobenzene
$(i)$ Bromocyclohexane
(ii) $3-$Bromocyclohexene
(iii) $3-$Bromocyclohexa$-1,4-$diene
(iv) Bromobenzene
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View SolutionThe major product obtained in the reaction of $1$-phenyl-$2$-bromobutane with $NaOMe$ is:
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View SolutionPure $CHCl_3$ and pure $CHI_3$ can be distinguished by which of the following?
Medium
View SolutionIn the given reaction:
$CH_3-CH_2-\ddot{S}-CH_2^*-CH_2-Br \xrightarrow{HOH} [X]$
$[X]$ will be:
$CH_3-CH_2-\ddot{S}-CH_2^*-CH_2-Br \xrightarrow{HOH} [X]$
$[X]$ will be:
Match the $CH_{3}X$ compounds in Column-$(I)$ with their $C-X$ bond length in Column-$(II)$ and dipole moment in Column-$(III)$.
$(I)$ Compound $(II)$ $C-X$ Bond Length $(pm)$ $(III)$ Dipole Moment $(D)$ $(A)$ $H_3C-F$ $(i)$ $214$ $(1)$ $1.636$ $(B)$ $H_3C-Cl$ $(ii)$ $193$ $(2)$ $1.830$ $(C)$ $H_3C-Br$ $(iii)$ $178$ $(3)$ $1.847$ $(D)$ $H_3C-I$ $(iv)$ $139$ $(4)$ $1.860$
| $(I)$ Compound | $(II)$ $C-X$ Bond Length $(pm)$ | $(III)$ Dipole Moment $(D)$ |
| $(A)$ $H_3C-F$ | $(i)$ $214$ | $(1)$ $1.636$ |
| $(B)$ $H_3C-Cl$ | $(ii)$ $193$ | $(2)$ $1.830$ |
| $(C)$ $H_3C-Br$ | $(iii)$ $178$ | $(3)$ $1.847$ |
| $(D)$ $H_3C-I$ | $(iv)$ $139$ | $(4)$ $1.860$ |
Difficult
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