Which of the following reactants will not fav | vedclass

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(D) The correct answer is $D$. All the given compounds do not readily undergo nucleophilic substitution reactions ($S_N1$ or $S_N2$):$1$. $1$-Bromobic

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(D) The correct answer is $D$. All the given compounds do not readily undergo nucleophilic substitution reactions ($S_N1$ or $S_N2$):$1$. $1$-Bromobicyclo$[2.2.1]$heptane: The bridgehead carbon is sterically crowded,preventing $S_N2$ attack. Furthermore,the formation of a carbocation at the bridgehead is forbidden by Bredt's rule,preventing $S_N1$ reaction.$2$. $Ph-Br$ (Bromobenzene): The lone pair on the bromine atom is in resonance with the benzene ring,giving the $C-Br$ bond partial double bond character. This makes the bond stronger and shorter,preventing nucleophilic substitution.$3$. $(CH_3)_3C-CH_2-Br$ (Neopentyl bromide): This is a primary alkyl halide with significant steric hindrance at the $\beta$-carbon,which blocks the backside attack required for $S_N2$. Additionally,the formation of a primary carbocation is energetically unfavorable,preventing $S_N1$.

Which of the following reactants will not favour nucleophilic substitution reaction?

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At normal temperature,iodoform is:

Which of the following,upon treatment with $tert-BuONa$ followed by addition of bromine water,fails to decolourize the colour of bromine?

Assertion $(A)$: $S_N1$ hydrolysis of optically active $2-$bromooctane results in the formation of $(\pm)-$octan$-2-$ol.
Reason $(R)$: The reaction proceeds through a planar carbocation which can be attacked by the nucleophile from either side.
The correct answer is

Among the following rules,the one which is applied in the given reaction is:
$CH_3-CH(Br)-CH_2-CH_3 \xrightarrow{Alc. KOH}$
$I. CH_3-CH=CH-CH_3$ (major product)
$II. CH_2=CH-CH_2-CH_3$ (minor product)

Explain the mechanism of a unimolecular nucleophilic substitution $(S_N1)$ reaction with a suitable example.

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