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In a $S_N2$ substitution reaction of the type
$R-Br + Cl^- \xrightarrow{DMF} R-Cl + Br^-$
Which one of the following has the highest relative rate ?
$R-Br + Cl^- \xrightarrow{DMF} R-Cl + Br^-$
Which one of the following has the highest relative rate ?
A
$(CH_3)_3C-CH_2Br$
B
$CH_3CH_2Br$
C
$CH_3-CH_2-CH_2Br$
D
$(CH_3)_2CH-CH_2Br$
Detailed Solution
In an $S_N2$ reaction the nucleophile attacks the carbon from the side opposite to the leaving group in a single step, through a crowded transition state having five groups around the carbon.
So the rate depends on steric hindrance at and near the carbon carrying the halogen: less the crowding, faster the reaction.
All four halides are primary, so the crowding at the $\beta$-carbon decides the rate.
$CH_3CH_2Br$ has no alkyl substituent on the $\beta$-carbon, so it is the least hindered.
$CH_3CH_2CH_2Br$ has one methyl group, $(CH_3)_2CHCH_2Br$ has two and $(CH_3)_3CCH_2Br$ (neopentyl bromide) has three methyl groups on the $\beta$-carbon, which block the backside attack increasingly.
Relative rate: $CH_3CH_2Br \gt CH_3CH_2CH_2Br \gt (CH_3)_2CHCH_2Br \gt (CH_3)_3CCH_2Br$
So $CH_3CH_2Br$ has the highest relative rate.
So the rate depends on steric hindrance at and near the carbon carrying the halogen: less the crowding, faster the reaction.
All four halides are primary, so the crowding at the $\beta$-carbon decides the rate.
$CH_3CH_2Br$ has no alkyl substituent on the $\beta$-carbon, so it is the least hindered.
$CH_3CH_2CH_2Br$ has one methyl group, $(CH_3)_2CHCH_2Br$ has two and $(CH_3)_3CCH_2Br$ (neopentyl bromide) has three methyl groups on the $\beta$-carbon, which block the backside attack increasingly.
Relative rate: $CH_3CH_2Br \gt CH_3CH_2CH_2Br \gt (CH_3)_2CHCH_2Br \gt (CH_3)_3CCH_2Br$
So $CH_3CH_2Br$ has the highest relative rate.